CBSE Class 10

These NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.2

Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
Solution:
The day are given tues, wed, thr, fri, sat Sample space of the days.

Total number of sample space = 5 x 5
n(S) = 25
(i) Let E be the event on the same day
E = (t, t), (w, w), (th, th), (f, f), (s, s)
Number of events E occured the same day = 5
P(E) = \(\frac { 5 }{ 25 }\) = \(\frac { 1 }{ 2 }\) [P(E) = \(\frac { n(E) }{ n(S) }\)]

(ii) Let F be the event on the consecutive days
n( F) = 8
n(S) = 25
P(F) = \(\frac { n(E) }{ n(S) }\) = \(\frac { 8 }{ 25 }\)

(iii) Let G be the event of different days
n(G) = 20
P(G) = \(\frac { 20 }{ 25 }\) = \(\frac { 4 }{ 25 }\)

Question 2.
A die is numbered in such a way that its faces show the number 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

What is the probability that the total score is at least 6?
(i) even
(ii) 6
(iii) at least 6
Solution:

Question 3.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is doubles that of a red ball, determine the number of blue balls in the bag.
Solution:
Let the number of blue balls be x
Given that the number of red balls is 5
The total number of balls in bag = x + 5
P (getting a red bails) = \(\frac { x+5 }{ 5 }\)
P (getting a blue ball) = \(\frac { 5 }{ x+5 }\)
According to question
2(\(\frac { x+5 }{ 5 }\)) = \(\frac { 5 }{ (x+5) }\)
10 (x + 15) = x² + 5x
x² – 5x – 50 = 0
x² – 10x + 5x – 50 = 0
x[x – 10] – 5 (x – 10] = 0
(x – 10) (x + 5) = 0
x = – 5 which is not possible.
x = 10
Hence, the number of blue balls is 10.

Question 4.
A box contains 12 balls out of which x i are black. If one ball is drawn at random from the box, what is the probability that it will be black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is how double of what it was before. Find x.
Solution:
The total number of balls containing in a box = 12
The total number of black bails = x ;
P (getting a black ball) = \(\frac { x }{ 12 }\)
Now,
When 6 more black balls are put in the box then
The total number of balls in the box = 12 + 6 = 8
and The total number of black balls = x + 6
P (getting a black ball after putting 6 more balls) = \(\frac { x+6 }{ 18 }\)
According to question
2(\(\frac { x }{ 12 }\)) = \(\frac { x+6 }{ 18 }\)
36x = 12x + 72
24x = 72
x = 3

Question 5.
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \(\frac { 2 }{ 3 }\). Find the number of blue balls in the jar.
Solution:
The total number of marbles in the jar = 24
Let the total number of green marbles in the jar be x
Then the total number of blue marbles = 24 – x
P (getting a green marble) = \(\frac { x }{ 24 }\)
According to question
P (getting a green marble) = \(\frac { 2 }{ 3 }\)
\(\frac { x }{ 24 }\) = \(\frac { 2 }{ 3 }\)
x = \(\frac { 24×2 }{ 3 }\)
The total number of blue marbles = 34 – x
= 24 – 16 = 8
Hence, the jar contains 8 blue marbles.

These NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.1

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

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
(i) The outcome is not equally likely because the car starts normally only when there is some defect, the car does not start.
(ii) The outcome is not equally likely because the outcome depends on the training of the player.
(iii) The outcome in the trial of true-false question is, either true or false. Hence, the two outcomes are equally likely.
(iv) A baby can be either a boy or a girl and both the outcomes have equally likely chances.

Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the bail at the beginning of a football game?
Solution:
When we toss a coin, the outcomes head and tail are equally likely. So, the result of an individual coin toss is completely unpredictable.

Question 4.
Which of the following cannot be the probability of an event?
(A) \(\frac { 2 }{ 3 }\)
(B) -1.5
(C) 15%
(D) 0.7
Solution:
We know that probability of an event cannot be less than 0 and greater than 1.
Correct option is (B).

