TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.1

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 5 Measures of Lines and Angles Ex 5.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Exercise 5.1

Question 1.
Give any five examples of line segments observed in your classroom.
Eg: edge of black board.
Answer:
Edge of blackboard, edge of a book, edge of threshold, edge of window, height of the room, edge of table, edge of bench.

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.1

Question 2.
Why is it better to use a divider than a ruler, while comparing two line segments ?
Answer:
While finding the length of any object, the eye should be correctly positioned (i.e.) just vertically above the marks concerned. Otherwise there may be an error due to angular viewing. To avoid this problem a better way is to use a divider. In the same way, it would be better to use a divider while comparing two line segments.

Question 3.
Measure all the line segments in the figure given below and arrange them in the ascending order of their lengths.
TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.1 1
Line segments \(\overline{\mathbf{A B}}, \overline{\mathbf{A C}}, \overline{\mathbf{A D}}, \overline{\mathbf{A E}}, \overline{\mathbf{B C}}, \overline{\mathbf{B D}}, \overline{\mathbf{B E}}, \overline{\mathbf{C D}}, \overline{\mathbf{C E}}, \overline{\mathbf{D E}}\)
Answer:
\(\overline{\mathrm{A}} \overline{\mathrm{B}}\) = 1 cm; \(\overline{\mathrm{A}} \overline{\mathrm{C}}\) = 2 cm; \(\overline{\mathrm{A}} \overline{\mathrm{D}}\) = 3 cm; \(\overline{\mathrm{A}} \overline{\mathrm{E}}\) = 4 cm
\(\overline{\mathrm{B}} \overline{\mathrm{C}}\) = 1 cm; \(\overline{\mathrm{B}} \overline{\mathrm{D}}\) = 2.0 cm; \(\overline{\mathrm{B}} \overline{\mathrm{E}}\) = 3.0 cm
\(\overline{\mathrm{C}} \overline{\mathrm{D}}\) = 1 cm; \(\overline{\mathrm{C}} \overline{\mathrm{E}}\) = 2.0 cm; \(\overline{\mathrm{D}} \overline{\mathrm{E}}\) = 1.0cm
Ascending order: \(\overline{\mathrm{AE}}, \overline{\mathrm{AD}}, \overline{\mathrm{BE}}, \overline{\mathrm{CE}}, \overline{\mathrm{BD}}, \overline{\mathrm{AC}}, \overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{DE}}\)

Question 4.
Mid point of \(\overline{\mathrm{AB}}\) is located by Swetha and Reshrna like this.
TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.1 2
Which one do you feel correct ? Measure the lengths of \(\overline{\mathbf{A C}}, \overline{\mathbf{C B}}\) and verify.
Answer:
By observation, we can say that the mid point of \(\overline{\mathrm{AB}}\) located by Reshrna is correct. (∵ AC = CB)
AC = 1 cm; CB = 2 cm (Swetha)
AC = 1.5 cm; CB = 1.5 cm (Reshrna)

Question 5.
Each of these figures given along side has many line segments. For the almirah we have shown one line segment along the longer edge. Identify and mark all such line segments in these figures.
TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.1 3
Answer:
(i) \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{DA}}\) and \(\overline{\mathrm{AC}}\) are the line segments.
TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.1 4

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.1

(ii) \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{DA}}, \overline{\mathrm{AE}}, \overline{\mathrm{BF}}, \overline{\mathrm{CG}}, \overline{\mathrm{EF}}\) and \(\overline{\mathrm{FG}}\) are the line segments.
TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.1 5

(iii) \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{BE}}, \overline{\mathrm{AF}}, \overline{\mathrm{FE}},\) and \(\overline{\mathrm{ED}}\) are the line segments.
TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.1 6

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra Ex 9.2

Students can practice TS SCERT Class 6 Maths Solutions Chapter 9 Introduction to Algebra Ex 9.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra Exercise 9.2

Question 1.
Write the expressions for the following statements.
(i) q is multiplied by 5.
Answer:
q × 5 = 5q

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra Ex 9.2

(ii) y is divided by 4.
Answer:
\(\frac{y}{4}\)

(iii) One fourth of the product of numbers p and q.
Answer:
The product of p and q = p × q = pq
One fourth of the product = \(\frac{\mathrm{pq} \times 1}{4}\) = \(\frac{\mathrm{pq}}{4}\)

(iv) 5 is added to the three times z.
Answer:
3 times z = 3 × z = 3z
If 5 is added to 3z, then we get 3z + 5

(v) 9 times ‘n’ is added to ’10’.
Answer:
9 times ‘n’ = 9 × n = 9n;
If 10 is added 9n, then it becomes 9n + 10

(vi) 16 is subtracted from two times ‘y’.
Answer:
2 times ‘y’ = 2 × y = 2y;
If 16 is subtracted from 2y, it becomes (2y – 16)

(vii) ‘y’ is multiplied by 10 and then x is added to the product.
Answer:
y × 10 + x = 10y + x

Question 2.
Write two statements each for the following expressions.
(i) y – 11
Answer:
(a) In a class of ‘y’ students, 11 students failed in the examination. Find the number of passed students.
(b) The sum of two numbers is y. If one number is 11, then what is the other number ?

(ii) 10a
Answer:
(a) The cost of one pen is Rs. 10. What is the cost of ‘a’ such pens ?
(b) What is the value of 10 times ‘a’ ?

(iii) \(\frac{x}{5}\)
Answer:
(a) The cost of 5 books is .Rs. ’x’.
Then what is the cost of one book?
(b) What is the value of one fifth of x?

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra Ex 9.2

(iv) 3m + 11
Answer:
(a) Add 11 to 3 times of m.
(b) The length of a room is 11 more than 3 times of its breadth (say m).

(v) 2y – 5
Answer:
(a) Subtract 5 from 2 times of ‘y’.
(b) The age of A is y and the age of B is 5 less than 2 times of the age of A. What is the age of B ?

Question 3.
Peter has ‘p’ number of balls. Number of balls with David is 3 times the balls with Peter. Write this as an expression.
Answer:
Number of balls that Peter has = p
Number of balls with David is 3 times the balls with Peter.
∴ Number of balls with David is 3p.

Question 4.
Sita has 3 more notebooks than Githa. Find the number of books that Sita has ? Use any letter for the number of books that Githa has.
Answer:
Let the number of books that Githa be x.
Sita has 3 more books than Githa.
∴ The number of books that Sita has = x + 3

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra Ex 9.2

Question 5.
Cadets are marching in a parade. There are 5 cadets in each row. What is the rule for the number of cadets, for a given number of rows ? Use ‘n’ for the number of rows.
Answer:
The number of cadets in each row is 5. Let the number of rows be ‘n’.
Total number of cadets in ‘n’ rows = 5 × n = 5n
∴ The required rule is 5n.

TS 10th Class Maths Solutions Chapter 2 Sets Ex 2.4

Students can practice TS 10th Class Maths Solutions Chapter 2 Sets Ex 2.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 2 Sets Exercise 2.4

Question 1.
State which of the following sets are empty and which are not ?
a) The set of straight lines passing through a point.
b) Set of odd natural numbers divisible by 2.
c) {x : x is a natural number, x < 5 and x > 7}
d) {x : x is a common point to any two parallel lines}
e) Set of even prime numbers.
Answer:
a) The number of straight lines passing through a point is infinite. So, the given set is non-empty.
b) We know the odd natural numbers are 1, 3, 5, 7,…. are not divisible by 2. Hence the given set is empty.
c) There is no natural number satisfying the given condition. Hence the given set is empty.
d) There is no common point to any two parallel lines because they do not meet when produced on either side. Hence the given set is empty.
e) 2 is the only even prime number. Therefore the set contains one element. Hence the given set is non-empty.

Question 2.
Which of the following sets are finite or infinite ?
a) The set of months in a year.
b) {1, 2, 3,…….., 99, 100}
c) The set of prime numbers less than 99.
Answer:
a) There are 12 months in a year. The set of months in a year contains 12 elements. Hence the set is finite.
b) Obviously, the given set contains 100 elements. Hence the set is finite.
c) We can count the prime numbers less than 99. Hence the set is finite.

