Class 10

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 13 Probability Ex 13.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 13 Probability Exercise 13.1

Question 1.
Complete the following statements : (AS₃)
i) Probability of an event E + Probability of the event ‘not E’ = ____________
ii) The probability of an event that cannot happen is _________ Such an event is called __________
iii) The probability of an event that is certain to happen is _________ Such an event is called _________
iv) The sum of the probabilities of all the elementary events of an experiment is _________
v) The probability of an event is greater than or equal to _________ and less than or equal to.
Solution:
i) 1
ii) 0, impossible event
iii) 1, sure (or) certain event
iv) 1
v) 0, 1

Question 2.
Which of the following experiments have equally likely outcomes ? Explain. (AS₃)
i) A driver attempts to start a car. The car starts or does not start.
Solution:
Equally likely. Since both have the some probability = \(\frac{1}{2}\)

ii) A player attempts to shoot a basket ball. She! he shoots or misses the shot.
Solution:
Equally likely. Since both have the same probability = \(\frac{1}{2}\)

iii) A trial ¡s made to answer a true-false question. The answer is right or wrong.
Solution:
Equally likely. Since both have the same probability = \(\frac{1}{2}\)

iv) A baby is born. It is a boy or a girl.
Solution:
Equally likely. Since both the events have the same probability = \(\frac{1}{2}\)

Question 3.
If P(E) = 0.05, what is the probability of “not E” ? (AS₁)
Solution:
Given that P(E) = 0.05
Probability of “not E” is denoted by \(\overline{\mathrm{E}}\)
P(\(\overline{\mathrm{E}}\)) = 1 – P(E)
= 1 – 0.05
= 0.95

Question 4.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out.
i) an orange flavoured candy ? (AS₄)
ii) a lemon flavoured candy ?
Solution:
i) The bag contains lemon flavoured candies only.
(i.e.,) It does not contains orange flavoured candies.
So, the probability that she takes out an orange flavoured candy is ‘O’.

ii) Probability that she takes out lemon flavoured candy is 1 because the bag contains lemon flavoured candies only.

Question 5.
Rahim removes all the hearts from the cards. What is the probability of she takes out
i) Getting an ace from the remaining pack. (AS₄)
ii) Getting a diamond.
iii) Getting a card that is not a heart.
iv) Getting the Ace of hearts.
Solution:
All the hearts are taken out of 52 cards.
Therefore, the remaining cards will be 52 – 13 = 39
i) The probability of getting an ace from the remaining pack.
= \(\frac{\text { Number of outcomes favourable to the event }}{\text { Total number of all possible outcomes }}\)
= \(\frac{4}{52}\) = \(\frac{1}{13}\)

ii) The probability of picking out a diamond
= \(\frac{\text { Number of outcomes favourable to the event }}{\text { Total number of all possible outcomes }}\)
= \(\frac{13}{39}\) = \(\frac{1}{3}\)

iii) The probability of getting a card that is not a heart.
= \(\frac{\text { Number of outcomes favourable to the event }}{\text { Total number of all possible outcomes }}\)
= \(\frac{39}{39}\) = 1

iv) The probability of getting the Ace of hearts.
= \(\frac{\text { Number of outcomes favourable to the event }}{\text { Total number of all possible outcomes }}\)
= \(\frac{0}{39}\) = 0

Question 6.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday ? (AS₄)
Solution:
Probability that the 2 students have the same birthday.
= 1 – Probability that the 2 students do not have the same birthday.
= 1 – 0.992 = 0.008

Question 7.
A die is rolled once. Find the probability of getting
(i) a prime number ;
(ii) a number lying between 2 and 6 ;
(iii) an odd number. (AS₁, AS₄)
Solution:
Suppose we roll a die once. The possible out-comes are 1, 2, 3, 4, 5 and 6. Each number has the same possibility of showing up. So the equally likely outcomes of rolling a dice are 1, 2, 3, 4, 5 and 6.

i) Let E be the event of getting a prime number. Then the outcomes favourable to E are 2, 3 and 5 (prime numbers).
Therefore, the number of outcomes favourable to E is 3.
So, P(E)
= \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

ii) The numbers lying between 2 and 6 are 3, 4 and 5.
Let E be the event of getting a number lying between 2 and 6.
Then, the outcomes favourable to E are 3, 4 and 5.
Therefore the number of outcomes favourable to E is 3.
So, P (E)
= \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

iii) The odd numbers in the first 6 natural numbers is 1, 3 and 5.
Let E be the event of getting an odd number.
Then, the outcomes favourable to E are 1, 3 and 5.
Therefore, the number of outcomes favourable to E is 3.
So, P(E)
= \(\frac{\text { No. of outcomes favourable to E }}{\text { No. of all possible outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 8.
What is the probability of selecting out a red king from a deck of cards ?
Solution:
A deck has 52 cards.
Since the total number of cards is 52.
Number of all possible outcomes = 52
Let E be the event of getting a red king.
Then, the outcomes favourable to E are king of diamond and heart. Therefore, the number of outcomes favourable to E is 2.
So, P(E)
= \(\frac{\text { No. of outcomes favourable to E}}{\text { No. of all possible outcomes }}\)
= \(\frac{2}{52}\) = \(\frac{1}{26}\)

Question 9.
Make 5 more problems of this kind using die, cards or birthdays and discuss with friends and teacher about their solutions. (AS₃)
Solution:

1. A die is thrown once. Find the probability of getting a number less than 3.
2. From a well shuffled pack of cards is drawn at random. Find the probability of getting a black queen.
3. A die is thrown once. What is the probability of
   a) Getting a number greater than 2 ?
   b) Getting an even number ?
   c) Getting a number between 3 and 6 ?
4. Two coins are tossed simultaneously. Find the probability of getting exactly one head.
5. Raghu and Siva are friends. What is the probability that both will have
   1. Different birthdays ?
   2. The same birthday ? (ignoring a leap year)
6. The king, queen and jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of
   1. Heart
   2. Queen

