TS 10th Class Maths Notes Chapter 4 Pair of Linear Equations in Two Variables

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TS 10th Class Maths Notes Chapter 4 Pair of Linear Equations in Two Variables

→ An Equation of the form ax + by + c = 0, where a, b, c e R and a and b are not both zero, is called a linear equation in two variables x and y.

→ A pair of linear equations in two variables x and y can be represented as follows :
a1x + b1y + C1 = 0
a2x + b2y + c2 = 0
Where a1, a2, b1, b2, c1, c2 are real numbers such that a12 + b12 ≠ 0; a22 + b22 ≠ 0.

→ A pair of linear Equations in two variables forms a system of simultaneous linear equations.
Example : 3x – 4y = 2
2x + 5y = 9

→ A pair of values of the variables x and y satisfying each one of the equations that are given is called a solution of the system.
x = 2, y = 1 is a solution of the system of simultaneous linear equations.
3x-4y = 2 …………… (1)
2x + 5y = 9 …………… (2)
Putting x = 2 and y = 1 in equation (1), we get
L.H.S. = 3 × 2 – 4 × 1 = 6 – 4 = 2
R.H.S = 2
L.H.S = R.H.S
Similarly, put x = 2 and y = 1 in equation (2), we get
L.H.S = 2 × 2 + 5 × 1 = 4 + 5 = 9
R.H.S = 9
∴ L.H.S = R.H.S

→ A pair of linear equations in two variables can be solved using

  • Graphical method
  • Model method
  • Algebraic method :
    (a) Substitution method
    (b) Elimination method
    (c) Cross-Multiplication method

TS 10th Class Maths Notes Chapter 4 Pair of Linear Equations in Two Variables

→ Graphical method: The graph of a pair of linear equations in two variables is represented by two straight lines.

  • If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is consistent.
  • If the lines coincide, then there are infinitely many solutions – each point on the line being a solution. In this case, the pair of equations is dependent (consistent).
  • If the lines are parallel, then the pair of equations has no solution.
    In this case, the pair of equations is inconsistent.

→ The relation that exists between the coefficients and nature of system of equations.

→ If a pair of linear equations is given by a1x + b2y + c2 = 0 and a2x + b2y + c2 = 0, then

  • The pair of linear equations is consistent if \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
  • The pair of linear equation is inconsistent if \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
  • The pair of linear equation is dependent and consistent if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

Important Formulas:

  • The pair of linear equations is consistent if \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
  • The pair of linear equation is inconsistent if \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
  • The pair of linear equation is dependent and consistent if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

TS 10th Class Maths Notes Chapter 4 Pair of Linear Equations in Two Variables

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TS 10th Class Maths Notes Chapter 4 Pair of Linear Equations in Two Variables 1

William George Horner(1786 – 1832):

  • A William George Horner was a British mathematician; he was a schoolmaster; headmaster and schoolkeeper, proficient in classics as well as mathematics.
  • A He wrote extensively on functional equations, number theory and approximation theory.
  • A His contribution to approximation theory is honoured in the designation Horner’s method. The modern invention of the zoetrope, under the name Daedaleum in 1834 has been attributed to him.

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Students must practice these TS Inter 2nd Year Maths 2A Important Questions Chapter 6 Binomial Theorem to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 1.
Find the largest binomial coefficient(s) in the expansion of i) (1+x)19 (ii) (1 +x)24
Solution:
i) Here n = 19, an odd integer. Therefore, by corollary 6.1.19. the largest binomial coefficients are \({ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{\mathrm{n}-1}{2}\right)}\) and
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 1
ii) Here n 24 ¡s an even integer. Hence there is only one largest binomial coefficient, that is \({ }^n C_{\left(\frac{n}{2}\right)}={ }^{24} C_{12}\)

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 2.
If 22Cr is the largest bínomial coefficient in the expansion of (1+ x)22 find the value of 13Cr
Solution:
Here n = 22 is an even integer. Therefore, there is only one largest binomial coefficient
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 2

Question 3.
Find the 7th term in the expansion of
\(\left(\frac{4}{x^3}+\frac{x^2}{2}\right)^{14}\)
Solution:
The general term in the expansion of (X + a)n is given by
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 4

Question 4.
Find the 3rd term from the end in the expension of  \(\left(x^{\frac{-2}{3}}-\frac{3}{x^2}\right)^8\)
Solution:
Comparing the given expansion with (x + a), we get
\(X=x^{\frac{-2}{3}}, a=\frac{-3}{x^2}, n=8\)
The expansion has (n + 1) = 9 terms.
Hence the 3 term from the end is 7th term from the beginning and
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 5
Question 5.
Find the coefficients of x9 and x10 in the expansion of  \(\left(2 x^2-\frac{1}{x}\right)^{20}\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 6
To find the coefficient of x9 put 40 – 3r = 9.
Then we get r = \(\frac{31}{3}\)
Since r is a positive integer this is not possible. This means that the expansion of \(\left(2 x^2-\frac{1}{x}\right)^{20}\) doesn’t posess x9 term. This means that the coefficient of x9 in the expansion of \(\left(2 x^2-\frac{1}{x}\right)^{20}\)  is 0.
To find the coefficient of x10 put 40 –  3r = 10.
We get r = 10
Now, on substituting r 10 in (1), we get that
\(T_{11}=(-1)^{10} \cdot{ }^{20} \mathrm{C}_{10} \cdot 2^{10} \cdot x^{10}\)
Hence, the coefficient of x10 in the expansion of \(\left(2 x^2-\frac{1}{x}\right)^{20} \text { is }{ }^{20} \mathrm{C}_{10} \cdot 2^{10}\)

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 6.
Find the term independent of x (that is the constant term) in the expansion of \(\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^2}\right)^{10}\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 7

Question 7.
If the coefficient of x10 in the expansion \(\left(a x^2+\frac{1}{b x}\right)^{11}\) is equal to the coefficient of
x-10 In the expansion of \(\left(a x-\frac{1}{b x^2}\right)^{11}\) find the ration between a and b where a and b are real numbers.
Solution:
Step – 1: The general term in the expansion
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 8
To find the coefficient of x10 in this expansion, we should consider 22 – 3r = 10 or r = 4. Hence, the coefficient of x10 in the expansion of
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 9
To find the coefficient of x-10 in this expansion
we should consider 11 – 3r = – 10 or r = 7.
Thus the coefficient of x-10 in the expansion
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 10

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 8.
If the kth term is the middle term in the expansion of \(\left(x^2-\frac{1}{2 x}\right)^{20}\), find Tk and Tk+3
Solution:
The general term in the expansion of \(\left(x^2-\frac{1}{2 x}\right)^{20}\) is given by
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 11

Question 9.
If the coefficients of (2r + 4)th and (r – 2)nd terms in the expansion of (1 + x)18 are equal, find r.
Solution:
The rth term in the given expansion of
\((1+x)^{18} \text { is } \mathrm{T}_{\mathrm{r}}={ }^{18} \mathrm{C}_{(\mathrm{r}-1)} \cdot \mathrm{x}^{\mathrm{r}-1}\)
Thus, the coefficient of \({ }^{18} C_{r-1}\)
Given that the coefficient of (2r + 4)th term = the coefficient of (r -2)nd term.
That is \({ }^{18} C_{2 r+3}={ }^{18} C_{r-3}\)
⇒ 2r+ 3r – 3 or (2r + 3) +(r-3) 18
⇒ r = – 6 or r 6
Since r is a positive integer, we get r z 6

Question 10.
Prove that 2.C0 +7.C1 + 12C2 + …. + (5n + Z)Cn (5n + 4)2n-1
Solution:
First method:
The coefficieint of C0, C1, C2, ………………. Cn in LH.S.
are 2, 7, 12 , (5n + 2) which are in A.P. with first term a = 2 and common difference d=5
Hence from example 6.1.14 (1), we get that
2C0 + 7C1 + 12C2 + (5n+2).Cn
=(2a + nd) . 2n-1
= (4 + 5n) 2n-1

Second method:
The general term ((r +1)th term) in LH.S (5r + 2) Cr Therefore,
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 12

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 11.
Prove that
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 13
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 14
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 15

Question 12.
For r = 0, 1, 2 ……………….. n, prove that
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 16
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 17
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 18

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 13.
Prove that
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 19
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 20

Question 14.
Find the numerically greatest term(s) in the expansion of
(i) (2+3x)10 when x= \(\frac{11}{8}\)
(ii) (3 x-4y)14 when x=8, y=3
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 21
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 22
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 24
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 25
Therefore the numerically greatest terms in the expansion of (3x – 4y)14 are T5 and T6. They are
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 26

Question 15.
Prove that 62n – 35n -1 is divisible by 1225 for all natural numbers n.
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 30
i.e., 62n-35n – 1 = 1225 (k) for sorne integer k(if n≥2)
If n = 1, then 62n-35n – 1 = 62-35-1 = 0, which is trivially divisible by 1225. Hence, for all natural numbers n, 62I – 35n – 1 is divisible by 1225.
Note: The above problem can also be proved by induction.

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 16.
Suppose that n is a natural number and I, F are respectively the Integral part and fractional part of \((7+4 \sqrt{3})^n\). Then show that
(i) I Is an odd integer
(ii) (1 + F)(1 – F) = 1.
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 31
2k where k is a positive integer ………………. (1)
Thus, I + F + f is an even integer.
Since I is an integer, we get that
F + f is an integer. Also, since 0 < F < 1 and 0 < f < 1.
we get 0 < F + f < 2.
Since F + f is an integer, we get
F + f = 1 i.e., 1 – F = f

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

(i) From (1), I + F+ f= 2k
⇒ 1+ 1 = 2k ⇒ 1= 2k – I, an odd integer.

(ii) (1 +F) (1-F) (1 + F) f from (2)
= \((7+4 \sqrt{3})^n(7-4 \sqrt{3})^n=(49-48)^n=1\)

Question 17.
Find the coefficient of x6 in (3 + 2x + x2)6
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 32
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 33
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 34

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 18.
If n is a positive integer, then prove that
\(C_0+\frac{C_1}{2}+\frac{C_2}{3}+\ldots+\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 35

Question 19.
If n is a positive Integer and x is any non zero real number, then prove that
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 36
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 37
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 38

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 20.
Prove that
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 39
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 40
Now, we calculate the term independent of x in the L.H.S of equation (1). From (1)
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 41
Observe that the expansion in the numerator of (2) contains only even powers of x. Therefore, if n is odd, then there is no constant term in (2). In other words, the term independent of x in \((1-x)^n\left(1+\frac{1}{x}\right)^n\) is zero. Now, suppose n is an even integer say n=2 k. Then, from (2) we get
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 42
To get the term independent of x in (3), put 2 r-2 k=0. Then r=k and hence the term independent of x in
\((1-x)^n\left(1+\frac{1}{x}\right)^n\) is
\({ }^{2 \mathrm{k}} C_k \cdot(-1)^{\mathrm{k}}={ }^n C_{\frac{n}{2}}(-1)^{\frac{\mathrm{n}}{2}}\)
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 43

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 21.
Find the set E of the values of x for which the binomial expansions for the following are valid.
(i) \((3-4 x)^{\frac{3}{4}}\)
(ii) \((2+5 x)^{\frac{-1}{2}}\)
(iii) (7 – 4x)-5
(iv) \((4+9 x)^{\frac{-2}{3}}\)
(v) (a+bx)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 44
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 45

Question 22.
Find the
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 46
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 47
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 48
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 49
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 50
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 51
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 52

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 23.
Write the first 3 terms in the expansion of
(i) \(\left(1+\frac{x}{2}\right)^{-5}\)
(ii) \((3+4 x)^{\frac{-2}{3}}\)
(iii) \((4-5 x)^{\frac{-1}{2}}\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 53
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 54
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 55

Question 24.
Write the general term in the expansion of
(i) \(\left(3+\frac{x}{2}\right)^{\frac{-1}{3}}\)
(ii) \(\left(2+\frac{3 x}{4}\right)^{\frac{4}{5}}\)
(iii) (1 – 4x)-3
(iv) \((2-3 x)^{\frac{-1}{3}}\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 56
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 57
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 58
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 59

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 25.
Find the coefficient of x12 in \(\frac{(1+3 x)}{(1-4 x)^4}\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 60

Question 26.
Find the coefficient of x6 in the expansion of \((1-3x)^{\frac{-2}{5}}\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 61

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 27.
Find the sum of the infinite series
\(1+\frac{2}{3} \cdot \frac{1}{2}+\frac{2 \cdot 5}{3 \cdot 6}\left(\frac{1}{2}\right)^2+\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\left(\frac{1}{2}\right)^3+\ldots \infty\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 62

Question 28.
Find the sum of the series
\(\frac{3 \cdot 5}{5 \cdot 10}+\frac{3 \cdot 5 \cdot 7}{5 \cdot 10 \cdot 15}+\frac{3 \cdot 5 \cdot 7 \cdot 9}{5 \cdot 10 \cdot 15 \cdot 20}+\ldots ……….. \infty\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 63
Question 29.
If x = \(\frac{1}{5}+\frac{1 \cdot 3}{5 \cdot 10}+\frac{1 \cdot 3 \cdot 5}{5 \cdot 10 \cdot 15}+\ldots \ldots \infty\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 64
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 65

Question 30.
Find an approximate value of \(\sqrt[6]{63}\) correct to 4 decimal places.
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 66

Question 31.
If \(|\mathbf{x}|\) is so small that x2 and higher powers of x may be neglected, then find an approximate value of \(\frac{\left(1+\frac{3 x}{2}\right)^{-4}(8+9 x)^{\frac{1}{3}}}{(1+2 x)^2}\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 67

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 32.
If |x| is so small that x4 and higher powers of x may be neglected, then find an approximate value of
\(\sqrt[4]{x^2+81}-\sqrt[4]{x^2+16}\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 68

Question 33.
Suppose that x and y are positive and x is very small when compared to y. Then find an approximate value of
\(\left(\frac{y}{y+x}\right)^{\frac{3}{4}}-\left(\frac{y}{y+x}\right)^{\frac{4}{5}}\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 69

TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions

Question 34
Expand \(5 \sqrt{5}\) increasing powers of \(\frac{4}{5}\)
Solution:
TS Inter 2nd Year Maths 2A Binomial Theorem Important Questions 70

TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding

Telangana SCERT TS 10th Class Physical Science Study Material Pdf 8th Lesson Chemical Bonding Textbook Questions and Answers.

TS 10th Class Physical Science 8th Lesson Questions and Answers Chemical Bonding

Improve Your Learning
I. Reflections on concepts

Question 1.
Explain the difference between the valence electrons and the covalency of an element. \
Answer:

Valence ElectronsCovalency
1. The electrons present in the outermost orbital of an atom are called valence electrons.1. The total number of covalent bonds that an atom of an element forms is called covalency of the element
2. Valence electrons depend upon the number of electrons present in that atom.2. Covalency depends upon the valence electrons.

 

Question 2.
A chemical compound has the following Lewis notation:
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 1
(a) How many valence electrons does element Y have?
(b) What ¡s the valency of element Y’ x . X
(c) What is the valency of element X?
(d) How many covalent bonds are there in the molecule? H
(e) To which groups the elements X and Y belong?
Answer:
(a) Six electrons.
(b) Two. because it has combined with two elements namely X and H.
(c) One
(d) Two covalent bonds one in Y – X and another one is Y – H.
(e) X is Hydrogen and Y is oxygen. I suggest the molecule is H2O (water).

Question 3.
How bond energies and bond lengths of molecule help us in predicting their chemical properties? Explain with examples.
Answer:
1. Bond length: Bond length or bond distance ¡s the equilibrium distance between the nuclei of two atoms which form a covalent bond.

2. Bond energy: Bond energy or bond dissociation energy ¡s the energy needed to break a covalent bond between two atoms of a diatomic covalent compound in its gaseous state.

3. If the nature of the bond between the same two atoms changes, the bond length also changes. For example, the bond lengths between two carbon atoms are C-C>C=C>C ≡ C.

4. Thus the various bond lengths between the two carbon atoms are ¡n ethane 1.54 Å, ethylene 1.34 Å. acetylene 1.20 Å.

5. The bond lengths between two oxygen atoms are in H2O2 (O-O) Is 1.48 A° and in O2 (O = O) is 1.21 Å.

6. Observe the table.

BondBond length(Å)Bond (dissociation) energy (KJ mol-1’)
H -H0.74436
F-F1.44159
Cl-Cl1.95243
Br-Br2.28193
I-I2.68151
H-F0.918570
H-Cl1.27432
H-Br1.42366
H -I1.61298
H-O(of H2O)0.96460
H-N(of NH3)1.01390
H-C(of CH4)1.10410

7. When bond length decreases, then bond dissociation energy Increases.
8. When bond length increases, then bond dissociation energy decreases.
9. Bond length of H -H in H2 molecule is 0.74 Å and its bond dissociation energy is 436 KJ/mol, whereas bond length of F – F In F2 molecule is 1.44 Å and its bond dissociation energy Is 159 K/mol.
10. Melting and boiling points of substances also can be determined by these bond energies and bond lengths.

Question 4.
Draw simple diagrams to show how electrons are arranged in the following covalent molecules:
(a) Calcium oxide (CaO)
(b) Water (H2O)
(c) Chlorine (Cl2)
Answer:
(a) Calcium oxide (CaO)
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 2

(b) Water (H2O):
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 4
One oxygen and two hydrogen atoms form a water molecule, H2O

(c) Chlorine (Cl2):
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 5

Question 5.
Represent each of the following atoms using Lewis notation.
(a) Beryllium
(b) Calcium
(c) Lithium
(d) Bromine gas(Br2)
(e) Calcium Chloride (CaCl2)
(f) Carbon dioxide (CO2)
Answer:
(a) Beryllium:
Beryllium atomic number = 4
Be – Valency electrons = 2

(b) Calcium:
Calcium atomïc number = 20
Ca – Valency electrons = 2

(C) Lithium:
Lithium atomic number = 3
Li – Valency electron = 1

(d) Bromine gas (Br2):
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 7

(e) Calcium Chloride (CaCl2):
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 8

(f) Carbon dioxide (CO2):
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 9

Question 6.
Why do only valence electrons involve in bond formation? Why not electron of Inner shells? Explain.
Answer:

  1. When two atoms come sufficiently close together the valence electrons of each atom experience the attractive force of the nucleus in the other atom.
  2. The nucleus and the electrons ¡n the inner shell remain unaffected when atoms come close together.
  3. only the electrons in outermost shell of an atom get affected.
  4. Thus electrons in valence shell are responsible for the formation of bond between atoms.

Question 7.
List the factors that determine the type of bond that will be formed between two atoms.
Answer:
There are several factors that determine the type of bond which will be formed between two atoms. They are

  • The force of attraction or repulsion between the electrons and protons.
  • Number of valence electrons present in the valence shell of the atom.
  • Electronegative difference between the atoms.
    If the E.N difference between the two atoms is > 1.9, ionic bond is formed.
    If the E.N difference between the two atoms is <1.9, covalent bond is formed.
  • Atomic size
  • Ionisation potential
  • Electron affinity.

Question 8.
Represent the molecule H2O using Lewis notation.
Answer:
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 10

Question 9.
What is octet rule? How do you appreciate the role of the ‘Octet rule’ in explaining the chemical properties of elements?
Answer:
Octet rule: “The atoms of element tend to undergo chemical changes that help to leave their atoms with eight outer-shell electrons”.

→ Role of ‘Octet Rule :

  • ‘Octet rule’ helps to explain the chemical activities of atoms of many elements.
  • It explains why some elements are more reactive towards chemical reaction and some are not.
  • It can explain the high reactivity of Alkali, Alkaline earth metals.
  • It can also explain the high reactivity of halogens.

Question 10.
What is hybridization? Explain the formation of the following molecules using hybridization.
(a) BeCl2
(b) BF3
Answer:
(i) Hybridisation is a phenomenon of intermixing of atomic orbitals of almost equal energy which are present in the outer shells of the atom and their reshuffling or redistribution into the same number of orbitals but with equal properties like energy and shape.

(a) Formation of BeCl2 (Beryllium chloride) molecule:
(ii) The atomic number of Beryllium is 4
(iii) The electronic configuration of Beryllium atom in its ground state is 1s22s2.
(iv) The electronic configuration of Beryllium atom in its excited state is 1s2 2s12p1.
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 11
(i) In the excited Beryllium atom its ‘2s’ and ‘2px‘ orbitals intermix to give two equivalent ‘ sp ‘ hybrid orbitals.
(ii) The electronic configuration of Be is 1S22s12px1. It has one half-filled ‘p’ orbital.
(iii) The electronic configuration of 17 Cl is 1s2 2 s22p63s23px23py23pz1
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 12

(iv) The half-filled 3pz orbitals of two chlorine atoms overlap with ‘sp’ hybrid orbitals of beryllium atom in their axes to form two σ sp-p bonds.
(v) BeCl2 molecule so formed has linear shape. The bond angle in BeCl2 is 180°.
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 13

(b) Formation of Boron Trifluoride BF3:
(i) The central atom in BF3 is boron.
(ii) The electronic configuration of boron atom in its excited state is 1s2 2s1 2p2
B(Z=5)is
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 14
(iii) In the excited boron atom 2s’ orbital and two ‘2p’ orbitals intermix to give three equivalent sp2 hybrid orbitals.
(iv) In the formation of BF3 molecule, three sp2 hybrid orbitals of boron overlap with half-filled 2pz orbitals of three fluorine atoms. in their axes to give three bonds.
(v) BF3 molecule so formed has trigonal planar structure.
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 15
(iv) The bond angle ¡n BF3 is 120°.

Application of Concepts

Question 1.
Explain the formation of sodium chloride and calcium oxide on the basis of the concept of electron transfer from one atom to another atom.
Answer:
1. Formation of sodium chloride (NaCl): Sodium chloride is formed from the elements sodium and chlorine. It can be explained as follows.
(a) Formation of Cation: When sodium atom loses one electron to get octet electron configuration it forms a cation (Na+) and gets electron configuration that of Neon (Ne)
Na → Na++e
E.C: 2, 8, 1 2,8 +1e

(b) Formation of anion: Chlorine has shortage of one electron to get octet in its valence shell. So It gains the election that was lost by Na to form anion (Cl) and gets electron configuration of Argon (Ar)
Cl + e → Cl
E.C: 2,8,7 2,8,8

(c) Formation of the compound NaCl from ions: Transfer of electrons take place between ‘Na’ and ‘Cl’ atoms while they form Na+ and Cl ions. These oppositely charged ions get attracted towards each other due to
electrostatic forces and form the compound sodium chloride (NaCl).
Na+(g) + Cl(g) → NaCl(s)

2. FormatIon of calcium oxide (CaO) Calcium Oxide Is formed from the elements Calcium and Oxygen. It can be explained as follows :
(a) Formation of Cation: When Calcium atom loses two electrons to get octet electronic configuration it forms a cation (Ca+2) and gets electron configuration of Argon (Ar)
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 16

(b) Formation of anion: Oxygen has shortage of two electrons to get octet in its valence shell. So it gains the electrons that were lost by Ca to form anion (O-2) and gets electron configuration of Neon (Ne)
O + 2e → O-2
E.C: 2, 6 → 2,8

(c) Formation of the compound CaO from ions : Transfer of electrons between ‘Ca’ and ‘O’ atoms takes place while they form Ca+2 and O-2 ions. These oppositely charged ions get attracted towards each other due to
electrostatic forces and form the compound calcium oxide (CaO).
Ca+2(g) + O-2(g) →CaO(s)

Question 2.
A, B and C are three elements with atomic numbers 6, 11. and 17 respectively.
(i) Which of these cannot form Ionic bond? Why?
(ii) Which of these cannot form covalent bond? Why?
(iii) Which of these can form Ionic as well as covalent bonds?
Answer:
(i) ‘A’ cannot form ionic bond. Its valence electrons are 4. It is difficult to lose or gain 4e to get octet configuration. So it forms covalent bond [Z of A is 6 so it is carbon (c)].

(ii) ‘B’ cannot form covalent bond. Its valence electrons are 1 only. So it is easy to donate the electron for other atom and become an ion. So it can form ionic bond [Z of B is 11, so it is sodium (Na)].

(iii) Element C can form ionic as well as covalent bonds. The element with atomic number 17 is Cl. It is able to participate with Na in ionic bond forming NaCl and with hydrogen in HCl molecule as covalent bond.

Question 4.
How Lewis dot structure helps in understanding bond formation between atoms?
Answer:

  1. The valence electrons in an atom are represented by putting dots (.)on the symbol of the element, one dot for each valence electron.
  2. By knowing the valence electrons of two different atoms by Lewis dot structure, we can understand which type of bond is going to establish between them and forms corresponding molecule.