TS 10th Class Maths Solutions Chapter 1 Sets Ex 2.4

Question 3.
State whether each of the following sets is finite or infinite.
a) The set of letters in the english alphabet.
b) The set of lines which are parallel to the x – axis.
c) The set of numbers which are multiples of 5.
d) The set of circles passing through the origin (0, 0).
Answer:
a) The set of letters in the english alphabet contains 26 elements. Hence, the set is finite.
b) We cannot count the number of parallel lines drawn to the x – axis. Hence the set is infinite.
c) The set of numbers which are multiples of 5 is {5, 10, 15, 20, 25, …}. Hence the set contains infinite number of elements. Hence, the set is infinite.
d) The number of circles that can be drawn through the origin (0, 0) is countless. Hence the set is infinite.

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Students must practice this TS Intermediate Maths 2B Solutions Chapter 6 Integration Ex 6(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Exercise 6(a)

I. Evaluate the following integrals.

Question 1.
∫(x3 – 2x2 + 3) dx on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q1

Question 2.
∫2x√x dx on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q2

Question 3.
\(\int \sqrt[3]{2 x^2}\) dx on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q3

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Question 4.
\(\int\left(\frac{x^2+3 x-1}{2 x}\right)\) dx, x ∈ I ⊂ R – {0}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q4
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q4.1

Question 5.
\(\int \frac{1-\sqrt{x}}{x}\) dx on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q5

Question 6.
\(\int\left(1+\frac{2}{x}-\frac{3}{x^2}\right) d x\) on I ⊂ R – {0}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q6

Question 7.
\(\int\left(x+\frac{4}{1+x^2}\right) d x\) on R
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q7

Question 8.
\(\int\left(e^x-\frac{1}{x}+\frac{2}{\sqrt{x^2-1}}\right) d x\) on I ⊂ R – [-1, 1]
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q8
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q8.1

Question 9.
\(\int\left(\frac{1}{1-x^2}+\frac{1}{1+x^2}\right) d x\) on (-1, 1)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q9

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Question 10.
\(\int\left(\frac{1}{\sqrt{1-x^2}}+\frac{2}{\sqrt{1+x^2}}\right) d x\) on (-1, 1). [May ’11]
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q10

Question 11.
\(\int e^{\log \left(1+\tan ^2 x\right)} d x\) on I ⊂ R – {\(\frac{(2 n+1) \pi}{2}\) : n ∈ Z}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q11

Question 12.
\(\int \frac{\sin ^2 x}{1+\cos 2 x} d x\) on I ⊂ R – {(2n ± 1)π : n ∈ Z}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q12
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) I Q12.1

II. Evaluate the following integrals.

Question 1.
∫(1 – x2)3 dx on (-1, 1).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q1

Question 2.
\(\int\left(\frac{3}{\sqrt{x}}-\frac{2}{x}+\frac{1}{3 x^2}\right) d x\) on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q2

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Question 3.
\(\int\left(\frac{\sqrt{x}+1}{x}\right)^2 d x\) on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q3
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q3.1

Question 4.
\(\int \frac{(3 x+1)^2}{2 x} d x\), x ∈ I ⊂ R – {0}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q4

Question 5.
\(\int\left(\frac{2 x-1}{3 \sqrt{x}}\right)^2 d x\) on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q5

Question 6.
\(\int\left(\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x^2-1}}-\frac{3}{2 x^2}\right) d x\) on (1, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q6

Question 7.
∫(sec2x – cos x + x2) dx, x ∈ I ⊂ R – {\(\frac{n \pi}{2}\) : n is an odd integer}.
Solution:
∫(sec2x – cos x + x2) dx
= ∫sec2x dx – ∫cos x dx + ∫x2 dx
= tan x – sin x + \(\frac{x^3}{3}\) + c

Question 8.
∫(sec x tan x + \(\frac{3}{x}\) – 4] dx, x ∈ I ⊂ R – ({\(\frac{n \pi}{2}\) : n is an odd integer} ∪ {0})
Solution:
∫(sec x tan x + \(\frac{3}{x}\) – 4] dx
= ∫sec x + tan x dx + 3 ∫\(\frac{dx}{x}\) – 4∫dx
= sec x + 3 log|x| – 4x + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Question 9.
\(\int\left(\sqrt{x}-\frac{2}{1-x^2}\right) d x\) on (0, 1)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q9
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q9.1

Question 10.
\(\int\left(x^3-\cos x+\frac{4}{\sqrt{x^2+1}}\right) d x\), x ∈ R
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q10

Question 11.
\(\int\left(\cosh x+\frac{1}{\sqrt{x^2+1}}\right) d x\), x ∈ R
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q11

Question 12.
\(\int\left(\sinh x+\frac{1}{\left(x^2-1\right)^{\frac{1}{2}}}\right) d x\), x ∈ I ⊂ (-∞, -1) ∪ (1, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q12
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q12.1

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Question 13.
\(\int \frac{\left(a^x-b^x\right)^2}{a^x b^x} d x\), (a > 0, a ≠ 1 and b > 0, b ≠ 1) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q13

Question 14.
∫sec2x cosec2x dx on I ⊂ R – ({nx : n ∈ Z} ∪ {(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}) (May ’09)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q14
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q14.1

Question 15.
\(\int\left(\frac{1+\cos ^2 x}{1-\cos 2 x}\right) d x\) on I ⊂ R – {nπ : n ∈ Z} (Mar. ’13)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q15

Question 16.
\(\int \sqrt{1-\cos 2 x} d x\) on I ⊂ [2nπ, (2n + 1)π], n ∈ Z. (Mar. ’09)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a) II Q16

Question 17.
\(\int \frac{1}{\cosh x+\sinh x} d x\) on R.
Solution:
\(\int \frac{1}{\cosh x+\sinh x} d x\)
= \(\int \frac{\cosh x-\sinh x}{\cosh ^2 x-\sinh ^2 x} d x\)
= ∫(cos hx – sin hx) dx (∵ cosh2x – sinh2x = 1)
= ∫cosh x dx – ∫sinh x dx
= sinh x – cosh x + c

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(a)

Question 18.
\(\int \frac{1}{1+\cos x} d x\) on I ⊂ R – [(2n + 1)π : n ∈ Z].
Solution:
\(\int \frac{1}{1+\cos x} d x\)
= \(\int \frac{1-\cos x}{(1+\cos x)(1-\cos x)} d x\)
= \(\int \frac{1-\cos x}{\sin ^2 x} d x\)
= ∫cosec2x dx – ∫cot x cosec x dx
= -cot x + cosec x + c
= cosec x – cot x + c

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Questions

Students can practice TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Exercise InText Questions

Do This

Question 1.
Are 953, 9534, 900, 452 divisible by 2? Also check by actual division.
Answer:
(i) The given number is 953.
This number is divisible by 2 because it has not any one of the digits 0, 2, 4, 6 or 8 in its units place.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Quesions 1

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Questions

(ii) The given number is 9534.
This number is divisible by 2 because it has 4 in its units place.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Quesions 2

(iii) The given number is 900.
This number is divisible by 2 because it has ‘0’ in its units place.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Quesions 3

(iv) The given number is 452.
This number is divisible by 2 because it has 2 in its units place.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Quesions 4

Do This

Check whether the following numbers are divisible by 3 ?
(i) 45986
Answer:
The given number is 45986.
Sum of the digits = 4 + 5 + 9 + 8 + 6 = 32
32 is not a multiple of 3.
So the given number is not divisible by 3.

(ii) 36129
Answer:
The given number is 36129.
Sum of the digits = 3 + 6 + 1 + 2 + 9 = 21
21 is a multiple of 3.
So the given number is divisible by 3.

(iii) 7874
Answer:
The given number is 7874.
Sum of the digits = 7 + 8 + 7 + 4 = 26
26 is not a multiple of 3.
So the given number is not divisible by 3.

Try These

Question 1.
Is 7224 divisible by 6 ? Why ?
Answer:
The given number has 4 in its units place.
So it is divisible by 2.
The sum of the digits of the given number is 7 + 2 + 2 + 4 = 15.
It is a multiple of 3.
So the given number is divisible by 3.
If a number is divisible by both 2 and 3 then it is divisible by 6 also.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Questions

Question 2.
Give two examples of 4 digit numbers which are divisible by 6.
Answer:
9648 and 3756.