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Optional Exercise

Question 1.
Verify that the numbers given along side the cubic polynomials below are their zeroes. Also verify the relationship between the zeros and the coefficients in each case :
i) 2x³ + x² – 5x + 2;(\(\frac{1}{2}\), 1, -2)
ii) x³ + 4x² + 5x – 2; (2, 1, 1)
Solution:
i) Given polynomial is 2x³ + x² – 5x + 2
Comparing the given polynomial with ax³ + bx² + cx + d,
We get a = 2, b = 1, c = -5 and d = 2
p(\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\))³ + (\(\frac{1}{2}\))³ – 5(\(\frac{1}{2}\)) + 2
= \(\frac{1}{4}\) + \(\frac{1}{4}\) – \(\frac{5}{2}\) + \(\frac{2}{1}\)
= \(\frac{1+1-10+8}{4}\) = \(\frac{0}{4}\) = 0
P(1) = 2(1)³ + (1)² – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
p(-2) = 2(-2)³ + (-2)² – 5(-2) + 2
= 2(-8) + 4 + 10 + 2
= -16 + 16 = 0
∴ \(\frac{1}{2}\), 1, and -2 are the zeroes of 2x³ + x² – 5x + 2
So, α = \(\frac{1}{2}\), β = 1 and γ = -2
Therefore,
α + β + γ = \(\frac{1}{2}\) + 1 + (-2)
= \(\frac{1+2-4}{2}\) = \(\frac{-1}{2}\) = \(\frac{-b}{a}\)
αβ + βγ + γα = (\(\frac{1}{2}\))(1) + (1)(-2) + -2(\(\frac{1}{2}\))
= \(\frac{1}{2}\) – 2 – 1
= \(\frac{1-4-2}{2}\) = \(\frac{-5}{2}\) = \(\frac{c}{a}\)
And αβγ = \(\frac{1}{2}\) × 1 × (-2) = -1 = \(\frac{-2}{2}\) = \(\frac{-\mathrm{d}}{\mathrm{a}}\)

ii) Given polynomial is x³ – 4x² + 5x – 2
Comparing the given polynomial with ax³ + bx² + cx + d, We get a = 1, b = -4, c = 5 and d = -2
p(2) = (2)³ – 4(2)² + 5(2) – 2
= 8 – 16 + 10 – 2 = 0
p(1) = (1)³ – 4(1)² + 5(1) – 2
= 1 – 4 + 5 – 2 = 0
∴ 2, 1 and 1 are the zeroes of x³ – 4x² + 5x – 2.
So, α = 2, β = 1 and γ = 1
Therefore,
α + β + γ = 2 + 1 + 1
= 4 = \(\frac{-(-4)}{1}\) = \(\frac{-b}{a}\)
αβ + βγ + γα = 2(1) + (1)(1) + (1) (2)
= 2 + 1 + 2 = 5
= \(\frac{5}{1}\) = \(\frac{c}{a}\)
and
αβγ = (2) (1) (1) = 2 = \(\frac{-(-2)}{1}\) = \(\frac{-\mathrm{d}}{\mathrm{a}}\)

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and the product of its zeroes as 2, -7, -14 respectively.
Answer:
Let the cubic polynomial be
ax³ + bx² + cx + d and its zeroes be α, β and γ.
Then
α + β + γ = 2 = \(\frac{-(-2)}{1}\) = \(\frac{-b}{a}\)
αβ + βγ + γα = -7 = \(\frac{-7}{1}\) = \(\frac{c}{a}\)
and αβγ = -14 = \(\frac{-14}{1}\) = \(\frac{-d}{a}\)
∴ a = 1, b = -2, c = -7 and d = 14.
So, one cubic polynomial which satisfies the given conditions will be
x³ – 2x² – 7x + 14

Question 3.
If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.
Answer:
Given polynomial is x³ – 3x² + x + 1
Since, (a – b), a, (a + b) are the zeroes of the polynomial x³ – 3x² + x + 1.
Therefore, sum of the zeroes = (a – b) + a + (a + b)
⇒ 3a = 3 ⇒ a = 1
= \(\frac{-(-3)}{1}\) = 3
∴ Sum of the products of its zeroes taken two at a time.
= a(a – b) +a(a + b) + (a + b) (a – b)
= \(\frac{1}{1}\) = 1
⇒ a² – ab + a² + ab + a² – b² = 1
⇒ 3a² – b² = 1
So, 3(1)² – b² = 1
⇒ 3 – b² = 1
⇒ b² = 2 ⇒ b = \(\sqrt{2}\) = ±\(\sqrt{2}\)
Here, a = 1 and b = ±\(\sqrt{2}\)

Question 4.
If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± \(\sqrt{3}\) , find other zeroes.
Answer:
Given polynomial is
x⁴ – 6x³ – 26x² + 138x – 35
We have, 2 ± \(\sqrt{3}\) are two zeroes of the polynomial
p(x) = x⁴ – 6x³ – 26x² + 138x – 35
Let x = 2 ± \(\sqrt{3}\)
So, x – 2 = ±\(\sqrt{3}\)
On squaring, we get x² – 4x + 4 = 3, i.e., x² – 4x + 1 = 0
Let us divide p(x) by x² – 4x + 1 to obtain other zeroes

p(x) = x⁴ – 6x³ – 26x² + 138x – 35
= (x² – 4x + 1) (x² – 2x – 35)
= (x² – 4x + 1) (x² – 7x + 5x – 35)
= (x² – 4x + 1) [x(x – 7) +5(x – 7)]
= (x² – 4x + 1) (x + 5) (x – 7)
So, (x + 5) and (x – 7) are other factors of p(x)
So, -5 and 7 are other zeroes of the given polynomial.

Question 5.
If the polynomial
x⁴ – 6x³ – 16x² + 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
Answer:
Given polynomial is
x⁴ – 6x³ + 16x² – 25x + 10 and another polynomial is x² – 2x + k
Remainder is x + a
Let us divide
x⁴ – 6x³ + 16x² – 25x + 10 by x² – 2x + k.

∴ Remainder = (2k – 9)x – (8 – k)k + 10
But the remainder is given as x + a.
On comparing their coefficients we have
2k – 9 = 1 ⇒ 2k = 10
⇒ k = 5 and -(8 – k) k + 10 = a
So, a = -(8 – 5) 5 + 10 = -3 × 5 + 10
= -15 + 10 = -5
Hence, k = 5 and a = -5

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics InText Questions

Think – Discuss

Question 1.
The mean value can be calculated from both ungrouped and grouped data. Which one do you think is more accurate? Why? (Page No. 327) (AS₂. AS₃)
Solution:
Mean calculated from ungrouped data is more accurate than, mean calculated from the grouped data. Since its calculation takes the observations in the data into consideration.

Question 2.
When it Is more convenient to use grouped data for analysis? (AS₃) (Page No. 327)
Solution:
Grouped data is convenient when the values of f_i and x_i are low.

Question 3.
Is the result obtained by all the three methods the same? (AS₃) (Page No. 331)
Solution:
Yes, mean obtained by all the three methods is same.

Question 4.
If x_i and f_i are sufficiently small, then which method is an appropriate choice? (AS₃) (Page No. 331)
Solution:
Direct method.

Question 5.
If x_i and f_i are numerically large numbers then which methods are appropriate to use? (AS₃) (Page No. 331)
Solution:
Assumed mean method and step deviation method.