Question 5.
Explain the formation of the following molecules using valence bond theory.
(a) N2 molecule,
(b) O2 molecule
Answer:
(a) Formation of N2 molecule:
7N has electronic configuration 1s2 2s2 2px1 2py12pz1 Suppose that ‘px’orbital of one ‘N’ atom overlaps the ‘px’ orbital of the other ‘N’ atom giving σpx-px bond along the inter-nuclear axis. The py and pz orbitals of one ‘N’ atom overlap the py and pz orbital of other ‘N’ atom laterally, respectively perpendicular to internuclear axis giving 2π py-py and pz – pz bonds. Therefore N2 molecule has a triple bond between two nitrogen atoms. (N ≡ N)
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 17

b) Formation of O2 molecule:

  • 8O has electronic configuration 1s22s22px22py12pz1
  • If the ‘py’ orbital of one ‘O’ atom overlaps the ‘py’ orbital of other ‘O’ atom along the internuclear axis, a sigma py– py bond ((σpy- py) is formed.
  • pz orbital of one ‘O’ atom overlaps the pz orbital of other ‘O’ atom laterally, perpendicular to the inter-nuclear axis giving a Πpz-pz bond.
  • O2 molecule has a double bond between two oxygen atoms. (O=O)

TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 18

Question 6.
Predict the reasons for low melting points for covalent compounds when compared with ionic compounds.
Answer:

  1. The melting point is low due to the weak Vander Waal’s forces of attraction between the covalent molecules.
  2. The force of attraction between the molecules of a covalent compound is very weak.
  3. Only a small amount of heat energy is required to break these weak molecular forces, due to which covalent compounds have low melting points and low boiling points. :
  4. But some of the covalent solids like diamond and graphite have, however very high melting points and boiling points.

Higher Order Thinking questions

Question 1.
Two chemical reactions are described below.
(i) Nitrogen and hydrogen react to form ammonia.
(ii) Carbon and hydrogen bond together to form a molecule of methane (CH4)
For each reaction, give
a) The valence of each of the atoms involved in the reaction.
b) The Lewis structure of the product that ¡s formed.
Answer:
i) Nitrogen reacts with hydrogen to form Ammonia. The reaction is
N2 + 3H2 → 2NH3
a) The valency of each atom involved in the reaction.
Valence of Nitrogen = 3
Valence of Hydrogen = 1

(b) The Lewis structure of the product that is formed
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 19

(i) Carbon reacts with hydrogen to form a molecule of methane. The reaction is
C + 2H2 → CH4
(a) The valency of each atom involved in the reaction
Valence of Carbon = 4
Valence of Hydrogen = 1

(b) The Lewis structure of the product
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 20

Multiple choice questions

Question 1.
Which one of the following four elements is more electronegative? [ ]
(a) Sodium
(b) Oxygen
(c) Magnesium
(d) Calcium
Answer:
(b) Oxygen

Question 2.
An element, 11X23 forms an ionic compound with another element ‘Y’. Then the charge on the ior formed by X is [ ]
(a) +1
(b)+2
(c) -1
(d)-2
Answer:
(a) +1

Question 3.
An element ‘A’ forms a chloride ACl4. The number electrons in the valence shell of ‘A’ is [ ]
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 4.
The inert gas element which does not have octet electronic configuration in its outermost orbit is [ ]
(a) Helium
(b) Argon
(c) Krypton
(d) Radon
Answer:
(a) Helium

Question 5.
Number of covalent bonds in methane molecule [ ]
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 6.
The concept hybridisation of orbitals of an atoms was introduced by  [ ]
(a) Lives pooling
(b) Mosley
(c) Lewis
(d) Kossel
Answer:
(a) Lives pooling

Question 7.
The value of bond angle in Berileum chloride molecule is [ ]
(a) 180°
(b) 120°
(c) 110°
(d) 104°.31′
Answer:
(a) 180°

Suggested Projects

Question 1.
Collect the information about properties and uses of covalent compounds and prepare a report?
Answer:
A. Properties of covalent compounds:

  1. Covalent compounds are usually liquids or gases, only some of them are solids.
  2. They are usually liquids or gases due to the weak force of attraction between their molecules.
  3. They have usually low melting and low boiling points.
  4. They are usually insoluble in water but they are soluble in organic solvents.
  5. They do not conduct electricity.
  6. They show molecular reactions.

Uses of covalent compounds:

  1. Covalent compounds form 99% of our body.
  2. Water is a covalent compound. We know many uses of water.
  3. Sugars, food substances, tea, and coffee are all covalent compounds.
  4. Air we breathe in contains covalent molecules of oxygen and nitrogen.
  5. Almost everything on earth other than most simple inorganic salts are covalent compounds.

TS 10th Class Physical Science Chemical Bonding Intext Questions

Page 150

Question 1.
How do they(elements)usually exist?
Answer:
Elements usually exist as group of atoms.

Question 2.
Do they exist as a single atom or as a group of atoms?
Answer:
Except inert elements, all others exist as group of atoms. Inert elements exist as single atoms.

Question 3.
Are there elements which exist as atoms?
Answer:
Inert elements exist freely as atoms.

Question 4.
Why do some elements exist as molecules and some as atoms?
Answer:
Inert elements exist as atoms as they won’t form any bond other atoms form bonds and exist as molecules.

Question 5.
Why do some elements and compounds react vigorously while others are inert?
Answer:
Elements which do not have octet conflgurtion in their valence shelf react vigorously with other elements to form stable entities and which have octet configuration in their valence shell are chemically inert in nature.

Question 6.
Why is the chemical formula for water H20 and for sodium chloride NaCl, why not HO2 and NaCl2’?
Answer:
In water, oxygen atom bonds with two hydrogen atoms where as sodi + ion forms single bond with chloride in NaCl – ion.

Question 7.
Why do some atoms combine while others do not?
Answer:
Elements which do not have octet configuration in their valence shell combine with other elements and which have octet in their valence shell are chemically inert in nature.

Question 8.
Are elements and Compounds simply made up of separate atoms Individually arranged?
Answer:
No. They are arranged

Question 9.
Is there any attraction between atoms?
Answer:
Yes, there is attraction chemical bond.

Question 10.
What is that holdlng them together?
Answer:
Force of attraction called chemical bond.

Page 152

Question 11.
Why there is absorption of energy in certain chemical reactions and release of energy In other reactions?
Answer:
The absorption of energy in chemical reactions occurs when the reactants ‘‘ less chemical energy than the product where as release of energy in chemical reactions occurs when the reactants have higher chemical energy than the products.

Question 12.
Where the absorbed energy goes?
Answer:
The energy absorbed by the molecules makes electrons to reach excited s e and Increases kinetic energy of the molecule.

Question 13.
Is there any relation to energy and bond formation between atoms?
Answer:
The interacting energy is the potential energy between the atoms. It is negative if the atoms are bound and positive if they can move away from each other. The interaction energy is the integral of the force over the separation distance so these two quantities are directly related. The interaction energy is turning at the equilibrium position. This value of the energy Is called the bond energy and is the energy needed to separate completely to infinity (the work that needs to be done to overcome the attractive force.)

Question 14.
What could be the reason for the change in reactivity of elements?
Answer:
Number of valence electrons In the atoms of the element.

Question 15.
What could be the reason for the less reactivity of noble gases?
Answer:
All the noble gases have eight electrons in the outermost shell, except Helium (He). Thus they have no valency electrons and are less reactive or not at all reactive.

Page 155

Question 16.
What have you observed from tSe above conclusions about the main groups?
Answer:
Main group elements lose or gain electrons to get noble gas electronic configuration.

Question 17.
Why do atoms of elements try to combine and form molecules?
Answer:
To get stable electronic configuration In their valence shell.

Page 156

Question 18.
Is it accidental that IA to VIlA main group elements durIng chemical reactions get eight electrons In the outermost shells of their ions, similar to noble gas atoms?
Answer:
No. It cannot be simply accidental, Eight electrons in the outermost shell definitely gives stability to the ion or atom. Based on the above observations a statement known as The Octet Rule” Is framed.

Page 157

Question 19.
Explain the formation of ionic compounds NaCl, MgCl2, Na2O and AlCl3 through Lewis electron-dot symbols (formulae).
Answer:
(1) Lewis electron-dot symbol for NacI :
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 21
Formation of sodium chloride (NaCl) :
Sodium chloride Is formed from the elements sodium and chlorine. It can be explained as follows.
Na(s) + 1/2 Cl2(g) → Nacl(s)

Cation formation: When Sodium (Na) atom loses one electron to get octet electron configuration it forms a cation (Na+) and gets electron configuration that of Neon (Ne) atom.
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 22
Anion formation: Chlorine has shortage of one electron to get octet in its valence shell. So It gains the electron from Na atom to form anion and gets electron configuration as that of argon (Ar).
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 23
Formation of compound NaCl from its ions: Transfer of electrons between Na and cl atoms, results in the formation of Na+, and Cl- ions. These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound sodium chloride (NaCl).
Na+(g)+ Cl(g) → Na+(g)Cl(a) or NaCl

2. Lewis electron-dot symbol for MgCl2 :
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 24
Formation of magnesium chloride (MgCl2) :
Magnesium chloride is formed from the elements magnesium and chlorine. The bond formation MgCl2 in brief using chemical equation is as follows :
Mg(a)+Cl2(g) → MgCl2(a)
Cation Formation :
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 25
Anion formation :
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 26
The compound MgCl2 formation from its ions :
Mg2+ gets Ne configuration and Each Cl- gets Ar configuration.
Mg2+(g) + 2Cl(g) → MgCl2(a)
One ‘Mg’ atom transfers two electrons one each to two ‘Cl’ atoms and so formed Mg2+ and 2Cl- attract to form MgCl2.

3. Lewis electron-dot symbol for (Na2O) :
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 27
Formation of di-sodium monoxide (Na2O) :
Di-sodium monoxide formation can be explained as follows :
Cation formation (Na+ formation) :
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 28
Anion formation (O2-, the oxide formation) Electronic configuration
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 29
The compound Na20 formation from Its ions is as shown.
2Na+(g) + O2-(g) → Na2O(g)
Two ‘Na; atoms transfer one electrons each to one oxygen atom to form 1Na+ and 02-.
Each Na+ gets ‘Ne’ configuration and 02- gets ‘Ne’ configuration.
These ions (2Na+ and 02-) attract to form Na2O.

4. Lewis electron-dot symbol for (AlCl3) :
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 30
Formation of aluminium chloride (AlCl3) :
Aluminium chloride formation can be explained as follows :
Formation of aluminium ion (Al3+)0 the cation :
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 31
Formation of chloride ion (Cl-) the anIon :
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 32
Each aluminium atom loses three electrons and three chlorine atoms gain them, one electron each. The compound AlCl3 is formed from its component Ions by the electrostatic forces of attraction.
Al+ 3(g)+3Cl(g)

Page 160

Question 20.
How do cations and anions of an ionic compound exIst in its solid state?
Answer:
Cations and anions are surrounded themselves In three-dimensional lattices to give properly shaped crystals.

Question 21.
Do you think that pairs of Na+ Cl as units would be present in the solid crystal?
Answer:
No, electrostatic forces are non-directional. Therefore, it is not possible for one Na+ to be attracted by one Cl and vice-versa. Depending upon the size and charge of a particular ion, number of oppositely charged Ions get attracted by it, but, in a definite number. In sodium chloride crystal each Na is surrounded by 6 Cl and each Cl by six Na ions. Ionic compounds in the crystalline state consist of orderly arranged cations and anions held together by electrostatic forces of attractions in three dimensions

Page 161

Question 22.
Can you explain the reasons for all these?
Answer:
An ionic bond is formed between atoms of elements with electronegativity difference equal to or greater than 1.9.

Page 162

Question 23.
Can you say what type of bond exists between atoms of nitrogen molecules?
Answer:
Triple Bond

Page 164

Question 24.
What do you understand from bond lengths and bond energies?
Answer:
Bonds formed between two atoms in different molecules have different bond lengths and bond energies

Page 165

Question 25.
Are the values not different for the bonds between different types of atoms?
Answer:
Different for different molecules.

Page 167

Question 26.
What is the bond angle in a molecule?
Answer:
It is the angle subtended by two imaginary lines that pass from the nuclei of two atoms which form the covalent bonds with the central atom through the nucleus of the central atom at the central atom.

Page 169

Question 27.
How is HCl molecule formed?
Answer:
The ‘is’ orbital of ‘H’ atom containing unpaired electron overlaps with ‘3p’ orbital of chlorine atom containing unpaired electron of opposite spin.

TS 10th Class Physical Science Chemical Bonding Activities

Activity 1

Question 1.
Write the Lewis structure of the given elements ¡n the table. Also, consult the periodic table and fill in the group number of the element.
Answer:
TS 10th Class Physical Science Solutions Chapter 8 Chemical Bonding 33

Question 2.
Look at the periodic table. Do you see any relation between the number of valence electrons and group numbers’?
Answer:
For groups 2-6 the number of valence electrons is its group number. Group 1 has one outer electron, group 2 has two, where groups 13-17 number of valence electrons is (Group number-10). 3, 2, 1 respectively. (ie, 8-5=3;8-6=2;8-7= 1)

Question 3.
What did you notice in Lewis dot structure of noble gases and electronic configurations of the atoms of these elements shown in table – 1.
Answer:
It was found that the elements get octet or ns2 np6 configuration except helium, duplet.

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a)

I.
Question 1.
A p.d.f of a discrete random variable h zero except at the point to x = 0, 1, 2. At these points it has the value P(0) = 3c3 P(1)= 4c – 10c2, P(2)= 5c – 1 for same c > 0. Find the value of c.
Solution:
Given P(0) = 3c3,
P(1) = 4c – 10c2,
P(2) = 5c – 1
we have \(\sum_{i=0,1,2}\) P(i) = 1
⇒ P(0) + P(1) + P(2) = 1
⇒ 3c3 + 4c – 10c2 + 5c – 1 = 1
⇒ 3c3 – 10c2 + 9c – 2 = 0
c = 1 satisfy the equation by inspection and for c = 1, we have
P(0) = 3,
P(1) = 4 – 10 = 6,
P(2) = 4
Hence c = 1 does not satisfy 0 ≤ P (E) ≤ 1.
So we try for other solutions by synthetic division method.

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a) 1

∴ 3c2 – 7c + 2 = 0
⇒ 3c2 – Gc – c . 2 0
⇒ 3c (c – 2) – 1 (c – 2 )= 0
(3c – 1) (c – 2 ) = 0
c – 2 = 0
⇒ c = 2 is not admissible;
∴ c = \(\frac{1}{3}\) suit the solutions P (0), P (1) and P (2)
∴ c = \(\frac{1}{3}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a)

Question 2.
Find the constant c, so that F(x)= c\(\frac{2}{3}\), x = 1, 2, 3, …………… is the p.d.f. of a discrete random variable X.
Solution:
Since F (x) is the p.d.f. oF discrete random variable x.
we have \(\sum_{x=1}^{\infty}\) F(x) = 1

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a) 2

Question 3.

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a) 3

is the probability distribution of a random variable X. Find the value of k and the variance of X.
Solution:
Sum of the probability = 1
⇒ 0.1 + k + 0.2 + 2k + 0.3 + k = 1
⇒ 4k + 0.6 = 1
⇒ 4k = 0.4
⇒ k = 0.1

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a) 4

Mean = Σxi P (X = xi)
= (- 2 ) × 0.1 + (- 1 ) × 0.1 + 0 × 0.2 + 1 × 0.2 + 2 × 0.3 + 3 × 0.1 = 0.8
Variance σ2 = Σx2 P (X = x ) – μ2
= ( – 2 )2 × 0.1 + ( – 1)2 × 0.1 + 02 × 0.2 + 12 × 0.2 + 22 × 0.3 + 32 × 0.1 – ( 0.8 )2 = 2.16.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a)

Question 4.

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a) 5

is the probability distribution of a random variable X. Find the value of k and the variance of X.
Solution:

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a) 6

Question 5.
A random variable X has the following probability distribution.

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a) 7

Find i) k ii) the mean and iii) P(0 < X < 5).
Solution:
Sum of the probability = 1
⇒ 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
⇒ 10k2 + 9k – 1 = 0
⇒ 10k2+ 10k – k – 1 = 0
⇒ 10k(k + 1) -1 (k + 1) = 0
⇒ (10k – 1) (k + 1) = 0
⇒ k = \(\frac{1}{10}\); (k = – 1 is not admissible since probability is non-negative).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a)

II.
Question 1.
The range of a random variable X Is (0, 1, 2). Given that P (X = 0) = 3c3,
P (X = 1) = 4c – 10c2, P (X = 2) = 5c – 1.
Find (i) the value of c
ii) P(X < 1), P (1 < X ≤ 2) and P (0 < X ≤ 3).
Solution:
i) Sum of the probabilities = 1
⇒ P( X = 0) + P (X = 1) + P (X = 2) = 1
⇒ 3c3 + 4c – 10c2 + 5c – 1 = 1
⇒ 3c3 – 10c2 + 9c – 2 = 0
⇒ (3c – 1) (c – 1) (c – 2) = 0
⇒ c = \(\frac{1}{3}\) [∵ c ≠ 1, 2].

ii) P(X < 1) = P(X = 0)
= 3c3
= 3 × (\(\frac{1}{27}\)) = \(\frac{1}{9}\)

P(1 < X ≤ 2) = P(X = 2)
= 5c – 1
= 5(\(\frac{1}{27}\)) – 1
= \(\frac{2}{3}\).

P(0 < X ≤ 3) = P(X = 1) + P(X = 2)
= 4c – 10c2 + 5c – 1
= – 10c2 + 9c – 1
= \(-\frac{10}{9}+\frac{9}{3}-1=\frac{8}{9}\).

Question 2.
The range of a random variable X is {1, 2, 3, …………} and P (X = k) = \(\frac{c^{\mathbf{k}}}{\mathbf{k} !}\). Find the value of c and P (0 < X < 3).
Solution:
∵ Σ P(X = k) = 1

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(a) 8

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 9 Probability Ex 9(c) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

I.
Question 1.
Three screws are drawn at random from a lot of 50 screws, 5 of which are defective. Find the probability of the event that all 3 screws are non-defective assuming that the drawing is
a) with replacement
b) without replacement
Solution:
Total number of screws = 50
Number of defective screws = 5
Number of non-defective screws = 45
∴ n (S) = 50C3
a) With replacement:
P(E) = \(\frac{45}{50} \times \frac{45}{50} \times \frac{45}{50}=\frac{9}{10} \times \frac{9}{10} \times \frac{9}{10}\)
= \(\left(\frac{9}{10}\right)^3\)

b) Without replacement:
P(E) = \(\frac{45}{50} \times \frac{44}{49} \times \frac{43}{48}=\frac{1419}{1960}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 2.
If A, B, C are three independent events of an experiment such that
p(A ∩ B ∩ C) = \(\frac{1}{4}\), P\(\left(A^c \cap B \cap C^c\right)\) = \(\frac{1}{8}\), P(C) = \(\frac{1}{4}\) then find P(A), P(B) and P(C).
Solution:

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 1

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 2

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 3.
There are 3 black and 4 white balls in one bag, 4 black and 3 white balls In the second bag. A die is rolled and the first bag is selected if the die shows up 1 or 3, and the second bag for the rest Find the probability of drawing a black ball from the bag thus selected.
Solution:
Let E1, E2 be the events of selecting first and second bag respectively.
P(E1) = \(\frac{2}{6}\)
P(E2) = 1 – \(\frac{2}{6}\)
= \(\frac{4}{6}\)
Let B be the event of getting black ball from the selected bag.
\(P\left(\frac{B}{E_1}\right)=\frac{3}{7}\),
\(P\left(\frac{B}{E_2}\right)=\frac{4}{7}\)
∴ Required probability = P(E1 ∩ B) + P(E2 ∩ B)
= P(E1) . \(P\left(\frac{B}{E_1}\right)\) + P(E2). \(P\left(\frac{B}{E_2}\right)\)
= \(\frac{2}{6} \times \frac{3}{7}+\frac{4}{6} \times \frac{4}{7}\)
= \(\frac{6}{42}+\frac{16}{42}=\frac{22}{42}=\frac{11}{21}\)
Probability of drawing a black ball from the bag selected = \(\frac{11}{21}\).

Question 4.
A, B, C are aiming to shoot a balloon. A will succeed 4 times out of 5 attempts. The chance of B to shoot the balloon is 3 out of 4 and that of C is 2 out of 3. If the three aim the balloon simultaneously, then find the probability that atleasi two of them hit the balloon.
Solution:
Given that
P(A) = \(\frac{4}{5}\); P(B) = \(\frac{3}{4}\); P(C) = \(\frac{2}{3}\)
At least two of them will hit the balloon means if A, B hits the balloon then C will not hit or A, C hits the balloon then B will not hit or B, C hits the balloon then A will not hit or all the three will hit the balloon.
Required probability,

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 3

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 5.
If A, B are two events, then show that P(A/B) P(B) + P(A/\(\mathbf{B}^c\)) P(\(\mathbf{B}^c\)) = P(A).
Solution:

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 4

Question 6.
A pair of dice are roiled. What is the probability that they sum to 7 given that neither dic shows a 2?
Solution:
Given that neither die shows a ‘2’, when the two dice are rolled, the number of sample points.
n(S) = 36 – 11 = 25
(Faces showing ‘2’ will be excluded)
If E is the event of not getting ‘2’ on both the dice and sum is 7, then
E = {( 1, 6), (3, 4), (4, 3), (6, 1)}
∴ n(E) = 4
∴ Required probability = \(\frac{4}{25}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 7.
A pair of dice are rolled. What is the probability that neither die shows a 2 gIven that they sum to 7?
Solution:
Since S consists of points whose sum of points on faces is 7.
S = {(1, 6) (2, 5), (3, 4), (4, 3),(5, 2), (6, 1)}
∴ n(S) = 6
Let E be the event of a die not showing ‘2’. then
E = { (1, 6), (3, 4), (4, 3), (6, 1)}
n(E) = 4
∴ Required probability = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{4}{6}=\frac{2}{3}\).

Question 8.
If A, B are any two events in an experiment, and P (B) ≠ 1. Show that
P(A/\(\mathbf{B}^c\)) = \(\frac{P(A)-P(A \cap B)}{1-\mathbf{P}(\mathbf{B})}\)
Solution:
\(P\left(\frac{A}{B^{\prime}}\right)=\frac{P(A \cap \bar{B})}{P(\bar{B})}\)
= \(\frac{P(A)-P(A \cap B)}{1-P(B)}\)

Question 9.
An urn contains 12 red balls and 12 green balls. Suppose two balls are drawn one after another wlt.hout replacement Find the probability that the second ball drawn; is green, given that the first ball drawn is red.
Solution:
Let E1 be the event of drawing a red ball
n (S) = Total number of balls = 24
∴ n(E1) = 12C1 = 12
∴ P(E) = \(\frac{\mathrm{n}\left(\mathrm{E}_1\right)}{\mathrm{n}(\mathrm{S})}=\frac{12}{24}=\frac{1}{2}\)
Now there are 23 balls remaining.
∴ n(S) = 23
Let \(\frac{E_2}{E_1}\) be the event of drawing a green ball in the second attempt.
∴ n(\(\frac{E_2}{E_1}\)) = 12C1 = 12
∴ P(\(\frac{E_2}{E_1}\)) = \(\frac{12}{23}\)
P(E1 ∩ E2) = P(E1) . P(\(\frac{E_2}{E_1}\))
= \(\frac{1}{2} \cdot \frac{12}{23}=\frac{6}{23}\)
Hence the probability that the second ball drawn is green, given that the first ball drawn is red is \(\frac{6}{23}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 10.
A single die is rolled twice in succession. What is the probability that the number on the second toss is greater than that on the first rolling?
Solution:
n(S) = 36
Let E be the event that the number on the second toss is greater than that on the first rolling when a single die is rolled twice in succession.
Then E = {(1, 2), ( 1, 3), ( 1, 4),(1, 5), (1, 6 ), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)}
n(E) = 15
Hence the probability oF required event
P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{15}{36}\).

Question 11.
If one card is drawn at random from a pack of cards then show that the event of getting an ace and getting a heart are independent events.
Solution:
Let E1 be the event of getting an ace and E2 be the event of getting a heart.
P(E1) = \(\frac{4}{52}\)
P(E2) = \(\frac{13}{52}\)
P(E1) . P(E2) = \(\frac{4}{52} \times \frac{13}{52}=\frac{1}{52}\)
Common to E1 and E2 is an ace from hearts
∴ P(E1 ∩ E2) = \(\frac{1}{52}\)
∴ P(E1 ∩ E2) = P ( E1) . P (E2)
∴ E1 and E2 are independent events.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 12.
The probability that a boy A will get a scholarship is 0.9 and that another boy B will get is 0.8. What is the probability that atleast one of them will get the scholarship?
Solution:
Let E1 be the event that A will get scholarship and E2 be the event that B will get scholarship.
P(E1) = 0.9;
P(E2) = 0.8
Required probability
P(E1 ∪ E2) = P(E1) + P(E2) – P(E1 ∩ E2)
= P(E1) + P(E2) – P(E1) . P(E2)
= 0.9 + 0.8 – 0.72
= 1.7 – 0.72 = 0.98.