Question 5.
If P (E) = 0.05, what is the probability of ‘not E’?
Solution:
We have, P (E) + P (not E) = 1
Given: P(E) = 0.05
P (not E) = 1 – 0.05 = 0.95

Question 6.
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?
(ii) a lemon flavoured candy?
Solution:
(i) A bag contains only lemon flavoured candies.
P (an orange flavoured candy) = 0
(ii) P (a lemon flavoured candy) = 1

Question 7.
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?
Solution:
We have, P (E) + P (not E) = 1
⇒ P (E) + 0.992 = 1
⇒ P (E) = 1 – 0.992 = 0.008

Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is
(i) red?
(ii) not red?
Solution:
(i) Number of red balls = 3
Number of black balls = 5
Total number of balls = 3 + 5 = 8
∴ Probability (getting red ball) = \(\frac { 3 }{ 8 }\)

(ii) P (not red) = 1 – P(E)
= 1 – \(\frac { 3 }{ 8 }\)
= \(\frac { 8 – 3 }{ 8 }\)
= \(\frac { 5 }{ 8 }\)

Question 9.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red?
(ii) white?
(iii) not green?
Solution:
Total number of marbles = 5 + 8 + 4 = 17
(i) P (red marble) = \(\frac { 5 }{ 17 }\)
(ii) P (white marble) = \(\frac { 8 }{ 17 }\)
(iii) P (not a green marble) = \(\frac { 13 }{ 17 }\)

Question 10.
A piggy bank contains hundred 50 p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin?
(ii) will not be a ₹ 5 coin?
Solution:
(i) The total coin contain in piggy bank = 100 + 50 + 20 + 10 = 180
The number of 50 p coin = 100
P (getting a 50 p coin) = \(\frac { 100 }{ 180 }\)
= \(\frac { 10 }{ 18 }\)
= \(\frac { 5 }{ 9 }\)

(ii) Let F be the event “will not be a ₹ coin”.
Number of outcomes favourable to the event F = 180 – 10 = 170 [ ∵ There are ten ₹ 5 coins]
∴ The number of possible outcomes = 180
∴ P(F) = \(\frac { 170 }{ 180 }\) = \(\frac { 17 }{ 18 }\).

Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see figure). What is the probability that the fish taken out is a male fish?
Solution:
Number of male fish = 5
Number of female fish = 8
Total number of fish = 5 + 8 = 13
P (a male fish) = \(\frac { 5 }{ 13 }\)

Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see figure.), and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Solution:
(i) P (getting 8) = \(\frac { 1 }{ 8 }\)
(ii) P (an odd number) = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\) ( odd numbers are 1, 3, 5, 7)
(iii) P (a number greater than 2) = \(\frac { 6 }{ 8 }\) = \(\frac { 3 }{ 4 }\)
(iv) P (a number less than 9) = \(\frac { 8 }{ 8 }\) = 1

Question 13.
A die is thrown once. Find the probability of getting
(i) a prime number
(ii) a number lying between 2 and 6
(ill) an odd number
Solution:
(i) Prime numbers on a die = 2, 3, 5
P (a prime number) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

(ii) Number lying between 2 and 6 = 3, 4, 5
P(a number lying between 2 and 6) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

(iii) Odd numbers = 1, 3, 5
P (an odd number) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

Question 14.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds
Solution:
Number of cards in a well-shuffled deck = 52.
(i) P (a king of red colour) = \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26 }\)
(ii) P (a face card) = \(\frac { 12 }{ 52 }\) = \(\frac { 3 }{ 13 }\)
(iii) P (a red face card) = \(\frac { 6 }{ 52 }\) = \(\frac { 3 }{ 26 }\)
(iv) P (the jack of hearts) = \(\frac { 1 }{ 52 }\)
(v) P(a spade) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
(vi) P (the queen of diamonds) = \(\frac { 1 }{ 52 }\)

Question 15.
Five cards – the ten, jack, queen, king and ace of diamonds, are well shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is
(a) an ace?
(b) a queen?
Solution:
Out of 5 cards there is only one queen.
(i) P (getting queen) = \(\frac { 1 }{ 5 }\) [when queen is drawn, four cards are left]
(ii) (a) P (an ace) = \(\frac { 1 }{ 4 }\)
(b) P (a queen) = \(\frac { 0 }{ 4 }\) = 0

Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Number of defective pens = 12
Number of good pens = 132
Total number of pens = 12 + 132 = 144
P (the pen is good one) = \(\frac { 132 }{ 144 }\) = \(\frac { 11 }{ 12 }\)

Question 17.
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
(i) Total number of bulbs = 20
Number of defective bulbs = 4
P (bulb drawn is defective) = \(\frac { 4 }{ 20 }\) = \(\frac { 1 }{ 5 }\)
(ii) Remaining bulbs = 19
P (bulb drawn is not defective) = \(\frac { 15 }{ 19 }\)

Question 18.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two digit number.
(ii) a perfect square number.
(iii) a number divisible by 5.
Solution:
Total numbers of discs = 90
(i) P (a two digit number) = \(\frac { 81 }{ 90 }\) = \(\frac { 9 }{ 10 }\)

(ii) Here, perfect square numbers are 1, 4, 9, 16, 25, 36, 49, 64, 81
P (getting a perfect square number) = \(\frac { 9 }{ 90 }\) = \(\frac { 1 }{ 10 }\)

(iii) Numbers divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
P (getting a number divisible by 5) = \(\frac { 18 }{ 90 }\) = \(\frac { 1 }{ 5 }\)

Question 19.
A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting
(i) A?
(ii) D?
Solution:
(i) P (getting A) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)
(ii) P (getting D) = \(\frac { 1 }{ 6 }\)

Question 20.
Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m?

Solution:
Area of rectangular region = 1 x b
= 3 x 2 = 6m²
Area of circle whose diameter is 1 = πr²

Question 21.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) she will buy it?
(ii) she will not buy it?
Solution:
Total number of ball pens = 144
Number of defective pens = 20
Number of good pens = 144 – 20 = 124

Question 22.
Two dice, one blue and one grey, are thrown at the same time. Now
(i) Complete the following table:

(ii) A student argues that-there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability \(\frac { 1 }{ 11 }\). Do you agree with this argument? Justify your answer.
Solution:
(i) Total number of possible outcomes = 36
(1, 2) and (2, 1) are the favourable events of getting the sum 3.
P(sum 3) = \(\frac { 2 }{ 36 }\) = \(\frac { 1 }{ 18 }\)
(1, 3) , (2, 2) and (3, 1) are the favourable events of getting the sum 4.
P(sum 4) = \(\frac { 3 }{ 36 }\) = \(\frac { 1 }{ 12 }\)
(1, 4) , (2, 3), (3, 2) and (4, 1) are the favourable events of getting the sum 5.
P(sum 5) = \(\frac { 4 }{ 36 }\) = \(\frac { 1 }{ 9 }\)
(1, 5) , (2, 4), (3, 3), (4, 2) and (5, 1) are the favourable events of getting the sum 6.
P (sum 6) = \(\frac { 5 }{ 36 }\)
(1, 6) , (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) are the favourable events of getting the sum 7.
P(sum 7) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)
(3, 6) , (4, 5), (5, 4) and (6, 3) are the favourable events of getting the sum 9.
P(sum 9) = \(\frac { 4 }{ 36 }\) = \(\frac { 1 }{ 9 }\)
(4, 6) , (5, 5) and (6, 4) are the favourable events of getting the sum 10.
P(sum 10 = \(\frac { 3 }{ 36 }\) = \(\frac { 1 }{ 12 }\)
(5,6) and (6,5) are the favourable events of getting the sum 11.
P(sum 11) = \(\frac { 2 }{ 36 }\) = \(\frac { 1 }{ 18 }\)

(ii) No, because the outcomes as 11 different sum are not equally likely.