Question 3.
Can you give an example of a number which is divisible by 6 but not by 2 and 3, why ?
Answer:
A number is divisible by 6 only when it is divisible by 2 and 3.
So, it is not possible to give an example for such number.

Do This

Question 1.
Test whether 9846 is divisible by 9 ?
Answer:
Number = 9846
Sum of the digits = 9 + 8 + 4 + 6 = 27 27
\(\frac{27}{9}\) = 3 9
∴ 9846 is divisible 9.

Question 2.
Without acutal division, find whether 8998794 is divisible by 9 ?
Answer:
Number = 8998794
Sum of the digits = 8 + 9 + 9 + 8 + 7 + 9 + 4 = 54
\(\frac{54}{9}\) = 6
∴ 8998794 is divisible by 9.

Question 3.
Check whether 786 is divisible by both 3 and 9 ?
Answer:
Number = 786
Sum of the digits = 7 + 8 + 6 = 21
\(\frac{21}{3}\) = 7
So 786 is divisible by 9.
But 21 is not divisible by 9.
So 786 is not divisible by 9.

Do This

Question 1.
Find the factors of 80.
Answer:
Factors of 80 are 1, 2, 4, 5, 8, 10, 16, 20, 40, 80.

Question 2.
Do all the factors of a given number divide the number exactly ? Find the factors of 28 and verify by division.
Answer:
Yes, all the factors of a given number divide the number exactly.
Factors of 28 are 1, 2, 4, 7, 14, 28.

Verification:
\(\frac{28}{28}\) = 1, \(\frac{28}{14}\) = 2, \(\frac{28}{7}\) = 4, \(\frac{28}{4}\) = 7, \(\frac{28}{2}\) = 14

Question 3.
3 is a factor of 15 and 24. Is 3 a factor of their difference also ?
Answer:
Yes. (The difference between 15 and 24 is 9 and 9 is a multiple of 3.)

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Questions

Try These

Question 1.
What is the smallest prime number?
Answer:
The smallest prime number is 2.

Question 2.
What is the smallest composite number?
Answer:
The smallest composite number is 4.

Question 3.
What is the smallest odd composite number?
Sol.
The smallest odd composite number is 6.

Question 4.
Give 5 odd and 5 even composite numbers.
Answer:
The odd composite numbers are
9, 15, 21, 25, 27 etc.
The even composite numbers are
4, 6, 8, 10, 12 etc.

Question 5.
Is 1 prime or composite and why?
Answer:
The number 1 has only one factor i.e. (itself). So, 1 is neither prime nor composite.

Question 6.
Can you guess a prime number which when on reversing Its digits, gives another prime number?
(Hint : Take a 2 digit prime number)
Answer:
13 is a prime number. On reversing its digits it becomes 31, which is also a prime number.

Question 7.
You know 311 is a prime number. Can you find the other two prime numbers just by rearranging the digits?
Answer:
Given prime number is 311.
The other two prime numbers just by rearranging the digits are 113 and 131.

Do This

From the following numbers identify different pairs of co-primes. 2, 3, 4, 5, 6, 7, 8, 9 and 10.
Answer:
(2, 3); (2, 5); (2, 7); (3, 5); (3, 7); (5, 7)

Do This

Question 1.
Write the prime factors of 28 and 36 through division method.
Answer:
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Quesions 5
∴ Prime factors of 36 is 2 × 2 × 3 × 3

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Questions

Question 2.
Write the prime factors of 42 by factor tree method.
Answer:
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Quesions 6
∴ Prime factors of 42 is 2 × 3 × 7

Do This

Find the HCF of 12, 16 and 28 by prime factorization method.
Answer:
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Quesions 7
The common factors of 12, 16 and 28 is 2 × 2 = 4
Hence, HCF of 12, 16 and 28 is 4.

Do This

Find the HCF of 28, 35 and 49 by division method^
Answer:
First find the HCF of any two numbers.
Let us find the HCF of 28 and 35
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Quesions 8
Last divisor is 7
∴ HCF of 28 and 35 is 7.
Then find the HCF of the third numbern and the HCF of first two numbers.
Let us find the HCF of 49 and 7.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Quesions 9
HCF of 49 and 7 is 7.
∴ The HCF of 28, 35 and 49 is 7.

Think, Discuss and Write

What is the HCF of any two
(i) Consecutive numbers ?
Answer:
The HCF of any two consecutive numbers is 1.

(ii) Consecutive even numbers ?
Answer:
2

(iii) Consecutive odd numbers ?
Answer:
1 (one)

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Questions

What do you observe ? Discuss with your friends.
Answer:
It was observed that the HCF of two consecutive numbers and consecutive odd numbers is same, i.e., 1.

Try This

Question 1.
find the LCM of
(i) 3, 4
(ii) 10, 11
(iii) 5, 6, 7
(iv) 10, 30
(v) 4, 12, 24
(vi) 3, 12
What do you observe ?
Answer:
(i) LCM of 3 and 4 = 3 × 4 = 12
(ii) LCM of 10 and 11 = 10 × 11 = 110
(iii) LCM of 5, 6 and 7 = 5 × 6 × 7 = 210
(iv) LCM of 10 and 30 = 10 × 1 × 3 = 30
(v) LCM of 4, 12 and 24 = 4 × 3 × 2 = 24
(vi) LCM of 3 and 12 = 3 × 4 = 12
It is observed that the LCM of two numbers will be their product, if the given numbers have no common factor except 1.

Think, Discuss and Write

When will the LCM of two or more numbers be their own product ?
Answer:
If the numbers are co-primes or relatively prime numbers then the LCM of two or more numbers be their own product.

Think, Discuss and Write

Question 1.
What is the LCM and HCF of twin prime numbers?
Answer:
Let the twin primes may be (3, 5)
LCM of 3, 5 is their product 3 × 5 = 15
HCF of 3, 5 is 1.
(for any type of twin prime)

Question 2.
Interpret relationship between LCM and HCF of any two numbers?
Answer:
Consider the two numbers be 14 and 21.
Now find LCM of 14 and 21.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Quesions 10
∴ LCM of 14 and 21 = 7 × 2 × 3 = 42
Now find HCF of 14 and 21.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Quesions 11
∴ HCF of 14 and 21 is 7.
Relation between LCM and HCF of 14 and 21:
42 × 7 = 14 × 21 = 294
Product of LCM and HCF of two numbers = Product of two numbers.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Questions

Do This

Divisibility Rule for 4:

Question 1.
Is 100000 is divisible by 4? Why?
Answer:
100000 = 1000 × 100
The given number is a multiple of 100.
We know, 100 is divisible by 4.
∴ The given number (i.e., 100000) is divisible by 4.

Question 2.
Give an example of a 2 digit number that is divisible by 2 but not divisible by 4?
Answer:
22, 26, 30, 34, 38 98 .
All the above two digit numbers are divisible by 2 but not divisible by 4.

Do This

Question 1.
Is 76104 divisIble by 8?
Answer:
The number formed by the last three digits is 104. It is divisible by 8.
Hence, the given number is divisible by 8.

Question 2.
Write the numbers that are divisible by 8 and lie between 100 and 200.
Answer:
104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192 are ail divisible by 8.
They lie between 100 and 200.

Page No. 92 (45)

Divisibility Rule for 11:

Using the division rule of ’11’. Fill the following table.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Quesions 12
Answer:
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Quesions 13

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers InText Questions

1221 is a Palindrome number, which on reversing their digits gives the same number.
Thus, every Palindrome number with even number of digits, is always divisible by 11.

Write a Palindrome number of 6 digits and verify whether it is divisible by 11 or not.
Answer:
Palindrome number which on reversing their digits gives the same number.
Every Palindrome number with even number is always divisible by 11.

∴ The 6 digited Palindrome number is 123321. It is divisible by 11.

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.3

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 5 Measures of Lines and Angles Ex 5.3 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Exercise 5.3

Question 1.
Which of the following are models for parallel lines, perpendicular lines and which are neither of them.
(i) The vertical window bars.
(ii) Railway lines (Track)
(iii) The adjacent edges of door.
(iv) The letter ‘V’ in English alphabet.
(v) The opposite edges of Blackboard.
Answer:
(i) For parallel lines —— The vertical window bars.
(ii) For parallel lines —— Railway lines (Track)
(iii) For perpendicular lines —— The adjacent edges of door.
(iv) Neither parallel lines nor perpendicular lines —— The letter ‘V’ in English alphabet.
(v) For parallel lines The opposite edges of Blackboard.