Question 13.
If A, B are two events with P(A ∪ B) = 0.65, P (A ∩ B) = 0.15, then find the value of P(\(A^{\mathrm{C}}\)) + P(\(B^{\mathrm{C}}\)).
Solution:
Given that P (A ∪ B) = 0.65;
P (A ∩ B ) = 0.15
From addition theorem on probabilities
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ P(A) + P(B) = P(A ∪ B) + P(A ∩ B)
= 0.65 + 0.15 = 0.8
∴ \(P(\bar{A})+P(\bar{B})\) = 1 – P(A) + 1 – P(B)
= 2 – [P(A) + P(B)]
= 2 – 0.8 = 1.2.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 14.
If A, B, C are independent events then show that A B and C are also independent events.
Solution:
P[(A ∪ B) ∩ C] = P(A ∩ C) ∪ (B ∩ C)]
= P(A ∩ C) + P(B ∩ C) – P[(A ∩ C) ∪ (B ∩ C)]
= P(A ∩ C) + P(B ∩ C) – P(A ∩ B ∩ C)
= P(A) . P(C) + P(B) . P(C) – P(A)P(B)P(C)
(∵ A, B, C are independent events)
= [P (A) + P ( B ) – P (A) . P (B)] P( C)
= [P(A) P(B) – P(A ∩ B) ] . P(C)
= P(A ∪ B) . P(C)
= (A ∪ B ) and C are independent events.

Question 15.
A, B are two independent events such that, the probability of both the events to occur is \(\frac{1}{6}\) and the probability of both the events do not occur \(\frac{1}{3}\). Find P(A).
Solution:
Given P(A ∩ B) = \(\frac{1}{6}\) …………..(1)
\(\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\frac{1}{3}\) ………….(2)
∴ \(\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\overline{\mathrm{B}})=\frac{1}{3}\)
⇒ [1 – P(A)] [1 – P(B)] = \(\frac{1}{3}\)
⇒ 1 – P(A) – P(B) + P(A) . P(B) = \(\frac{1}{3}\)
⇒ 1 – [P(A) + P(B)] + P(A ∩ B) = \(\frac{1}{3}\)
⇒ 1 – [P(A) + P(B)] + \(\frac{1}{6}\) = \(\frac{1}{3}\)
⇒ P(A) + P(B) = \(\frac{5}{6}\)
From (1) P(A) . P(B) = \(\frac{1}{6}\)
P(B) = \(\frac{1}{6 \mathrm{P}(\mathrm{A})}\)
Substituting this in (3)
P(A) + \(\frac{1}{6 \mathrm{P}(\mathrm{A})}\) = \(\frac{5}{6}\)
⇒ 6[P(A)]2 + 1 = 5P(A)
⇒ 6[P(A)]2 – 5P(A) + 1 = 0
⇒ [3P(A) – 1] [2P(A) – 1] = 0
⇒ P(A) = \(\frac{1}{3}\) or P(A) = \(\frac{1}{2}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 16.
A fair die is rolled. Consider the events A = {1, 3, 5}, B = {2, 3) and C = {2, 3, 4, 5}.
Find
i) P (A ∩ B), P (A ∪ B)
ii) P(A/B), P(J3/A)
iii) P(A/C), P(C/A)
iv) P(B/C), P(C/B)
Solution:
n(S) = 6
P(A) = \(\frac{3}{6}=\frac{1}{2}\);
P(B) = \(\frac{2}{6}=\frac{1}{3}\) and
P(C) = \(\frac{4}{6}=\frac{2}{3}\)

i) A ∩ B = {3}
∴ n(A ∩ B) = 1
∴ P(A ∩ B) = \(\frac{1}{6}\)
A ∪ B = {1, 2, 3, 4, 5}
n(A ∪ B) = 4,
∴ P(A ∪ B) = \(\frac{4}{6}=\frac{2}{3}\)

ii)

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 5

iii)

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 6

iv)

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 7

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 17.
If A, B, C are three events in a random experiment, prove the following
i) P(A/A) = 1
ii) P(Φ / A) = 0
iii) A ⊆ B ⇒ P(A/C) ≤ P(B/C)
iv) P(A – B) = P(A) – P(A ∩ B)
v) If A, B are mulnally exclusive and P(B) > 0 ihen P (A/B) = 0.
vi) If A, C are mutually exclusive then \(\mathbf{P}\left(\mathbf{A} / \mathbf{B}^{\mathrm{C}}\right)=\frac{\mathbf{P}(\mathbf{A})}{1-\mathbf{P}(\mathbf{B})}\) where P(B) ≠ 1
vii) If A and B are mutually exclusive and P (A ∪ B) ≠ 0 then
P(A/A ∪ B) = \(\frac{\mathbf{P}(\mathbf{A})}{\mathbf{P}(\mathbf{A})+\mathbf{P}(\mathbf{B})}\)
Solution:
i) A ∩ A is the common sample points of A and A is A.
\(P\left(\frac{A}{A}\right)=\frac{P(A \cap A)}{P(A)}=\frac{P(A)}{P(A)}\)
= 1 = R.H.S.

ii) \(P\left(\frac{\phi}{A}\right)=\frac{P(\phi \cap A)}{P(A)}\)
= \(\frac{\mathrm{P}(\phi)}{\mathrm{P}(\mathrm{A})}=\frac{\theta}{\mathrm{P}(\mathrm{A})}\)
= 0 (∵ P(Φ) = 0)

iii) A ⊆ B
⇒ A ∩ C ≤ B ∩ C
⇒ P(A ∩ C) ≤ P(B ∩ C)
\(P\left(\frac{A}{C}\right)=\frac{P(A \cap C)}{P(C)} \leq \frac{P(B \cap C)}{P(C)} \leq P\left(\frac{B}{C}\right)\)
∴ \(\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{C}}\right) \leq \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{C}}\right)\)

iv) P(A – B) P(A ∩ \(\))
= P(A ∩ (S – B))
= P[(A ∩ S) – (A ∩ R)]
= P(A) – P(A ∩ B)

v) Given A, B are mutually exclusive.
⇒ A ∩ B = Φ
⇒ P(A ∩ B) = P(Φ) = 0
∴ \(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{0}{P(B)}\) = 0
(∵ P(B) > 0).

vi) Given A, B are mutually exclusive.

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 8

vii) Given A, B are mutually exclusive.

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 9

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 18.
Suppose that a coin is tossed three times. Let event A be “getting three heads” and B be the event of “getting a head on the first toss”. Show that A and B are dependent events.
Solution:
When a coin is tossed 3 times,
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)
⇒ n(S) = 6, n(A) = 1,
P(A) = \(\frac{1}{6}\)
When event B occurred, the possibilities are HHH, FIHT, HTI-I, HTT in which HHH is favourable for the event \(\frac{A}{B}\).
P(\(\frac{A}{B}\)) = \(\frac{1}{4}\)
Hence P(A) ≠ P(\(\frac{A}{B}\))
∴ A, B are dependent events.

Question 19.
Suppose that an unbiased pair of dice is rolled. Let A denote the event that the same number shows on each die. Let B denote the event that the sum is greater than 7. Find
(i) P(\(\frac{A}{B}\))
(ii) P(\(\frac{B}{A}\))
Solution:
n(S) = 36
Let A be the event of getting the same number on two dice.
n(A) = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} = 6
P(A) = \(\frac{6}{36}\)
Let B be the event at getting the sum greater then 7.
n(B) = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)} = 15
P(B) = \(\frac{15}{36}\)
n(A ∩ B) = {(4, 4), (5, 5), (6, 6)} = 3
P(A ∩ B) = \(\frac{3}{36}\)

i) \(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}\)
= \(\frac{\frac{3}{36}}{\frac{15}{36}}=\frac{3}{15}=\frac{1}{5}\)

ii) \(P\left(\frac{B}{A}\right)=\frac{P(B \cap A)}{P(A)}\)
= \(\frac{\frac{3}{36}}{\frac{1}{6}}=\frac{3}{6}=\frac{1}{2}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 20.
Prove that A and B are independent events if and only if \(P\left(\frac{A}{B}\right)=\mathbf{P}\left(\frac{A}{B^C}\right)\).
Solution:
Let A and B are independent.

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 10

∴ L.H.S = R.H.S.

II.
Question 1.
Suppose A and B are independent events with P(A)= 0.6, P (B)= 0.7 thencompute
i) P(A ∩ B)
ii) P(A ∪ B)
iii) P(B/A)
iv) \(\mathbf{P}\left(\mathbf{A}^{\mathrm{C}} \cap \mathbf{B}^{\mathrm{C}}\right)\).
Solution:
i) Given P(A) = 0.6, P(B) = 0.7
and A, B are independent events.
i) P(A ∩ B) = P(A) . P(B)
= 0.6 × 0.7 = 0.42.

ii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.6 + 0.7 – 0.42
= 1.3 – 0.42 = 0.88.

iii) \(P\left(\frac{B}{A}\right)=\frac{P(B \cap A)}{P(A)}=\frac{0.42}{0.6}\) = 0.7

iv) \(P(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\overline{\mathrm{B}})\)
= [ 1 – P(A)] [1 – P(B)]
= ( 1 – 0.6) ( 1 – 0.7)
= (0.4) (0.3) = 0.12.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 2.
The probability that Australia wins a match against India in a ciickel gaine is given to be if India and Australia play 3 matches, what is the probability that
i) Australia will loose all the three matches?
ii) Australia will win atleast one match?
Solution:
Let E1, E2, E3 be the events that Australia wins a cricket match against India in the 1st, 2nd and 3rd matches respectively.
P(E1) = P(E2) = P(E3) =
\(\mathrm{P}\left(\overline{\mathrm{E}}_1\right)=\mathrm{P}\left(\overline{\mathrm{E}}_2\right)=\mathrm{P}\left(\overline{\mathrm{E}}_3\right)\)
= 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\).

i) Australia will loose all the three matches is \(\mathrm{P}\left(\overline{\mathrm{E}}_1 \cap \overline{\mathrm{E}}_2 \cap \overline{\mathrm{E}}_3\right)\)
= \(P\left(\bar{E}_1\right) P\left(\overline{\mathrm{E}}_2\right) P\left(\overline{\mathrm{E}}_3\right)\)
= \(\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3}=\frac{8}{27}\).

ii) Australia will win atleast one match is P(E1 ∪ E2 ∪ E3)
= 1 – P\(\left(\overline{\mathrm{E}}_1 \cap \overline{\mathrm{E}}_2 \cap \overline{\mathrm{E}}_3\right)\)
= 1 – \(\frac{8}{27}\) = \(\frac{19}{27}[latex].

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 3.
Three boxes numbered I, II, III contain the balls as Follows:

WhiteBlackRed
I123
II211
II453

 

One box is randomly selected and a ball is drawn from it. If the ball is red, then find the probability that it is from box II.
Solution:
Let E1, E2, E3 be the events of selecting 1st, 2nd and 3rd boxes.
P(E1) = P(E2) = P(E3) = [latex]\frac{1}{3}\)
Let us denote the event of drawing a red ball by R.

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 11

∴ The probability for the ball to he red that it is from box II is \(\frac{1}{4}\).

Question 4.
A person secures a job in a construction company in which the probability that the workers go ou strike is 0.65 and the probability that the construction job will be completed on time if there is no strike is 0.80. If the probability that the construction job will be completedon tinte even if there is a strike is 0.32, determine the probability that the construction job will be completed on time.
Solution:
Given that the probability that the workers go on strike = 0.65
∴ P(S) = 0.65
P\((\overline{\mathrm{S}})\) = 1 – P(S)
= 1 – 0.65 = 0.35.
The probability that the construction job will be completed on time, if there is no strike = 0.80.
\(\mathrm{P}\left(\frac{A}{\overline{\mathrm{S}}}\right)\) = 0.80
The probability that the onstruction job will be completed on time even if there is a strike = 0.32
P\(\left(\frac{A}{S}\right)\) = 0.32
∴ P(A) = P(A ∩ S) + P(A ∩ \(\overline{\mathrm{S}}\))
= P(S) . P\(\left(\frac{A}{S}\right)\) + P\((\overline{\mathrm{S}})\) . \(\mathrm{P}\left(\frac{A}{\overline{\mathrm{S}}}\right)\)
= 0.65 × 0.32 + 0.35 × 0.80 = 0.488.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 5.
For any two events A, B show that P(A ∩ B) – P(A) P(B) = \(P\left(A^C\right) P(B)-P\left(A^C \cap B\right)\)
= \(\mathbf{P}(\mathbf{A}) \mathbf{P}\left(\mathbf{B}^{\mathrm{C}}\right)-\mathbf{P}\left(\mathbf{A} \cap \mathbf{B}^{\mathrm{C}}\right)\)
Solution:
\(P\left(A^C\right) P(B)-P\left(A^C \cap B\right)\)
= [1 – P(A)] P(B) – P[(S – A) ∩ B]
= P(B) – P(A P(B) – P [B – (A ∩ B)]
= P(B) – P(A) P(B) – P(B) + P(A ∩ B)
= P(A ∩ B) – P(A)P(B) ……………………(1)
\(\mathbf{P}(\mathbf{A}) \mathbf{P}\left(\mathbf{B}^{\mathrm{C}}\right)-\mathbf{P}\left(\mathbf{A} \cap \mathbf{B}^{\mathrm{C}}\right)\)
= P(A)[1 – P(B)] – P(A ∩ (S – B))
= P(A) – P(A) P(B) – P[A – (A ∩ B)]
= P(A) – P(A) P(B) – P(A) + P(A ∩ B)
= P(A ∩ B) – P(A) P(B) ………….(2)
From (1) and (2) we get
P(A ∩ B) – P(A) P(B) = \(P\left(A^C\right) P(B)-P\left(A^C \cap B\right)\)
= \(\mathbf{P}(\mathbf{A}) \mathbf{P}\left(\mathbf{B}^{\mathrm{C}}\right)-\mathbf{P}\left(\mathbf{A} \cap \mathbf{B}^{\mathrm{C}}\right)\).

III.
Question 1.
Three urns have the following composition of balls.
Urn I: 1 white, 2 black
Urn II: 2 white 1 black
(ini III: 2 white, 2 black
One of the urns is selected at random and a ball is drawn. It turns out to he while. Find the probability that it came from uni III.
Solution:
Let E1, E2, E3 be the events of selecting Urns B1, B2, B3 respectively.
Then P(E1) = P( E2) = P(E3) = \(\frac{1}{3}\)
Let W denote the event of white ball chosen from the Urn III.
Then we have to find P\(\left(\frac{E_3}{\mathrm{~W}}\right)\)
Now
\(P\left(\frac{\mathrm{W}}{\mathrm{E}_1}\right)=\frac{1}{3}, \mathrm{P}\left(\frac{\mathrm{W}}{\mathrm{E}_2}\right)=\frac{2}{3}, \mathrm{P}\left(\frac{\mathrm{W}}{\mathrm{E}_3}\right)=\frac{2}{4}\)
∴ Probability for the ball to be white selected from Urn III by Bayes theorem is

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 12

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 2.
In a shooting lest the probability of A, B, C hitting the targets are \(\frac{1}{2}\), \(\frac{1}{3}\), \(\frac{1}{4}\) and respectively. If all of them fire at the same target, find the probability that (i) only one of them hits the target (ii) atleast one of them bits the target.
Solution:
Given P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{3}\) and P(C) = \(\frac{3}{4}\)
i) The probability that only one of them hits the target

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 13

Question 3.
In a certain college, 25% of the boys and 10% of the girls are studying mathematics. The girls constItute 60% of student strength. If a student selected at random is found studying mathematics, find the probability that the student is a girl.
Solution:
Given girls constitute 60% of student strength, we have 40% students are boys probability of an event being girl.

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 14

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c)

Question 4.
A person is known to speak truth 2 out of 3 times. He throws a die and reports that it is 1. Find the probability that it is actually 1.
Solution:
Given that the probability of a person to speak truth = \(\frac{2}{3}\)
P(T) = \(\frac{2}{3}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(c) 15

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

These TS 10th Class Maths Chapter Wise Important Questions Chapter 8 Similar Triangles given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Previous Exams Questions

Question 1.
Construct a triangle of sides 4.2 cm, 5.1 cm and 6 cm. Then construct a triangle similar to it, whose sides are \(\frac{2}{3}\) of corresponding sides of the first triangle. (A.P. Mar. ’15)
Solution:
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 19

  1. Draw a triangle ABC, with sides AB = 4.2 cm, BC = 5.1 cm, CA = 6 cm.
  2. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
  3. Locate 3 points B1, B2, B3 on BX so that BB1 = B1B2 = B2B3
  4. Join B3, C and draw a line through B2 parallel to B3C intersecting BC and C1.
  5. Draw a line through C1 parallel to CA intersect AB at A1.
  6. ∆A1BC1 is required triangle.

Question 2.
Is a square similar to a rectangle ? Justify your answer. (A.P. Mar.’15)
Solution:
In a square and rectangle, the corresponding angles are equal. But the corresponding sides are not proportional.
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 20
∴ A square and a rectangle are not similar.

Question 3.
In a ∆DEF, A, B and C are mid points of EF, FD and DE respectively. If the area of ∆DEF is 14.4 cm2 then find the area of ∆ABC. (T.S. Mar. ’15)
Solution:
Area of ∆ABC = 1/4 of area of ∆DEF
= 1/4 (14.4) = 3.6 cm2

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Question 4.
Observe the below diagram and find the values of x and y. (T.S. Mar. ’15)
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 21
Solution:
∆ABC ~ AEFC
\(\frac{\mathrm{x}}{6}\) = \(\frac{9}{15}\)
x = \(\frac{9 \times 6}{15}\) = \(\frac{18}{5}\) = 3.6 cm.
∆BDC ~ ABEF
\(\frac{\mathrm{x}}{9.6}\) = \(\frac{\mathrm{y}}{15}\)
⇒ y = \(\frac{3.6 \times 15}{9.6}\) = \(\frac{45}{8}\) = 5.625 cm

Question 5.
Observe the figure given below in ∆FQR if XY // PQ, \(\frac{\mathrm{PX}}{\mathrm{XR}}\) = \(\frac{5}{3}\) and QR = 7.2. Then find the length of RY. (T.S. Mar. ’15)
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 22
Solution:
XY || PQ
⇒ \(\frac{\mathrm{PX}}{\mathrm{XR}}\) = \(\frac{\mathrm{QY}}{\mathrm{YR}}\)
⇒ \(\frac{5}{3}\) + 1 = \(\frac{\mathrm{QY}}{\mathrm{RY}}\) + 1
⇒ \(\frac{8}{3}\) = \(\frac{\mathrm{QR}}{\mathrm{RY}}\)
⇒ RY = 7.2 × \(\frac{3}{8}\) = 2.7 cm.

Question 6.
Find the value of ‘x’ in the given figure where ∆ ABC ~ ∆ ADE. (T.S. Mar. ’15)
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 23
Solution:
∆ABC ~ ∆ADE
∴ \(\frac{\mathrm{AB}}{\mathrm{AD}}\) = \(\frac{\mathrm{BC}}{\mathrm{DE}}\) = \(\frac{\mathrm{AC}}{\mathrm{AE}}\) ;
BC = x, DE = 5, AE = 3, AC = 9
By substituting \(\frac{\mathrm{BC}}{\mathrm{DE}}\) = \(\frac{\mathrm{AC}}{\mathrm{AE}}\)
∴ \(\frac{\mathrm{x}}{5}\) = \(\frac{9}{3}\) ⇒ x = \(\frac{9 \times 5}{3}\) = 15
∴ x = 15 cm

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Question 7.
Construct a triangle of sides 5 cm, 6 cm, and 7 cm then construct a triangle similar to it, whose sides are \(\frac{2}{3}\) of the corresponding sides of the triangle. (T.S. Mar. ’15)
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 24
Solution:
Construction steps :

  1. Draw a triangle ∆ABC with sides AB = 5 cm, BC = 6 cm and CA = 7 cm.
  2. Draw a ray \(\overrightarrow{\mathrm{BX}}\) making an acute angle with BC on the side opposite to vertex A.
  3. Locate 3 points B1, B2, B3 on BX. So that BB1 = B1B2 = B2B3.
  4. Join B3, C and draw a line through B2 parallel to B3 C intersecting BC at C’.
  5. Draw a line through C’ parallel to CA intersect AB at A’.
  6. ∆A’B’C’ is required triangle.

Additional Questions

Question 1.
In a ∆ ABC, DE || BC and AD = \(\frac{1}{3}\) BD. If BC = 5.6 cm, find DE.
Solution:
Given DE || BC, AD = \(\frac{1}{3}\) BD, BC = 5.6 cm
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 1
AD = \(\frac{1}{3}\) BD ⇒ BD = 3 AD …………… (1)
From similar triangles, ABC and ADE
We have \(\frac{\mathrm{AD}}{\mathrm{AB}}\) = \(\frac{\mathrm{DE}}{\mathrm{BC}}\)
⇒ \(\frac{\mathrm{AD}}{\mathrm{AD}+\mathrm{DB}}\) = \(\frac{\mathrm{DE}}{5.6}\)
⇒ \(\frac{\mathrm{AD}}{\mathrm{AD}+3\mathrm{DB}}=\) = \(\frac{\mathrm{DE}}{5.6}\)
[∵ from (1) DB = BD = 3AD]
⇒ \(\frac{\mathrm{AD}}{4 \mathrm{AD}}\) = \(\frac{\mathrm{DE}}{5.6}\)
⇒ \(\frac{1}{4}\) = \(\frac{\mathrm{DE}}{5.6}\)
⇒ DE = \(\frac{5.6}}{4\)
∴ DE = 1.4 cm

Question 2.
In the adjacent figure ∆ABC ~ ∆AHK. If AK = 8 cm, BC = 4.5 cm and HK = 9 cm find AC.
Solution:
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 2
Given AK = 8 cm, BC 4.5 cm, HK from similar ∆les, ABC and AHK
We have \(\frac{\mathrm{AC}}{\mathrm{AK}}\) = \(\frac{\mathrm{BC}}{\mathrm{HK}}\)
AC = \(\frac{\mathrm{AK} \times \mathrm{BC}}{\mathrm{HK}}\)
= \(\frac{8 \times 4.5}{9}\)
= 8 × 0.5 = 4
∴ AC = 4 cm

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Question 3.
In the below given figure P and Q are points on the sides AB and AC respectively of ∆ABC such that AQ = 3 cm, QC = 5 cm and PQ || BC. Find the ratio of areas of ∆APQ and ∆ABC.
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 3
Solution:
Given AQ = 3 cm, QC = 5 cm and PQ || BC.
We know that by theorem,
Ratio of the areas of two similar triangles = Ratio of the squares of their corresponding sides
Now AC = AQ + QC = 3 + 5 = 8 cm
By Theorem \(\frac{\text { area of } \triangle \mathrm{APQ}}{\text { area of } \triangle \mathrm{ABC}}\) = \(\frac{(\mathrm{AQ})^2}{(\mathrm{AC})^2}\)
= \(\frac{3^2}{8^2}\)
= \(\frac{9}{64}\)
∴ Ratio of areas of ∆APQ and ∆ABC = 9 : 64

Question 4.
What value of (5) of x will make DE || AB, in the given figure ?
AD = 5x + 5, CD = x + 2 BE = 3x + 3, CE = x
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 4
Solution:
Given in ∆ ABC, DE || AB
AD = 5x + 5, CD = x + 2
BE = 3x + 3, CE = x
y Basic proportional theorem,
If DE || AB, then we have
\(\frac{\mathrm{CD}}{\mathrm{DA}}\) = \(\frac{\mathrm{CE}}{\mathrm{EB}}\)
⇒ \(\frac{x+2}{5 x+5}\) = \(\frac{x}{3 x+3}\)
⇒ (5x + 5) x = (x +2) (3x + 3)
⇒ 5x2 + 5x = 3x2 + 3x + 6x + 6
⇒ 5x2 – 3x2 + 5x – 3x – 6x – 6 = 0
⇒ 2x2 – 4x – 6 = 0
⇒ 2x2 – 6x + 2x – 6 = 0
⇒ 2x (x – 3) + 2(x – 3) = 0
⇒ (x – 3) + (2x + 2) = 0
⇒ x-3 = 0 or 2x + 2 = 0
⇒ x = 3 or 2x = -2
⇒ x = –\(\frac{2}{2}\) = -1
∴ The value of x = 3 will make DE || AB.

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Question 5.
∆ABC ~ ∆PQR and their areas are respectively 81 cm2 and 144 cm2. If QR = 16cm then find BC.
Solution:
\(\frac{\text { area of } \triangle \mathrm{ABC}}{\text { area of } \triangle \mathrm{DEF}}\) = \(\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)^2\)
⇒ \(\frac{81}{144}\) = \(\left(\frac{\mathrm{BC}}{16}\right)^2\)
⇒ \(\frac{9}{12}\) = \(\frac{\mathrm{BC}}{16}\)
⇒ BC = \(\frac{9}{12}\) × 16 = \(\frac{144}{12}\) = 12
∴ BC = 12 cm.