Do This

Question 1.
Find the mode of the following data. (AS₁) (Page No. 334)
a) 5, 6, 9, 10, 6, 12, 3, 6, 11, 10, 4, 6, 7
Solution:
Mode : 6 (Most repeated value of the data)

b) 20, 3, 7, 13, 3, 4, 6, 7, 19, 15, 7, 18, 3
Solution:
Mode : 3, 7 (Most repeated values)

c) 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6
Solution:
Mode : 2, 3, 4, 5, 6 (Most repeated values)

Question 2.
Is the mode always at the center of the data ? (AS₃) (Page No. 334)
Solution:
No, Mode may not beat the centre always.

Question 3.
Does the mode change ? If another observation is added to the data in Example ? Comment. (Page No. 334)
Solution:
The data given in example 4 is as follows :
0, 1, 2, 2, 2, 3, 3, 4, 5, 6.
Except 2 and 3, the other observations occur only once. 2 occurs three times while 3 occurs two times. Any other observation except 2 and 3 is added, the mode of the data will not be affected. If 3 is added, then 3 occurs three times. There are already three 2’s in the data. Hence, the data has two modes (i.e.,) 3 and 2. If 2 is added, then 2 occurs four times. So, the mode of the data will be 2, (i.e.,) the mode of the data is not affected because on observing the data prior to the addition of 2, we can say that the mode is 2.

Question 4.
If the maximum value of an observation in the data in example 4 is changed to 8, would the mode of the data be affected ? Comment. (Page No. 334)
Solution:
If the maximum value is altered to 8, the mode remains the same. Mode doesn’t consider the values but consider their frequencies only.

Think – Discuss

Question 1.
It depends upon the demand of the situation whether we are interested in finding the average marks obtained by the students or the marks obtained by most of the students. (AS₃) (Page No. 336)
a) What do we find in the first situation ?
Solution:
We find A.M.

b) What do we find in the second situation?
Solution:
We find the mode.

Question 2.
Can mode be calculated for grouped data which unequal class sizes ? (Mar. ’15 (AP)) (Page No. 336)
Solution:
Yes. mode can be calculated for grouped data with unequal class sizes.

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.4

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following.
i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x
iii) p(x) = x⁴ – 5x + 6 and g(x) = 2 – x²
Solution:
i) p(x) = x³ – 3x² + 5x – 3 and
g(x) = x² – 2
The given polynomials are in standard form.

The degree of x² – 2 is 2.
The degree of 7x – 9 is 1.
∴ We stop here since the degree of (7x – 9) < degree of (x² – 2)
So, the quotient is x – 3 and the remainder is 7x – 9.

ii) p(x) = x⁴ – 3x² + 4x + 5,
g(x) = x² + 1 – x
p(x) = x⁴ – 3x² + 4x + 5, it is in standard form.
g(x) = x² + 1 – x, it is not in standard form. Writing it in standard form, we have x² – x + 1. Now we apply the division algorithm to the given polynomials.

The degree of x² – x + 1 is 2
The degree of 8 is 0.
∴ We stop here since the degree of (8) < degree of (x² – x + 1)
So, the quotient is x² + x – 3 and the remainder is 8.

iii) p(x) = x⁴ – 5x + 6 and g(x) = 2 – x²
p(x) = x⁴ – 5x + 6 → it is in standard form
g(x) = 2 – x² → it is not in standard form writing it in standard form, we have -x² + 2.
Now we apply the division algorithm to the given polynomials

The degree of -x² + 2 is 2 and that of (-5x + 10) is 1
The degree of (-x² + 2) > degree of (-5x + 10)
So, we stop here
So, the quotient is -x² – 2 and the remainder is -5x + 1o.

Question 2.
Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1
Solution:
Divide the second polynomial by the first polynomial. If the remainder is zero, then the first polynomial is a factor of the second one. The given polynomials are in standard form.
i)

Since the remainder is ‘0’, t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

ii)

Since the remainder is ‘0’, x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

iii)

Here, the remainder is 2.
∴ x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x + 1.
x⁵ – 4x³ + x² + 3x + 1

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are
\(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\).
Solution:
The given polynomial is
3x⁴ + 6x³ – 2x² – 10x – 5.
Two of its zeroes are \(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\).
∴ [x – \(\sqrt{\frac{5}{3}}\)][x + \(\sqrt{\frac{5}{3}}\)] = [x² – \(\sqrt{\frac{5}{3}}\)] is a factor of the given polynomial.
Now, we apply the division algorithm to the given polynomial & x² – \(\frac{5}{3}\).

3x⁴ + 6x³ – 2x² – 10x – 5 = (x² – \(\frac{5}{3}\)) (3x² + 6x + 3)
3x² + 6x + 3 = 3(x² + 2x + 1) = 3(x + 1)²
So, the zeroes of 3(x + 1)² are -1 and -1.
Therefore, the zeroes of the given polynomial are and \(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\), -1 and -1

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4 respectively. Find g(x).
Solution:
By division algorithm, we have
p(x) = g(x) × q(x) + r(x)
g(x) × q(x) = p(x) – r(x)
g(x) × (x – 2) = x³ – 3x² + x + 2 – (-2x + 4)
= x³ – 3x² + x + 2 + 2x – 4
= x³ – 3x² + 3x – 2
∴ g(x) = (x³ – 3x² + 3x – 2) ÷ (x – 2)

∴ g(x) = x² – x + 1

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
i) deg p(x) = deg q(x)
ii) deg q(x) = deg r(x)
iii) deg r(x) = 0
Solution:
i) p(x) = 6x² – 12x + 15, g(x) = 3,
q(x) = 2x² – 4x + 5, r(x) = 0
ii) p(x) = x³ + 2x² + x – 6, g(x) = x² + 2
q(x) = x + 2, r(x) = -x – 10
iii) p(x) = x³ + 5x² – 3x – 10, g(x) = x² – 3
q(x) = x + 5, r(x) = 5

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year :

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. (AS₃, AS₄)
Solution:
Maximum number of patients joined in the age group 35 – 45
∴ Modal class is 35 – 45.
Lower limit of the modal class ‘l’ = 35
class size, h = 10
Frequency of modal class. f₁ = 23
Frequency of the class preceding the modal class f₀ = 21
Frequency of the class succeding the modal class f₂ = 14

∴ Mode = l + \(\frac{\left(f_1-f_0\right)}{2 f_1-f_0-f_2}\) × h
= 35 + \(\left[\frac{23-21}{2 \times 23-21-14}\right]\) × 10
= 35 + \(\left[\frac{2}{46-35}\right]\) × 10
= 35 + \(\frac{2}{11}\) × 10 = 35 + 1.81818 ….
= 36.8 years

Mean \(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) = \(\frac{2830}{80}\) = 35.37 years.

Interpretation : Mode age is 36.8 years, Mean age = 35.37 years.
Maximum number of patients admitted in the hospital are the age 36.8 years, while on an average the age of patients admitted to the hospital is 35.37 years. Mode is less than the mean.

Question 2.
The following data gives the information on the observed life times (in hours) of 225 electrical components : (AS₄)

Determine the modal life times of the components.
Solution:

Since the maximum frequency 61 is in the class 60 – 80, this is the required modal class.
Modal class frequency, f₁ = 61
Frequency of the class preceding the modal class f₀ = 52
Frequency of the class succeding the modal class f₂ = 38
lower boundary of the modal class l = 60
Height of the class, h = 20
∴ Mode (z) = l + \(\frac{\left(f_1-f_0\right)}{2 f_1-\left(f_0+f_2\right)}\) × h
= 60 + \(\left[\frac{61-52}{2 \times 61-(52+38)}\right]\) × 20
= 60 + \(\frac{9}{122-90}\) × 20
= 60 + \(\frac{9}{32}\) × 20
= 60 + 5.625 = 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of Gummadidala village. Find the modal monthly expenditure of the families. Also find the mean monthly expenditure. (AS₄)

Solution:

Assumed mean (a) = 3250
Σf_i = 200; Σf_iu_i = – 235
Mean monthly income = \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
= 3250 – \(\frac{235}{200}\) × 500
= 3250 – 587.5
= ₹ 2662.50
Since the maximum families 40 lies in the class 1500 – 2000, this is the required modal class.
Lower boundary of modal class (l) = 1500
Frequency of the modal class (f₁) = 40
Frequency of the class preceding the modal class f₀ = 24
Frequency of the class succeding the modal class f₂ = 33
Height of the class, h = 500
Mode (z) = l + \(\frac{\left(f_1-f_0\right)}{2 \times f_1-\left(f_0+f_2\right)}\) × h
⇒ 1500 + \(\frac{40-24}{2 \times 40-(24+33)}\) × 500
= 1500 + \(\frac{16 \times 500}{80-57}\) = 1500 + \(\frac{8000}{23}\)
⇒ 1500 + 347.826
= ₹ 1847.83
Hence, modal monthly income = ₹ 1847.83

Question 4.
The following distribution gives the state-wise, teachers, student ratio in higher second’ ary schools of India. Find the mode and mean of this data. Interpret the two measures. (AS₃, AS₄)

Solution:

Mean (\(\overline{\mathrm{x}}\)) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
a – assumed mean = 32.5; h – height of the class = 5
∴ \(\overline{\mathrm{x}}\) = 32.5 – \(\frac{22}{35}\) × 5 = 32.5 – 3.28 = 29.22
Since the maximum number of states ‘10′ lies in the class interval 30 – 35, this is the modal class.
Lower boundary of the modal class, l = 30
Frequency of the modal class, f₁ = 10
Frequency of the class preceding the modal class, f₀ = 9
Frequency of the class succeding the modal class, f₂ = 3
Height of the class h = 5
Mode (z) = l + \(\left[\frac{f_1-f_0}{\left(f_1-f_0\right)+\left(f_1-f_2\right)}\right]\) × h
⇒ 30 + \(\frac{10-9}{(10-9)+(10-3)}\) × 5
= 30 + \(\frac{1 \times 5}{1+7}\) = 30 + \(\frac{5}{8}\)
= 30 + 0625 ⇒ 30.625.
Mode of states have a teacher students ratio 30.625 and on an average of this ratio is 29.22.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one international cricket matches.

Find the mode of the data. (AS₄)
Solution:

Maximum no.of batsmen are in the class 4000 – 5000
Modal class is 4000 – 5000
Lower boundary of the modal class ‘l’ = 4000
frequency of the modal class, f₁ = 18
frequency of the class preceding the modal class f₀ = 4
frequency of the class succeeding the modal class f₂ = 9
Size of the class, h = 1000
Mode(z) = l + \(\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{\left(\mathrm{f}_1-\mathrm{f}_0\right)+\left(\mathrm{f}_1-\mathrm{f}_2\right)}\right]\) × h
Mode (z) = 4000 + \(\frac{18-4}{(18-4)+(18-9)}\) × 1000
= 4000 + \(\frac{14}{14+9}\) × 1000
= 4000 + \(\frac{14000}{23}\) = 4000 + 608.695
= 4608.69
≅ 4608.7 runs

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods, each of 3 minutes and summarised this in the table given below.

Find the mode of the data. (AS₄)
Solution:

Since, the maximum frequency is 20, the modal class is 40 – 50.
Lower boundary of the modal class, ‘l’ = 40
Frequency of the modal class, f₁ = 20
Frequency of the class preceding the modal class f₀ = 12
Frequency of the class succeeding the modal class f₂ = 11
Height of the class, h = 10;
Mode (z) = l + \(\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{\left(\mathrm{f}_1-\mathrm{f}_0\right)+\left(\mathrm{f}_1-\mathrm{f}_2\right)}\right]\) × h
= 40 + \(\frac{(20-12)}{(20-12)+(20-11)}\) × 10
= 40 + \(\frac{8}{8+9}\) × 10
= 40 + \(\frac{80}{17}\)
= 40 + 4.70588
= 44.705
≅ 44.7 cars.

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.4

Question 1.
The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive. (AS₅)
Solution:
For less than ogive, we take the upper class limits on X-axis and their corresponding cumulative frequencies on Y – axis, choosing a convenient scale.

The points to be plotted are (300, 12) (350, 26) (400, 34) (450, 40) and (500, 50)
Scale :
On X-axis : 1 cm = 50 units
On Y-axis : 1 cm = 5 units

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows : (AS₅)

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:
The points to be plotted on a graph paper are : (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35)
Scale :
On X-axis : 1 cm = 2 units
On Y-axis : 1 cm = 2 units

Number of observations = 35
Here \(\frac{\mathrm{n}}{2}\) = \(\frac{35}{2}\) = 17.5
Locate the point 17.5 on the Y – axis. From the point, draw a line parallel to the X – axis cutting the curve at a point. From the point, draw a perpendicular to the X – axis.

The point of intersection of this perpendicular with the X – axis determines the median of the given data as 46.8 kg.

Number of observations = n = 35
\(\frac{\mathrm{n}}{2}\) = \(\frac{35}{2}\) = 17.5
17.5 belongs to the class 46 – 48
∴ Median class = 46 – 48
l – lower boundary of class = 46
f – frequency of the median class = 14
c.f. = 14
Class size = 2
Median = l + \(\frac{\left(\frac{n}{2}-c . f .\right)}{f}\) × h
= 46 + \(\frac{17.5-14}{14}\) × 2
= 46 + \(\frac{3.5}{14}\) × 2
= 46 + \(\frac{7}{14}\) = 46 + \(\frac{1}{2}\) = 46.5
Hence, median is 46.5 either ways.