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.3

Question 2.
Trace the copy of set squares (Geometry box) on a paper and mark the perpendicular edges.
Answer:
TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.3 1

TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.3

Question 3.
ABCD is a rectangle. \(\overline{\mathbf{A C}}\) and \(\overline{\mathbf{B D}}\) are diagonals. Write the pairs of parallel lines, perpendicular lines from the figure in symbolic form. Also write pairs of intersecting lines.
TS 6th Class Maths Solutions Chapter 5 Measures of Lines and Angles Ex 5.3 2
Answer:
(a) Parallel lines : AB || CD, AD || BC.
(b) Perpendicular lines: AD ⊥ DB, AB ⊥ BC, BC ⊥ CD, CD ⊥ DA
(c) Pair of intersecting lines: AC, BD.

TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.1

Students can practice 10th Class Maths Solutions Telangana Chapter 3 Polynomials Ex 3.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.1

Question 1.
If p(x) = 5x7 – 6x5 + 7x – 6, find
(i) Coefficient of x5
(ii) degree of p(x)
(iii) constant term.
a) If P(x) = 5x7 – 6x5 + 7x – 6
Solution:
i) coefficient of x5 is -6
ii) degree of p(x) = highest degree of x = 7
iii) constant term is -6

Question 2.
State which of the following statements are true and which are false ? Give reasons for your choice.

i) The degree of the polynomial
\(\sqrt{2}\)x2 – 3x + 1 is \(\sqrt{2}\).
Solution:
The given statement is false because \(\sqrt{2}\) is the coefficient of x2 but not its degree. The degree of the polynomial is 2.

ii) The coefficient of x2 in the polynomial p(x) = 3x3 – 4x2 + 5x + 7 is 2.
Solution:
The given statement is false because the coeffi-cient of x2 in the polynomial is -4 but not 2.

iii) The degree of a constant term is zero.
Solution:
The given statement is true. Because 3x0 = 3 the degree is 0

TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.1

iv) \(\frac{1}{x^2-5 x+6}\) is a quadratic polynomial.
Solution:
The given statement is false because the variable ‘x’ appears in the denominator.

v) The degree of a polynomial is one more than the number of terms ¡n it.
Solution:
The given statement is false. There is a no relationship between the degree of the polynomial and the number of terms in it.

Question 3.
If p(t) = t3 – 1, find the values of p(1), p(-1), p(0), p(2). p(-2) (A.P.Mar. ’15)
Solution:
Given that p(t) = t3 – 1
∴ p(1) = (1)3 – 1 = 1 – 1 = 0
p(-1) = (-1) – 1 = -1 – 1 = -2
p(0) = (0) – 1 = 0 – 1 = -1
p(2) = (2) – 1 = 8 – 1 = 7
= (-2) – 1 = -8 – 1 = -9

Question 4.
Check whether -2 and 2 are the zeroes of the polynomial x4 – 16.
Solution:
p(x) = x4 – 16
p(-2) = (-2)4 – 16 = 16 – 16 = 0
p(2) = (24) – 16 = 16 – 16 = 0
Yes, -2 and 2 are zeroes of the polynomial x4 – 16.

TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.1

Question 5.
Check whether 3 and -2 are the zeroes of the polynomial p(x) when p(x) = x2 – x – 6
Solution:
p(x) = x2 – x – 6
= (3)2 – 3 – 6 = 9 – 3 – 6 = 9 – 9 = 0
p(-2) = (-2)2 – (-2) – 6 = 4 + 2 – 6 = 6 – 6 = 0
Yes, 3 and -2 are zeroes of the polynomial
p(x) = x2 – x – 6

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Students can practice TS 10th Class Maths Solutions Chapter 2 Sets InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Do This

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions 1

Question 1.
List the teeth under each of the following type.
i) Incisors
Solution:
Central incisors = 4
Lateral incisors = 4
TS 10th Class Maths Solutions Chapter 2 Sets InText Questions 2
Total incisors = 8

ii) Canines
Solution:
Total Canines = 4

iii) Pre-molars
Solution:
First premolars = 4
Second premolars = 4
TS 10th Class Maths Solutions Chapter 2 Sets InText Questions 3
Total premolars = 8

iv) Molars
Solution:
First molars = 4
Second molars = 4
Third molars = 4
TS 10th Class Maths Solutions Chapter 2 Sets InText Questions 4
Total molars = 12

Question 2.
Identify and write the “common property” of the following collections. (Page No. 26)

1) 2, 4, 6, 8 ….
2) 3, 5, 7, 11, ….
3) 1, 4, 9, 16, ………
4) January, February, March, April,………..
5) Thumb, index finger, middle finger, ring finger, pinky.
Solution:
1) For given integers 2, 4, 6, 8 …….. 2n, n = 1, 2, 3, 4,…. is any positive integer.
2) 2, 3, 5, 7, 11, …. prime numbers.
3) For given integers 1, 4, 9, 16, …. n2, where ‘n’ is any positive integer and square of the numbers.
4) January, February, March, April,…. months of the every year.
5) Thumb, index finger, middle finger, ring finger, pinky fingers of the human hand.

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 3.
Write the following sets. (Page No. 27)
1) Set of the first five positive integers.
2) Set of multiples of 5 which are more than 100 and less than 125.
3) Set of first five cubic numbers.
4) Set of digits in the Ramanujan number.
Solution:
1) { + 1, +2, +3, +4, +5}
2) {105, 110, 115, 120}
3) {1, 8, 27, 64,125} = {13, 23, 33, 43, 53}
4) Ramanujan Number = 1729. So the set = {1, 2, 7, 9}

Question 4.
Some numbers are given below. Decide the numbers to which number sets they belong to and does not belong to and express with correct symbols. (Page No. 27)
i) 1
ii) 0
iii) -4
iv) \(\frac{5}{6}\)
v) \(\text { 1. } \overline{3}\)
vi) \(\sqrt{2}\)
vii) log 2
viii) 0.03
ix) π
x) \(\sqrt{-4}\)
Solution:
Set of natural numbers = N
set of integers = Z
Set of rational numbers = Q
Set of real numbers = R
i) 1 ∈ (N, Z, Q, R}
ii) 0 ∉ N, 0 ∈ {Z, Q, R}
iii) -4 ∉ N, -4 ∈ (Z, Q, R}
iv) \(\frac{5}{6}\) ∉ (N, Z} But \(\frac{5}{6}\) ∈ {Q, R}
v) \(1 . \overline{3}\) ∉ {N, Z} But \(1 . \overline{3}\) ∈ {Q, R}
vi) \(\sqrt{2}\) ∉ {N, Z} But \(\sqrt{2}\) ∈ {Q, R}
vii) log2 ∉ N,Z But log 2 ∈ {Q, R}
viii) 0.03 ∉ {N, Z}But 0.03 ∈ {Q, R}
ix) π ∉ {N, Z} But π ∈ (Q, R}
x) \(\sqrt{-4}\) ∉ (N, Z, Q} But \(\sqrt{-4}\) ∈ R.

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 5.
List the elements of the following sets.
i) G = {all the factors of 20}
ii) F = {the multiples of 4 between 17 and 61 which are divisible by 7}
iii) S = {x : x is a letter in the word MADAM’}
iv) P = {x : x is a whole number between 3.5 and 6.7} (Page No. 29)
Answer:
i) G = {1, 2, 4, 5, 10, 20}
ii) Multiples of 4 between 17 and 61.
x = {20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60}
F = {28, 56}
iii) S = {M, D, A}
iv) P = {4, 5, 6}

Question 6.
Write the following sets in the roaster form. (Page No. 29)
i) B is the set of all months in a year having 30 days.
ii) P is the set of all prime numbers less than 10.
iii) X is the set of colours of the rainbow.
Answer:
i) B = {April, June, September, November}
ii) P = {2, 3, 5, 7}
iii) X = {Violet, Indigo, Blue, Green, Yellow, Orange, Red}

Question 7.
A is the set of factors of 12. Which one of the following is not a member of A ? (Page No. 29)
A) 1
B) 4
C) 5
D) 12
Answer:
[C]

Think – Discuss

Question 1.
Observe the following collections and prepare as many as generalised statements you can describing their more properties. (Page No. 26)
i) 2, 4, 6, 8, ….
ii) 1, 4, 9, 16
Answer:
i) 3, 6, 9, 12, ….
Property : multiples of 3.

ii) 1, 8, 27, 64,….
Property : cube of the numbers, i.e., 13, 23, 33, 43, ….

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 2.
Can you write the set of rational numbers listing elements in it ? (Page No. 28)
Solution:
No

Try This

Question 1.
Write some sets of your choice, involving algebraic and geometrical ideas. (Page No. 29)
Answer:
1) The natural number greater than three and less than twelve.
2) A two digit number such that the sum of its digits is 8.
3) The set of quadrilaterals.
4) The set of all triangles in a plane.

Question 2.
Match roaster forms with the set builder form.
i) {P, R, I, N, C, A, L}
ii) {0}
iii) {1, 2, 3, 6, 9, 18}
iv) {3, -3} (Page No. 29)
a) {x : x is a positive integer and is a divisor of 18}
b) {x : x is an integer and x2 – 9 = 0}
c) {x : x is an integer and x + 1 = 1}
d) {x: x is a letter of the word PRINCIPAL}
Answer:
i) d
ii) c
iii) a
iv) b

Do This

Question 1.
A = {1, 2, 3, 4}, B = {2, 4}, C = {1, 2, 3, 4, 7}, F = { } (Page No. 33)
Fill in the blanks with ⊂ or ⊄.
i) A………B
ii) C……..A
iii) B……..A
iv) A……..C
v) B……..C
vi) F……B
Answer:
i) AB
ii) CA
iii) BA
iv) AC
v) BC
vi) FB

Question 2.
State which of the following statements are true. (Page No. 33)
i) { } = ϕ
ii) ϕ = 0
iii) 0 = {ϕ}
Answer:
i) True (T)
ii) False (F)
iii) False (F)

Question 3.
Let A = {1, 3, 7, 8} and B = {2, 4, 7, 9}. Find A∩B.(Page No. 37)
Solution:
Given sets
A = {1, 3, 7, 8} and B = {2, 4, 7, 9}
A ∩ B = {1, 3, 7, 8} ∩ {2, 4, 7, 9}
= {7}

Question 4.
If A = {6, 9, 11}; ϕ = { }, find A ∪ ϕ. (Page No. 37)
Solution:
Given sets A = {6, 9, 11} and B = {2, 4, 7, 9}
A ∪ ϕ = {6, 9, 11} ∪ {ϕ}
= {6, 9, 11} = A
∴ A ∪ ϕ = A

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 5.
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
B = {2, 3, 5, 7}. Find A ∩ B. (Page No. 37) (June ’15(AP))
Solution:
Given sets A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = {2, 3, 5, 7}
A ∩ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∩ {2, 3, 5, 7}
= {2, 3, 5, 7} = B
∴ A ∩ B = B

Question 6.
If A = {4, 5, 6}; B = {7, 8}, then show that A ∪ B = B ∪ A. (Page No. 37)
Solution:
Given sets are
A = (4, 5, 6} and B = (7, 8}
A ∪ B = {4, 5, 6} ∪ {7, 8}
= {4, 5, 6, 7, 8}
B ∪ A = {7, 8} ∪ {4, 5, 6}
= {4, 5, 6, 7, 8}
∴ A ∪ B = B ∪ A

Question 7.
If A = {1, 2, 3, 4, 5}; B = {4, 5, 6, 7} then find A-B and B-A. Are they equal ? (Page No. 38)
Solution:
Given sets are A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7}
A – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7} = {1, 2, 3}
B – A = {4, 5, 6, 7} – {1, 2, 3, 4, 5} = {6, 7}
No, A – B ≠ B – A

Question 8.
If V = {a, e, i, o, u} and B = {a, i, k, u}, find V – B and B – V. (Page No. 38)
Solution:
Given sets are
V = {a, e, i, o, u} and B = {a, i, k, u}
V – B = {a, e, i, o, u} – {a, i, k, u}
= {e, o}
B – V = {a, i, k, u} – {a, e, i, o, u}
= {k}

Try This

Question 1.
A = {set of quadrilaterals}, B = {square, rectangle, trapezium, rhombus}
State whether
A ⊂ B or B ⊂ A. Justify your answer. (Page No. 33)
Answer:
A ⊄ B
B ⊂ A every element of B is also an element of A.

Question 2.
If A = {a, b, c, d}. How many subsets does the set A have? (Remember null set and equal sets). (Page No. 33)
A) 5
B) 6
C) 16
D) 65
Solution:
A = {a, b, c, d}
Subsets of A = {a}, {b}, {c}, {d};
n(A) = 4
Number of subsets for a set, which is having ‘n’ elements is 2n.
So n(A) = 4
Number of subsets for A is 24 =16.
Answer:
(C)

Question 3.
P is the set of factors 5, Q is the set of factors of 25 and R is the set of factors of 125.
Which one of the following is false?
A) P ⊂ Q
B) Q ⊂ R
C) R ⊂ P
D) P ⊂ R (Page No. 33)
Solution:
P = {1, 5}
Q = {1, 5, 25}
R = {1, 5, 25, 125}
Answer:
(C)

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 4.
A is the set of prime numbers less than 10, B is the set of odd numbers < 10 and C is the set of even numbers < 10. How many of the following statements are true ? (Page No. 33)
i) A ⊂ B
ii) B ⊂ A
iii) A ⊂ C
iv) C ⊂ A
v) B ⊂ C
vi) X ⊂ A
Solution:
A = {2, 3, 5, 7}
B = {1, 3, 5, 7, 9}
C = {2, 4, 6, 8}
The given all statements are false.

Question 5.
List out some sets A and B and choose their elements such that A and B are disjoint. (Page No. 37)
Solution:
A and B are disjoint sets.
i) A = {2, 3, 5} B = {4, 6, 8}
ii) A = {1, 2, 3} B = {4, 5, 6}
iii) A = {1, 3, 5, 7} B = {2, 4, 6, 8}

Question 6
If A = {2, 3, 5}, find A∪ϕ and ϕ∪A and compare. (Page No. 37)
Solution:
A = {2, 3, 5}; ϕ = { }
A∪ϕ = {2, 3, 5} ∪ { } = {2, 3, 5}
ϕ∪A = { } ∪ {2, 3, 5} = {2, 3, 5}
∴ A∪ϕ = ϕ∪A

Question 7.
If A = {1, 2, 3, 4}; B = {1, 2, 3, 4, 5, 6, 7, 8} then find A ∪ B, A ∩ B. What do you notice about the result ? (Pg. No. 37)
Solution:
A = {1, 2, 3, 4};
B = {1, 2, 3, 4, 5, 6, 7, 8}
A∪B = {1, 2, 3, 4} ∪ {1, 2, 3, 4, 5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}
The common elements in both A and B sets are 1, 2, 3, 4.
A∩B = {1, 2, 3, 4}
Result :
i) A∪B = B
ii) A∩B = A
iii) A ∩ B ⊂ A ∪ B

Question 8.
A = {1, 2, 3, 4, 5, 6}; B = {2, 4, 6, 8, 10}. Find the intersection of A and B. (Page No. 37)
Solution:
A = {1, 2, 3, 4, 5, 6}
B = {2, 4, 6, 8, 10}
The common elements in both A and B are 2, 4, 6.
∴ A ∩ B = {2, 4, 6}

Think : Discuss

Question 1.
Is empty set subset to every set? (Page No. 34)
Solution:
Yes.

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 2.
Is any set subset to itself? (Page No. 34)
Solution:
Yes.

Question 3.
You are given two sets such that a set is not a subset of the other. If you have to prove this, how do you prove ? Justify your answers. (Page No. 34)
Solution:
The above statement is true only in the two sets does not have common elements.
Ex : A = {1, 2, 3}; B = {4, 5, 6}
So, AB.

Question 4.
The intersection of any two disjoint sets is a null set. Justify your answer. (Page No. 37)
Answer:
Yes, this statement is true because A ∩ B = ϕ when A and B are disjoint sets.

Question 5.
The sets A – B, B – A and A ∩ B are mutually disjoint sets. Use examples to observe if this is true. (Page No. 38)
Solution:
Let the sets are
A = {1, 2, 3, 4} and B = {5, 6, 7, 8}
A – B = {1, 2, 3, 4} – {5, 6, 7, 8} = {1, 2, 3, 4}
B – A = {5, 6, 7, 8} – {1, 2, 3, 4} = {5, 6, 7, 8}
A ∩ B = {1, 2, 3, 4} ∩ {5, 6, 7, 8} = { } = ϕ
∴ A – B, B – A and A ∩ B are disjoint sets.

Do This

Question 1.
Which of the following are empty sets? Justify your answer. (Page No. 44)
i) Set of integers which lie between 2 and 3.
ii) Set of natural numbers that are less than 1.
iii)Set of odd numbers that have remainder zero, when divided by 2.
Answer:
i) This is null set. We know that there is no integer that lie between 2 and 3.
ii) This is also a null set. We know that there is natural number less than ‘1’.
iii) This is a null set. We know that odd numbers do not leave remainder zero when divided by 2.

Question 2.
State which of the following sets are finite and which are infinite. Give reasons for your answers.
i) A = {x: x ∈ N and x < 100}
ii) B = {x : x ∈ N and x ≤ 5}
iii)C = {12, 22, 32, ………… )
iv)D = {1, 2, 3, 4)
v) {x : x is a day of the week) (Page No. 44)
Answer:
i) A = {1, 2, 3, ………., 98, 99}
This set is finite, because there are 99 numbers possible to count.

ii) B = {1, 2, 3. 4, 5}
This set is finite, because there are 5 numbers possible to count.

iii) C = {12, 22, 32,………)
This set is infinite, because there are infinite numbers.

iv) D = {1, 2, 3, 4}
This set is finite because there are 4 numbers that are possible to count.

v) E = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday)
This set is finite, because there are 7 days in a week possible to count.

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 3.
Tick the set which is infinite
A) The set of whole numbers < 10
B) The set of prime numbers < 10
C) The set of integers < 10
D) The set of factors of 10 (Page No. 44)
Answer:
[C]
The set of integers < 10
{……., -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Try This

Question 1.
Which of the following sets are empty sets? Justify your answer.
i) A = {x : x2 = 4 and 3x = 9}
ii) The set of all triangles in a plane having the sum of their three angles less than 180. (Page No. 44)
Answer:
i) Empty set.
To satisfy these equations same x value is not possible.
ii) Empty set.
The sum of the three angles of a triangle is equal to 180°.

Question 2.
B = {x : x + 5 = 5} is not an emptyset. Why? (Page No. 44)
Solution:
x + 5 = 5
x = 5 – 5
x = 0
For x = 0 it is true
Only one element is there.
So it is not an empty set.

Think — Discuss

Question 1.
An empty set is a finite set. Is this statement true or false? Why? (Page No. 44)
Answer:
Yes, it is a finite set because there is finite number i.e., ‘0’ elements it consists.

Question 2.
What is the relation between n(A), n(B), n(A ∩ B) and n (A ∪ B)? (Page No. 45)
Solution:
n(A) = elements in set A, n(B) = elements in set B
n(A ∩ B) = set of all elements which are common to both A and B.
n(A ∪ B) = elements in set A and set B (or) union of sets A and B.
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

TS 10th Class Maths Solutions Chapter 2 Sets InText Questions

Question 3.
If A and B are disjoint sets, then how can you find n(A ∪ B)? (Page No. 45)
Solution:
n(A) = elements in set A.
n(B) = elements in set B.
n(A ∩ B) = elements in set A and set B
Here it is ‘0’ (∴ A and B are disjoint sets)
n(A ∪ B) = elements in set A or set B.
∴ n(A ∪ B) = n(A) ÷ n(B) – n(A ∩ B)
= n(A) + n(B) – 0
= n(A) + n(B).

TS 10th Class Maths Solutions Chapter 2 Sets Ex 2.3

Students can practice TS 10th Class Maths Solutions Chapter 2 Sets Ex 2.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 2 Sets Exercise 2.3

Question 1.
Which of the following sets are equal ?
a) A = {x : x is a letter in the word FOLLOW’}
b) B = {x : x is a letter in the word ‘FLOW’}
c) C = {x : x is a letter in the word ‘WOLF’}
Answer:
a) Writing the given set in the roaster form, we have A = {F, O, L, W}
b) Writing the given set in the roaster form, we have B = {F, L, O, W}
c) Writing the given set in the roaster form, we have C = {W, O, L, F}
Therefore, A, B, C are equal sets.
[∴ The sets A, B, C have exactly the same elements]

Question 2.
Consider the following sets and fill up the blank in the statement given below with = or ≠ so as to make the statement true.
A = {1, 2, 3};
B = {The first three natural numbers}
C = {a, b, c, d};
D = {d, c, a, b}
E = {a, e, i, o, u};
F = {set of vowels in English Alphabet}
i) A ……… B
ii) A …….. E
iii) C ……. D
iv) D …… F
v) F ……. A
vi) D …… E
vii) F ……. B
Answer:
i) A = B
ii) A ≠ E
iii) C=D
iv) D ≠ F
v) F ≠ A
vii) D ≠ E
viii) F ≠ B

TS 10th Class Maths Solutions Chapter 1 Sets Ex 2.3

Question 3.
In each of the following, state whether
A = B or not.
i) A = {a, b, c, d}; B = {d, c, a, b}
ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18}
iii) A = (2, 4, 6, 8, 10)
B = {x: x is a positive even integer and x ≤ 10}
iv) A = {x: x ¡s a multiple of 10};
B = {10, 15, 20, 25, 30,……. }
Answer:
i) A = B because A and B have exactly the same elements i.e., a, b, c, d.
ii) A ≠ B because A and B have not exactly the same elements.
iii) A = B because A and B have exactly the same elements.
Writing B in roaster form, we have
B = {2, 4, 6, 8, 10}
iv) A = {10, 20, 30, 40,……..}
B = {10, 15, 20, 25,……..}
A ≠ B because A and B have not exactly the same elements.

Question 4.
State the reasons for the following:
i) {1,2, 3,…, 10} ≠ {x : x ∈ N and 1 < x < 10}
ii) {2, 4, 6, 8, 10} ≠ {x : x = 2n + 1 and x ∈ N}
iii) {5, 15, 30, 45} ≠ {x : x is a multiple of 15
iv) {2, 3, 5, 7, 9} ≠ {x : x is a prime number}
Solution:
The first set is {1, 2, 3, ……, 10}
Writing the second set in roaster form, we have {2, 3, 4, ……, 9}
The first set and the second set have not exactly the same elements.
∴ {1, 2, 3,……10} ≠ {x : x ∈ N and 1 < x < 10}

ii) The first set is {2, 4, 6, 8, 10}
Writing the second set in roaster form, we have {3, 5, 7, 9, ….}
∴ {2, 4, 6, 8, 10} ≠ {3, 5, 7, 9, ….}
x = 2n + 1 means x is odd.

iii) The first set is {5, 15, 30, 45}
Writing the second set in roaster form, we have {15, 30, 45, 60, …}
∴ {5, 15, 30, 45} ≠ {15, 30, 45, 60,…}
5 does not exist, since x is multiple of 15.

iv) The first set is {2, 3, 5, 7, 9}
Writing the second set in roaster form, we have {2, 3, 5, 7, 11, 13,…}
∴ {2, 3, 5, 7, 9} ≠ {2, 3, 5, 7, 11, 13 }
9 is not a prime number.

TS 10th Class Maths Solutions Chapter 1 Sets Ex 2.3

Question 5.
List all the subsets of the following sets.
i) B = {p, q}
ii) C = {x, y, z}
iii) D = {a, b, c, d}
iv) E = {1, 4, 9, 16}
v) F = {10, 100, 1000}
Solution:
i) {p}, {q}, {p, q}, { ϕ }
ii) {x}, {y}, {z}, {x, y}, {y, z}, {x, z}, {x, y, z}, {ϕ}
iii) {a}, {b}, {c}, {d}, {a, b}, {b, c}, {c, d}, {a, c}, [a, d}, {b, d}, {a, b, c}, {b, c, d}, {a, b, d}, {a, c, d}, {a, b, c, d}, {ϕ}
iv) {1}, {4}, {9}, {16}, {1, 4}, {4, 9}, {9, 16}, {1, 9}, {1, 16}, {4, 16}, {1, 4, 9}, {4, 9, 16}, {1, 4, 16}, {1, 9, 16}, {1, 4, 9, 16}, {ϕ}
v) {10}, {100}, {1000}, {10, 100}, {100, 1000}, {10, 1000}, {10, 100, 1000}, {ϕ}

TS 10th Class Maths Solutions Chapter 2 Sets Ex 2.2

Students can practice TS 10th Class Maths Solutions Chapter 2 Sets Ex 2.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 2 Sets Exercise 2.2

Question 1.
If A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, then find A ∩ B and B ∩ C. Are they same ?
Solution:
A = {1, 2, 3, 4}; B = {1, 2, 3, 5, 6}
∴ A ∩ B = {1, 2, 3, 4} ∩ {1, 2, 3, 5, 6}
= {1, 2, 3}
B ∩ A = {1, 2, 3, 5, 6} ∩ {1, 2, 3, 4)
= {1, 2, 3}
Yes, A ∩ B and B ∩ A are same.

Question 2.
A = {0, 2, 4}, find A ∩ ϕ and A ∩ A. Comments. (June 15(A.P.))
Solution:
A = {0, 2, 4}
A ∩ ϕ = {0, 2, 4} ∩ ϕ
= ϕ
A ∩ A = {0, 2, 4} ∩ {0, 2, 4}
= {0, 2, 4} = A

TS 10th Class Maths Solutions Chapter 1 Sets Ex 2.2

Question 3.
If A = {2, 4, 6, 8, 10} and B = {3, 6, 9, 12, 15}, find A – B and B – A.
Solution:
A = {2, 4, 6, 8, 10);
B = {3, 6, 9, 12, 15)
A – B = {2, 4, 8, 10}
B – A = {3, 9, 12, 15}

Question 4.
If A and B are two sets such that A ⊂ B then what is A ∪ B?
Solution:
Let us consider A ⊂ B.
Set A = {1, 2, 3}
Set B = {1, 2, 3, 4, 5}
A ∪ B = {1, 2, 3} ∪ {1, 2, 3, 4, 5)
= {1, 2, 3, 4, 5}
= B
A ∪ B = B

Question 5.
If A = {x : x is a natural number}
B = {x : x is an even natural number}
C = {x : x is an odd natural number}
D = {x : x is a prime number
Find A∩B, A∩C, A∩D, B∩C, B∩D, C∩D.
Solution:
A = {1, 2, 3, 4, …….. }
B = {2, 4, 6, 8, ……….}
C = {1, 3, 5,7, ……….}
D = {2, 3, 5, 7, ……….}
A ∩ B = {1, 2, 3, 4,………. } ∩ {2, 4, 6, 8,………. }
= {2, 4, 6,…… }
= {even natural numbers)
A ∩ C = {1, 2, 3, 4,……} ∩ {1, 3, 5,……}
= {1, 3, 5,………}
= {odd natural numbers}
A ∩ D = {1, 2, 3, 4,……} ∩ {2, 3, 5, 7, ……..}
= {2, 3, ………}
= {prime natural numbers}
B ∩ C = {2, 4, 6, 8,…..} ∩ {1, 3, 5, 7, …….}
= ϕ
= Null set (Or) empty set
B ∩ D = {2, 4, 6, 8,…….. } ∩ {2, 3, 5, 7,…….. }
= {2} = {even natural number}
C ∩ D = {1, 3, 5, 7,……. } ∩ {2, 3, 5, 7,…….}
= {3, 5, 7,…….. }
= {All odd prime numbers)

Question 6.
If A = {3, 6, 9, 12, 15, 18, 21);
B = {4, 8, 12, 16, 20};
C = {2, 4, 6, 8, 10, 12, 14, 16};
D = {5, 10, 15, 20}, find
i) A – B
ii) A – C
iii) A – D
iv) B – A
v) C – A
vi) D – A
vii) B – C
viii) B – D
ix) C – B
x) D – B
Solution:
i) A – B = {3, 6, 9, 12, 15, 18, 21} – {4, 8, 12, 16, 20}
= {3, 6, 9, 15, 18, 21}

ii) A – C = {3, 6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10, 12, 14, 16)
= {3, 9, 15, 18, 21}

iii) A – D = {3, 6, 9, 12, 15, 18, 21} – {5, 10, 15, 20}
= {3, 6, 9, 12, 18, 21}

iv) B – A = {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21}
= {4, 8, 16, 20}

v) C – A = {2, 4, 6, 8, 10, 12, 14, 16} – {3, 6, 9, 12, 15, 18, 21}
= {2, 4, 8, 10, 14, 16}

vi) D – A = {5, 10, 15, 20} — {3, 6, 9, 12, 15, 18, 21)
= (5, 10, 20)

vii) B – C = {4, 8, 12, 16, 20) – {2, 4, 6, 8, 10, 12, 14, 16}
= {20}

viii) B — D = {4, 8, 12, 16, 20} – {5, 10, 15, 20}
= {4, 8, 12, 16)

ix) C – B = {2, 4, 6, 8, 10, 12, 14, 16} – {4, 8, 12, 16, 20}
= {2, 6, 10, 14)

x) D – B = {5, 10, 15, 20} – {4, 8, 12, 16, 20}
= {5, 10, 15}

TS 10th Class Maths Solutions Chapter 1 Sets Ex 2.2

Question 7.
State whether each of the following statement is true or false. Justify your answers.
i) {2, 3, 4, 5} and {3, 6) are disjoint sets.
ii) {a, e, i, o, u) and {a, b, c, d) are disjoint sets.
iii){2, 6, 10, 14) and {3, 7, 11, 15) are disjoint sets.
iv) (2, 6, 10) and {3, 7, 11) are disjoint sets.
Answer:
Two sets are said to be disjoint sets when they have no elements in common.
i) In the given two sets. ‘3’ is common. So, they are not disjoint sets. Hence, the given statement is false.
ii) In the given two sets, ‘a’ is common. So, they are not disjoint sets. Hence, the given statement is false.
iii) There are no elements common in the given two sets. So they are disjoint sets. Hence, the given statement is true.
iv) There are no elements common in the given two sets, So they are disjoint sets. Hence. the given statement is true.

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3

Students can practice Telangana 6th Class Maths Textbook Solutions Chapter 7 Fractions and Decimals Ex 7.3 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Exercise 7.3

Question 1.
Write shaded portion as fraction. Arrange them in ascending or descending order using sign ‘<‘, ‘=’, ‘>’ between the fractions.
(i)
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 1
Answer:
\(\frac{3}{8}, \frac{6}{8}, \frac{4}{8}, \frac{1}{8}\); Arranging them in ascending order, we get \(\frac{1}{8}<\frac{3}{8}<\frac{4}{8}<\frac{6}{8}\)
Descending order: \(\frac{6}{8}>\frac{4}{8}>\frac{3}{8}>\frac{1}{8}\)

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3

(ii)
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 2
Answer:
\(\frac{8}{9}, \frac{4}{9}, \frac{3}{9}, \frac{6}{9}\); Arranging them in ascending order, we get \(\frac{3}{9}<\frac{4}{9}<\frac{6}{9}<\frac{8}{9}\)
Descending order: \(\frac{8}{9}>\frac{6}{9}>\frac{4}{9}>\frac{3}{9}\)

Question 2.
Show \(\frac{2}{6}, \frac{4}{6}, \frac{8}{6}, \frac{5}{6}\) and \(\frac{6}{6}\) on the number line. Also arrange them in ascending order.
Answer:
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 3

Question 3.
Look at the figures and write ‘<‘ or’>’, ‘=‘ between the given pairs of fractions:
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 4
(i) \(\frac{1}{6}\) ________ \(\frac{1}{3}\)
(ii) \(\frac{3}{4}\) ________ \(\frac{2}{6}\)
(iii) \(\frac{2}{3}\) ________ \(\frac{2}{4}\)
(iv) \(\frac{6}{6}\) ________ \(\frac{3}{3}\)
(v) \(\frac{5}{6}\) ________ \(\frac{5}{5}\)
Make five more such problems and ask your friends to solve them.
Answer:
(i) \(\frac{1}{6}\) < \(\frac{1}{3}\)
(ii) \(\frac{3}{4}\) > \(\frac{2}{6}\)
(iii) \(\frac{2}{3}\)> \(\frac{2}{4}\)
(iv) \(\frac{6}{6}\) = \(\frac{3}{3}\)
(v) \(\frac{5}{6}\) < \(\frac{5}{5}\)

Question 4.
Fill with the appropriate sign. (‘<‘, ‘=’, ‘>’)
(i) \(\frac{1}{2}\) _______ \(\frac{1}{5}\)
Answer:
\(\frac{1}{2}\) > \(\frac{1}{5}\)

(ii) \(\frac{2}{4}\) _______ \(\frac{3}{6}\)
Answer:
\(\frac{2}{4}\) = \(\frac{3}{6}\)

(iii) \(\frac{3}{5}\) _______ \(\frac{2}{3}\)
Answer:
\(\frac{3}{5}\) < \(\frac{2}{3}\)

(iv) \(\frac{3}{4}\) _______ \(\frac{2}{8}\)
Answer:
\(\frac{3}{4}\) > \(\frac{2}{8}\)

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3

(v) \(\frac{3}{5}\) _______ \(\frac{6}{5}\)
Answer:
\(\frac{3}{5}\) < \(\frac{6}{5}\)

(vi) \(\frac{7}{9}\) _______ \(\frac{3}{9}\)
Answer:
\(\frac{7}{9}\) > \(\frac{3}{9}\)

Question 5.
Answer the following. Also write how you solved them.
(i) Is \(\frac{5}{9}\) equal to \(\frac{4}{5}\)?
Answer:
We write the fractions \(\frac{5}{9}\) and \(\frac{4}{5}\) having the same denominators
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 5

(ii) Is \(\frac{9}{16}\) equal to \(\frac{5}{9}\)?
Answer:
We write the fractions \(\frac{9}{16}\) and \(\frac{5}{9}\) having the same denominators
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 6

(iii) Is \(\frac{4}{5}\) equal to \(\frac{16}{20}\)?
Answer:
We write the fractions \(\frac{4}{5}\) and \(\frac{16}{20}\) having the same denominators
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 7

(iv) Is \(\frac{1}{15}\) equal to \(\frac{4}{30}\)?
Answer:
We write the fractions \(\frac{1}{15}\) and \(\frac{4}{30}\) having the same denominators
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 8

Question 6.
Varshith read 25 pages of a story book containing 100 pages. Lalitha read \(\frac{2}{5}\) of the same story book. Who read less? Give reason.
Answer:
Total number of pages in the story book = 100
Number of pages that Varshith read = 25
Number of pages that Lalitha read
= \(\frac{2}{5}\) of 100
= \(\frac{2}{5}\) × 100 = 40
So, Varshith read less pages.

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3

Question 7.
Fill the appropriate (+ or -) sign in the blank space.
(i)
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 9
Answer:
+

(ii)
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 10
Answer:

(iii)
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 11
Answer:
+

Question 8.
simplify:
(i) \(\frac{1}{18}+\frac{1}{18}\)
Answer:
\(\frac{1}{18}+\frac{1}{18}=\frac{1+1}{18}=\frac{2}{18}=\frac{1}{9}\)

(ii) \(\frac{8}{15}+\frac{3}{15}\)
Answer:
\(\frac{8}{15}+\frac{3}{15}=\frac{8+3}{15}=\frac{11}{15}\)

(iii) \(\frac{7}{7}-\frac{5}{7}\)
Answer:
\(\frac{7}{7}-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}\)

(iv) \(\frac{1}{22}+\frac{21}{22}\)
Answer:
\(\frac{1}{22}+\frac{21}{22}=\frac{1+21}{22}=\frac{22}{22}\) = 1

(v) \(\frac{12}{15}-\frac{7}{15}\)
Answer:
\(\frac{12}{15}-\frac{7}{15}=\frac{12-7}{15}=\frac{5}{15}=\frac{1}{3}\)

(vi) \(\frac{5}{8}+\frac{3}{8}\)
Answer:
\(\frac{5}{8}+\frac{3}{8}=\frac{5+3}{8}=\frac{8}{8}\) = 1

(vii) 1 – \(\frac{2}{3}\)
Answer:
1 – \(\frac{2}{3}=\frac{3}{3}-\frac{2}{3}=\frac{3-2}{3}=\frac{1}{3}\) [∵ 1 = \(\frac{3}{3}\)]

(viii) \(\frac{1}{4}+\frac{0}{4}\)
Answer:
\(\frac{1}{4}+\frac{0}{4}=\frac{1+0}{4}=\frac{1}{4}\)

(ix) 3 – \(\frac{2}{15}\)
Answer:
3 – \(\frac{12}{5}=\frac{3 \times 5}{1 \times 5}-\frac{12}{5}=\frac{15}{5}-\frac{12}{5}=\frac{15-12}{5}=\frac{3}{5}\)

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3

Question 9.
Fill in the missing fractions.
(i) \(\frac{7}{10}\) – ___________ = \(\frac{3}{10}\)
Answer:
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 12

(ii) ____________ – \(\frac{3}{21}\) = \(\frac{5}{21}\)
Answer:
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 13

(iii) ___________ – \(\frac{3}{3}\) = \(\frac{3}{6}\)
Answer:
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 14

(iv) ___________ + \(\frac{5}{27}\) = \(\frac{12}{27}\)
Answer:
TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3 15

Question 10.
Narendra painted \(\frac{2}{3}\) area of the wall in his room. His brother Ritesh helped and painted \(\frac{1}{3}\) area of the wall. How much did they paint together?
Answer:
Area of the wall painted by Narendra = \(\frac{2}{3}\)
Area of the wall painted by Ritesh = \(\frac{1}{3}\)
Area of the wall painted by both Narendra and Ritesh = \(\frac{2}{3}+\frac{1}{3}\)
= \(\frac{2+1}{3}\) = \(\frac{3}{3}\) = 1
∴ Narendra and his brother Ritesh painted the complete wall.

Question 11.
Neha was given \(\) of a basket of bananas. What fraction of bananas was left in the basket?
Answer:
The part of a basket of bananas given to Neha = \(\frac{5}{7}\)
The, part of bananas left in the basket
= 1 – \(\frac{5}{7}=\frac{1 \times 7}{1 \times 7}-\frac{5}{7}=\frac{7}{7}-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}\)

Question 12.
A piece of rod \(\frac{7}{8}\) metre long is broken into two pieces. One piece was \(\frac{1}{4}\) metre long. How long is the other piece?
Answer:
Length of a piece of rod = \(\frac{7}{8}\) metre
Length of one broken piece of rod = \(\frac{1}{4}\) metre.
Length of the other piece = \(\frac{7}{8}-\frac{1}{4}\)
= \(\frac{7}{8}-\frac{1 \times 2}{4 \times 2}=\frac{7}{8}-\frac{2}{8}=\frac{7-2}{8}\) = \(\frac{5}{8}\)m
∴ \(\frac{5}{8}\)m long is the other piece.

TS 6th Class Maths Solutions Chapter 7 Fractions and Decimals Ex 7.3

Question 13.
Renu takes 2\(\frac{1}{5}\) minutes to walk around the school ground. Snigdha takes \(\frac{7}{4}\) minutes to do the same. Who takes less time and by what fraction?
Answer:
Time taken by Renu to walk around the school ground = 2\(\frac{1}{5}\) minutes
= \(\frac{11}{5}\)
Time taken by Snigdha to walk around the school ground = \(\frac{7}{4}\)minutes
To find the person who takes less time to do the same, we write the fractions \(\frac{11}{5}\) and \(\frac{7}{4}\) having the same denominators.
\(\frac{11}{5} \times \frac{4}{4}=\frac{44}{20}\); \(\frac{7}{4} \times \frac{5}{5}=\frac{35}{20}\)
We know that \(\frac{35}{20}\) < \(\frac{44}{20}\)
Therefore, Snigdha takes \(\frac{9}{20}\) minutes less time to walk around the school ground.
(∵ \(\frac{44}{20}-\frac{35}{20}=\frac{44-35}{20}=\frac{9}{20}\))