Question 6.
∆ABC ~ ∆DEF, BC = 5 cm, EF = 6 cm and area of ∆DEF = 72 cm2. Determine the area of ∆ABC.
Solution:
Given ∆ABC ~ ∆DEF
and BC = 5cm, EF = 6cm
Area of ∆DEF = 72 cm2
We know that areas of two similar triangles are in the ratios of the squares of the corresponding sides.
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 5

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Question 7.
A ladder 13m long reaches a window of building 12cm above the ground. Determine the distance of the foot of the ladder from the building.
Solution:
In ∆ ABC, B = ∠90°
⇒ AC2 = AB2 + BC2 (By Pythagoras theorem)
Let AC = 13m, AB = 12m, BC = ?
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 6
⇒ 132 = 122 + BC2
⇒ 169 = 144 + BC2
⇒ BC2 = 169 – 144 = 25
⇒ BC = \(\sqrt{25}\) = 5m
Hence, the foot of the ladder from the building is at a distance of 5m.

Question 8.
The hypotenuse of a right angled triangle is 3 m more than twice of the shortest side. If the third side is 1 m less than the hypotenuse find the sides of the triangle.
Solution:
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 7
Let the shortest side = x m = BC
Then hypotenuse = 2x + 3 m = AC
The third side = (2x + 3) – 1
= (2x + 2) m = AB
By phythagoras theorem, we have
AC2 = AB2 + BC2
⇒ (2x + 3)2 = (2x + 2)2 + x2
⇒ 4x2 + 9 + 12x = 4x2 + 4 + 8x + x2
⇒ x2 + 8x – 12x + 4 – 9 = 0
⇒ x2 – 4x – 5 = 0
⇒ x2 – 5x + x – 5 = 0
⇒ x (x – 5) + 1 (x – 5) = 0
⇒ (x – 5) (x + 1) = 0
⇒ x – 5 = 0 or x + 1 = 0
⇒ x = 5 or x = -1
But x can’t be negative x = 5
Hence, the sides of the triangle are
x, 2x + 3, 2x + 2
⇒ x = 5, 2 × 5 + 3, 2 × 5 + 2
= 5, 10 + 3, 10 + 2
i.e. 5, 13, 12

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Question 9.
A ladder 25 m long reaches a window which is 15 m above the ground an one side of the road. Keeping its foot at the same point, the ladder is turned to other side of the road to reach a window 20 m high. Find the width of the road.
Solution:
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 8
Let A and D be the windows on the either sides of the road
From phythagoras theorem,
AC2 = AB2 + BC2
⇒ 252 = 152 + BC2
⇒ BC2 = 252 – 152 = 625 – 225 = 400
BC = \(\sqrt{400}\) = 20m ……………. (1)
Also CD2 = CE2 + DE2
252 = CE2 + 202
CE2 = 252 – 202 = 625 – 400 = 225
CE = \(\sqrt{225}\) = 15m
Width of the road = BE = BC + CE
= 20 + 15
= 35 m

Question 10.
In the given figure below, If AD ⊥ BC, prove that AB2 – BD2 = AC2 – CD2.
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 9
Solution:
Given in ∆ ABC, AD ⊥ BC
R.T.P : AB2 – BD2 = AC2 – CD2
Proof : ∆ ABD is a right angled triangle.
We have AB2 = AD2 + BD2 (By pythagaros theorem)
⇒ AB2 – BD2 = AD2 ……………… (1)
∆ ACD is a right angled triangle.
We have AC2 = AD2 + CD2
⇒ AC2 – CD2 = AD2 ………………. (2)
From (1) and (2), we have
AB2 – BD2 = AC2 – CD2

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Question 11.
In an equilateral triangle ABC, if AD is the altitude prove that 3AB2 = 4 AD2.
Solution:
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 10
Given ABC is an equilateral triangle.
Here, AD ⊥ BC and AB = BC = CA
In triangles ADB and ADC
∠ADB = ∠ADC = 90°
Hypotenuse AB = Hypotenuse AC
AD is common
∴ ∆ ADB ≅ ∆ ADC
∴ BD = DC
In ∆ ADB, ∠ADB = 90°(By Phythagoras theorem)
AB2 = AD2 + BD2
= AD2 + \(\left(\frac{\mathrm{BC}}{2}\right)^2\) (∵ BD = CD)
= AD2 + \(\frac{\mathrm{BC}^2}{4}\)
AB2 = \(\frac{4 A D^2+B C^2}{4}\)
⇒ 4AB2 = 4AD2 + BC2
⇒ 4AB2 – BC2 = 4AD2
⇒ 4AB2 – AB2 = 4AD2 (∵ BC = AB)
∴ 3AB2 = 4AD2

Question 12.
A wire attached to a vertical pole of height 15 m is 25 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?
Solution:
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 11
Let AB = Vertical pole
AC = Wire
B = Base
C = Stake
By Pythagoras theorem
AC2 = AB2 + BC2
252 = 152 + BC2
BC2 = 252 – 152
BC2 = 622 – 225 = 400
BC = \(\sqrt{400}\) = 20 m
∴ The stake should be driven 20 m for the base of the pole so that the wire will be taut.

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Question 13.
The larger of two complimentary angles is double the smaller. Find the angles.
Solution:
Let first complementary angle = x
Second complementary angle = 2x
Sum of the two complementary angles = 90°
∴ x + 2x = 90
3x = 90
x = \(\frac{90}{3}\)
x = 30
First angle = 30°, second angle = 60°.

Question 14.
In ∆ ABC, DE || BC and \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{5}\). If AE = 2.1 cm, then find AC.
Solution:
Here, Given DE || BC and \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{5}\)
AE = 2.1
∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{CE}}\)
\(\frac{3}{5}\) = \(\frac{2.1}{\mathrm{CE}}\)
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 12
CE = \(\frac{5 \times 2.1}{3}\)
CE = \(\frac{10.5}{3}\) = 3.5
CE = 3.5
Then AC = AE + CE
AC = 2.1 + 3.5
AC = 5.6

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Question 15.
What can you say about the ratio of areas of two similar triangles ?
Solution:
The ratio of areas of two similar triangles is( equal to the ratio of the squares of their corresponding sides.
Here
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 13

Question 16.
Construct an isosceles triangle whose base is 8 cm and altitude is 4 cm. Then, draw another similar triangle whose side are 1 \(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Solution:
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 14

  1. Draw a lines segment BC = 8 cm
  2. Draw the perpendicular bisector PQ of BC intersecting BC at ‘O’
  3. Mark a point ‘A’ on PQ such that \(\overline{\mathrm{OA}}\) = 4 cm
  4. Join AB and AC to the isosceles triangle ABC
  5. Extend BC on either side O’C’ = 1 \(\frac{1}{2}\) times; BC = 12 times.
  6. Similarly extend OA So that OA’ = 1 \(\frac{1}{2}\) times A = \(\frac{12}{2}\) = 6
  7. Join A’ B’ and A’ C’
  8. Now A’ B’ C’ – The required triangle

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Question 17.
Give two different examples of pair of 3 similar figures and non similar figures.
Solution:
Similar figures :

  1. Any two circles
  2. Any two squares

Non-similar figures :

  1. A square and a rhombus
  2. A square and a rectangle

Question 18.
∆ ABC ~ ∆ DEF and their areas are respectively 64 cm2 and 121 cm2. If EF = 15.4 cm then find BC.
Solution:
Given area of ∆ABC = 64 cm2
area of ∆DEF = 121 cm2
EF = 15.4 cm, BC = ?
We know that \(\frac{\text { area of } \triangle \mathrm{ABC}}{\text { area of } \triangle \mathrm{DEF}}\)=\(\frac{\mathrm{BC}^2}{\mathrm{EF}^2}\)
⇒ \(\frac{64}{121}\) = \(\frac{\mathrm{BC}^2}{(15.4)^2}\)
⇒ BC2 = \(\frac{64}{121}\) × (15.4)2
= \(\frac{64}{121}\) × 15.4 × 15.4
BC2 = 125.44
∴ BC = \(\sqrt{125.44}\) = 11.2 cm.

Question 19.
In D ABC, DE || BC and \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{5}\) ; AC = 5.6 find AE.
Solution:
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 15
But \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{5}\) , So \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = \(\frac{3}{5}\)
given AC = 5.6
from (1) \(\frac{3}{5}\) = \(\frac{\mathrm{AE}}{\mathrm{AC – AE}}\)
\(\frac{3}{5}\) = \(\frac{\mathrm{AE}}{5.6-\mathrm{AE}}\) (By cross-multiplication)
5 AE = 3 (5.6 – AE)
5 AE = 3 × 5.6 – 3AE
5AE + 3 AE = 3 × 5.6
8AE = 16.8
AE = \(\frac{16.8}{8}\) = 2.1 cm

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Question 20.
State and prove basic proportionality theorem.
Solution:
Preposition : If a line is drawn parallel to one side of the triangle to intersect the other sides in distinct points, then the other two sides are divided in the same ratio.

Given : A triangle ABC in which DE || BC and DE intersects AB in D and AC in E.
To Prove : \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 16
Construction : Join BE, CD and draw EF ⊥ AB. DG ⊥ AC.
Proof: In ∆ EAD and ∆ EDB, Here as EF ⊥ AB
therefore EF is the height for both of triangles EAD and EDB.
Now, Area of ∆ EAD = \(\frac{1}{2}\) (base × height)
= \(\frac{1}{2}\) × AD × EF
Area of EDB = \(\frac{1}{2}\) (base × height)
= \(\frac{1}{2}\) × DB × EF
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 17
∴ ∆ DBF, ECD are on the same base DE and between the same parallels DE || BC,
We have area of ∆DBE = area of ∆ECD
Hence (1) = (2)
i.e, \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\) (Q.E.D)

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Question 21.
Construct a triangle of sides 4 cm, 5 cm and 6 cm then construct a triangle. Similar to it. Whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
Solution:
TS 10th Class Maths Important Questions Chapter 8 Similar Triangles 18
Steps of construction :

  1. Draw ∆ ABC with AB = 4 cm, BC = 5 cm and CA = 6 cm.
  2. Draw a Ray \(\overrightarrow{\mathrm{BX}}\) making an acute angle with BC on the side opposite to vertex A.
  3. Mark three points B1, B2 and B3 on \(\overrightarrow{\mathrm{BX}}\) Such that BB1 = B1B2 = B2B3.
  4. Join B3, C.
  5. Draw a line parallel B3, C through B2 meeting BA at A’.
  6. ∆ A’ BC’ is required triangle.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Telangana TSBIE TS Inter 1st Year English Study Material 2nd Lesson A Red Red Rose Textbook Questions and Answers.

TS Inter 1st Year English Study Material 2nd Lesson A Red Red Rose

Annotations (Section – A, Q.No. 2, Marks : 4)

Question 1.
O my Luve’s like a red, red rose.
That’s newly sprung in June.
Answer:
Introduction:
This couplet is taken from the poem, A Red Red Rose written by Robert Burns. It is one of the best lyrics of English poetry. It blends the eternity of love with the mortality of life.

Context & Explanation:
The poet begins by using a simile to compare his love to a rose. In other words, his love is like a flower that has just bloomed in June. His love is fresh and is bursting with life. His feelings are very profound.

Critical Comment:
It is an address to the speaker’s lover to whom he swears eternal love and loyalty.

కవి పరిచయం :
రాబర్ట్ బర్న్స్ అంత్యానుప్రాసయుక్తముగా వ్రాసిన ‘ఒక ఎర్ర ఎర్ర గులాబి’ పద్యం నుండి ఈ పంక్తులు తీసుకోబడినవి. ఇతను 18వ శతాబ్దపు స్కాటిష్ కవి. ఈ పద్యంలో ఒక యువకుడి భావోద్వేగ తీవ్రత వర్ణింపబడినది.

సందర్భం :
ఇది పద్యము యొక్క ప్రారంభ ద్విపద. ఒక యువప్రేమికుడు తన ప్రియురాలికి తన హృదయాన్ని వివరిస్తున్నాడు. అతను తన ప్రేమ ఎర్ర గులాబి అంత అందంగాను మరియు కోమలంగాను ఉందని ప్రకటించుకుంటున్నాడు. తన ప్రేమ జూన్లో వికసించిన గులాబీ అంత తాజాగా ఉందని. గులాబి పువ్వు విశ్వవ్యాప్తంగా ఆమోదించబడిన ప్రేమ చిహ్నము. గులాబి యొక్క సౌకుమార్యము మరియు సౌందర్యము ప్రేమ లక్షణాలను ప్రముఖంగా ప్రతిబింబిస్తాయి. ఉపమానములు కవిత యొక్క కళాసంపదను పెంచుతాయి. వివరణ : ఇక్కడ ఒక యువ ప్రేమికుడు తన ప్రియురాలికి తన హృదయాన్ని వివరిస్తున్నాడు.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Question 2.
O my Luve’s like the melodie
That’s sweetly pla’d in tune.
Answer:
Introduction:
This couplet is taken from the beautiful lyric A Red Red Rose, written by Robert Burns. It is one of the best lyrics of English poetry. It blends the eternity of love with the mortality of life.

Context & Explanation:
The poet compares his love to a melody that is sweetly played in tune. His love is a song that is sung just so right in fact that it’s kind of sweet. His feelings are very profound.

Critical Comment:
Here, the poet compares his beloved to a sweet melody which is nice to hear.

కవి పరిచయం :
రాబర్ట్ బర్న్స్ అంత్యానుప్రాసయుక్తముగా వ్రాసిన ‘ఒక ఎర్ర ఎర్ర గులాబి’ పద్యం నుండి ఈ పంక్తులు తీసుకోబడినవి. ఇతను 18వ శతాబ్దపు స్కాటిష్ కవి. ఈ పద్యంలో ఒక యువకుడి భావోద్వేగ తీవ్రత వర్ణింపబడినది.

సందర్భం :
‘ఒక ఎర్ర ఎర్ర గులాబి’ ఒక మధుర ప్రేమ గేయము. లోతైన ప్రేమలో పడ్డ ఒక యువకుడు తన ప్రియురాలికి తన హృదయాన్ని చిత్రించుతున్నాడు. ఆమె పట్ల తన ప్రేమ ఎంత గాఢమైనదో తెలపాలని కోరుకుంటున్నాడు. ఆ బలమైన కోరిక అతనిని ఉపమానముల వాడుక వైపు నడిపిస్తుంది. తన ప్రేమ ఎర్ర గులాబి లాంటిది అంటాడు అతను. ఆయన ఇంకా అదనంగా (ఇక్కడ) తన ప్రేమ తీయగా పాడబడిన మధుర గీతిక అంటున్నారు. ప్రేమకు, గులాబీలకు మరియు సంగీతానికి మధ్య బంధం బలమైనది. అది ఇక్కడ మరింత బలోపేతం చేయబడింది.

వివరణ :
ఇక్కడ కవి తన ప్రేమ తియ్యగా పాడబడిన మధురగీతికగా పోలుస్తున్నాడు.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Question 3.
And I will Luve thee still, my dear,
Till a’ the seas gang dry :
Answer:
Introduction:
This couplet is extracted from the beautiful lyric A Red Red Rose, written by Robert Burns. It is one of the best lyrics of English poetry. It blends the eternity of love with the mortality of life.

Context & Explanation :
The speaker says he will love his bonny lass until all the seas dry up. The word ‘a’ is the shortened form of all. It is very common in Scots English. Gang does not refer to a group of people. It is an old word that means ‘go or a walk. The seas will probably never gang dry. So, the speaker seems to be saying that he will love his lass forever.

Critical Comment:
Here, the poet makes several promises to love his beloved forever.

కవి పరిచయం :
రాబర్ట్ బర్న్స్ అంత్యానుప్రాసయుక్తముగా వ్రాసిన ‘ఒక ఎర్ర ఎర్ర గులాబి’ పద్యం నుండి ఈ పంక్తులు తీసుకోబడినవి. ఇతను 18వ శతాబ్దపు స్కాటిష్ కవి. ఈ పద్యంలో ఒక యువకుడి భావోద్వేగ తీవ్రత వర్ణింపబడినది.

సందర్భం :
“ప్రేమ విచిత్రము” అన్నారు విజ్ఞులు. ప్రేమకు తర్కం తెలియదు. ఇక్కడ యువ ప్రేమికుడికి ఒకే ఒక లక్ష్యము ఉంది. తన ప్రియురాలికి తను ఆమెను ఎంత గాఢంగా ప్రేమిస్తున్నాడో నొక్కి చెప్పాలని అతను కోరుకుంటున్నాడు. ఆ విషయాన్ని నిరూపించటానికి అతను ఉపమానాలు మరియు అతిశయోక్తులు వాడుతున్నాడు. ఇచ్చిన ద్విపదలో ఆయన అంటున్నారు ఇలా : సముద్రాలన్నీ ఎండిపోయినా కూడా, నా ప్రేమ కొనసాగుతూనే ఉంది. సముద్రాలన్నీ నిజంగా ఎప్పటికైనా పూర్తిగా ఆవిరి అవుతాయా ? ఆ విషయం పట్ల అతనికి బాధ అసలు లేదు. అతని ఏకైక పని తన మనసును ఆమెకు చూపటం.

వివరణ :
ఇక్కడ కవి తన ప్రియురాలికి తన రకరకాల వాగ్దానాలను ప్రకటిస్తున్నాడు.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Question 4.
And fare thee weel, my only
Luve and fare thee weel a while!
Answer:
Introduction:
This couplet is taken from the beautiful lyric, ‘A Red Red Rose’ written by Robert Burns. It is one of the best lyrics of English poetry. It blends the eternity of love with the mortality of life.

Context & Explanation :
The speaker says that he will love his beloved forever. Even after the seas get dried up, all the rocks melt, and the sands of life exhaust their love stays alive. It will last forever. For the present, the speaker says good bye only to return soon, though the journey is to a far off place the poem blends the eternity of love with the mortality of life.

Critical Comment:
The poet makes several promises to love his beloved forever.

కవి పరిచయం :
రాబర్ట్ బర్న్ అంత్యానుప్రాసయుక్తముగా వ్రాసిన ‘ఒక ఎర్ర ఎర్ర గులాబి’ పద్యం నుండి ఈ పంక్తులు తీసుకోబడినవి. ఇతను 18వ శతాబ్దపు స్కాటిష్ కవి. ఈ పద్యంలో ఒక యువకుడి భావోద్వేగ తీవ్రత వర్ణింపబడినది.

సందర్భం :
మనసు మార్గాలు విచిత్రము. మరీ ప్రేమికుడి మనసు మార్గాలు మరింత విచిత్రము. ప్రేమలో పడ్డ యువకుడు తన ప్రియురాలి పట్ల తన ప్రేమ ఎంత లోతైనదో పదే, పదే నొక్కి చెబుతాడు. పద్యం యొక్క మొదటి మూడు స్టాంజాల నిండా అతని ప్రేమకు మద్దతుగా అతిశయోక్తులు మరియు ఉపమానాలే. సముద్రాలు ఎండిపోయినా, పర్వతాలు కరిగిపోయినా, జీవితాలు అనే ఇసుక అంతరించినా నా ప్రేమ కొనసాగుతూనే ఉంటుందని చెప్పాడు. ఇక ఇప్పుడు ఆ ప్రేమికుడు తన ప్రియురాలికి వీడ్కోలు పలుకుతున్నాడు, కొద్ది కాలానికి మాత్రమే అయినప్పటికి అతను ఒక వెయ్యి మైళ్ళ దూర ప్రయాణం చేపడుతున్నాడు. అతను ఆమెకు హామీ ఇస్తున్నాడు తను మళ్ళీ తప్పక ఆమె దగ్గరకు వస్తానని. “Fare thee well” అంటే “మీకు వీడ్కోలు” అని.

వివరణ :
ఇక్కడ కవి తన ప్రియురాలికి తన రకరకాల వాగ్దానాలను ప్రకటిస్తున్నాడు.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Paragraph Questions & Answers (Section – A, Q.No. 4, Marks : 4)

Question 1.
How is the feeling of love expressed in the poem A Red Red Rose ?
Answer:
The poem A Red Red Rose is written by Robert Burns. It is one of the best lyrics of English poetry. It blends the eternity of love with the mortality of life. It is an address to the speaker’s lover to whom he swears eternal love and loyalty.

The speaker shares his romantic love for his beloved in this poem. His feelings are very profound. He compares his beloved with a fresh and beautiful rose sprung in June and to a sweet melody as well. He also makes several promises to live his beloved forever. He makes a promise that he will return to her life after their temporary separation. He promises to be with her, no matter how long the journey takes.
A Rose speaks of Love silently

రాబర్ట్ బర్న్స్ అనే స్కాటిష్ కవి. ‘ఒక ఎర్ర ఎర్ర గులాబి’ అనే పద్యాన్ని రచించాడు. ఆంగ్ల పద్యసాహిత్యంలో ఇది ఒక ఉత్తమమైన పద్యం. ఇందులో కథకుడు ప్రియురాలితో తన ప్రేమను వ్యక్తం చేస్తున్నాడు.

ప్రియుడు తన ప్రేమను ప్రియురాలికి తెలుపుతున్నాడు. అతని అనుభూతులు పంచుకుంటున్నాడు. అతను తన ప్రేమను జూన్లో వికసించే ఒక తాజా ఎర్రని గులాబితో పోలుస్తున్నాడు. అతను తన ప్రేయసికి రకరకాల వాగ్దానాలు చేస్తున్నాడు. వారి ఎడబాటు తాత్కాలికమేనని మళ్ళీ తప్పక ఆమె దగ్గరకు వస్తానని వాగ్దానం చేస్తున్నాడు. ప్రస్తుతం వీడ్కోలు పలుకుతున్నానని తెలిపాడు.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Question 2.
Why is love compared to a Red Red Rose ?
Answer:
The poem, ‘A Red Red Rose’ is written by Robert Burns. He is one of the leading voices of Scotland in English literature. The present poem pictures his love for his beloved. His love is as beautiful as a fresh rose that has just bloomed in June. It is fresh and bursting with life. Here love is compared to a red rose because red rose has been an ancient symbol of love in almost all cultures. In this case, rose is newly spring in June. So, we can understand that his love is always at the starting point. Robert uses his rose with the meaning that it is very strong and passionate. It shows how strong is the speaker’s feeling.

రాబర్ట్ బర్న్స్ అనే స్కాటిష్ కవి. ‘ఒక ఎర్ర ఎర్ర గులాబి’ అనే పద్యాన్ని రచించాడు. ఆంగ్ల పద్యసాహిత్యంలో ఇది ఒక ఉత్తమమైన పద్యం. ఇందులో ప్రియుడు ప్రియురాలితో తన ప్రేమను వ్యక్తం చేస్తున్నాడు.

ప్రియుడు తన ప్రేమను ప్రియురాలికి తెలుపుతున్నాడు. అతని అనుభూతులు పంచుకుంటున్నాడు. అతను తన ప్రేమను జూన్లో వికసించే ఒక తాజా ఎర్రని గులాబితో పోలుస్తున్నాడు. ఇక్కడ తన ప్రేమను ఒక ఎర్ర ఎర్ర గులాబితో పోలుస్తూ అన్ని సంస్కృతులలో పురాతన కాలం నుంచి ఎర్రగులాబి అనేది ప్రేమకు చిహ్నంగా పేర్కొన్నాడు. ఇక్కడ గులాబి జూన్లో నూతనముగా వికసించినది. రాబర్ట్ గులాబిని బలమైన ప్రేమకు తార్కాణంగా చెప్పాడు. కథకుని భావాలు తన ప్రేమ పట్ల ఎంత బలంగా ఉన్నాయో ఈ పద్యంలో తెలుస్తున్నాయి.

Question 3.
What does the speaker promise in A Red Red Rose ? * (Imp, Model Paper)
Answer:
The poem, A Red Red Rose, is written by Robert Burns. It is one of the best lyrics of English poetry. The speaker shares his romantic lone for his beloved. He promises different things to his beloned. He vones to love his beloved until the seas have dried up, the fire of the sun has melted the ice, and human life is over. He uses these examples to express his feelings. Thus, promises his eternal love to his borny lass and that no matter how far he might go, he will always return to her aside.
TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose 2
రాబర్ట్ బర్న్స్న గొప్ప ప్రేమ కవిగా పరిగణిస్తారు. అత్యంత విస్తృతంగా ఆమోదించబడిన ప్రేమ చిహ్నం గులాబి. బర్న్స్ యొక్క కవిత ‘ఒక ఎర్ర ఎర్ర గులాబి’ పేరు నుండి చివరి పంక్తి వరకు అదొక ప్రేమ పద్యము అని నిరూపించుకుంటుంది. ఒక యువ ప్రేమికుడు ఈ కవితలో కథకుడు. అతను పదే పదే నొక్కి చెబుతాడు తన ప్రేమ ఒక ఎర్ర గులాబి లాంటిది మరియు ఒక మధుర గేయం లాంటిది అని.

అతను ఇంకా ఇంకా తన ప్రేమ కొనసాగుతుందని, సముద్రాలు ఎండిపోయినా, పర్వతాలు కరిగిపోయినా, జీవితం అనే ఇసుక అంతరించిపోయినా కూడా. ఆమెను చాలా గాఢంగా, లోతుగా ప్రేమిస్తున్నానని నొక్కి చెబుతాడు. గులాబీలు, సంగీతం ప్రేమకు ప్రతినిధులుగా చక్కగా ఉంటాయి. అందుకే ఇక్కడ ప్రేమికుడు తన ప్రేమ ఎర్ర గులాబి లాంటిది అంటాడు. కవిత పేరు కూడా ఈ విషయాన్ని నొక్కి చెబుతుంది.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Question 4.
Describe the speaker’s devotion to his beloved as expressed in the last two lines of A Red Red Rose.
Answer:
Robert Burns poem A Red Red Rose pictures a young speaker’s love for his beloved. The poem blends the eternity of love with the mortality of life. Especially, in the last two lines, the speaker is completely devoted to his beloved. The has promised his sweet heart that he will return to her wherever he goes, no matter what the distance. Even if her relationship is ten thousand miles away, his love will never die. He will continue to love her. All in all these lines represent the immortality of his love for his beloved.

రాబర్ట్ బర్న్స్ అనే స్కాటిష్ కవి. ‘ఒక ఎర్ర ఎర్ర గులాబి’ అనే పద్యాన్ని రచించాడు. ఆంగ్ల పద్యసాహిత్యంలో ఇది ఒక ఉత్తమమైన పద్యం. ఇందులో కథకుడు ప్రియురాలితో తన ప్రేమను వ్యక్తం చేస్తున్నాడు. చివరిగా ఆ ప్రేమికుడు తన ప్రియురాలికి వీడ్కోలు పలుకుతున్నాడు, కొద్దికాలానికి మాత్రమే అయినప్పటికి. అతను ఒక వెయ్యి మైళ్ళ దూర ప్రయాణం చేపడుతున్నాడు. అతను ఆమెకు హామీ ఇస్తున్నాడు తను మళ్ళీ తప్పక ఆమె దగ్గరకు వస్తానని. “Fare thee well” అంటే “మీకు వీడ్కోలు” అని.

A Red Red Rose Summary in English

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose 1

Robert Burns is a great lyrical poet. He became truly the national poet of Scotland. The last years of his life were fruitful in the lyrical songs that gave him not merely a national but a universal reputation. In the present poem “A Red Red Rose” he describes his love for his beloved. This is a simple, but sincere poem in which he pours out his intense love for his beloved. He describes his pious and ardent love in a heart-rending and picturesque manner.

The poet points out that love is newly emerging feeling fully bloomed like a pretty rose in lovely spring. It is filled with the warmth of June, the summer. He says that his love is fully grown, bloomed like a lovely rose, sprung in June. It is like a sweet melody played in a passionate tune.

He endows his emotion with a concrete form he sees vividly in his beloved’s image. He declares the immortality of his love. He says that his love will remain till the seas get dried, till the rocks melt with the sun, and till death snatches him away from his sweetheart.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Finally, he overcomes all his grief. It is because he realizes and convinces the beloved that this parting is not an end. And they will be united again after crossing the path of death covering a distance of ten thousand miles. Thus the ravages of time will fail to bring any change upon his pious feelings. He bids his beloved farewell physically. He bids it only temporarily as he is sure to get united with her immortally in the other world beyond the limits of life and death-the physical concepts. Burns very convincingly assures, his beloved that he will reach her through the distance between them were ten thousand miles, symbolically.

A Red Red Rose Summary in Telugu

Lyric అనగా గేయము అని అర్థం. దీనిని లయబద్ధముగా పాడవచ్చును. ఆంగ్ల భాషలో Robert Burns ఖ్యాతి గాంచిన గేయ రచయిత. ఇతను Scotland జాతీయ కవిగా కూడా గుర్తింపు పొందాడు. అతని చివరి కాలంలోని గేయరచనలు జాతీయ స్థాయిలోనే కాక ప్రపంచ స్థాయిలో కూడా పేరు తెచ్చాయి. ప్రస్తుత పద్యంలో కవి తన ప్రేయసి పట్లున్న తన ప్రేమను వివరిస్తున్నాడు. తన ప్రేయసిని గులాబీతో పోల్చి పద్య రచన చేశాడు.

తన ప్రేయసిని వసంత ఋతువులో అప్పుడే వికసించిన గులాబీతో పోల్చాడు. ఆ గులాబీ పరిమళాలు వెదజల్లుతూ ఆకర్షణీయంగా, సుందరంగా, కోమలంగా, లలితముగా ఎలా ఉందో తన ప్రేయసి కూడా అలాగే ఉందని వర్ణించినాడు. తన ప్రేమకు చావులేదని, సూర్యుడు, చంద్రుడు, భూమి, ఆకాశము, సముద్రములు ఉన్నంతకాలము వారి ప్రేమ కూడా చిరకాలమని వివరంగా వర్ణిస్తాడు.

తన ప్రేయసితో విడిపోయినప్పటికీ అది కేవలం తాత్కాలికమని, స్వర్గలోకములో ఆమెను కలుసుకుంటానని లేకుంటే మరియొక జన్మలోనైనా ఆమెను కలుస్తానని అతని ఆశ. మనస్సులు ఒకటైనవేళ, శరీరములు వేరైనప్పటికీ ప్రేమ శాశ్వతమైనదని అభివర్ణిస్తాడు.

A Red Red Rose Summary in Hindi

स्कॉट लैड – कवि रॉबर्ट बर्न्स से 18 वीं शती में लिखित प्रेम गीत है यह ‘एक लाल लाल गुलाब’ – ‘A Red Red Rose’ यह क्षाव्य गीत है, जिसमें झलकता है कि यौवन में प्रेम कितना भावोद्वेग भरित होता है । लयबद्ध रुप से बढ़नेवाली यह कविता पाठकों से अपार प्रेम पाती है । कथक अपनी प्रिया को अपना प्यार सोलह पंक्तियों में बताता है ।

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

“मेरा प्रेम जून में विकसित ताजा लाल गुलब जैसा है । जितना तेरा सौंदर्थ और आनंद है, तेरे प्रति मेरा प्रेम उतना गहरा है । तेरे प्रति मेरा प्रेम शाखत है । समुद्र सूख जाने पर भी, पर्वत गल जाने पर भी जीवन की रेत मिट जाने पर भी मेरा पेर प्रेम तो जारी रहता है। फिर भी, कुछ समय के तुझे विदा करता हूँ जरुर वापस आता हूँ। मेरी यात्रा दस हज़ार मील की होने पर भी लौट है । आता तब तक विदाई ।

Meanings and Explanations

Melody (n) / mel / adi/ (మెలడీ) : sweet sounds, మధురగీతం , मधुर गीत
Bonny (adj) / bani / (బోని) : healthy looking, with glow of health, ఆరోగ్యాంగా అందంగా ఉన్న , कमनीय, रमणीय
Lass (n)/aæs/(ల్యాస్) : girl, యువతి , युवती, किशोरी
Art : are, ఉన్నారు, है, हैं, हो
Thou (pron)/ ðau / (దౌ) : you, నీవు, మీరు,, आप, तुम
Gang dry : get dried, ఎండిపోవుట , अदुश्य होना
Fare thee well : Good-bye, may god bless you, వీడ్కోలు , तुझे विदा
Sprung(v-pp) / spraŋ / స్స్పంగ్) : bloomed, వికసించిన, పూసెను,, खिला हुआ
Weel : well, ఆరోగ్యవంతమైన , स्वस्थ
Luve (n)/lav / (లవ్) : love, ప్రేమ, प्यार, प्रेम, मुहबत
Play’d : played, ఆడటం, खेलना

TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table

Telangana SCERT TS 10th Class Physical Science Study Material Pdf 7th Lesson Classification of Elements- The Periodic Table Textbook Questions and Answers.

TS 10th Class Physical Science 7th Lesson Questions and Answers Classification of Elements- The Periodic Table

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I. Reflections on concepts

Question 1.
What are the limitations of Mendeleeff’s periodic table? How could the modern periodic table overcome the limitations of Mendeleeff’s table?
Answer:
Limitations of Mendeleeff’s periodic table: –

  1. Position of hydrogen: The position of hydrogen in table is not certain because it can be placed in group IA as well as in group VII A as it resembles both with alkali metals of IA group and halogens of VITA group.
  2. Anomalous pair of elements: Certain elements of higher atomic mass precede those with lower atomic mass. Eg: Tellurium proceeds Iodine, potassium placed after Argon. (Atomic mass 40) .
  3. DissimIlar elements placed together: Elements with dissimilar properties were placed in same groups as sub-group A and sub-group B.
    Eg: U, Na, and K of the IA group have little resemblance with Cu, Ag, Au of IB group.
  4. Some similar elements are separated: Some similar elements like copper and mercury ‘silicon and thallium’ etc., are placed ¡n different groups of the periodic table.

Question 2.
Define the modern periodic law. Discuss the construction of the long form of the periodic table.
Answer:
Modern Periodic Law: The physical and chemical properties of elements are the periodic function of the electronic configurations of their atoms.

  1. The modern periodic table has eighteen vertical columns known as groups and seven horizontal rows known as periods.
  2. The horizontal rows of elements In a periodic table are called periods. The elements in a period have consecutive atomic numbers. The vertical columns of elements In the table are called groups.
  3. Based on which subshell the differentiating electron enters In the atom of the given element the elements are classified as s, p. d, and f block elements.

s-Block elements :- The elements with valence shell configuration ns1 and ns2 are s- Block elements.
p-Block elements: The elements with electronic configuration from ns2np1 to ns2 np6 are p-block elements.
s and s block elements together are known as representative elements. d-Block elements or Transition elements:
The elements with valency electronic configuration ns2np6 nd1 to ns2np6 (n-1) d10 are called d-Block elements. These move from left to right in periodic table and we observe a transition of elements from metals to non-metals. Hence these are called transition elements.

f-Block elements or Inner transition elements: The elements In which f-orbitals are being filled in their atoms are called f-Block elements. These elements are called inner transition elements.

Inert Gases: Helium, Neon, Argon, Krypton, Xenon, Radon do not involve In any reaction and they are called Inert gases or noble gases. These are placed In zero group and every period ends with noble gas.

The long periodic table Is divided into 7 periods and 18 groups. First period contains 2 elements second period contains 8 elements. Third period contains 8 elements 4th and 5th periods contain 18 elements each and sixth period contains 32 elements. Seventh period is incomplete.

The 4f elements are called Lanthanolds. Elements from 58Ce to 71Lu. The 5f elements are called Actinoids from 90Th58 to 103 Lr 71 These are shown separately at the bottom of the periodic table.

TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table

Question 3.
Explain how the elements are classIfied Into s, p, d, and f-block elements In the periodic table and give the advantage of this kind of classification.
Answer:
Depending on the valency shell electronic configuration elements are classified into s, p, d, and f block elements.
s-Block elements: The elements with valence shell electronic configuration ns1 and ns2 are called s-block elements.

p-Block elements: The elements with valence shell electronic configuration ns2 np6 are called p-block elements. s and p block elements together known as Representative elements.

d-Biock elements : The elements with valence electronic configuration from ns2np6 (n -1)d1 to ns2np6(n -1)d10 are called d-block elements. These are also called as Transition elements.

f-Block elements: The elements In which f-orbitaIs are being filled in their atoms are called f – block elements. These are also known as Inner Transition Elements.

Advantage: This division of elements into blocks Is useful to divide the elements into groups. Every group has the elements with same valence electronic configuration. So they have similar chemical properties.
s-block elements – I A and II A groups
p-block elements – III A to VITI A or zero group.
d-block elements – I B to VIII B groups
f-block elements – Lanthanoids and Actinoids.

Question 4.
Write down the characteristics of the element having atomic number 17.
Electronic configuration …………………….
Period number …………………….
Group number …………………….
Element family …………………….
No. of valence electrons …………………….
Valency …………………….
Metal or Non-metal …………………….
Answer:
Electronic configuration: 1s22s22p63s23p5
Period number : 3
Group number: 17
Element family: Halogen (VII A) or 17th group
No. of valency electrons : 2+5 = 7 ( seven )
Valency: 1
Metal or Non-metal : Non-metal

Question 5.
Complete the following table using the periodic table.
TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table 1
Answer:
TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table 2

Question 6.
Complete the following table using the periodic table.
TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table 3
Answer:
TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table 4

Question 7.
Newlands proposed the law of octaves. Mendeleeff suggested eight groups for elements in his table. How do you explain these observations in terms of modern periodic classification?
Answer:

  • Newlands proposed the law of octaves.
  • According to Newlands law of octaves, every eight element starting from a given one be a repetition of the first with regard to its properties.
  • Mendeleeff suggested eight groups for elements in his table.
  • The elements in the same group have same properties that means every eight element starting from a given element have same property with that.
  • In terms of modern concepts, after completion of one shell the properties of elements are repeated.
  • After completion of ns2 np6 configuration the properties of elements are repeated.
  • So Newlands law of octaves and Mendeleeff’s suggestion of eight groups for elements are also reliable according to modem concepts.

Question 8.
Why was the basis of classification of elements changed from the atomic mass to the atomic number?
Answer:

  1. The first attempt to classify elements was made by Dobereiner.
  2. Dobereiner’s attempt gave a clue that atomic masses could be correlated with properties of elements.
  3. Newlands’ law of octaves also followed the same basis for classification but this law Is not valid for the elements that had atomic masses higher than calcium.
  4. Mendeleeff’s classification also was based on the atomic masses of elements, but it lead to some limitations like Anomalous pair of elements and Dissimilar elements placed together.
  5. Mosley by analyzing the X-ray patterns of different elements was able to calculate the number of positive charges in the atoms of respective elements.
  6. With this analysis, Mosley realized that the atomic number is more fundamental characteristic of an element than its atomic weight.
  7. So, he arranged the elements in the periodic table according to the increasing order of their atomic number.
  8. This arrangement eliminated the problem of anomalous senes and dissimilar elements placed together In same group as done by Mendeleeff’s classification.

TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table

Question 9.
(a) What is a periodic property? How do the following properties change In a group and period? Explain.
(a) Atomic radius
(b) Ionization energy
(c) Electron affinity
(d) Electronegativity.
Answer:
Periodic property: Periodicity means that when elements are arranged according to their electronic configuration, the periodic properties of elements in the periodic table reoccur at regular intervals – such properties which reoccur are called as periodic properties.

Change of properties In a group and a period.

Type of Periodic PropertyTrends in
Groups (from top to bottom)Periods (from left to right)
Atomic radiusIncreasingDecreasing
Ionization energyDecreasingIncreasing
ElectronegativityDecreasingIncreasing
Electron AffinityDecreasingIncreasing

Explanation:
(a) Atomic radius: Atomic radius of an element Is not possible to measure in its isolated state, because it is not possible to determine the location of the electron that surrounds the nucleus. Hence the distance between nucleus and outermost orbit of an element is taken as atomic radius measured In Pm (Pico Meters).

Variation In group: In a group as we go down, the atomic number of the element Increases, as a new shell Is added. As a result, the distance between nucleus and the outer shell Increases. This Is the reason why the atomic radius increases as we go down in a group.

Variation In period: AtomIc radii of elements decrease across a period from left to right. As we go from left to right, electrons enter into the same main shell or even Inner shell. Therefore there should be no change In the distance between nucleus and outer shell, but nuclear charge Increases. Hence the nuclear attraction on the outer shell increases. As a result the size of the atom decreases.

(b) Ionization enegry: The energy required to remove an electron from the outermost orbit or shell of a neutral gaseous atom is called ionization energy.

Variation in group: As we go down in a group, atomic radius increases. Hence the ionization energy decreases from top to bottom.

Variation In period: As we go from left to right in a period, the atomic radius decreases. Hence the ionization energy increases from left to right In a period but does not follow gradient order as the electron is attracted more towards the nucleus.

(c) Electron affinity: The electron affinity of an element is defined as the energy liberated when an electron Is added to its neutral gaseous atom. Electron affinity of arm atom is also called electron gain enthalpy of that element.

Variation in group: Electron gain enthalpy values decrease as we go down In a group, due to Increase In atomic radius.

Variation In period: Electron gain enthalpy values Increase as we go from left to right in a period, due to decrease in atomic radius.

(d) Electronegativity: The electronegativity of an element Is defined as the relative tendency of its atom to attract electrons towards itself when it is bounded to the atom of another element.

Variation in groups and periods: Electronegativity values of elements decrease as we go down in a group and increase along a period from left to right.

Question 10.
(b) Explain the ionization energy order In the following sets of elements.
(a) Na, Al, Cl
(b) Li, Be, B
(c) C, N, O
(d) F, Ne, Na
(e) Be, Mg, Ca.
Answer:
(a) Na, Al, Cl: All these three elements belong to same period. The order of their atomic size Is Na > Al> Cl. As we move from left to right in a period Ionization energy increases
∴ The order of ionization energy of these elements is Cl> Al> Na.

(b) Li, Be, B: U, Be, B belong to same period. The electronic configuration of Li-1S22S1, Be-1S22S2, B-1S22S22p1. The penetration power of 2P is less when compared to 2s. So it is easy to remove electrons from ‘2p’.
∴ The order of ionization energy of these elements is: Be > Li> B.

(c) C, N, O : C, N, O belong to same period. The electronic configuration of C – 1s22s22p2, N-1s22s22p3, O-1s22s22p4. Nitrogen has half filled degenerate orbital configuration. So, it Is more stable compared to Cl and ‘O’so Its Ionisation energy Is high.
∴ Order of ionization energy of these elements is: N > C> O.

(d) F, Ne, Na: Electronic configuration of F -1S22s22p5; Ne – 1S22S22p6; Na – 1s22s22p63s1. Ne is an Inert gas, so It has highest Ionization energy.
‘Na’ has smaller size compared to F. So ¡t ha high ionization energy.
∴ The order of ionization energy is: Ne>F>Na.

(e) Be, Mg, CaP: These elements belong to the same group the atomic size of these elements Is In the order of Ca > Mg > Be. As atomic size increases ionization energy decreases. The order of Ionization energy Is Be > Mg > Ca.

Application Of Concepts

Question 1.
Given below Is the electronic configuration of elements A, B, C, and D.
(A) 1s2 2s2 1. WhIch are the elements coming within the same period?
(B) 1s22s22p63s2 2. Which are the ones coming within the same group?
(C) 1s22s22p63s2 3p3 3. Which are the noble gas elements?
(D) 1s22s22p6 4. To which group and period does the clement C ‘belong?
Answer:
(1) A and D belong to same penned B and C belong to the same period.
(2) A, B coming in the same group.
(3) D is the noble gas.
(4) C’ belongs to 15th group and third period.

Question 2.
s – block, and p – block elements (except 18th group elements) are sometimes called as ‘Representative elements’ based on their abundant availability In the nature. Is It Justified? Why?
Answer:

  • s-block and p-block elements (except 18th group elements) are called ‘Representative elements’.
  • All these elements have incomplete outermost shells.
  • So they are chemically reactive to obtain stable electronic configuration of noble gases ns2np6.
  • So, they are abundant In nature n the form of compounds.

Question 3.
The electronic configuration of the elements X, Y, and Z are given below.
(A) X = 2 (B) Y = 2, 6 (C) Z =2, 8, 2
(i) Which element belongs to second period?
Answer:
‘Y’ belongs to the second period. Because differentiating electron enters into second shell.
(ii) Which element belongs to second group?
Answer:
Element ‘Z’ belongs to second group. Because its valence is ‘2’.
(iii) Which element belongs to 18 group?
Answer:
Element X belongs to 18th” group because its 1st shell is completely filled with electrons.

TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table

Question 4.
(a) State the number of valence electrons, the group number and the period number of each element given in the following table.
TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table 5
Answer:

ElementValence electronsGroup numberPeriod number
Sulphur616 (VIA)3
Oxygen616(VLA)2
Magnesium22 ( II A)3
Hydrogen1No group1
FluorIne717 ( VII A)2
Aluminium313 (III A)3

(b) State whether the following elements belong to a Group (G), Period (P) or Neither Group nor Period Indicating the letters G or P or N
TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table 6
Answer:

ElementsGroup / Period / Neither group nor period
Li, C, OP
Mg, Ca, BaG
Br, Cl, FG
C, S, BrN
Al, SI, ClP
Li, Na, KG
C, N, OP
K, Ca, BrN

 

Question 5.
Identify the element that has the larger atomic radius In each pair of the following and mark it with a symbol (✓).
(i) Mg, Ca
(ii) Li, Cs
(iii) N, P
(iv) B, AI
Answer:

  1. Mg, Ca: Calcium has large atomic radius. Since Mg and Ca are in same group Ca falling below Mg. Atomic radius increases from top to bottom In a group.
  2. Li, Cs: Cs has large atomic radius as Li and Cs are also in the same group and Cs lying below Li.
  3. N, P: Phosphorous has large atomic radius. As N and P are also In the same group and P lying below N.
  4. B, AI: Al has large atomic radius. B, Al are in same group III A and Al is below B.

Question 6.
Identify the element that has the lower Ionization energy in each pair of the following and mark it with a symbol (✓).
(i) Mg, Na
(ii) Li, O
(iii) Br, F
(iv) K, Br
Answer:

  1. Mg, Na: Na has low ionization energy. Since Mg, Na are in same period ionization energy increases from left to right. Na lies left to Mg. So Na has lower ionization energy.
  2. Li, O: Li has lower ionization energy. U lies left to ‘O’ in 2nd period.
  3. Br, F: Bromine has lower ionization energy; Br & F are in same group. IE decreases from top to bottom. Bromine is below F. So It has lower lE.
  4. K, Br: ‘K’ has lower ionization energy. Since it is left to Br in the same period.

Question 7.
How does metallic character change when we move
(i) Down in a group?
(ii) Across a period?
Answer:
(i) Down in a group?
Metallic character increases when we move down the group.
(ii) Across a period?
Answer:
Metallic character decreases when we move along a period.

Question 8.
On the bails of atomic numbers predict to which block the elements with atomic number 9, 37, 46 and 64 belong to?
Answer:

  1. Electronic configuration of element with atomic number 9 is 2, 7 – belongs to p-block.
  2. Electronic configuration of element with atomic number 37 is 2, 8, 8, 18, 1 – belongs to s- block.
  3. Electronic configuration of element with atomic number 46 is 2, 8, 8, 18, 10 – belongs to d – block.
  4. Electronic configuration of element with atomic number 64 Is 2, 8, 8, 18, 18, 10 -belongs tO f – block.

Question 9.
Using the periodic table, predict the formula of compound formed between an element X of group 13 and another element Y of group 16.
Answer:
Element X of group 13 Element Y of group 16
Valency of element X = 3 Valency of element Y
18-16=2
TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table 7
Compound: X2Y3

Question 10.
An element has atomic number 19. Where would you expect this element in the periodic table and why?
Answer:
The element with atomic number 19 is in 4th’ period and first group of periodic table.
Reason:

  1. Electronic configuration : Is2 2s2 2p6 3s2 3p6 4s1’.
  2. The differentiating electron enters Into 4t shell. Hence it belongs to 4th period.
  3. The differentiating electron is In ‘s’ orbital. So it belongs to ‘s’ block.
  4. The outermost orbital has only one electron. Hence it belongs to first group.

Question 11.
Elements In a group generally possess similar properties, but elements along a period have different properties. How do you explain this statement?
Answer:

  1. Physical and chemical properties of elements are related to their electronic configurations, particularly the outer shell configurations.
  2. The atoms of the elements in a group possess similar electronic configurations. Therefore, we expect all the elements in a group should have similar chemical properties and there should be a regular gradation in their physical properties from top to bottom.
  3. But in a period from left to right, elements get an increase in the atomic number by one unit between any two successive elements.
  4. Therefore, the electronic configuration of valence shell of any two elements in a given period is not same. Due to this reason elements along a period possess different chemical properties with regular gradation in their physical properties from left to right.

TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table

Question 12.
Name two elements that you would expect to have chemical properties similar to Mg. What is the basis for your choice?
Answer:

  • Calcium (Ca) and Stranúum (Sr) are the two elements, which are similar to Mg in chemical properties.
  • Because they belong to the same group lIA. Ail the elements which are in the same group have same electronic configuration and same chemical properties.

Question 13.
An element X belongs to 3rd perIod and group 2 of the periodic table. State (a) The no. of valence electrons in it (b) The valency (C) Whether It is metal or a non-metal.
Answer:
The element ‘X’ belongs to 3rd peñad and group 2 is Mg.
(a) The no.of valence electrons 2.
(b) The valency = 2.
(c) It is a metal.

Question 14.
How do you appreciate the role of electronic configuration of the atoms of elements In perIodic classification?
Answer:

  1. According to Modern Periodic Iaw the properties of elements are the functions of their atomic number or electronic configuration”.
  2. Elements having same number of valence electrons are placed in the same group based on their electronic configuration.
  3. So, we can easily locate the position of any element in the periodic table with Its electronic configuration.
  4. Hence electronic configuration plays a major role In the preparation of Modern period table. So its role is thoroughly appreciated.

Question 15.
Without knowing the electronic configurations of the atoms of elements Mendeleeff still could arrange the elements nearly close to the arrangements in the modern periodic table. How can you appreciate this?
Answer:

  1. Mendeleeff’s vision must be appreciable because he left some gaps predicting the existing of some elements, which are not available at that time.
  2. Atomic weights of some elements can be corrected by placing them In correct places.
  3. If we compare the long-form periodic table with Mendeleeff’s table, we find many elements with their places unchanged.
  4. One can see from these discussions, that the Mendeleeffs accurate prediction and their eventual success made him and his classification of elements famous, Even today his table of elements Is followed with minimum modifications.

Question 16.
Comment on the position of hydrogen in periodic table.
Answer:

  1. The configuration of Hydrogen (1s1) Is responsible for its dual nature.
  2. Either the electron can be lost behaving as electropositive elements (H+) like alkali metals or the electron can be gained as to complete, is subshell behaving as electronegative element (H) like halogens.
  3. The position of Hydrogen is not fixed in the periodic table sometimes it is placed with alkali metals and sometimes with halogens.
  4. However, Hydrogen exhibits other properties which differ from both alkali metals and halogens.

Higher Order Thinking Questions

Question 1.
How does the positions of elements In the periodic table help you to predict Its chemical properties? Explain with an example.
Answer:

  1. The physical and chemical properties of atoms of the elements depend on their electronic configuration, particularly the outer shell configurations.
  2. Elements are placed in the periodic table according to the increasing order of their electronic configuration.
  3. The elements In a group possess similar electronic configurations. Therefore all the elements in a group should have similar chemical properties.

Eg: Consider K

  • It is the element In 4th’ period 1st’ group.
  • It is on left side of the periodic table. Hence It Is a metal.
  • It is ready to lose an electron to get octet configuration. Hence its reactivity is more.
  • It is Alkali metal.
  • All alkali metals react with both acids and bases and release H2 gas.

Question 2.
In period 2, element X is to the right of element Y. Then find which of the elements have
(i) Low nuclear charge
(ii) Low atomic size
(iii) High Ionisation energy
(iv) High electronegativity
(v) More metallic character.
Answer:
X s to the right of Y. So they will be YX
(i) Low nuclear charge: Nuclear charge increases from left to right in a period. So Y has low nuclear charge.
(ii) Low atomic size: Atomic size decreases from left to right in a period. So X has low atomic size.
(iii) High ionization energy: Ionization energy increases across a period from left to right. So X has high ionization energy.
(iv) More metallic character: Metallic nature decreases from left to right. So Y has more metallic character.

TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table

Multiple choice questions

Question 1.
Number of elements present in period -2 of the long fom of periodic table ………………. [ ]
(a) 2
(b) 8
(c) 18
(d) 32
Answer:
(b) 8

Question 2.
Nitrogen (Z = 7) is the element of group VAofthe periodic table. Which of the following is the atomic number of the next element in the same group’? [ ]
(a) 9
(b) 14
(c) 15
(d) 17
Answer:
(c) 15

Question 3.
Electron configuration of an atom is 2,8,7. To which of the following elements would it be chemically similar? [ ]
(a) nitrogen(Z=7)
(b) fluorine(Z = 9)
(c) phosphorous(Z = 15)
(d) argon(Z=1 8)
Answer:
(b) fluorine(Z = 9)

TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table

Question 4.
Which of the following is the most active metal? [ ]
(a) lithium
(b) sodium
(c) potassium
(d) rubidium
Answer:
(d) rubidium

Suggested Experiments

Question 1.
Aluminium does not react with water at room temperature but reacts with both dil. HCl and NaOH solutions. Verify these statements experimentally. Write your observations with chemical equations. From these observations, can we conclude that Al is a metalloid?
Answer:
1. Aluminum reacts with dii. HCl and releases hydrogen gas with formation of Aluminium chloride.
TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table 8
2. Aluminium reacts with NaOH solution and releases hydrogen gas.
TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table 9
3. Aluminium does not react with water at room temperature. This concludes that the properties of Aluminium in between a metal and non-metal. So it behaves like a metalloid.

Suggested Projects

Question 1.
Collect the Information about reactivity of VIII A group elements (noble gases) from your school library and prepare a report on their special character when compared to other elements of periodic table.
Answer:

  1. The elements of Group VIII A (noble gases) are chemically inactive, because all of them have stable “Octet” configurations, except He.
  2. He has two electrons in the outermost orbital, i.e., is orbitalis is the maximum, the is orbital can accommodate. Hence “He” is also inactive.
  3. The losing or gaining an electron or sharing of electrons is difficult due to high I.E. and zero E.A value of inert gases.
  4. Under suitable conditions, the heavier elements of the group form compounds with F2 and O2 or XeF2 or XeO2F2.
  5. Ar forms coordination compounds with BF3 whose composition is Ar.n.BF3
  6. Xenon Hexa fluoro platinate (IV)(Xe[PtF6]) was prepared by N.Barltel. This was the first compounds of inert gases. After this compounds of Xenon with F2 and O2 were prepared.
  7. The compounds of inert gases have found varied applications in recent times.

Question 2.
Collect information regarding metallic character of elements of IA group and prepare report to support the idea of metallic character increases In a group as we move from top to bottom.
Answer:
Group IA elements are Li, Na, K, Rb, Cs, and Fr
1. Metallic character is the name given to the set of chemical properties associated with elements that are metals.

2. Chemical characteristics of metals include the following

  • form cations in ionic compounds with non-metals.
  • have ionic halides.
  • have ionic hydrides containing the H’ ion.
  • have basic oxides.

3. The elements with less electronegative character are metals.
4. All IA group elements have less electronegative characters because they are ready to lose one electron to get octet configuration.
5. If we observe this group, we can find that

Reaction with non-metals:
4M + O2 → 2M2O. In this cation is M+
Hydrides:
TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table 10
M stands for alkali metal.
Order of ionic nature of hydrides Is LiH < NaH < KH < RbH < CSH
Halides:
2M+X2 → 2MX
where M stands for alkali metal and X stands for halogen.
All the metal halides are ionic.

Oxides:
4M + O2 → 2M2O
But, 4 Li + O2 →‘ 2 Li2O2
All metals form peroxides.
6. From the above reactions we conclude that Group IA elements are metals and their metallic character increases as we go down in the group.

TS 10th Class Physical Science Classification of Elements- The Periodic Table Intext Questions

Page 124

Question 1.
Can you establish the same relationship with the set of elements given in the remaining rows?
Answer:
Yes, we can establish the same relationship with the set of elements given in the remaining rows.

Question 2.
Find average atomic weights of first and third elements ¡n each row and compare It with the atomic weight of the middle element.
Answer:
TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table 11

Question 3.
What do you observe?
Answer:
Dobereiner’s attempts gave a clue that atomic masses could be correlated with properties of elements.

Page 137

Question 4.
Find out the valencies of first 20 elements.
Answer:

ElementValency
1.Hydrogen1
2. Helium0
3. Lithium1
4. Berflium2
5. Boron3
6. Carbon4
7. Nitrogen3
8. Oxygen2
9. Fluorine1
10. Neon0
11. Sodium1
12. Magnesium2
13. Aluminium3
14. Silicon4
15. Phosphorous3, 5
16. Sulphur2, 6
17. Chlorine1
18. Argon0
19. Potassium1
20. Calcium2

(b) How does the valency vary in a period on going from left to right?
Answer:
The valency increases from 1 to 4 and then decreases to zero as we move from left to right in a period with respect to Hydrogen. The valency increases from 1 to 7 with respect to oxygen, in a period from left to right.

(c) How does the valency vary on going down a group?
Answer:
The valency remains same in a group.

Page 139

Question 5.
Do the atom of an element and Its Ion have same size?
Answer:
No, the atom of an element and its ion do not have the same size. The size of cation is less than its neutral atom and size of anion Is greater than its neutral atom.

TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table

Question 6.
Which one between Na and Na would have more size? Why?
Answer:

  1. The atomic number of sodium is ti and It has 11 protons and ii electrons with outer electron as3s’ whereas Na4 ¡on has il protons but only 10 electrons.
  2. The 3s1 shell of Na has no electron In It.
  3. So the outer shell configuration is 2S2 2p6.
  4. As proton number is moie than electrons the nudeus of Na+ ion attracts outer shell electrons with strong nudear force.
  5. As a result the Na lori shrinks In size.
  6. Therefore, the size of Na ion is less than Na atom.

Question 7.
Which one between Cl and Cl would have more size? Why?
Answer:
The electronic configuration of Cl is 1s22s22p63s23p5 whereas Cl is 1s22s22p63s2 3p6. In Cl , same no.of protons are attracting more number of electrons compared to Cl. Due to decrease in effective nuclear attraction atomic size of anion is more than its neutral atom.

Question 8.
Which one in each of the following pairs Is larger In size?
a) Na, AI
b) Na, Mg+2
c) S2, Cl
d) Fe+2, Fe+3
e) C-4, F
Answer:
a) Na
b) Na
c) S-2
d) Fe+2
e) C-4

Think And Discuss

Question 1.
What relation about elements did Doberelner wanted to establish?
Answer:
The relative atomic mass of the middle element En each triad is nearer to the average of the relative atomic masses of other two elements. This is what Doberelner wanted to establish.

Question 2.
The densities of Calcium (Ca) and Barium (Ba) are 1.55 and 3.51gm/ cm3 respectively. Based on Doberolner’s law of triads can you give the approximate density of strontium (Sr)?
Answer:
The density of strontium = (1.55 + 3.51 )/2 = 5.06 / 2 = 2.53gm/cm3

Question 3.
Do you know why Newland’s proposed the law of octaves? Explain your answer In terms of the structure of atom.
Answer:
Newlands proposed the law of octaves. According to this law, every 8” element starting from a given elements resembles in its properties to that of starting element with the elements with similar chemical properties are to be present along horizontal.

In modem periodic table, every period starts with a new shell and ends when the shell is filled. We know about octet configuration. Newlands law of octaves Is similar to octet. configuration.

Question 4.
Do you think that Newlands law of octaves is correct? JustIfy.
Answer:
Newlands law of octaves is corretc. Mendeleeff also suggested only 8 groups in his table. Later it was discovered that eighth element has the same properties as the first one.

Question 5.
Why Mendeleeff had to leave certain blank spaces In his periodic table. What is your explanation for this?
Answer:
Based on the arrangement of the elements in the table, he predicted that some elements were missing and left blank spaces at the appropriate places In the table. Mendeleeff believed that some new elements would be discovered definitely. He predicted the properties of these new additional elements in advance purely depending on his table. His predicted properties were almost the same as the observed properties of those elements after their discovery.

Question 6.
What is your understanding about Ea2O3 and EsO2?
Answer:
Ea2O3 is the predicted formula of oxide,of the predicted element In eka – Aluminium group, This is similar to Ga2O3, the observed property of oxide of Gallum, which was discovered in 1875. EsO2 is the predicted formula of oxide of predicted element In eka-silicon group. This is similar to GeO2, the observed property of oxide of Germanium, which was discovered in.1886.

Question 7.
All alkali metals are solids but hydrogen is a gas with diatomic molecules. Do you justify the inclusion of hydrogen in first group with alkali metals?
Answer:
The similarity of hydrogen with alkali metals is ns’ configuration and ability to exhibit +1 oxidation state. Hydrogen, because of its small size and high electronegativity, it is a non-metal with high I.P. It shares one electron with another hydrogen and forms diatomic molecules. Due to weak van der Waals forces of attraction It is a gas whereas in alkali metals, metallic bond is present hence they are solids. Hence, hydrogen cant be included in first group along with alkali metals.

Question 8.
Why lanthanoids and actinoids placed separately at the bottom of the periodic table?
Answer:
Lanthanolds and actinoids are placed separately in the periodic table to maintain its structure and to preserve the principle of classification with similar properties in a single column.

Question 9.
If they are inserted within the table imagine how the table would be?
Answer:
If f-block elements are inserted within the table, the table would be too long.

Question 10.
Second Ionization energy of an element is higher than Its first ionization energy. Why?
Answer:
The required to remove first electron from the outermost orbit or shell of a neutral gaseous atom is called first ionization energy (IE1).
The energy required to remove an electron from uni-positive ion is called the 2nd ionization energy (IE2).
The second ionization energy (IE2) is greater than the first ionization energy (IE1). On removing an electron from an atom, the uni-positive ion formed will have more effective nudear charge than the neutral atom. This decreases the repulsion between the electrons and at the same time increases the nuclear attraction on the electron in the outer shells. As a result, more energy is required to remove an electron from the uni-positive ion. Hence the second ionization
energy is greater than the first ionization energy.

Question 11.
The calculated electron gain enthalpy values for alkaline earth metals and noble gases are positive. How can you explain this?
Answer:
Due to completely filled s-subshell in alkaline earth metals and completely filled valence shell (i.e. octet) in noble gases, they are highly stable. Hence energy Is required to add an electron I.e their electron gain enthalpy values are positive.

Question 12.
The second-period element, for example, F has less electron gaIn enthalpy than the third-period element of the same group for example ‘Cl’. Why?
Answer:
The atomic size of second period elements decreases and electron number increases as we move from left to right. Because of this elements of N, O, F possess vert small size and high electron density. Hence incoming electron feels repulsion due to existing electrons. Because of this, electron gain enthalpy
is less for second-period element compared to third period of the same group (In 5th’ , 6th and 7th groups).

TS 10th Class Physical Science Classification of Elements- The Periodic Table Activities

Activity 1

Question 1.
Observe the following table and till It.
Answer:
TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table 12
Observations:
a) Can you establish the same relationship with the set of elements given In the remaining rows?
Answer:
Except in the first row, we cannot establish the relationship as said by law of triads.
b) Find average atomic weights of first and third elements in each row and compare It with the atomic weights of the middle element.
Answer:
From table we observe that. except Group A, in the remaining Groups, the average of atomic weights of first and third elements is roughly equal to the middle elements.

c) What do you observe?
Answer:
Dobereiner’s attempt gave a clue that atomic mass could be correlated with properties of elements.

TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table

Activity 2

Question 2.
Write the different chemical families and write involved, valency of the family.
Answer:
TS 10th Class Physical Science Solutions Chapter 7 Classification of Elements- The Periodic Table 13

Question 3.
(a) Find out the valencies of first 20 elements.
(b) How does the valency vary in a period on going from left to right?
(c) How does the valency vary on going down a group?
Answer:

ElementValency
1.Hydrogen1
2. Helium0
3. Lithium1
4. Berflium2
5. Boron3
6. Carbon4
7. Nitrogen3
8. Oxygen2
9. Fluorine1
10. Neon0
11. Sodium1
12. Magnesium2
13. Aluminium3
14. Silicon4
15. Phosphorous3, 5
16. Sulphur2, 6
17. Chlorine1
18. Argon0
19. Potassium1
20. Calcium2

(b) How does the valency vary in a period on going from left to right?
Answer:
The valency increases from 1 to 4 and then decreases to zero as we move from left to right in a period with respect to Hydrogen. The valency increases from 1 to 7 with respect to oxygen, in a period from left to right.

(c) How does the valency vary on going down a group?
Answer:
The valency remains same in a group.

TS 10th Class Maths Important Questions Chapter 2 Sets

These TS 10th Class Maths Chapter Wise Important Questions Chapter 2 Sets given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 2 Sets

Previous Exams Questions

Question 1.
Write roster and builder form of “The set of all natural numbers which divide 42”. (T.S. Mar. ’15)
Solution:
Factors of 42 = 1, 2, 3, 6, 7, 14, 21, 42
So roster form = {1, 2, 3, 6, 7, 14, 21, 42}
The builder form = (x/x ∈ N, x is a factor of 42 }

Question 2.
Write A = {1, 2, 3, 4} in set builder form. (A.P. June ’15)
Solution:
The given set A = {1, 2, 3, 4}
The set builder form is A = {x/x ∈ N, x < 5 }

TS 10th Class Maths Important Questions Chapter 2 Sets

Question 3.
Let A = { x/x is an even number }
B = { x/x is an odd number}
C = { x/x is a prime number}
D = { x/x is a multiple of 5 }
Find (A.P. June’15)
i) A ∪ B
ii) A ∩ B
iii) C – D
iv) A ∩ C
Solution:
A = { x : x is an even number }
= {2, 4, 6, 8, 10}
B = { x : x is an odd number}
= {1, 3, 5, 7, 9 }
C = { x : x is a prime number}
= { 2, 3, 5, 7, 11}
D = { x : x is a multiple of 5}
= { 5, 10, 15, 20, 25}

i) A ∪ B = { 2, 4, 6, 8, 10, ……..} ∪ {1, 3, 5, 7, 9, …….}
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …………}

ii) A ∩ B = { 2, 4, 6, 8, 10, ……….} ∩ {1, 3, 5, 7, 9, ………… }
= { } = Φ

iii) C – D = {2, 3, 5, 7, 11, ……….} – { 5, 10, 15, 20, 25, ………}
= {2, 3, 7, 11, …….}

iv) A ∩ C = { 2, 4, 6, 8, 10, ………….} ∩ {2, 3, 5, 7, 11, ……….}
= {2}

Question 4.
Write all the subsets of B = {p, q} (A.P. Mar ’16)
Solution:
{p}, {q}, {p, q} and { } are the subsets of the given set B = (p, q}
As then n(B) = 2 then number of all subsets = 2n = 22 = 4.

TS 10th Class Maths Important Questions Chapter 2 Sets

Question 5.
Write the roster form of the set A. (A.P. Mar. ’16)
A = {x : x = 2n + 1 ∀ n ∈ N}
Solution:
If n = 1 then 2n + 1 = 2(1) + 1
= 2 + 1 = 3
If n = 2 then 2n + 1 = 2(2) + 1
=4 + 1 = 5
If n = 3 then 2n + 1 = 2(3) + 1
= 6 + 1 = 7
So {3, 5, 7, 9, ………..} is the roster form of given set.

Question 6.
If A = {1, 2, 3, 4} and B = (1, 2, 3, 5, 6} then find
(i) A ∩ B
(ii) B ∩ A
(iii) A – B and
(iv) B – C then comment on the above. (A.P. Mar. ’16)
Solution:
Given A = {1, 2, 3, 4} and B = { 1, 2, 3, 5, 6} then
i) A ∩ B = {1, 2, 3, 4} ∩ { 1, 2, 3, 5, 6}
= {1, 2, 3}

ii) B ∩ A = {1, 2, 3, 5, 6} ∩ { 1, 2, 3, 4}
= {1, 2, 3}
So A ∩ B = B ∩ A

iii) A – B = {1, 2, 3, 4} – { 1, 2, 3, 5, 6}
= {4}

iv) B – A = { 1, 2, 3, 5, 6} – {1, 2, 3, 4}
= {5, 6}
So A – B ≠ B – A

TS 10th Class Maths Important Questions Chapter 2 Sets

Question 7.
Write the Set builder form of A – B where A = { x : x ∈ N and x < 20 } and B = { x : x ∈ N and x ≤ 5 } (T.S. Mar. ’16)
Solution:
Given A = { x : x ∈ N and x < 20 } is
A = {1, 2, 3, …….. 17, 18, 19} and
for B = {x : x ∈ N and x ≤ 5}
B = { 1, 2, 3, 4, 5} then
A – B = {1, 2, 3, ……… 17, 18, 19} – {1, 2, 3, 4, 5}
= {6, 7, 8, …… 18, 19} (T.S. Mar ’15)
The builder form of above A – B is
A – B = {x/x ∈ N and 6 ≤ x ≤ 19}
Or
A – B = {x : x ∈ N and 5 < x < 20}

Question 8.
If A = {x / x ∈ N, x < 6 } and B = { x : x ∈ N, 3 < x < 8} then show that A – B ≠ B – A with the help of venn – diagram.
Solution:
A = {x : x ∈ N, x < 6}
A = {1, 2, 3, 4, 5}
B = {x : x ∈ N, 3 < x < 8}
B = {4, 5, 6, 7}
∴ A – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7}
= {1, 2, 3}
TS 10th Class Maths Important Questions Chapter 2 Sets 5
and
B – A = {4, 5, 6, 7} – {1, 2, 3, 4, 5}
= {6, 7}
TS 10th Class Maths Important Questions Chapter 2 Sets 6
∴ A – B ≠ B – A

Question 9.
Write the set builder form of A = {1, \(\frac{1}{4}\), \(\frac{1}{9}\), \(\frac{1}{16}\), \(\frac{1}{25}\)} (T.S. Mar. ’16)
Solution:
{\(\frac{1}{1}\), \(\frac{1}{4}\), \(\frac{1}{9}\), \(\frac{1}{16}\), \(\frac{1}{25}\)} are in the form of \(\frac{1}{\mathrm{p}^2}\) whereas p < 6
So, A = { x : x = \(\frac{1}{\mathrm{p}^2}\), p ∈ N, and p < 6} is the set builder form.

TS 10th Class Maths Important Questions Chapter 2 Sets

Question 10.
If x is set of all factors of 24 and y is set all factors of 36 then find X ∪ Y and X ∩ Y using venn – diagrams. Comment. (T.S. Mar. ’16)
Solution:
X = Set of all factors of 24
= {1, 2, 3, 4, 6, 8, 12, 24}
Y = Set of all factors of 36
= {1, 2, 3, 4, 6, 9, 12, 18, 36}
Venn diagram of X ∪ Y
i)
TS 10th Class Maths Important Questions Chapter 2 Sets 7
∴ X ∪ Y = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}

ii)
TS 10th Class Maths Important Questions Chapter 2 Sets 8
∴ X ∩ Y = {1, 2, 3, 4, 6, 12}
∴ It is clear X ∪ Y ≠ X ∩ Y

Additional Questions

Question 1.
Which of the following are sets ? Justify your answer.
i) The collection of all months of a year beginning with letter ‘M’
ii) A team of eleven best cricket players of the world.
iii) The collection of all girls in your class.
iv) The collection of all odd integers.
Solution:
i) The months of a year which begin with the letter ‘M’ are March and May.
The required set is {March, May}.
∴ It is a well defined collection of objects.
So, it is a set.
ii) It is not a set, because we cannot determine the eleven best cricket players of the world.
iii) It is a set, because we can divide whether he / she belongs to the given set or not.
iv) It is a set, because we can divide whether the number belongs to the given set or not.

Question 2.
If A = (1, 2, 3, 4} B = (5, 6, 7} and C = (p, q, r, s}, then fill the appropriate symbol, ∈ or ∉ in the blanks.
i) 2 ………… A
ii) 6 …………. C
iii) 4 ………… B
iv) q …………. C
v) 5 …………. B
vi) 8 …………. A
Solution:
i) 2 ∈ A
ii) 6 ∉ C
iii) 4 ∉ B
iv) q ∈ C
v) 5 ∈ B
vi) 8 ∉ A

TS 10th Class Maths Important Questions Chapter 2 Sets

Question 3.
Express the following statements using symbols.
i) The element p does not belong to ‘A’
ii) ‘q’ is an element of the set ‘B’
iii) 4 belongs to the set of Natural numbers N
iv) 5 belongs to the set prime numbers P
Solution:
i) p ∉ A
ii) q ∈ B
iii) 4 ∈ N
iv) 5 ∈ p

Question 4.
Write the following sets in roaster form.
i) A = (x : x is a natural number less than 8}
ii) B = (x : x is a two digital natural number such that the sum of its digits is 5}
iii) C = (x : x is a prime number which is a divisor of 30}
iv) D = { the set of all letters in the word CRICKET}
Solution:
i) A = (1, 2, 3, 4, 5, 6, 7}
ii) B = (14, 41, 23, 32}
iii) C = (2, 3, 5}
iv) D = { C, E, I, K, R, T}

Question 5.
Write the following sets in the set-builder form.
i) A = {5, 10, 15, 20}
ii) B = {3, 9, 27, 81}
iii) C = {4, 16, 64, 256}
iv) D = {1, 8, 27, 64, 125, ……….. 1000}
Solution:
i) A = {x : x is a multiple of 5 and less than 21}
ii) B = {x : x is a power of 3 and x is less than 5}
iii) C = {x : x is a power of 4, and x is less than 5}
iv) D = {x : x is a cube of natural numbers and not greater than 10}

TS 10th Class Maths Important Questions Chapter 2 Sets

Question 6.
Match the roaster form with set builder
i) {1, 2, 4, 8} (a) (x : x is an even natural number less than 11}
ii) {3, 5} (b) (x : x is a natural number and divisor of 8}
iii) {L, E, T, R} (c) (x : x is a prime number and a divisor of 15}
iv) (2, 4, 6, 8, 10}(d) (x : x is a letter of the word ‘LETTER’}
Solution:
i) b
ii) c
iii) d
iv) a

Question 7.
If A = {1, 2, 3, 4, 5}; B = {3, 4, 5, 6}, then find A ∪ B and A ∩ B.
Solution:
Given A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6}
A ∪ B = {1, 2, 3, 4, 5, 6}
A ∩ B = {3, 4, 5}.

Question 8.
If A = {4, 5, 6, 7, 8}; B = {7, 8, 9, 10, 11} then find A – B and B – A.
Solution:
Given A = {4, 5, 6, 7, 8},
B = {7, 8, 9, 10, 11}
A – B = {4, 5, 6}
B – A = {9, 10, 11}

TS 10th Class Maths Important Questions Chapter 2 Sets

Question 9.
Which of the following pairs of sets are disjoint ?
i) A = {3, 4, 5, 6}; B = {4, 6}
ii) A = {a, e, i, o, u}; B = {i, j, k}
iii) A = {5, 10, 15, 20}; B = {8, 16, 24}
iv) A = {1, 3, 5, 7};B = {2, 4, 6, 8}
Solution:
Disjoint sets: Two sets are said to be disjoint sets when they have no elements in common.
i) A and B are not disjoint sets (∵ 4, 6 are common)
ii) A and B are not disjoint sets (∵ i is common)
iii) A and B are disjoint sets (∵ No elements in common)
iv) A and B are disjoint sets (∵ No elements in common)

Question 10.
If A = {3, 4, 5, 6, 7}, B = {2, 4, 6, 8} then find n (A ∪ B).
Solution:
n(A) = 5 (∵ A contains 5 elements)
n(B) = 4 (∵ B contains 4 elements)
But A ∪ B = {2, 3, 4, 5, 6, 7, 8}
Now n(A ∪ B) = 7 (∵ A ∪ B contains 7 elements)

Question 11.
If A and B are two sets and n(A) – 15, n(B) = 25 and n(A ∩ B) = 10, find n (A ∪ B)
Solution:
We Know that n (A ∪ B) = n (A) + n (B) – n(A ∩ B)
∴ n(A ∪ B) = 15 + 25 – 10
n(A ∪ B) = 40 – 10 = 30

Question 12.
Which of the following sets are equal ?
i) A = {3, 4, 5, 6}, B = {7, 8, 9, 10}
ii) C = {a, b, c, d}, D = {d, c, a, b} :
iii) E = {2, 4, 6, 8}, F = {x : x is a positive even number < 10}
iv) G = {5, 10, 15, 20, . . . }, H = {x : x is a multiple of 5}
Solution:
i) A and B are not equal sets
ii) C and D are equal sets
iii) E and F are equal sets
iv) G and H are equal sets

TS 10th Class Maths Important Questions Chapter 2 Sets

Question 13.
State the reason for the following.
i) {4, 5,6, ……….. 12} ≠ {x : x ∈ N and 4 < x < 12}
ii) {2, 4, 6, 8, 10} ≠ {x : x = 2n + 1 and x ∈ N}
iii) {3, 6, 9, 12} ≠ {x : x is a multiple of 6}
iv) {1, 3, 5, 7, 9} ≠ {x : x is an even number}
Solution:
The first set is {4, 5, 6, … ,12}
Writing the second set in roaster form is {5, 6, 7, …,11}
The 1st and 2nd set have not exactly the same elements.
i) ∴ {4, 5, 6, . . . , 12} ≠ {x : x ∈ N and 4 < x < 12}

ii) The first set is {2, 4, 6, 8, 10}
Writing the 2nd set in roaster form is {3, 5, 7, 9}
∴ {2, 4, 6, 8, 10} ≠ {3, 5, 7, 9}

iii) The first set is {3, 6, 9, 12}
Writing the 2nd set is roaster form is {6, 12, 18, 24, … }
∴ {3, 6, 9, 12} ≠ {6, 12, 18, 24,. . .}

iv) The first set is {1, 3, 5, 7, 9}
Writing the 2nd set in roaster form is {2, 4, 6, 8}
∴ {1, 3, 5, 7, 9} ≠ {2, 4, 6, 8,…}

Question 14.
List all the subsets of the following :
i) A = {a, b}
ii) B = {p, q, r}
iii) {1, 2, 3}
Solution:
i) {a}, {b}, {a, b}, {Φ}
ii) {p}, {q}, {r}, {p, q}, {p, r}, {q, r}, {p, q, r}, {Φ}
iii) {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1,2, 3}, {Φ}

TS 10th Class Maths Important Questions Chapter 2 Sets

Question 15.
State which of the following sets are empty.
i) Set of integers which lie between 5 and 6.
ii) Set of even natural numbers divisible by 2.
iii) The set of lines which are parallel to the Y-axis.
iv) {x : x is a natural number, x < 6 and x > 8}
Solution:
i) It is empty set, because there is no integer lying between 5 and 6
ii) It is not empty set, because there are even natural numbers which are divisible by 2.
iii) It is not empty set, because there are number of lines parallel to the Y-axis is infinite.
iv) It is empty set, because there is no natural number satisfying this condition.

Question 16.
Which of the following set is finite or infinite ?
i) A = {x : x ∈ N and x < 50}
ii) B = {x : x ∈ N and x ≤ 10}
iii) C = {13, 23, 33, ……}
iv) D = {x : x ∈ N and x is even}
Solution:
i) A = {1,2, 3, 4,…. 49}
This set is finite, because there are 49 numbers possible to count

ii) B = {1, 2, 3, 4, . . . , 10}
This set is finite, because there are 10 numbers possible to count

iii) C = {13, 23, 33, . . . }
This set is infinite, because there are infinite numbers.

iv) D = {2, 4, 6, 8, . . . }
This set is infinite, because there are infinite even numbers.

TS 10th Class Maths Important Questions Chapter 2 Sets

Question 17.
If A = {1, 2, 3, 4, 5}; B = {4, 5, 6, 7} then find B – A.
Solution:
Given A = {1, 2, 3, 4, 5}; B = {4, 5, 6, 7}
B – A = {4, 5, 6, 7} – {1, 2, 3, 4, 5}
B – A = {6, 7}

Question 18.
If A = {0, 1, 2} and B = {2, 4} then find n (A ∪ B)
Solution:
Given A = {0, 1, 2}, B = {2, 4}
A ∪ B = {0, 1, 2, 4}
∴ n(A ∪ B) = 4

Question 19.
If A’ is the set of all primes below ‘5’ and ‘B’ is the set of all prime factors of ’30’, then is A – B = B – A?
Solution:
Given A = Set of all primes below 5 = {2, 3}
B = Set of all prime factors of 30 = {2, 3, 5}
A – B = { } and B – A = { 5 }
∴ A – B ≠ B – A

Question 20.
Represent the following through Venn- diagram.
i) A – B
ii) B – A
iii) A ∪ B
iv) A ∩ B
Solution:
i) Venn-diagram of A – B is
TS 10th Class Maths Important Questions Chapter 2 Sets 1

ii) Venn-diagram of B – A is
TS 10th Class Maths Important Questions Chapter 2 Sets 2

iii) Venn-diagram of A ∪ B is
TS 10th Class Maths Important Questions Chapter 2 Sets 3

iv) Venn-diagram of A ∩ B is
TS 10th Class Maths Important Questions Chapter 2 Sets 4

Question 21.
If A = (x : Y is a Natural number below 10}
B = {x : Y is an even number below 10}
C = {x : Y is an odd number below 10}then
find (i) A – B
(ii) A – C
(iii) B ∪ C
(iv) Also mention the Mutually disjoint sets among (i), (ii) and (iii).
Solution:
Given A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
B = {2, 4, 6, 8}
C = {1, 3, 5, 7, 9}
i) A – B = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}

ii) A – C = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 3, 5, 7, 9}
= {2, 4, 6, 8}

iii) B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9}

iv) (i) and (ii) are disjoint sets because there is no element in common.
(ii) and (iii) are not disjoint sets because there are 2, 4, 6, 8 common elements.
(i) and (iii) are not disjoint sets because there are 1, 3, 5, 7, 9 common elements.

TS 10th Class Maths Important Questions Chapter 2 Sets

Question 22.
If A = {1, 4, 9, 16, 25, . . . } then write it in set builder form.
Solution:
A = {1, 4, 9, 16, 25, ……..}
= {12, 22, 32, 42, 52, . . . . }
= {x2/ x ∈ N}

Question 23.
If A = {x / x is a prime number and x < 20} then B = {x / 2x + 1, x ∈ w and x < 9} then Find
(i) A ∩ B
(ii) B ∩ A
(iii) A – B
(iv) B – A. What do you observe ?
Solution:
A = {x / x is a prime number and x < 20}
B = {2x + 1, x ∈ w and x < 9}
∴ A = {2, 3, 5, 7, 11, 13, 17, 19}
B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
1) A ∩ B = {2, 3, 5, 7, 11, 13, 17, 19} ∩ (1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
= {3, 5, 7, 11, 13, 17, 19}

2) B ∩ A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} ∩ {2, 3, 5, 7, 11, 13, 17, 19}
= {3, 5, 7, 11, 13, 17, 19}

3) A – B = {2, 3, 5, 7, 11, 13, 17, 19} – {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
= {2}

TS 10th Class Maths Important Questions Chapter 2 Sets

4) B – A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} – {2, 3, 5, 7, 11, 13, 17, 19}
= {1, 9, 15}
Note : i) A ∩ B = B ∩ A
ii) A – B ≠ B – A

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Students must practice these TS Inter 2nd Year Maths 2A Important Questions Chapter 3 Quadratic Expressions to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 1.
Find the roots of the equation
3x2 + 2x – 5 = 0.
Solution:
The roots of the quadratic equation
ax + bx + c = 0 are \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
Here a = 3, b = 2 and c = – 5.
Therefore the roots of the given equation are
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 1
Hence 1 and – \(\frac{5}{3}\) are the roots of the given equation.

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Another method:
We can also obtain these roots in the following way.
3x2 + 2x – 5 = 3x2 + 5x -3x – 5
= x(3x+5) – 1 (3x+5)
= (x-1) (3x + 5)
= 3 (x – 1) \(\left(x+\frac{5}{3}\right)\)
Since 1 and \(\frac{5}{3}\) are the zeros of 3x2 + 2x – 5, they are the roots of 3x2 + 2x – 5 = 0.

Question 2.
Find the roots of the equation
4x2– 4x + 17 = 3x2 – 10x – 17.
Solution:
Given equation can be rewritten as x2 + 6x + 34 = 0.
The roots of the quadratic equation, ax2 + bx + c = 0 are
\(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
Therefore the roots of the given equation are
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 2
Hence the roots of the given equation are
– 3 + 5i and -3-5i

Question 3.
Find the roots of the equation
\(\sqrt{3} x^2+10 x-8 \sqrt{3}=0\)
Solution:
The roots of the quadratic equation a+ b
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 3

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 4.
Find the nature of the roots of 4x2 – 20x + 25 = 0.
Solution:
Here a = 4, b = -20 and c = 25
Hence Δ = b2 – 4ac = (-20)2– 4 (4) (25) = 0
Since Δ is zero and a, b, c are real, the roots of the given equation are real and equal.

Question 5.
Find the nature of the roots of 3x2 + 7x + 2 = 0.
Solution:
Here a=3, b=7 and c=2
Hence = b2 – 4ac
= (7)2 4(3) (2) = 25 = 52  > 0.
Since Δ = 5 is a square number, the roots of the given equation are rational and unequal.

Question 6.
For what values of m, the equation x2 – 2 (1 + 3m) x + 7 (3 + 2m) = 0 will have equal roots?
Solution:
The given equation will have equal roots if its discriminant is 0.
Here Δ = {- 2(1+3m))}2 + 4(1) 7(3 + 2m)
= 4(1+3m)2 – 28 (3+2m) = 4(9m2 – 8m – 20)
= 4 (m – 2) (9m + 10) = 36 (m – 2) m + \(\left(m+\frac{10}{9}\right)\)
Hence Δ = 0 ⇔ m = 2 or m= \(-\frac{10}{9}\)
Therefore the roots of the given equation are equal \(\left\{-\frac{10}{9}, 2\right\}\)

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 7.
If α and β are the roots of ax2 +bx + c = 0, find the values of α2 + β2 and α3 + β3 in terms of a, b, c.
Solution:
From the hypothesis
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 4

Question 8.
Form a quadratic equation whose roots are 2\(\sqrt{3}\) -5 and -2\(\sqrt{3}\) – 5.
Solution:
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 5

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 9.
Let α and β be the roots of the quadratic equation ax2 + bx + c = 0. If c ≠ 0, then form the quadratic equation whose roots are \(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\)
Solution:
From the hypothesis we have
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 6
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 7

Question 10.
Find a quadratic equation, the sum of whose roots is I and the sum of the squares of the roots is 13.
Solution:
Let a and 3 be the roots of a required equation
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 8

Therefore x2 – a + x a = 0 becomes  – x  – 6 = 0. This is a required equation.
Equations reducible to quadratic equations: We now explain by some illustrations how to solve some equations which are reducible to quadratic equations by suitable substitutions.

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 11.
Solve \(x^{\frac{2}{3}}+x^{\frac{1}{3}}\) – 2 = 0.
Solution:
On taking \(\mathrm{x}^{\frac{1}{3}}\) = t, the given equation becomes
t2 + t – 2= 0, which is a quadratic equation in t.
Hence a complex number a is a solution of the equation.
\(x^{\frac{2}{3}}+x^{\frac{1}{3}}\) – 2 = 0, if there exists λ such that λ2 + λ -2 = 0 and λ3 = α.
Therefore the set of all solutions of the given equation is
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 9
Therefore the solution set of the given equation is { – 8, 1}.

Question 12.
Solve 71+x +71-x = 50 for real x.
Solution:
The given equation can be written as
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 10
X = – 1 or x = 1
Therefore the solution set of the given equation is (-1, 1).

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 13.
Solve \(\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}\)
Solution:
On taking \(\sqrt{\frac{x}{1-x}}=t\) the given equation becomes
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 11

Question 14.
Find all numbers which exceed their square root by 12.
Solution:
Let x be any such number.
Then \(x = \sqrt{\mathrm{x}}\) + 12 ie., x – 12 = \(x = \sqrt{\mathrm{x}}\)
On squaring both sides and simplifying we obtain (x – 12)2
i.e., x2  – 24x 144 = x
i.e., x2 – 25x + 144 = 0
i.e.,x (x – 16) – 9(x – 16) = 0
i.e., (x – 16) (x – 9) = 0
The roots of the equation (x – 16) (x – 9) = 0 are 9 and 16.
But x = 9 does not satisfy equation (1), while x = 16 satisfies (1). Therefore the required number is 16.

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 15.
Prove that there is a unique pair of consecutive positive odd integers such that the sum of their squares is 290 and find it.
Solution:
Since two consecutive odd integers differ by
2, we have to prove that there is a unique
positive odd integer x such that
⇔ x2+(x+2)2=290 ……………. (1)
⇔ x2+ (x + 2)2 = 290
⇔x2 + x2 + 4x + 4 = 290
⇔ 2x2 + 4x – 286 = 0
⇔ x2 + 2x – 143 = 0
⇔ x2+ 13x – 11x – 143 = 0
⇔ x(x + 13) – 11 (x + 13) = 0
⇔ (x+ 13)(x – 11)=0
⇔ x∈( – 13,11)
Hence 11 is the only positive odd integer satisfying equation (1).
Therefore (11, 13) is the unique pair of integers which satisfies the given condition.

Question 16.
The cost of a piece of cable wire is Rs. 35/-. If the length of the piece of wire is 4 meters more and each meter costs Rs. 1/- less, the cost would remain unchanged. What is the length of the wire?
Solution:
Let the length of the piece of wire be ‘l’ meters and the cost of each meter be Rs. x I –
By the given conditions lx = 35 ……………………. (1)
Also, (1+ 4) (x – 1) = 35
i.e., lx – 1 + 4x – 4 = 35 ……………………… (2)
From (1) and (2), 35 – l + 4x – 4 = 35
i.e., 4x = l + 4
Therefore \(x=\frac{l+4}{4}\)
On substituting this value of ‘x’ in (1) and simplifying, we get
\(l\left(\frac{l+4}{4}\right)=35\)
i.e., l2 + 4l- 140 = 0
i.e., l2 + 14l – 10l- 140 = 0
i.e., l(l+ 14)-10 (1+ 14)=0
i.e., (l+ 14) (1- 10) = 0
The roots of the equation (l + 14) (l- 10) = 0 are-14 and 10.
Since the length can not be negative, i = 10
Therefore the length of the piece of wire is 10 metres.

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 17.
One fourth of a herd of goats was seen in the forest. Twice the square root of the number in the herd had gone up the bill and the remaining l5goatswere on the bank of the flyer. Find the total number of goats
Solution:
Let the number of goats in the herd be ‘x’.
By the given conditions, the number of goats seen in the forest is \(\frac{x}{4}\) the number of goats gone up the hill is \(2 \sqrt{x}\) and the number of the remaining goats which were on the bank of the river is 15.
Therefore \(\frac{x}{4}\) + \(2 \sqrt{x}\) + 15 = X
i.e., x + \(8 \sqrt{x}\) + 60 = 4x
i.e., 3x – \(8 \sqrt{x}\) –  60 = 0.
On taking \(\sqrt{x}\) = t, this equation becomes
3t2 – 8t – 60 = 0
i.e., 3t2– 18t+ 10t – 60 = 0
i.e.,3t(t – 6)+ 10(t – 6)=0
i.e.,(t – 6)(3t+ 10)=0
The roots of the equation (t – 6) (3t + 10) = 0 are 6 and \(-\frac{10}{3}\)
Hence 3x – \(8 \sqrt{x}\) – 60 = 0 = \(\sqrt{x}\) = 36
(since \(\sqrt{x}\) is non-negative)⇔ x = 36
Therefore the total number of goats in the herd is 36

Question 18.
In a cricket match Anli took one wicket less than twice the number of wickets taken by Ravi. If the product of the number of wickets taken by them is 15, find the number of wickets taken by each of them.
Solution:
Let the number of wickets taken by Anil and
Ravi be x and y respectively.
Then x = 2y – 1 …………… (1)
xy = 15 ……………(2)
From (1) and (2), (2y – 1) y = 15
i.e., 2y2-y-15=0
i.e., 2y2-6y + 5y- 15 = 0
i.e., 2y(y-3) + 5 (y-3) = 0
i.e., (y-3)(2y + 5)= 0
The roots of the equation (y – 3) (2y + 5) = 0 are 3 and \(-\frac{5}{2}\)
Since the number of wickets must be positive integer y = 3.
From (2) we get 3x = 15, i.e., x = 5
Therefore the wickets taken by Anil and Ravi are 5 and 3 respectively.

Question 19.
Some points on a plane are marked and they are connected pairwise by line segments. if the total number of line segments formed is 10, find the number of marked points on the plane.
Solution:
Let the number of points marked on the plane be ‘x’. Since each point is joined to the remaining (x – 1) points, the number of line segments having a given point as an end point is (x – 1). Hence the total number of line segments formed appears to be x (x – 1).
But in this counting, each line segment is counted exactly twice at each of its end points. Hence the total number of line
segments actually formed is \(\frac{x(x-1)}{2}\)
Therefore by hypothesis \(\frac{x(x-1)}{2}\) = 10
i.e., x – x – 20 = 0
i.e., (x – 5) (x +4) = 0
The roots of the equation (x – 5) (x + 4) = 0 are – 4 and 5.
x can not be negative, so x = 5.
Therefore the number of points marked on the plane is 5.

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 20.
Suppose that the quadratic equations ax2 + bx +c = 0 and bx2 + cx + a = 0 have a common root. Then show that
a3 + b3 + C3 = 3abc.
Solution:
The condition for two quadratic equations
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 12

Question 21.
For what values of x the expression – 5x – 14 is positive?
Sol.
Since x2 – 5x  – 14 = (x + 2) (x – 7), the roots of the equation x2 – 5x – 14 = 0 are – 2 and 7.
Flere the coefficient of x2 is 1, which is positive.
Hence x2 – 5x – 14 is positive when x < – 2 or x> 7.

Question 22.
For what values of x the expression  – 6x2+2x – 3 is negative?
Solution:
– 6x + 2x – 3 = 0 can be written as
6x2 – 2x + 3= 0.
The roots of this equation are
\(\frac{2 \pm \sqrt{(-2)^2-4(6)(3)}}{2(6)}\)
Therefore the roots of – 6x2+2x – 3 = 0 are non-real complex numbers.
Here the coefficient of x2 is – 6 which is negative.
Hence – 6x2 + 2x – 3 < 0 for all x R

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 23.
Find the value of x at which the following expressions have maximum or minimum.

(i) x2+5x+6
(ii) 2x – x2+7
Solution:
(i) In the expression x2+5x+6, the coefficient of x2 is positive.
So x2 + 5x + 6 has absolute minimum at
x – (since b=5, a= 1).

(ii) In the expression 2x – x2 + 7, the coefficient of x2 is negative.
So 2x – x2 + 7 has an absolute maximum at
\(x=-\frac{2}{2(-1)}=1\) (since b=2, a-1).

Question 24.
Find the maximum or minimum value of the quadratic expression.
(i) 2x – 7 – 5 x 2
(ii) 3x2 + 2x + 11
Solution:
(i) Comparing the given expression with ax2 + bx + C,
we have a = -5, b = 2 and c = – 7.
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 13

Question 25.
Find the changes in the sign of 4x – 5x2 + 2 for x ∈ R and find the extreme value.
Solution:
Comparing the given expression with
ax2 + bx+c, we have a = – 5 < 0.
The roots of the equation 5x2 – 4x – 2 = 0 are \(\frac{2 \pm \sqrt{14}}{5}\)
Therefore, when \(\frac{2-\sqrt{14}}{5}<x<\frac{2+\sqrt{14}}{5}\) the sign of 4x – 5x2 + 2 is positive and when
\(x<\frac{2-\sqrt{14}}{5} \text { or } x>\frac{2+\sqrt{14}}{5}\) the sign of 4x – 5x2 + 2 is negitive
Since a < 0, the maximum value of the expression 4x – 5x2 + 2 is
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 15

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 26.
Show that none of the values of the function \(\frac{x^2+34 x-71}{x^2+2 x-7}\) over R lies between 5 and 9.
Solution:
Let y0 be a value of the given function.
Then ∋ x0 ∈ R such that
\(y_0=\frac{x_0^2+34 x_0-71}{x_0^2+2 x_0-7}\)
If y0 = 1, then clearly y0 ∉(5,9).
Suppose that y0 ≠ 1.
Then the equation y0 (x2 + 2x – 7) = x2 + 34x – 71 is a quadratic equation and x0 is a real root of it.
Therefore(y0 -1)x2+(2y0 – 34)x – (7y0 – 71)
= 0 is a quadratic equation having a real root x0.
Since all the coefficients of this quadratic equation are real, the other root of the equation is also real.
Therefore
Δ = (2y0 – 34)2 +4 (y0 -1) (7y0 – 71) ≥ 0.
On simplifyìng this we get
y20 – 14y0 + 45 ≥ 0
i.e., (y0-5) (y0– 9) ≥ 0
Therefore y0 ≤ 5 or y0 ≥ 9.
Hence y0 does not lie in (5, 9).
Hence none of the values of the given function over R lies between 5 and 9.

Question 27.
Find the maximum value of the function
\(\frac{x^2+14 x+9}{x^2+2 x+3}\) over R.
Solution:
Since the discriminant of x2 + 2x + 3 is negative, x2 + 2x + 3 is never zero on R.
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 16
The above inequality is true even when y0 = 1.
Hence the range of the given function on R is a subset of [- 5, 4].
For m = 4, the equation (1 – m) x2 + (14 – 2m) x+ 9 – 3m = 0 is a quadratic equation with real coefficients and discriminant zero and hence has only one root which is real. Let it be α.
Then for m = 4, we have (1 – m) α2 + (14 – 2m) α + 9 – 3m = 0.
Hence \(\frac{\alpha^2+14 \alpha+9}{\alpha^2+2 \alpha+3}=m=4\)
Hence 4 is in the range of the given function.
Since the range is a subset of [-5, 4] and 4 is in the range, 4 is the absolute maximum value given function over R.

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 28.
Find the solution set of x2 + x – 12 ≤ 0 by both algebraic and graphical methods.
Solution:
Algebraic Method:
We have x2 + x – 12 (x + 4) (x – 3)
Hence – 4 and 3 are the roots of the equation x2 + x – 12 = 0
Since the coefficient of x2 in the quadratic expression x2 + x – 12 = 0 is positive, x2 + x – 12 is negative if – 4 < x < 3 and positive if either x < – 4 or x > 3.
Hence x2 + x – 12 ⇔ -4 ≤ x ≤ 3
Therefore the solution set is
{ x ∈ R : – 4 ≤ x ≤ 3}
Graphical Method:
Let y = f(x) x2 – x – 12
The values of y at some selected values of x are given in the following table:
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 17
The graph of the function y = f(x) is drawn using the above tabulated values.
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 18
Therefore from the graph of y = f(x we observe that
y = x2 – x – 12 ≤ 0 if i – 4 ≤ x≤3
Hence the solution set is {x ∈ R : – 4 ≤ x ≤ 3}.

Question 29.
Find the set of values of x for which the inequalities x2-3x-10<0, 10x – x2-16>0 hold simultaneously.
Solution:
We have x2 -3x – 10 (x + 2) (x – 5)
Flence – 2 and 5 are the roots of the equation x2-3x-10= 0.
Since the coefficient of x2 in the quadratic 30. expression x2 – 3x -10 is positive.
x2-3x-10<0 =-2<x<5
We have 10x – x2-16 = – (x – 2)(x – 8)
Hence 2 and 8 are the roots of the equation
10x – x2 – 16 = 0.
Since the coefficient of x2 in the quadratic expression 10x – x2 -16 is negative.
10x- x2-16>0=2 < x< 8
Hence x2 -3x -10 < 0 and
⇔ X∈ (-2,5)∩(2,8)
Therefore the solution set is {x∈R : 2 < x < 5}.
10x – x2 – 16 >0

Question 30.
Solve the inequation \(\sqrt{x+2}>\sqrt{8-x^2}\)
Solution:
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 19
We have (x + 2) >(8-x2)
⇔ x2+ x – 6>0
Now x2 + x-6 = (x + 3)(x-2)
Since -3 and 2 are the roots of the equation
x2 + x – 6 = 0 and the coefficient of x2 in x2 + x – 6 is positive
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 20

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 31.
Solve the inequation
\(\sqrt{(x-3)(2-x)}<\sqrt{4 x^2+12 x+11}\)
Solution:
The given inequation is equivalent to the following two inequalities.
(x-3) (2-x) ≥ 0 and (x – 3)(2-x)4x2+ 12x+11
(x – 3)(2- x)≥ 0 (x-2)(x-3)≤0
– x2+ 5x- 6<4x2 + 12x+ 11
⇔ 5x2 + 7x + 17 > 0
The discriminant of the quadratic expression 5x2 + 7x + 17 is negative.
Hence 5x2 + 7x + 17 > 0 ∀x ∈ R.
Hence the solution set of the given inequation is {x ∈R :2 ≤ x ≤ 3}.

Question 32.
Solve the inequation
\(\frac{\sqrt{6+x-x^2}}{2 x+5} \geq \frac{\sqrt{6+x-x^2}}{x+4}\)
Solution:
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 21
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 22

TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 33.
Solve the inequation
\(\sqrt{x^2-3 x-10}>(8-x)\)
Solution:
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 23
TS Inter 2nd Year Maths 2A Quadratic Expressions Important Questions 24

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 9 Probability Ex 9(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

I.
Question 1.
If 4 fair coins are tossed simultaneously, then find the probability that 2 heads and 2 tails appear.
Solution:
If getting Head and Tail are denoted by H, T the number of sample points in the sample space n(S ) = 24 = 16.
If E is the event of getting 2 heads and 2 tails then E = {THHT, TTHH, HTHT, HHTT, TTHH, HTTH}
∴ n ( E ) = 6
∴ Probability of getting 2 heads and 2 tails.
P(E) = \(\frac{n(E)}{n(S)}=\frac{6}{16}=\frac{3}{8}\)

Question 2.
Find the probability that a non-leap year contains
(i) 53 Sundays
(ii) 52 Sundays only.
Solution:
A non-leap year contains 365 days in which there are 52 full weeks and one day.
52 full weeks contain 52 × 7 = 364 days and
left out of one day may be either Sun, Mon, Tues, Wed, Thu, Fri or Sat.
∴ n(S) = 7.

i) For a non-leap year to contain 53 Sundays, we have only on possibility to have 365th day a Sunday.
∴ n(E) = 1

ii) For a non-leap year to contain 52 Sundays, we have 6 possibilities for 365th day be a day other Sunday
∴ n ( E ) = 6
∴ Probability of getting 52 Sundays in a non leap year is
∴ p(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{6}{7}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

Question 3.
Two dice are rolled. What is the probability that none of the dice shows the number 2?
Solution:
n = 6, m = 2 i.e., n(S) = 6, n (E) = 2
P ( E ) = Probability of not getting 2 when a single die is rolled = \(\frac{5}{6}\).
∴ The probability of not getting 2 when two dice are rolled = \(\left(\frac{5}{6}\right) \times\left(\frac{5}{6}\right)=\left(\frac{5}{6}\right)^2\).

Question 4.
In an experiment of drawing a card at random from a pack, the event of getting a spade is denoted by A and getting a pictured card (King, Queen or Jack) is denoted by B. Find the probabilities of A, B, A ∩ B and A ∪ B.
Solution:
Total number of cards in the pack n ( S ) = 52
If A is the event of getting a spade card then n (A) = 13
∴ Probability of getting a spade card,
P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{13}{52}=\frac{1}{4}\).

If B is the event of getting a picture card (King, Queen or Jack) then n (B) = 12
(there are 4 kings, 4 queens and 4 jacks)
∴ Probability of getting a picture card = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{12}{52}=\frac{3}{13}\)
∴ P(B) = \(\frac{3}{13}\)
Common to the events A and B there is a spade king, spade queen and spade jack
∴ n(A ∩ B) = 3
∴ Probability of getting (A ∩ B) is
P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{3}{52}\)
By addition theorem,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{13}{52}+\frac{12}{52}-\frac{3}{52}=\frac{22}{52}=\frac{11}{26}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

Question 5.
In a class of 60 boys and 20 girls, half of the boys and half of the girls know cricket. Find the probability of the event that a person selected from the class is either a boy or a girl who knows cricket.
Solution:
Let E, be the event of selecting a boy and E, be the event of selecting a person who knows cricket.
P(E1) = \(\frac{60}{80}\);
P(E2) = \(\frac{40}{80}\)
(30 boys + 10 girls who knows cricket)
P (E1 ∩ E2) = \(\frac{30}{80}\)
(∵ 30 boys are common to E1 and E2)
∴ Probability of selecting a boy or a girl who knows cricket is .
P(E1 ∪ E2) = P(E1) + P(E2) – P(E1 ∩ E2)
= \(\frac{60}{80}+\frac{40}{80}-\frac{30}{80}=\frac{70}{80}=\frac{7}{8}\)
∴ ProbabilIty of the event that a person selected from the class is either a boy or a girl who knows cricket = \(\frac{7}{8}\).

Question 6.
For any two events A and B, show that P\(\left(A^c \cap B^c\right)\) = 1 + P(A ∩ B) – P(A) – P(B)
Solution:
\(\left(A^c \cap B^c\right)\) = \((A \cup B)^c\)
∴ P\(\left(A^c \cap B^c\right)\) = P\((A \cup B)^c\)
= 1 – P(A ∪ B) (∵ P(S) = 1)
= 1 – [P (A) + P (B ) – P (A ∩ B)]
= 1 + P(A ∩ B) – P(A) – P(B)
[From Demorgan’s law \(\left(A^c \cap B^c\right)\) = \((A \cup B)^c\)].

Question 7.
Two persons A and B are rolling a dice on the condition that the person who gets 3 will win the game. If A starts the game, then find the probabilities of A and B respectively to win the game.
Solution:
Two persons A and B toss a die.
The probability of throwing 3 with a die = \(\frac{1}{6}\)
The probafility of not throwing 3 with the die = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
If A is to win, A must throw 3 in the 1st or 2nd or 3rd rounds etc.
P(A) = \(\frac{1}{6}+\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}+\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}+\ldots .\)
which is an infinite G.P. with a = \(\frac{1}{6}\)
r = \(\frac{5}{6} \cdot \frac{5}{6}=\left(\frac{5}{6}\right)^2\)
∴ P(A) = S
= \(\frac{a}{1-r}=\frac{\frac{1}{6}}{1-\left(\frac{5}{6}\right)^2}=\frac{6}{11}\)
B wins when A fails to win
∴ P(B) = \(P(A)^c\)
= 1 – P(A)
= 1 – \(\frac{6}{11}\) = \(\frac{5}{11}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

Question 8.
A, B, C are 3 newspapers from a city, 20% of the population read A, 16% read B, 14% read C, 8% both A and B, 5% both A and C, 4% both B and C and 2% all the three. Find the percentage of the population who read atleast one newspaper.
Solution:
Given P(A) = \(\frac{20}{100}\),
P(B) = \(\frac{16}{100}\)
P(C) = \(\frac{14}{100}\)
P(A ∩ B) = \(\frac{8}{100}\)
P(A ∩ C) = \(\frac{5}{100}\)
P(B ∩ C) = \(\frac{4}{100}\) and
P(A ∩ B ∩ C) = \(\frac{2}{100}\)
∴ The percentage of the population who read atleast one newspaper.
∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)
= \(\frac{20}{100}+\frac{16}{100}+\frac{14}{100}-\frac{8}{100}-\frac{4}{100}-\frac{5}{100}+\frac{2}{100}\)
= \(\frac{35}{100}\)
∴ 35 % read as atleast one newspaper.

Question 9.
If one ticket is randomly selected from tickets numbered 1 to 30, then find the probability that the number on the ticket is
i) a multiple of 5 or 7
ii) a multiple of 3 or 5.
Solution:
Given n (S) = 30
i) Let E1 he the event of getting a number which is a multiple of 5.
Let E2 be the event of getting a number which is a multiple of 7..
n(E1) = {5, 10, 15, 20, 25, 30} = 6
and n(E2) = {7, 14, 21, 28} = 4
∴ P(E1) = \(\frac{n\left(E_1\right)}{n(S)}=\frac{6}{30}\)
P(E2) = \(\frac{\mathrm{n}\left(\mathrm{E}_2\right)}{\mathrm{n}(\mathrm{S})}=\frac{4}{30}\)
∴ P(E1 ∪ E2) = P(E1) + P(E2)
(∵ P(E1 ∩ E2) = 0, because E1 ∩ E2 = Φ)
= \(\frac{6}{30}+\frac{4}{30}=\frac{10}{30}=\frac{1}{3}\)

ii) Let E3 be the event of getting ticket which is a number multiple of 3.
Let E4 be the event of getting a number which is a multiple of 5.
n(E3) = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}
n(E3) = 10
∴ P(E3) = \(\frac{n\left(E_3\right)}{n(S)}=\frac{10}{30}=\frac{1}{3}\)
E4 = {5, 10, 15, 20, 25, 30} = 6
n(E4) = 6,
∴ P(E4) = \(\frac{\mathrm{n}\left(\mathrm{E}_4\right)}{\mathrm{n}(\mathrm{S})}=\frac{6}{30}=\frac{1}{5}\)
n(E3 ∩ E4) = {15, 30} = 2
∴ P(E3 ∩ E4) = \(\frac{\mathrm{n}\left(\mathrm{E}_3 \cap \mathrm{E}_4\right)}{\mathrm{n}(\mathrm{S})}=\frac{2}{30}\)
∴ P(E3 ∪ E4) = P(E3) + P(E4) – P(E3 ∩ E4)
= \(\frac{10}{30}+\frac{6}{30}-\frac{2}{30}=\frac{14}{30}=\frac{7}{15}\)
∴ Probability for the number on the ticket to be a multiple of 3 or 5 is \(\frac{7}{15}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

Question 10.
If two numbers are selected randomly from 20 consecutIve natural numbers, find the probability that the sum of the two numbers is
i) an even number
ii) an odd number.
Solution:
i) Let A be the event that the sum of the numbers is even when two numbers chosen out of 20 consecutive positive integers.
n(S) = 20C2 = 190
n (A) = 10C2 + 10C2
= 45 + 45 = 90
∴ Probability for the sum of two numbers to be even = \(\frac{90}{190}=\frac{9}{19}\)

ii) Probability for the sum of numbers to be odd = 1 – \(\frac{9}{19}\) = \(\frac{10}{19}\).

Question 11.
A game consists of tossing a coin 3 tintes and noting its outcome. A boy wins if all tosses give tite same outcome and loses otherwise. Find the probability that the boy loses the gante.
Solution:
If a coin is tossed 3 times then the total number of sample points n(S) = 23 = 8
For winning the game all tosses gives the same outcome.
Let E be such that event then E = {HHH, TTT}
∴ n(E) = 2
P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{2}{8}\)
∴ The probability for the boy to loose the game
P\((\overline{\mathrm{E}})\) = 1 – P(E)
= 1 – \(\frac{2}{8}\) = \(\frac{6}{8}\).

Question 12.
If E1, E2 are two events with E1∩ E2 = Φ, then show that
\(P\left(E_1^C \cap E_2^C\right)=P\left(E_1^C\right)-P\left(E_2\right)\).
Solution:
\(P\left(E_1^C \cap E_2^C\right)=P\left(E_1^C\right)-P\left(E_2\right)\)
= 1 – [P(E1 ∪ E2)]
= 1 – [P(E1) + P(E2)]
[∵ P(E1 ∩ E2) = P(Φ) = 0]
(∵ From Addition theorem)
= 1 – P(E1) – P(E2)
= \(\mathrm{P}\left(\mathrm{E}_1^{\mathrm{C}}\right)\) – P(E2).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

II.
Question 1.
A pair of dice is rolled 24 times. A person wins by not getting a pair of 6’s on any ot the 24 rolls. What is the probability of his winnIng ? (This is the problem proved by the French gambler Chevalier de Mere (1607 – 1685) to Blaise Pascal, who in turn discussed it with Pierre de Fermat and solved the saine).
Solution:
When a pair of dice are rolled at a time, we get n(S) = 36.
If \(\overline{\mathrm{E}}\) is the event of getting a pair of 6’s then
n(\(\overline{\mathrm{E}}\)) = 1
∴ P(\(\overline{\mathrm{E}}\)) = \(\frac{1}{36}\)
∴ Probability of getting a pair of 6’s when two dice are roiled at a time = \(\frac{1}{36}\)
∴ Probability of not getting a pair of 6’s when two dice are rolled = P (E)
= 1 – P (\(\overline{\mathrm{E}}\))
= 1 – \(\frac{1}{36}\) = \(\frac{35}{36}\)
∴ A person wins the game if the dice are rolled for 24 times without getting pair of sixes.
∴ The probability that the person wins the game = \(\left(\frac{35}{36}\right)^{24}\).

Question 2.
If P is a probability function, then show that for any two events A and B.
P(A ∩ B) ≤ P(A) ≤ P(A ∪ B) ≤ P(A) + P(B)
Solution:
Here P(A ∩ B) = P(A) . P\(\left(\frac{\mathrm{B}}{\mathrm{A}}\right)\) where
P\(\left(\frac{\mathrm{B}}{\mathrm{A}}\right)\) is the conditional probability of B relative to A.
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ P(A ∩ B) ≤ P(A) …………(i)
(∵ P\(\left(\frac{\mathrm{B}}{\mathrm{A}}\right)\) ≤ 1)
P(A) ≤ P(A ∪ B) …………..(ii)
(∵ A ⊆ A ∪ B)
∴ P(A ∪ B) ≤ P(A) + P(B) ………….(iii)
From (i), (ii) and (iii).
P(A ∩ B) ≤ P(A) ≤ P(A ∪ B) ≤ P(A) + P(B).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

Question 3.
In a box contaIning 15 bulbs, 5 are defective. If 5 bulbs are selected at random from the box, find the probabflky of the event, that
i) none of them is defective.
ii) only one of theni is defective.
iii) atleast omie of ihietim is defective.
Solution:
Total number of bulbs = 15
Defective bulbs = 5
Number of good bulbs = 10
n(S) = 15C5 = 3003.

i) None of theni is defective:
If E is the event of selecting all good bulbs from 10,
= 10C2 = 252
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{252}{3003}=\frac{12}{143}\).

ii) Only one of them is defective:
So selecting one defective from 5 and 4 good bulbs from 10.
n(E1) = 10C4 . 5C1 = 1050
∴ P(E1) = \(\frac{\mathrm{n}\left(\mathrm{E}_1\right)}{\mathrm{n}(\mathrm{S})}=\frac{1050}{3003}=\frac{50}{143}\)

iii) Atleast one of them is defective:
P(E) = 1 – \(\frac{12}{143}\)
= \(\frac{131}{143}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

Question 4.
A and B are seeldng admission into IIT. If the probability for A to be selected is 0.5 and that both to be selected is 0.3, then it is possible that, the probability of B to be selected is 0.9?
Solution:
Given that P(A) = 0.5; P (A ∩ B) = 0.3
and P(A ∪ B) ≤ 1
⇒ P (A) + P(B ) – P (A ∩ B) ≤ 1
⇒ 0.5 + P(B) – 0.3 ≤ 1
⇒ P( B) + 0.2 ≤ 1
⇒ P(B) ≤ 1 – 0.2 = 0.8
∴P (B) = 0.9 is not possible.

Question 5.
The probability for a contractor to get a road contract is \(\frac{2}{3}\) and to get a building contract is \(\frac{5}{9}\). The probability to get atleast one contract is \(\frac{4}{5}\). Find the probability that he gets both the contracts.
Solution:
Let A be the event that a contractor gets road contract and R he the event that a contractor gets a building contract.
Given P(A) = \(\frac{2}{3}\);
P(B) = \(\frac{5}{9}\)
P(A ∪ B) = \(\frac{4}{5}\)
We have to find P(A ∩ B).
From Addition theorem on probabilities, we have the probability that the contractor gets both the contracts is
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= \(\frac{2}{3}+\frac{5}{9}-\frac{4}{5}=\frac{19}{45}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

Question 6.
In a committee of 25 members, each member is proficient either in Mathematics or in Statistics or in both. If 19 of these are proficient in Mathematics, 16 in Statistics, find the probability that a person selected from the committee is proficient in both.
Solution:
Let A be the event that a person is proficient in Mathematics and B be the event that a person is proficient in Statistics.
n(S) = 25;
n(A) = 19;
n(B) = 16
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= \(\frac{19}{25}+\frac{16}{25}\) – 1
= \(\frac{35-25}{25}=\frac{10}{25}=\frac{2}{5}\)
10 members arc proficient in Mathematics and Statistics.

Question 7.
A, B, C are three horses in a race. The probability of A to win the race is twice that of B and probability of B is twice that of C. What are the probablilties of A, B and C to win the race?
Solution:
Let probability of C to win the race be ‘x’
i.e.. P(C) = x
Given that P(B) = 2P(C) = 2x
and P(A) = 2P(B) = 2 × 2x = 4x
Sum of the probabilities is ‘1’.
x + 2x + 4x = 1
⇒ 7x = 1
⇒ x = \(\frac{1}{7}\)
∴ P(A) = 2(\(\frac{1}{7}\)) = \(\frac{2}{7}\)
and P(C) = x = \(\frac{1}{7}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

Question 8.
A bag contains 12 two rupee coins, 7 one rupee coins and 4 half rupee coins. 1f three coins are selected at random, then find the probability that
i) the sum of three coins is maximum
ii) the sum of three is minimum
iii) each coin is of different value.
Solution:
Total number of coins 12 + 7 + 4 = 23 coins
n(S) = 23C3
i) The sum of the three coins is maximum when we select 3 two rupee coins from 12 coins in 12C3 ways.
∴ n(E) = 12C3
Probability that te sum of three coins is maximum.
P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} C_3}{{ }^{23} C_3}\)

ii) The sum of the three coins is minimum when we selecting 3 half rupee coins from 4 half a rupee coins which can be done in 4C3 ways.
∴ n(E) = 4C3
∴ Probability for the sum of three coins to be minimum P(E) = \(\frac{{ }^4 C_3}{{ }^{23} C_3}\)

iii) Three coins each of different value
n(E) = 12C1 . 7C1 . 4C1 and probability that the each coin has of different value = \(\frac{{ }^{12} C_1 \cdot{ }^7 C_1 \cdot{ }^4 C_1}{{ }^{23} C_3}\) ways.

Question 9.
The probabilities of three events A, B, C are such that P (A) = 0.3, P(B) = 0.4, P(C) = 0.8, P(A ∩ B) = 0.08, P(A ∩ C) = 0.28, P(A ∩ B ∩ C) = 0.09 and P(A ∩ B ∩ C) ≤ 0.75. Show that P(B ∩ C) lies in the interval [0.23, 0.48].
Solution:
Probability for any event is always less than or equal to ‘1’ and given that
P (A ∪ B ∪ C) ≥ 0.75
∴ 0.75 ≤ P(A ∪ B ∪ C) ≤ 1
⇒ 0.75 ≤ P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C) ≤ 1
⇒ 0.75 ≤ 0.3 + 0.4 + 0.8 – 0.08 – P(B ∩ C) – 0.28 + 0.09 ≤ 1
⇒ 0.75 ≤ 1.23 – P(B ∩ C) ≤ 1
⇒ – 1.23 + 0.75 ≤ – P(B ∩ C) ≤ 1 – 1.23
⇒ – 0.48 ≤ – P(B ∩ C) ≤ – 0.23
⇒ 0.23 ≤ P(B ∩ C) ≤ 0.48
⇒ P(B ∩ C) ∈ [0.23, 0.48].

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

Question 10.
The probabilities of three mutually exclusive events are respeclively given as \(\frac{1+3 p}{3}, \frac{1-\mathbf{p}}{4}, \frac{1-2 \mathbf{p}}{2}\). Prove that \(\frac{1}{3} \leq p \leq \frac{1}{2}\).
Solution:
Suppose A, B, C are exclusive events
such that P(A) = \(\frac{1+3 \mathrm{P}}{3}\)
P(B) = \(\frac{1-P}{4}\)
P(C) = \(\frac{1-2 P}{2}\)
We know that
0 ≤ P(A) ≤ 1
o ≤ \(\frac{1+3 \mathrm{P}}{3}\) ≤ 1
0 ≤ 1 + 3P ≤ 3
– 1 ≤ 3P ≤ 3 – 1
\(\frac{-1}{3} \leq P \leq \frac{2}{3}\) ……………(1)
0 ≤ P(B) ≤ 1
0 ≤ \(\frac{1-\mathrm{P}}{4}\) ≤ 4
0 ≤ 1 – P ≤ 4
– 1 ≤ – P ≤ 4 – 1
1 ≥ P ≥ – 3
– 3 ≤ P ≤ 1 …………(2)
0 ≤ P(C) ≤ 1
0 ≤ \(\frac{1-2 \mathrm{P}}{2}\) ≤ 1
0 ≤ 1 – 2P ≤ 2
– 1 ≤ – 2P ≤ 2 – 1
1 ≥ 2P ≥ – 1
\(\frac{1}{2} \geq \mathrm{P} \geq-\frac{1}{2}\)
\(\frac{-1}{2} \leq P \leq \frac{1}{2}\) ……….(3)
Since A, B, C are exclusive events,
0 ≤ P(A ∪ B ∪ C) ≤ 1
⇒ 0 ≤ P(A) + P(B) + P(C) ≤ 1
⇒ 0 ≤ \(\frac{4+12 P+3-3 P+6-12 P}{12}\) ≤ 1
⇒ 0 ≤ \(\frac{13-3 P}{12}\) ≤ 1
⇒ 0 ≤ 13 – 3p ≤ 12
⇒ – 13 ≤ – 3P ≤ 12 – 13
⇒ 13 ≥ 3P ≥ 1
⇒ \(\frac{13}{3} \geq \mathrm{P} \geq \frac{1}{3}\)
⇒ \(\frac{1}{3} \leq \mathrm{P} \leq \frac{13}{3}\) ……………(4)
Max.of \(\left\{\frac{-1}{3},-3, \frac{-1}{2}, \frac{1}{3}\right\}=\frac{1}{3}\)
Min.of \(\left\{\frac{2}{3}, 1, \frac{1}{2}, \frac{13}{3}\right\}=\frac{1}{2}\)
(1), (2), (3) and (4) holds if \(\frac{1}{3} \leq P \leq \frac{1}{2}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

Question 11.
On a festival day, a man plans to visit 4 holy temples A, B, C, D in a random order. Find the probability that he visits
(i) A before B
(ii) A before B and B before C
Solution:
Given that 4 holy temples are A, B, C and D.
Number of ways to visit 4 holy tempies in 4P4 ways.
∴ n(S) = 4P4
= 4! = 24

i) A before B:

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b) 1

ii) A before B and B before C:

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b) 2

Question 12.
From the employees of a company, 5 persons are selected to represent them in the managing committee  lite company. The particulars of 5 persons are as follows :

NameSexAge in years
1. HarishM30
2. RohanM33
3. SheetalaF46
4. AlisF28
5. SalimM41

 

A person is selected at random front this group to act as a spokesperson. Find the probability that the spokesperson will be either male or above 35 years.
Solution:
Let ‘A’ be the event of selecting a male
n(A) = 3
Let ‘B’ be the event of selecting a person whose age is above 35.
n(B) = 2
n(S) = 5, n(A ∩ B) = 1
P(A ∩ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A \cap B)}{n(S)}\)
= \(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}=\frac{4}{5}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

Question 13.
Out of 100 Students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, find the probability that
(ï) you both enter the same section
(ii) you both enter the different sections.
Solution:
n(S) = 100C40
i) You both enter (lie saine seclion:
n(A) = 98C38 + 98C58
P(A) = \(\frac{n(A)}{n(S)}\)
= \(\frac{{ }^{98} C_{38}+{ }^{98} C_{58}}{{ }^{100} C_{40}}=\frac{17}{33}\)

ii) You both enter the different sections :
n(A) = 98C39 + 98C59
P(A) = \(\frac{n(A)}{n(S)}\)
= \(\frac{{ }^{98} C_{39}+{ }^{98} C_{59}}{{ }^{100} C_{40}}=\frac{16}{33}\).