Question 23.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e. three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Possible outcomes are
HHH, TTT, HHT, HTH, THH, TTH, THT, HTT = 8
P (win the game) = \(\frac { 2 }{ 8 }\) = \(\frac { 1 }{ 4 }\)
P (lose the game) = 1 – \(\frac { 1 }{ 4 }\) = \(\frac { 3 }{ 4 }\)

Question 24.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment.]
Solution:
Total outcomes = 36
Number of outcomes in favour of 5 is (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 6) = 11
(i) P (5 will not come up either time) = \(\frac { 25 }{ 36 }\)
(ii) P (5 will come up at least once) = \(\frac { 11 }{ 36 }\)

Question 25.
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes- two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac { 1 }{ 3 }\).
(ii) If a die is thrown, there are two possible outcomes- an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac { 1 }{ 2 }\).
Solution:
(i) Argument is incorrect.
The possible outcomes are (HH), (HT), (TH), (TT)
P(HH) = \(\frac { 1 }{ 4 }\)
P(TT) = \(\frac { 1 }{ 4 }\)
P(HT or TH) = \(\frac { 2 }{ 4 }\) = \(\frac { 1 }{ 2 }\)

(ii) Argument is correct.
Possible outcomes = 1, 2, 3, 4, 5, 6
Odd numbers are = 1, 3, 5
P (an odd number) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\).

These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5

Question 1.
A copper wire, 3 mm in diameter, is wound about in cylinder whose length is 12 cm, and diameter 10 cm, so as the cover the curve surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm³.
Solution:
One round of wire covers 3 m = \(\frac { 3 }{ 10 }\) cm in thickness of the surface of the cylinder.
Length of the cylinder = 12 cm
Number of the rounds to cover 12 cm

Diameter of the cylinder = 10 cm.
Radius = 5 cm
Length of the wire in completing one round
= 2 x π x cm = 10π cm
Length of the wire in covering the whole surface = Length of the wire in completely 40 rounds
(10π x 40) cm = 400π cm
= (400 x 3.14) cm.
= 1256 cm
= 125.6 m
Radius of copper wire = \(\frac { 3 }{ 20 }\) cm
Volume of wire = π x r²h
= π x \(\frac { 3 }{ 20 }\) x \(\frac { 3 }{ 20 }\) x 400π
= 9π² cm³.
So, mass of the wire = 9π² x 8,88 per cm³.
= 787.98 gm
Length of wire = 12.56 m and mass of the wire = 787.98 gm.

Question 2.
A right triangle, whose sides 3 cm and 4 cm (other than hypotenuse) is made to resolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate),
Solution:
Let ΔABC be the right angle triangled at B whose sides AB and BC are 3 cm, 4 cm respectively.
Length of hypotenuse = \(\sqrt{(3)^{2}+(4)^{2}}\) = \(\sqrt{25}\) = 5 cm.

BO and DO are the common base of double cone formed by revolving the right triangle about AC,
Slant height = 3 cm (for cone ABD)
Slant height = 4 cm (for cone BCD)
(∵ AOD = 90 = ADC and DAC is common)
\(\frac { AO }{ 3 }\) = \(\frac { 3 }{ 4 }\) = \(\frac { OD }{ 4 }\)
⇒ AO = \(\frac { 9 }{ 5 }\) = 1.8 cm
h = 1.8 cm
CO = AC – OA
⇒ CO = 5 – 18 = 3.2 cm (H = 3.2 cm)
Similarly OD = \(\frac { 12 }{ 5 }\) 2.4 cm = r
Now, volume of double cone

Curved surface area of double cone
= πr(l₁) + πr(l₂)
= πr(3) + πr(4)
– 3.14 x 2.4 (7)
= 52.75 cm².

Question 3.
A cistern, internally measuring 150 cm x 120 cm x 110 cm, has 129600 cm³ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one – seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each being 22.5 cm x 7.5 cm x 6.5 cm
Solution:
Area of cistern = 150 cm x 120 cm x 110 cm – 1980000 cm³.
Empty volume of cistern = (1980000 – 129600)cm³.
Let number of bricks dropped in the water be n
So the volume of bricks = (22.5 7.5 x 6.5) n = 1096.875 cm³.
The volume of water absorbed by each brick = \(\frac { 1 }{ 17 }\) x (1096.S95) cm³.
Volume of total bricks = (1096.875) \(\frac { n }{ 17 }\) cm³.
So, 1850400 cm³ = 1096.875
n = (1096.875) \(\frac { n }{ 17 }\)

∴ Number of bricks = 1792.