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village. (AS₅)

Change the distribution to a more than type distribution and draw its ogive.
Solution:
More than type distribution

Points to be plotted on the graph paper are:
(50, 100), (55, 98), (60, 90), (65. 78), (70, 54) and (75, 16)
Scale:
On X-axis: 1 cm = 5 units
On Y-axis: 1 cm = 10 units

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. (AS₄)

Median :
Solution:

Here n = 68; \(\frac{\mathrm{n}}{2}\) = \(\frac{68}{2}\) = 34
The median lies in the class 125 – 145.
Lower limit (l) of the median class = 125
Frequency of the median class (f) = 20 of (cumulative frequency of the class 105 – 125) = 22
class size (h) = 20
∴ Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 125 + \(\frac{\left[\frac{n}{2}-c f\right]}{f}\) × 20
= 125 + 12
= 137 units.

Mean :

Mean (\(\overline{\mathrm{X}}\)) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
Here a = 135, h = 20, Σf_i = 68, Σf_iu_i = 7
= 135 + \(\left[\frac{7}{68}\right]\) × 20
= 135 + \(\frac{35}{17}\)
= 135 + 2.05
= 137.05

Mode :
Since the maximum number of consumers have their monthly consumption (in units) in the interval 125 – 145
∴ lower limit of the modal class (l) = 125
class size (h) = 20
Frequency of the modal class (f₁) = 20
Frequency of the class preceding the modal class (f₀) = 13
Frequency of the class succeeding the modal class (f₂) = 14
Modal = l + \(\frac{\left(f_1-f_0\right)}{\left(2 f_1-f_0-f_2\right)}\) × h
= 125 + \(\frac{(20-13)}{(2 \times 20-13-14)}\) × 20
= 125 + \(\frac{7}{(40-13-14)}\) × 20
= 125 + \(\frac{7 \times 20}{13}\) = 125 + \(\frac{140}{13}\)
= 125 + 10.76 = 135.76 units

Comparison :
In this case, the three measures i.e., mean, median and mode are approximately equal.

Question 2.
If the median of 60 observations, give below is 28.5, find the value of x and y. (AS₁)

Solution:

Hence, n = 60 (given)
\(\frac{\mathrm{n}}{2}\) = \(\frac{60}{2}\) = 30
⇒ 45 + x + y = 60
⇒ x + y = 60 – 45 = 15 ……………… (1)
The median is given as 28.5.
If lies in the class 20 – 30.
So, l = 20;
Frequency of the median class (f) = 20
cf (cumulative frequency of the class preceding the median class 10 – 20) = 5 + x
∴ Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
⇒ 28.5 = 20 + \(\left[\frac{30-(5+x)}{20} \times 10\right]\)
⇒ 28.5 = 20 + \(\frac{25-x}{2}\)
⇒ \(\frac{25-x}{2}\) = 28.5 – 20 = 8.5
⇒ 25 – x = 2 × 8.5
⇒ 25 – x = 17
⇒ x = 25 – 17 = 8 ……………… (2)
from (1) and (2) : we get
8 in x + y = 15, 8 + y = 15
∴ y = 15 – 8 = 7
Hence, x = 8 and y = 7

Question 3.
A life insurance agent found the following data about distribution of ages of 100 policy holders. Calculate the median age. (policies are given only to persons having age 18 years onwards but less than 60 years.) (AS₄)

We find the class intervals and their corresponding frequencies to calculate the median age.
Solution:

Hence, n = 100: \(\frac{\mathrm{n}}{2}\) = \(\frac{100}{2}\) = 50
The median lies in the class 35 – 40. So l = 35
frequency of the median class (f) = 33
cf (cumulative frequency of the class preceeding the median dass 30 – 35) = 45
class size(h) = 5
∴ Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 35 + \(\frac{(50-45)}{33}\) × 5
= 35 + \(\frac{5 \times 5}{33}\) = 35 + \(\frac{25}{33}\)
= 35 + 0.76 = 35.76
Hence, the median age = 35.76 years.

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre and the data obtained Is represented in the following table: (AS₄)

Find the median length of the leaves (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5 ………. 171.5 – 180.5)
Solution:

Here, n = 40; \(\frac{\mathrm{n}}{2}\) = \(\frac{40}{2}\) = 20
The median class is 144.5 – 153.5
Lower limit (l) of the median class = 144.5
cf (cumulative frequency of class preceding the median class 144.5 – 153.5) = 17
f (frequency of the median class) = 12
h (class size) = 9
using the formula, median
= l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 144.5 + \(\frac{(20-17) \times 9}{12}\) × 9
= 144.5 + \(\frac{3 \times 9}{12}\) = 144.5 + \(\frac{9}{4}\)
= 144.5 + 2.25
= 146.75
∴ Median length = 146.75 mm.

Question 5.
The following table gives the distribution of the life-time of 400 neon lamps

Find the median life time of a lamp. (AS₄)
Solution:

Here, n = 400; \(\frac{\mathrm{n}}{2}\) = \(\frac{400}{2}\) = 200
The median class is 3000 – 3500
lower limit (1) of the median class = 3000
cf (cumulative frequency of the class preceding the median class 2500 – 3000) = 130
f (frequency of the median class) = 86
h(class size) = 500
using the formula, Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 3000 + \(\frac{(200-130) \times 500}{86}\)
= 3000 + \(\frac{70 \times 500}{86}\)
= 3000 + 406.98
= 3406.98 hours
∴ Median life = 3406.98 hours.

Question 6.
loo surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters ¡n the English alphabet in the surnames was obtained as follows:

Determine the median number of letters In the surnames. Find the mean number of letters in the surnames P Also, find the modal size of the surnames. (AS₄)
Solution:
Median:

Here, n = 100; \(\frac{\mathrm{n}}{2}\) = \(\frac{100}{2}\) = 50
So, the median lies in the class 7 – 10
lower limit (1) of the median class = 7
cf (cumulative frequency of the class preceding median class 7 – 10) = 36
f (frequency of the median class) = 40
h (class size) = 3
using the formula. Median
= l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h × h
= 7 + \(\frac{(50-36) \times 3}{40}\)
= 7 + \(\frac{14 \times 3}{40}\) = 7 + \(\frac{21}{20}\)
= 7 + 1.05 = 8.05
Hence, the median of letters in the surnames = 8.05

Mean :

Mean (\(\overline{\mathrm{x}}\)) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}} \times \mathrm{h}\right]\)
Here a = 8.5, h = 3, Σf_i = 100, Σf_iu_i = -6
= 8.5 – \(\frac{6 \times 3}{100}\)
= 8.5 – \(\frac{18}{100}\)
= 8.5 – 0.18
= 8.32
Hence, the mean of the surnames = 8.32

Mode:

Since, the maximum number of surnames have number of letters in the interval 7 – 10, the modal class is 7 – 10
Lower limit (1) of the modal class = 7
frequency of the modal class (f₁) = 40
Frequency of the class preceding the modal class (f₀) = 30
frequency of the class succeeding the modal class (f₂) = 16
class size(h) = 3
∴ Mode = l + \(\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{2 \mathrm{f}_1-\mathrm{f}_0-\mathrm{f}_2}\right]\) × h
= 7 + \(\frac{(40-30)}{(2 \times 40-30-16)}\) × 3
= 7 + \(\left[\frac{10 \times 3}{80-30-16}\right]\)
= 7 + \(\frac{30}{34}\)
= 7 + \(\frac{15}{17}\)
= 7 + 0.88 = 7.88
Hence, the modal size of the surnames = 7.88

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students. (AS₄) (Mar ’16 (A.P))

Solution:

Hence, n = 30; \(\frac{\mathrm{n}}{2}\) = \(\frac{30}{2}\) = 15
So, the median lies in the class 55 – 60
∴ l = 55
frequency of the median class (f) = 6
cf (cumulative frequency, of the class 50 – 55) = 13
class size (h) = 5
∴ Median = l + \(\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{c.f}\right]}{\mathrm{f}}\) × h
= 55 + \(\frac{(15-13) \times 5}{6}\)
= 55 + \(\frac{2 \times 5}{6}\)
= 55 + \(\frac{5}{3}\) = 55 + 1.67
= 56.67
Hence, the median weight of the students = 56.67 kgs.

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 14 Statistics Ex 14.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 14 Statistics Exercise 14.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Solution:

∴ \(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) = \(\frac{162}{20}\) = 8.1
!! Since f_i and x_i are of small values we use direct method.

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory. (AS₄)

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:

Here, the x_i are of large numerical values in 250 – 300. So, a = 275.
So we use Assumed Mean method then,
\(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
Here, the Assumed mean is taken as 275.
∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) = 275 + \(\frac{1900}{50}\) = 275 + 38 = 313

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f. (AS₄)

Solution:

\(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
\(\overline{\mathrm{x}}\) = 18 (given)
⇒ 18 = \(\frac{752+20 f}{(44+f)}\)
⇒ 18(44 + f) = 752 + 20f ⇒ 20f – 18f = 792 – 752
⇒ 2f = 40
∴ f = \(\frac{40}{2}\) = 20

Question 4.
Thirty women were examined in a hospital by a doctor and their of heart beats per minute were recorded and summarised as shown. Find the mean heart beats per minute for these women, choosing a suitable method. (AS₄)

Solution:

\(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 75.5 + \(\frac{12}{30}\) = 75.5 + 0.4 = 75.9

Question 5.
In a retail market, fruit vendors were selling oranges kept in packing baskets. These baskets contained varying number of oranges. The following was the distribution of oranges. (AS₄)

Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose ?
Solution:

Here, we use step division method where a = 22, h = 5
\(\overline{\mathrm{x}}\) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right]\) × h
⇒ 22 + \(\frac{25}{400}\) × 5
⇒ 22 + 0.31 = 22.31

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method. (AS₄)
Solution:

Here, a = 225, h = 50
\(\overline{\mathrm{x}}\) = a + \(\left[\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right]\) × h
= 225 + \(\frac{(-7)}{25}\) × 50 = 225 – 14 = 211
The average daily expenditure on food = ₹ 211

Question 7.
To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collect for 30 localities in a certain city and is presented below.

Find the mean concentration of SO₂ in the air. (AS₄)
Solution:

∴ \(\overline{\mathrm{x}}\) = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{2.96}{30}\) = 0.0986666 …….
≅ 0.099 ppm

Question 8.
A class teacher has the following attendence record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term. (AS₄)

Solution:

Here, a = 51.5
∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 51.5 – \(\frac{99}{40}\)
⇒ 51.5 – 2.475 = 49.025 ≅ 49 days

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. (AS₄)

Solution:

Here a = 70, h = 10
∴ \(\overline{\mathrm{x}}\) = a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\) × h
⇒ \(\overline{\mathrm{x}}\) = 70 – \(\frac{2}{35}\) × 10
= 70 – \(\frac{20}{35}\) = 70 – 0.5714
= 69.4285 ≅ 69.43%

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.3

Question 1.
Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and coefficients.

i) x² – 2x – 8
ii) 4s² – 4s + 1
iii) 6x² – 3 – 7x
iv) 4u² + 8u
v) t² – 15
vi) 3x² – x – 4
Solution:
i) Let p(x) = x² – 2x – 8
= x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x + 2) (x – 4)
The zeroes of p(x) are given by P(x) = 0
⇒ (x + 2) (x – 4) = 0
⇒ x + 2 = 0 (or) x – 4 = 0
⇒ x = -2 (or) x = 4
Hence the zeroes of x² – 2x – 8 are -2 and 4.
Sum of the zeroes = -2 + 4 = 2

Product of the zeroes = -2 × 4 = -8

ii) Let p(s) = 4s² – 4s + 1
= (2s)² – 2(2s) (1) + (1)²
= (2s – 1)²
= (2s – 1) (2s – 1)
The zeroes of p(s) are given by p(s) = 0
⇒ (2s – 1) (2s – 1) = 0
⇒ 2s – 1 =0 (or) 2s – 1 = 0
⇒ s’ = \(\frac{1}{2}\) (or) s = \(\frac{1}{2}\)
Hence the zeroes of 4s² – 4s + 1 are \(\frac{1}{2}\), \(\frac{1}{2}\)
Sum of the zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1;

(iii) Let p(x) = 6x² – 7x – 3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1 (2x – 3)
= (3x + 1) (2x – 3)
The zeroes of p(x) are given by p(x) = 0
⇒ (3x + 1) (2x – 3) = 0
⇒ 3x + 1 = 0 (or) 2x – 3 = 0
⇒ 3x = -1 (or) 2x = 3
⇒ x = \(\frac{-1}{3}\) (or) x = \(\frac{3}{2}\)
Hence the zeroes of 6x – 7x – 3 are \(\frac{-1}{3}\) and \(\frac{3}{2}\)
Sum of the zeroes = \(\frac{-1}{3}\) + \(\frac{3}{2}\)

iv) Let p(u) = 4u² + 8u
= 4u(u + 2)
The zeroes of p(u) are given by
⇒ p(u) = 0
⇒ 4u(u + 2) = 0
⇒ 4u = 0 (or) u + 2 = 0
⇒ u = 0 (or) u = -2
Hence the zeroes of 4u + 8u are 0 and -2.
Sum of the zeroes = 0 + (-2) = -2

v) Let p(t) = t² – 15
The zero of p(t) is given by p(t) = 0
⇒ t² – 15 = 0 ⇒ t² = 15

vi) Let P(x) = 3x²
The zero of p(x) is given by P(x) = 0
⇒ 3x² – x – 4 = 0
⇒ 3x² + 3x – 4x – 4 = 0
⇒ 3x(x + 1) – 4(x + 1) = 0
⇒ (x + 1) (3x – 4) = 0
⇒ x + 1 = 0 (or) 3x – 4 = 0
⇒ x = -1 (or) x = 4/3
Hence the zeroes of 3x² – x – 4 are -1 (or) 4/3
Sum of the zeroes of 3x² – x – 4 are -1 (or) 4/3.
Sum of the zeroes =

Question 2.
Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

i) \(\frac{1}{4}\), -1
ii) \(\sqrt{2}\), \(\frac{1}{3}\)
iii) 0, \(\sqrt{5}\)
iv) 1, 1
v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
vi) 4, 1 (A.P. Mar. ’15)
Solution:
i) \(\frac{1}{4}\), -1
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = \(\frac{1}{4}\)
Product of the zeroes = αβ = -1
The required quadratic polynomial will be
k[x² – x (α + β) + αβ] where k is a constant
⇒ k[x² – x(\(\frac{1}{4}\)) + (-1)]
⇒ k(x² – \(\frac{x}{4}\) – 1)
If k = 4, then the polynomial will be 4(x² – \(\frac{x}{4}\) – 1) = 4x² – x – 4

ii) \(\sqrt{2}\), \(\frac{1}{3}\)
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = \(\sqrt{2}\)
Product of the zeroes = αβ = \(\frac{1}{3}\)
∴ The required quadratic polynomial will be
k[x² – x(α + β) + αβ] where k is a constant
⇒ k[x² – x(\(\sqrt{2}\)) + \(\frac{1}{3}\)]
⇒ k[x² – \(\sqrt{2}\)x + \(\frac{1}{3}\)]
when k = 3, then the polynomial will be
3[x² – \(\sqrt{2}\)x + \(\frac{1}{3}\)] (or) 3x² – 3\(\sqrt{2}\)x + 1

iii) 0, \(\sqrt{5}\)
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = 0
Product of the zeroes = αβ = \(\sqrt{5}\)
The required quadratic polynomial will be k[x² – x(α + β) + αβ where k is a constant
⇒ k[x² -x(0) + \(\sqrt{5}\)]
⇒ k[x² + \(\sqrt{5}\)].
Where k = 1, the polynomial will be x² + \(\sqrt{5}\).

iv) 1, 1
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = 1
Product of the zeroes = αβ = 1
∴ The required quadratic polynomial will be
k[x² – x(α + β) + αβ] where k is a constant
⇒ k[x² – x(α – β) + 1]
⇒ k[x² – x + 1]
Where k = 1, the polynomial will be x² – x + 1.

v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = –\(\frac{1}{4}\)
Product of the zeroes = αβ = \(\frac{1}{4}\)
The required quadratic polynomial will be k[x² – x(α + β) + αβ] where k is a constant
⇒ k[x² – x(-\(\frac{1}{4}\)) + \(\frac{1}{4}\)]
⇒ k[x² + \(\frac{x}{4}\) + \(\frac{1}{4}\)]
Where k = 4, the polynomial will be
4[x² + \(\frac{x}{4}\) + \(\frac{1}{4}\)] = 4x² + x + 1

Question 3.
Find the quadratic polynomial for the zeroes α, β given in each case. (A.P.Mar.’16)

i) 2, -1
ii) \(\sqrt{3}\), –\(\sqrt{3}\)
iii) \(\frac{1}{4}\), -1
iv) \(\frac{1}{2}\), \(\frac{3}{2}\)
Solution:
i) 2,-1
Let the quadratic polynomial be ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = 2, β = -1
Sum of the zeroes = α + β = 2 + (-1) = 1
Product of the zeros = αβ = 2 × (-1)
= -2
∴ The required quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant
⇒ k [x² – x(1) + (-2)]
⇒ k[x² – x – 2]
Where k = 1, the quadratic polynomial will be x² – x – 2.

ii) \(\sqrt{3}\), –\(\sqrt{3}\)
Let the quadratic polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = \(\sqrt{3}\) , β = –\(\sqrt{3}\)
Sum of the zeroes = α + β
= \(\sqrt{3}\) + (-\(\sqrt{3}\)) = o
Product of the zeros
= αβ = \(\sqrt{3}\) × –\(\sqrt{3}\) = -3
Therefore the quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant
⇒ k [x² – x(0) + (-3)]
⇒ k[x² – 3]
Where k = 1, the quadratic polynomial will be [x² – 3].

iii) \(\frac{1}{4}\), -1
Let the quadratic polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = \(\frac{1}{4}\), β = -1
Sum of the zeroes = α + β
= \(\frac{1}{4}\) + (-1) = \(\frac{-3}{4}\)
Product of the zeroes = αβ = \(\frac{1}{4}\) × (-1)
= –\(\frac{1}{4}\)
Therefore, the quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant
Where k = 4, the quadratic polynomial will be [4x² + 3x – 1]

iv) \(\frac{1}{2}\), \(\frac{3}{2}\)
Let the quadratic polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = \(\frac{1}{2}\), β = \(\frac{3}{2}\)
Sum of the zeroes = α + β
= \(\frac{1}{2}\) + \(\frac{3}{2}\) = \(\frac{4}{2}\) = 2
Product of the zeroes = αβ
= \(\frac{1}{2}\) × \(\frac{3}{2}\) = \(\frac{3}{4}\)
Therefore the quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant

Where k = 4, the quadratic polynomial will be [4x² – 8x + 3].

Question 4.
Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x³ + 3x² – x – 3 and check the relationship between zeroes and the coefficients. (A.P. Mar. ’15)
Solution:
The given polynomial is x³ + 3x² – x – 3
Comparing the given polynomial with ax³ + bx² + cx + d
We get a = 1, b = 3, c = -1, d = -3
Let p(x) = x³ + 3x² – x – 3
Then p(1) = (1)³ + 3(1)² – 1 – 3
= 1 + 3 – 1 – 3
= 4 – 4 = 0
p(-1) = (-1)³ + 3(-1)² – (-1) – 3
= -1 + 3 + 1 – 3 = 0
p(-3) = (-3)³ + 3(-3)² – (-3) – 3
= -27 + 27 + 3 – 3 = 0
Therefore 1, -1 and -3 are the zeroes of
x³ + 3x² – x – 3.
So α = 1, β = -1, γ = -3
α + β + γ = 1 – 1 – 3 = -3 = -3/1 = \(\frac{-b}{a}\)
αβ + βγ + αγ = 1(-1) + (-1)(-3) + (-3)(1)
= -1 + 3 – 3 = -1 = -1/1 = c/a
αβγ = (1)(-1)(-3) = – \(\left(\frac{-3}{1}\right)\) = \(\frac{-\mathrm{d}}{\mathrm{a}}\)

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.2

Question 1.
The graphs of y = p(x) are given in the figure below, for some polynomials p(x). In each case, find the number of zeroes
of p(x).
Solution:

Solution:
Six figures are given and they are numbered.
In the first figure, there are no zeroes.
In the second figure, there is one zero.
In the third figure, there are three zeroes.
In the fourth figure, there are two zeroes.
In the fifth figure, there are four zeroes.
In the sixth figure, there are three zeroes.

Question 2.
Find the zeroes of the given polynomials.
(i) p(x) = 3x
(ii) p(x) = x² + 5x + 6
(iii) p(x) = (x + 2) (x + 3)
(iv) p(x) = x⁴ – 16
Solution:
i) p(x) = 3x
p(0) = 3 × 0 = 0
∴ The zero of p(x) = 3x is 0

ii) p(x) = x² + 5x + 6
= x² + 3x + 2x + 6
= x(x + 3) + 2(x + 3) = (x + 2) (x + 3)
To find zeroes, let p(x) = 0
⇒ (x + 2) (x + 3) = 0
So, x + 2 = 0 (or) x + 3 = 0
⇒ x = -2 (or) x = -3
Therefore, the zeroes of x + 5x + 6 are – 2 and -3.

iii) p(x) = (x + 2) (x + 3)
⇒ x² + 2x + 3x + 6
⇒ x² + 5x + 6
To find zeroes, let p(x) = 0
⇒ (x + 2) (x + 3) = 0
So, x + 2 = 0 (or) x + 3 = 0
⇒ x = -2 (or) x = -3
∴ The zeroes of (x + 2) (x + 3) are -2 and -3.

iv) p(x) = x⁴ – 16
= (x²)² – (4)² = (x² + 4) (x² – 4)
= (x² + 4)(x + 2)(x – 2)
To find zeroes, let p(x) = O
⇒ (x² + 4) (x + 2)(x – 2) = 0
(i.e.,) x² + 4 = 0 (or) x + 2 = 0 (or) x – 2 = 0
If x² + 4 = 0, then x² = -4 ⇒
x = ± \(\sqrt{-4}\)
If x + 2 = 0, then x = -2
If x – 2 = 0, then x = 2
∴ The zeroes of the polynomial x⁴ – 16 are -2, 2 and ± \(\sqrt{-4}\).

Question 3.
Draw the graphs of the given polynomials and find the zeroes. Justify the answers. (A.P.June’ 15)
(i) p(x) = x² – x – 12
(ii) p(x) = x² – 6x + 9
(iii) p(x) = x² – 4x + 5
(iv) p(x) = x² + 3x – 4
(v) p(x) = x² – 1
Solution:
i) p(x) = x² – x – 12

∴ The zeros of x² – x – 12 are 4 and -3.

Scale:
On X-axis: 1 cm = 1 unit
OnY-axis: 1 cm = 2 units

The graph intersects the X-axis at (4, 0) and (-3, 0)

ii) p(x) = x² – 6x + 9


The graph intersects one the X-axis at only point (3, 0)

iii) p(x) = x² – 4x + 5

Scale:
On X-axis: 1 cm = 1 unit
OnY-axis: 1 cm = 2 units

The graph does not intersect at the X-axis.
There are no zeroes of the polynomial x² – 4x + 5

iv) p(x) = x² + 3x – 4


∴ The graph intersects the X-axis at (1, 0) and (-4, 0)
∴ The zeroes of the polynomial x² + 3x – 4 are 1 and -4.

v) p(x) = x² – 1


The graph intersects the X-axis at (-1, 0) and (1, 0)
∴ The zeroes of the polynomial x² – 1 are -1 and 1.

Question 4.
Why are \(\frac{1}{4}\) and -1 zeroes of the polynomial p(x) = 4x² + 3x – 1?
Solution:
p(x) = 4x² + 3x – 1
∴ p(\(\frac{1}{4}\)) = 4(\(\frac{1}{4}\))² + 3(\(\frac{1}{4}\)) – 1
= \(\frac{1}{4}\) + \(\frac{3}{4}\) + \(\frac{1}{1}\)
= \(\frac{1+3-4}{4}\) = \(\frac{0}{4}\) = 0
∴ p(-1) = 4(-1)² + 3(-1) – 1
= 4(1) – 3 – 1
= 4 – 3 – 1
= 4 – 4 = 0
Since p(\(\frac{1}{4}\)) and p(-1) are equal to zero.
\(\frac{1}{4}\) and -1 are the zeroes of the polynomial.

Students can practice 10th Class Maths Textbook SSC Solutions Telangana Chapter 12 Applications of Trigonometry InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 12 Applications of Trigonometry InText Questions

Do This

Question 1.
Draw diagram for the following situations :
i) A person is flying a kite at an angle of elevation a and the length of thread from his hand to kite is ‘l’.
ii) A person observes two banks of a river at angles of depression θ₁ and θ₂ (θ₁ < θ₂) from the top of a tree of height ‘h’ which is at a side of the river. The width of the river is ‘d’. (AS₅) (Page No. 297)
Solution:
i)

In the figure
A is the position of the person.
B is the position of the kite.
AB is the length of thread.

ii)

In the figure,
‘D’ is the position of person
CD is the height of the tree
AB is the width of the river
Angles of depression are θ₁ and θ₂.

Think – Discuss

Question 1.
You are observing top of your school building at an angle of elevation a from a point which is at ‘d’ meter distance from foot of the building. Which trigonometric ratio would you like to consider to find the height of the building.
(AS₃) (Page No. 297)
Solution:
Rough sketch

I would like to use the “Tangent” to find the height of the building by the “Rough sketch”.

Question 2.
A ladder of length ‘x’ meter is leaning against a wall making angle θ with the ground. Which trigonometric ratio would you like to consider to find the height of the point on the wall at which the ladder is touching ? (AS₃) (Page No. 297)
Solution:

I would like to use the “sin θ” to find the height of the point on the wall at which the ladder is touching.