Question 4.
In one fortnight of a given month, there was rainfall of 10 cm in a river valley. If the area of one valley is 97280 km², show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Solution:
Volume of a river = 1072 km x 75 m x 3 m
= 1072000 m x 75 m 3 m
= 241200000 m³
Volume of 3 rivers
= 3 x 241200000
= 723600000 m³.
Now value of rainfall in the valley
= height of rainfall x Area
= 10 cm x 97280 km².
= \(\frac { 10 }{ 100 }\) m x 97280 x (100)² m².
= \(\frac { 10 }{ 100 }\) x 97280000000
= 9728000000 m³.
Now 9728000000 m³ > 73360000 m³.
Hence, the additional water i,e. of the 3 river cannot be equivalent to the rainfall.

Question 5.
An oil funnel made of tin sheet consist of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (See the adjoining figure).

Solution:
Let l be the slant height of the frustum part of the funnel.
Then, l = \(\sqrt{(9-4)^{2}+12^{2}}\) = \(\sqrt{25+144}\) cm
= \(\sqrt{169}\) cm = 13 cm
Now, Tin required = Curved surface area of cylindrical portion + Curved surface area of frustum potion
= 2πr₁h + π(r₁ + r₂)h
= [2π x 4 x 10 + π (4 + 9) x 13] cm²
= (80π + 169π) cm² = 249π cm²
= \(\left(249 \times \frac{22}{7}\right)\) cm²
= \(\frac { 5478 }{ 7 }\)
= 782 \(\frac { 4 }{ 7 }\) cm².

Question 6.
Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in section 13.5, using the symbols as explained.
Solution:
APO’B and APQR are similar

Now curved surface area of the fur stum is given by
= curved surface area of cone PQR (bigger)- curved surface and of cone PAB (smaller).
= πr₁l₁ – πr₂l₂ … (3)
∴ From (2) and (3)
Curved surface area of the frustum

= π(r₁ + r₂)l [Using (2)]
∴ Total surface area of the frustum curved surface area of the frustum + area of both circular bases.
= π(r₁ + r₂) l + πr²₁ + πr²₂
= π(r₁ + r₂) l + r²₁ + r²_2.

Question 7.
Derive the formula for the volume of the frustum of a cone, given to you in section 13.5, using the symbols as explained.
Solution:
Since ΔPO’B and ΔPQR are similar

Now volume of the frustum = Volume of cone PQR (bigger) – Voolume of cone PAQ (smaller)

These NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths 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.
Solution:

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

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:

\(\frac { n }{ 2 }\) = \(\frac { 35 }{ 2 }\) = 17.5, therfore median class is 46 – 48.

By using formula,

where l₁, = 46 (lower limit)
h = 2 (class size)
c = 14 (cumulative frequency)
f = 14
then from the graph, median = 46.5
Thus, we see that two value i.e., from graph and formula are same.

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

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

These NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths 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.

Which method did you use for finding the mean, and why?
Solution:

Since the values of x_i and f_i are small, so we have used direct method to find the mean.

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

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Here h = 20

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.

Solution:

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

Solution:

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:

∴ The mean number of mangoes kept in packing box is 57.19
We have used direct method because the numerical values of x_i and f_i are small.

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.
Solution:

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

Find the mean concentration of SO₂ in the air.
Solution:
Here h = 0.04

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:

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

Solution:
Here h = 10

These NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.2

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

Solution:
Here the maximum class frequency is 23, and the class corresponding this frequency is 35-45. So, the modal class 35 – 45.
Here, modal class = 35 – 45, l = 35, class size = (h) = 10

∴ The mode of the given data is 36.8 years.
Table for mean

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

Determine the modal lifetimes of the components.
Solution:
Here the maximum class frequency is 61, and the class corresponding to this frequency is 60 – 80. So, the modal class is 60 – 80.
Now, lower limit (l) of modal class = 60, class size (h) = 20, frequency (f) = 61
Frequency (f₀) of class preceding the modal class = 52
Frequency (f₂) of class succeeding the modal class = 38

∴ Modal lifetimes of the components = 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:

Question 4.
The following distribution gives the state-wise teacher- student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures

Solution:

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

Find the mode of the data.
Solution:

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 it in the table given below. Find the mode of the data:

Solution: