## PSEB 10th Class Maths Solutions Chapter 15 Probability Ex 15.1

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 15 Probability Ex 15.1

Question 1.
Complete the following statements:

(i) Probability of an event E + Probability of the event ‘not E’ = _________.
Solution:
Probability of an event E +
Probability of the event ‘not E’ = 1

(ii) The probability of an event that cannot happen is ___________. Such an event is called _________.
Solution:
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 _________. Such an event is called ________.
Solution:
The probability of an event that is certain to happen is 1. Such event is called sure event.

(iv) The sum of the probabilities of all the elementary events of an experiment is __________.
Solution:
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 _________ and less than or equal to _________.
Solution:
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.
Solution:
When a dnver attempts to start a car the car starts normally. Only when there is some defects the car does not start. So the outcome is not equally likely.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Solution:
When a player attempts to shoot a basketball the outcome in this situation is not equally likely because the outcome depends on many factors such as the training of the player, quality of the gun used etc.

(iii) A trial is made to answer a true – false question. The answer is right or wrong.
Solution:
Since for a question there are two possibilities either right or wrong the outcome in this trial of true-false question is either true or false i.e. one out of the two and both have equal chances to happen. Hence, the two outcomes are equally likely.

(iv) A baby is born. It is a boy or a girl.
Solution:
A new baby (i.e. who took birth at a moment) can be either a boy or a girl and both the outcome have equally likely chances.

Question 3.
Why is tossing a coin considered to he a fair way of deciding which team should get the ball at the beginning of a football game?
Solution:
When a coin is tossed there are only two possibilities i.e. Head or tail both are equally likely to happen. Result of the toss of a fair coin 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:
As we know probability of event cannot be less than O and greater than 1
i.e. 0 ≤ P ≤ 1
∴ (B) – 1.5 is not possible.

Question 5.
If P(E) = 0.05, what is the probability of not E.
Solution. As we know P (E) + P $$(\overline{\mathrm{E}})$$ = 1
P$$(\overline{\mathrm{E}})$$ = 1 – P(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) Since bag contains only lemon flavoured candies
∴ There is no orange candies
∴ It is impossible event.
∴ Probability of getting orange flavoured = 0.

(ii) Since there are only lemon flavoured candies, it is sure event
∴ Probability, of getting lemon flavoured candy = $$\frac{1}{1}$$ = 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 studenís have the same birthday?
Solution:
Let A is event that two students have same birthday
∴ $$(\overline{\mathrm{A}})$$ is event that 2 students not having same birthday is 0.992
∴ P $$(\overline{\mathrm{A}})$$ = 0.992
∴ P (A) = 1 – P (A) (P (A) + P $$(\overline{\mathrm{A}})$$ = 1)
= 1 – 0.992 = 0.008
∴ Probability that two students have saine birthday = 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 probabifity that the ball drawn is
(i) red?
(ii) not red?
Solution:
Number of Red balls = 3
Number of Black balls = 5
Total number of balls = 3 + 5 = 8
One hail is drawn at random
(i) Probability of getting Red ball = $$\frac{\text { Number of favourable cases }}{\text { Total number of cases }}$$
P (Red ball) = $$\frac{3}{8}$$.

(ii) Probability of getting not red ball = 1 – P (Red ball)
= 1 – $$\frac{3}{8}$$ = $$\frac{3}{8}$$ [P $$(\overline{\mathrm{A}})$$ = 1 – P(E)].

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:
Number of red marbles = 5
Number of white marbles = 8
Number of green marbles = 4
Total number of marbles = 5 + 8 + 4 = 17
Since, one marble is taken out
(i) There are 5 Red marbles
Probability of drawing Red marble = $$\frac{\text { Number of favourable cases }}{\text { Total number of cases }}$$
= $$\frac{5}{17}$$

(ii) Since there are 8 white marbles
Probability of drawing white marble = $$\frac{\text { Number of favourable cases }}{\text { Total number of cases }}$$
= $$\frac{8}{17}$$

(iii) There are 4 green bails
Probability of drawing green ball = $$\frac{\text { Number of favourable cases }}{\text { Total number of cases }}$$
= $$\frac{4}{17}$$

∴ Probability of not drawing green ball = 1 – Probability of green ball
= 1 – $$\frac{4}{17}$$ = $$\frac{17-4}{17}$$ = $$\frac{13}{17}$$.

Question 10.
A piggy bank contains hundred 50p 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;
Number of 50 coins = 100
Number of ₹ 1 coins = 50
Number of ₹ 2 coins =20
Number of ₹ 5 coins = 10
∴ Total number of coins = 100 + 50 + 20 + 10 = 180

(i) Since there are 100 ; 50’ p coin
Probability of getting 50p coin = $$\frac{\text { Number of favourable cases }}{\text { Total number of outcomes }}$$

= $$\frac{100}{180}$$

P (50 p coins) = $$\frac{5}{9}$$.

(ii) Number of ₹ 5 coins = 10
∴ Probability of getting ₹ 5 coin = $$\frac{\text { Number of favourable cases }}{\text { Total number of cases }}$$

P (₹ 5 coins) = $$\frac{10}{180}$$ = $$\frac{1}{18}$$
Probability of getting not ₹ 5 coin = 1 – P (₹ 5 coins)
= 1 – $$\frac{1}{18}$$
= $$\frac{18-1}{18}$$ = $$\frac{17}{18}$$.

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. 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 in the tank = 5 + 8 = 13
Probability of getting a male fish = $$\frac{\text { Number of favourable cases }}{\text { Total number of cases }}$$
P(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 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) Total number of outcomes = {1, 2, 3, 4, 5, 6, 7, 8)
Probability of getting ‘8’ = $$\frac{1}{8}$$

(ii)Odd numbers are = {1, 3, 5, 7)
Probability of getting odd number = $$\frac{4}{8}=\frac{1}{2}$$

(iii) Numbers greater than 2 are {3, 4, 5, 6, 7, 8)
∴ Probability of getting number greater than 2 = $$\frac{6}{8}=\frac{3}{4}$$
P (number greater than 2) = $$\frac{3}{4}$$.

(iv) Numbers less than 9 are: {1, 2, 3, 4, 5, 6, 7, 8)
∴ Probability of getting number less than 9 = $$\frac{8}{8}$$
P(a numher less than 9) = 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;
(iii) an odd number.
Solution:
When dice is thrown number of possible outcomes
S = {1, 2, 3, 4, 5, 6)
(i) Prime numbers are {2, 3, 5)
∴ Probability of getting prime number = $$\frac{3}{6}=\frac{1}{2}$$

(ii) Numbers lying between 2 and 6 = {3, 4, 5}
Probability of getting number between 2 and 6 = $$\frac{3}{6}=\frac{1}{2}$$.

(iii) The odd numbers are = {1, 3, 5}
Probability of getting an odd number = $$\frac{3}{6}=\frac{1}{2}$$
P (odd number) = $$\frac{1}{2}$$.

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
(vi) the queen of diamonds.
Solution:
There are 52 cards in a pack
(i) There are two red kings i.e. king of heart and king of diamond
Probability of getting red king = $$\frac{2}{52}=\frac{1}{26}$$
P(Red king) = $$\frac{1}{26}$$

(ii) There are 12 face cards
i.e. 4 Jack, 4 Queens and 4 kings
Probability of getting face card = $$\frac{12}{52}$$
∴ P (A face card) = $$\frac{3}{13}$$.

(iii) Since there are 6 Red face cards i.e 2 Jacks; 2 Queens and 2 Kings
∴ Probability of getting 6 Red face cards = $$\frac{6}{52}$$
P (Red face card) = $$\frac{3}{26}$$.

(iv) There is only one Jack of Heart
∴ Probability of getting Jack of Heart = $$\frac{1}{52}$$
P (A Jack card) = $$\frac{1}{52}$$

(v) Since there are 13 spade cards
∴ Probability of getting a spade card = $$\frac{13}{52}$$
P (A spade card) = $$\frac{1}{4}$$.

(vi) Since there is only one queen of diamonds
∴ Probability of getting queen of spade card = $$\frac{1}{52}$$
P (A queen of spade) = $$\frac{1}{52}$$.

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:
Five cards are ten, jack, queen, king and ace
(i) Probability of getting queen = $$\frac{1}{5}$$
∴ P (A queen) = $$\frac{1}{5}$$.

(ii) If the queen is drawn and put aside then there are 4 cards left – Ten, a Jack, a king and an ace.
(a) Probability of getting an ace = $$\frac{1}{4}$$
P (An Ace) = $$\frac{1}{4}$$.
There’s no queen left

(b) Probability of getting a queen = $$\frac{0}{4}$$ = 0
P (a queen) = 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
Probability of getting good pen = $$\frac{132}{144}=\frac{11}{12}$$
P (a good pen) = $$\frac{11}{12}$$.

Question 17.
(i) A lot of 20 bulbs contains 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 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) Number of defective bulbs 4
Number of good bulbs (Not defective) = 16
Total number of bulbs = 4 + 16 = 20
Probability of getüng defective bulb = $$\frac{4}{20}=\frac{1}{5}$$.

(ii) When a defective bulb drawn is not being replaced, we are left with 19 bulbs
Now probability of getting not defective bulb = $$\frac{15}{19}$$
∴ P (Not defective bulb) = $$\frac{15}{19}$$

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:
From 1 to 90 there are 90 numbers in all and 81 two – digit numbers from 10 to 90
(i) Probability of getting two digit number = $$\frac{81}{90}$$
∴ P (two digit number) = $$\frac{81}{90}=\frac{9}{10}$$.

(ii) Perfect square numbers are (1, 4, 9, 16, 25, 36, 49, 64, 81 } there are 9 perfect square numbers between 1 to 90
Probability of getting perfect square = $$\frac{9}{90}=\frac{1}{10}$$
∴ P (Perfect square) = $$\frac{1}{10}$$

(iii) Numbers divisible by 5 are (5, 10, 15, 20, 25, 30, 35, 40, 45. 50, 55. 60, 65, 70, 75, 80, 85, 90}
There are 18 numbers divisible by 5
∴ Probability of number getting divisible by 5 = $$\frac{18}{90}=\frac{1}{5}$$
∴ Required probability = $$\frac{1}{5}$$.

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

The die is thrown. What is the probability of getting
(i) A ?
(ii) D?
Solution:
Number of faces of a die = 6
S = {A, B, C, D, E, A}
n(S) = 6
(i) Since there are two A’s
∴ Probability of getting A = $$\frac{2}{6}=\frac{1}{3}$$
P(A) = $$\frac{1}{3}$$

(ii) Since there is only one face with D
Probability of getting D = $$\frac{1}{6}$$
∴ P(D) = $$\frac{1}{6}$$

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

Solution:
Length of rectangle (l) = 3 m
Width of rectangle (b) = 2 m
∴ Area of rectangle = 3 m × 2 m = 6m2
Diameter of circle = 1 m
Radius of circle (R) = $$\frac{1}{2}$$ m
∴ Area of circle = πR2 = π($$\frac{1}{6}$$)2
= $$\frac{\pi}{4}$$ m2.

Probability of die to land on a circle = $$\frac{\text { Area of circle }}{\text { Area of rectangle }}$$
= $$\frac{\frac{\pi}{4} \mathrm{~m}^{2}}{6 \mathrm{~m}^{2}}=\frac{\pi}{24}$$
∴ Required Probability = $$\frac{\pi}{24}$$.

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

(i) Let ‘A’ is event showing she buy the pen
∴ Probability that she buy a Pen = $$\frac{124}{144}$$
P(A) = $$\frac{31}{36}$$

(ii) $$\bar{A}$$ is event showing that she will not buy the pen
P $$(\bar{A})$$ = 1 – P(A)
= 1 – $$\frac{31}{36}$$ = $$\frac{36-31}{36}$$
∴ P (Not buy the pen) = $$\frac{5}{36}$$.

Question 22.
Two dice, one blue and one grey are thrown at the same time. Write down all the possible outcomes
(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:
When two dices are thrown total number of possible outcomes
{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
S (4, 1) (4, 2) (4, 3) (4, 4) (4,5) (4,6)
(5, 1) (5, 2) (5.3) 5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}
n(S) = 36
Let A is event of getting sum as 3
∴ A = {(1,2) (2, 1)}
n(A) = 2
∴ Probability of getting sum as 3 = $$\frac{2}{36}=\frac{1}{18}$$
P(A) = $$\frac{1}{18}$$

Let B is event of getting sum as 4 B = ((1, 3), (3, 1), (2, 2))
n(B) = 3
∴ P(B) = $$\frac{3}{36}=\frac{1}{12}$$

Let C is event of getting sum as 5.
C = {(1, 4) (4, 1) (2, 3) (3, 2)}
n(C) = 4
P(C) = $$\frac{4}{36}=\frac{1}{9}$$

Let D is event of getting sum as 6
D = {(1, 5) (5, 1)(2, 4) (4,2) (3, 3)}, n (D) = 5
∴ P(6) = $$\frac{5}{36}$$

Let E is event of getting sum as 7
E = {(1, 6) (6, 1) (2, 5) (5,2) (4, 3) (3, 4)}
∴ P (E) = P (Sum as 7) = $$\frac{6}{36}=\frac{1}{6}$$

Let F is event of getting sum as 8
F = {(2, 6) (6, 2) (3, 5) (4, 4) (5, 3)}
∴ n(F) = 5
P(F) = P(sum as 8) = $$\frac{5}{36}$$

Let G is event of getting sum as 9 when two dices are thrown
G = {(4, 5) (5, 4) (3, 6) (6, 3))
n(G) = 4
∴ P (G) P (Sum as 8) = $$\frac{4}{36}=\frac{1}{9}$$

Let H is event of getting sum as 10
H= {(6, 4) (4, 6) (5, 5)}
n(H) = 3
∴ P (H) = P (sum as 10) = $$\frac{3}{36}=\frac{1}{12}$$

Let I is event of getting sum as 11
I = ((5,6) (6, 5))
n(I) = 2
∴ P(D) = $$\frac{2}{36}=\frac{1}{18}$$

Let J is event of,getting sum as 12
J = {(6,6)}; n(J) = 1
∴ P(J) = $$\frac{1}{36}$$

(ii) No, here all 11 possible outcomes are not equally likely
∴ Three probabilites are different.

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:
When a coin tossed three times, then possible out comes are
S = {HHH, HHT HTH, THH, HTF, THT, TTH, TTT)
n(S) = 8
Let A is event of getting all the three same results i.e., {HHH, TTT}
∴ P(A) = $$\frac{2}{8}=\frac{1}{4}$$
Probability of lossing the game = 1 – P (A)
P $$(\bar{A})$$ = 1 – $$\frac{1}{4}$$
= $$\frac{4-1}{4}$$ = $$\frac{3}{4}$$
∴ Probability of losing the game = $$\frac{3}{4}$$.

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?
Solution:
When a die is thrown twice all possible outcomes are
S = {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) 5) (4, 6) (5, 1) (5, 2) (5, 3) (5,4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}
n(S) = 36
Ler A is event that 5 will come up either time
A = {(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 6)}
n(A) = 11
∴ $$(\bar{A})$$ is event that 5 will not come up either time.
n$$(\bar{A})$$ = 36 – 11 = 25.

(i) ∴ Probability of not getting 5 up either time = $$\frac{25}{36}$$
P $$(\bar{A})$$ = $$\frac{25}{36}$$
Probability that 5 will come up at least once = $$\frac{11}{36}$$
∴ P(A) = $$\frac{11}{36}$$.

Question 25.
Which of the following arguments are correct ? Give reasons for your answer:
(i) 1f 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) When two coins are tossed the possible outcomes are S = {HH, HT, TH, TT}
Probability of getting 2 Heads = $$\frac{1}{4}$$
P(HH) = $$\frac{1}{4}$$
Probability of getting two tails = $$\frac{1}{4}$$
P(TT) = $$\frac{1}{4}$$
Probability of getting one head and one tail = $$\frac{2}{4}=\frac{1}{2}$$
∴ (i) argument is incorrect.

(ii) When a die is thrown possible outcomes are S = (1, 2, 3, 4, 5, 6)
n(S) = 6
Odd numbers are 1, 3, 5
∴ Probability of getting odd number = $$\frac{3}{6}=\frac{1}{2}$$
Even numbers are 2, 4, 6
∴ Probability of getting even number = $$\frac{3}{6}=\frac{1}{2}$$
(ii) argument is correct.

## PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.4

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 14 Statistics Ex 14.4 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 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:

Now, by drawing the points on the graph
i.e. (120, 12); (140, 26); (160, 34); (180, 40); (200, 50).
We get graph of less than type cumulative frequency.

Scale chosen:
On x-axis 10 units = Rs. 10
On y-axis 10 units = 5 workers.

Question 2.
During the medial 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 verWy the result by using the formula.
Solution:

Now, By drawing the points on the graph i.e., (38, 0); (40, 3); (42, 5); (44, 9); (46, 14); (48, 28) ; (50, 32) ; (52, 35) we get graph of less than type cumulative frequency.

Scale Chosen:
On x-axis, 10 units = 2 kg
On y-axis units = 5 students

From above graph, it is clear that
Median = 46.5 kg ; which lies in class interval 46 – 48.
Now, in the given table
$$\Sigma f_{i}$$ = n = 35

∴ $$\frac{n}{2}=\frac{35}{2}$$ = 17.5 ; which lies in the interval 46 – 48.

∴ Median class = 46 – 48
So, l = 46; n = 35; f = 14; cf = 14 and h = 2

Using formula, Median = l + $$\left\{\frac{\frac{n}{2}-c f}{f}\right\}$$ × h

Median = 46 + $$\left\{\frac{\frac{35}{2}-14}{14}\right\}$$ × 2

= 46 + $$\left\{\frac{\frac{35-28}{2}}{14}\right\}$$

= 46 + $$\frac{7}{2} \times \frac{1}{14}$$ = 46 + $$\frac{1}{2}$$

= 46 + 0.5 = 46.5
From above discussion and graph; it is clear that median is same in both cases. Hence, Median weight of students is 46.5 kg.

Question 3.
The following table gives production yield per hectare of wheat of 1(X) farms of a village.

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

Solution:

Now, by drawing the points on the graph i.e. (50, 100); (55, 98); (60, 90); (65, 78); (70, 54); (75, 16)
we get graph of more than type cumulative frequency.

Scale chosen:
On x-axis 10 units = 5 kg/ha
On y-axis 10 units = 10 forms

## PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter Statistics Ex 14.3 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 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.

Solution:

Here, $$\Sigma f_{i}$$ = 68 then $$\frac{n}{2}=\frac{68}{2}$$ = 34
Which lies in interval 125 – 145
Median class = 125 – 145
So, l = 125; n = 68; f = 20; çf = 22 and h = 20

Using formula, Median = l + $$\left[\frac{\frac{n}{2}-c f}{f}\right]$$ × h

= 125 + $$\left\{\frac{\frac{68}{2}-22}{20}\right\}$$ × 20

=125+ $$\frac{34-22}{20}$$ × 20

= 125 + 12 = 137

For mean:

From above data, assumed mean (a) = 135
Width of class (h) = 20
∴ $$\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{7}{68}$$ = 0.102
Using formula, Mean $$(\overline{\mathrm{X}})=a+h \bar{u}$$
$$\overline{\mathrm{X}}$$ = 135 + 20 (0.102)
= 135 + 2.04 = 137.04.

For Mode:
In the given data,
Maximum frequency is 20 and it correspond to 125 – 145.
∴ Modal class = 125 – 145
So l = 125; f1 = 20; f0 = 13; f2 = 14and h = 20
Using formula, Mode = l + $$\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)$$ × h

Mode = 125 + $$\left(\frac{20-13}{2(20)-13-14}\right)$$ × 20

= 125 + $$\frac{7}{40-27}$$ × 20

= 125 + $$\frac{140}{13}$$
= 125 + 10.76923
= 125 + 10.77 = 135.77.
Hence. median, mean and mode of given data is 137 units: 137.04 units and 135.77 units.

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.

Solution:

In thegiven data, $$\Sigma f_{i}$$ = n = 60
∴ $$\frac{n}{2}=\frac{60}{2}$$ = 30
Also, median of the distribution = 28.5 ………….(Given)
which lies in the class interval 20 – 30
Median class = 20 – 30
So, l = 20; f = 20; cf = 5 + x; h = 10
From table, it is clear that 45 + x + y = 60
x + y = 60 – 45 = 15
or x + y = 15 ……………….(1)
Now, using formula, Median = l + {$$\left\{\frac{\frac{n}{2}-c f}{f}\right\}$$

28.5 = 2o + $$\left\{\frac{30-(5+x)}{20}\right\}$$

or 28.5 = 20 + $$\frac{30-5-x}{2}$$

or 28.5 = 

or 2(28.5) = 65 – x
or 57.0 = 65 – x
or x = 65 – 57 = 8
∴ x = 8
Substitute this value of x in (1), we get
8 + y = 15
Hence, values of x and y is 8 and 7.

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

Solution:

Here, $$\Sigma f_{i}$$ = n = 100
then, $$\frac{n}{2}=\frac{100}{2}$$ = 50, which lies in the interval 35 – 40
∴ Median class = 35 – 40
So, l = 35; n = 100; f = 33; cf = 45 and h = 5
Using formula, Median = l + $$\left\{\frac{\frac{n}{2}-c f}{f}\right\}$$ × h

= 35 + $$\left\{\frac{\frac{100}{2}-45}{33}\right\}$$ × 5

= 35 + $$\frac{50-45}{33}$$ × 5

= 35 + $$\frac{25}{33}$$
= 35 + 0.7575 = 35 + 0.76 (approx.) = 35.76
Hence, median age of given data is 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:

Find the median length of the leaves.
Solution:
Since the frequency distribution is not continuous, so firstly we shall make it continuous.

Here, $$\Sigma f_{i}$$ = n = 40
then, $$\frac{n}{2}=\frac{40}{2}$$ = 20, which lies in the interval 144.5 – 153.5
∴ Median class = 144.5 – 153.5
So, l = 144.5; f = 12; cf = 17; h = 9
Using formula, Median = l + $$\left\{\frac{\frac{n}{2}-c f}{f}\right\}$$ × h

Median = 144.5 + $$\left\{\frac{20-17}{12}\right\}$$ × 9

= 144.5 + $$\frac{3 \times 9}{12}$$
= 144.5 + 225 = 146.75
Hence, median length of the leaves is 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.

Solution:

Here, $$\Sigma f_{i}$$ = n = 400
∴ $$\frac{n}{2}=\frac{400}{2}$$ = 200; which lies in the interval 3000 – 3500.
∴ Median class = 3000 – 3500
So, l = 3000; n = 400; f = 86; cf = 130 and h = 500
Using formula, Median = l + $$\left\{\frac{\frac{n}{2}-c f}{f}\right\}$$ × h

Median = 3000 + $$\left\{\frac{\frac{400}{2}-130}{86}\right\}$$ × 500

= 3000 + $$\left(\frac{200-130}{86}\right)$$ × 500

= 3000 + $$\frac{70 \times 500}{86}$$ + 406.9767441

= 3000 + 406.98 (approx.) = 3406.98
Hence, median life time of a lamp is 3406.98 hours.

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets 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 ? Also, find the modal size of the surnames.
Solution.
For Median:

Here, Here, $$\Sigma f_{i}$$ = n = 100
∴ $$\frac{n}{2}=\frac{100}{2}$$ = 50, which lies in interval 7 – 10.
∴ Median class = 7 – 10
So, l = 7; n = 100; f = 40; cf = 36 and h = 3
Using formula, Median = l + $$\left\{\frac{\frac{n}{2}-c f}{f}\right\}$$ × h

Median = 7 + $$\left\{\frac{\frac{100}{2}-36}{40}\right\}$$ × 3

= 7 + $$\left\{\frac{50-36}{40}\right\}$$ × 3

= 7 + $$\frac{14 \times 3}{40}$$

= 7 + $$\frac{21}{20}$$ = 7 + 1.05 = 8.05
Hence, the median of letters in the surnames is 8.05.

For Mean:

From above data, Assumed Mean (a) = 8.5
Width of class (h) = 3
∴ $$\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\bar{u}=\frac{-6}{100}$$ = – 0.06

Using formula, Mean $$(\overline{\mathrm{X}})=a+h \bar{u}$$
$$\bar{X}$$ = 8.5 + 3 (- 0.06) = 8.5 – 0.18 = 8.32
Hence, mean number of letters in the surnames is 8.32.

For Modal:
In the given data Maximum frequency is 44 and it corresponds to interval 7 — 10
∴ Modal class = 7 – 10
So l = 7; f1 = 40; f0 = 30; f2 = 16 and h = 3
Using formula, Mode = l + $$\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)$$ × h

Mode = 7 + $$\left(\frac{40-30}{2(40)-30-16}\right)$$ × 3

= 7 + $$\frac{10}{80-46}$$ × 3

= 7 + $$\frac{30}{34}$$ = 7 + 0.882352941

= 7 + 0.88 (approx.) = 7.88.
Hence. modal size of the surnames is 7.88 letters.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Solution:

Here, $$\Sigma f_{i}$$ = n = 30
∴ $$\frac{n}{2}=\frac{30}{2}$$ = 15; which lies in the interval 55 – 60.
∴ Median class = 55 – 60
So, l = 55; n = 30; f = 6; cf = 13 and h = 5
Using formula, Median = l + $$\left\{\frac{\frac{n}{2}-c f}{f}\right\}$$ × h

Median = 55 + $$\left\{\frac{\frac{30}{2}-13}{6}\right\}$$ × 5

= 55 + $$\left\{\frac{15-13}{6}\right\}$$ × 5

= 55 + $$\frac{2 \times 5}{6}$$

= 55 + $$\frac{5}{3}$$ = 55 + 1.66666
= 55 + 1.67 (approx.) = 56.67
Hence, median weight of the students are 56.67 kg.

## PSEB 10th Class Maths Solutions Chapter 15 Probability Ex 15.2

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 15 Probability Ex 15.2 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 15 Probability Ex 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:
When Shyam and Ekta visit a particular shop in the same week. Possible outcomes are:
S = {(T, T) (T, W) (T, Th) (T, F) (T, S) (W, T) (W, W) (W, Th) (W, F) (W, S) (Th, T) (Th, W) (Th, Th) (Th, F) (Th, S) (F, T) (F, W) (F, Th) (F, F) (F, S) (S, T) (S, W) (S, Th) (S, F) (S, S)}

Here T stands For Tuesday
W stands For Wednesday
Th stands For Thursday
F stands For Friday
S stands For Saturday
n(S) = 25
(i) Let A is event that Shyam and Ekta visit the shop on the same day
A = {(T, T) (W, W) (Th, Th) (F, F) (S, S)}
n(A) = 5
Probability that both will visit the shop on same day = $$\frac{5}{25}$$
∴ P(A) = $$\frac{1}{5}$$.

(ii) Let B is event that both will visit consecutive days particular shop
B = {(T, W) (W, T) (W, Th) (Th, W) (Th, F) (F, Th) (F, S) (F, S)}
n(B) = 8
∴ Probability that both will visit particular shop on consecutive days = $$\frac{8}{25}$$.

(iii) Probability that both will visit the shop on different days = 1 – Probability that both will visit the shop
on same day.
= 1 – $$\frac{1}{5}$$ [∵ P $$(\bar{A})$$ = 1 – P(A)]
= $$\frac{5-1}{5}$$
P $$(\bar{A})$$ = $$\frac{4}{5}$$.

Question 2.
A die, is numbered in such a way that its faces show the numbers 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
(i) even?
(ii) 6?
(iii) at least 6?
Solution:
The complete table is

Number of all possible out comes = 6 × 6 = 36
(i) Let A is event of getting total as even
A = {2, 4, 4, 4, 4, 4, 4, 4, 4, 6, 6, 6, 6, 8, 8, 8, 8, 12}
n (A) = 18
∴ Probability of getting an even number = $$\frac{18}{36}=\frac{1}{2}$$
P (Even Number) = $$\frac{1}{2}$$.

(ii) Let B is event of getting sum as 6 B = {6, 6, 6, 6)
n(B) = 4
Probability of getting an even number = $$\frac{4}{36}$$
∴ P(B) = $$\frac{1}{9}$$.

(iii) Let C is event of getting sum at least 6
C = (6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 12}
n(C) = 15
∴ Probability of getting at least 6 = $$\frac{15}{36}=\frac{5}{12}$$
∴ P(C) = $$\frac{5}{12}$$.

Question 3.
A bag contains 5 red balls and some blue balls. lithe probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Solution:
Number of red balls = 5
Let number of blue balls = x
∴ Total number of balls = 5 + x
According to question,
Probability of drawing blue ball = 2 Probability of Red ball
$$\frac{x}{5+x}=2\left[\frac{5}{5+x}\right]$$
x = 10
∴ Number of blue balls = 10.

Question 4.
A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?
If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Solution:
Total number of balls in bag = 12
Number of black balls x
∴ Probability of getting black ball = $$\frac{x}{12}$$
If 6 more balls put in the box then total number of balls in the box = 12 + 6 = 18
Number of black balls = x + 6
Probability of getting black ball = $$\frac{x+6}{18}$$
According to Question,
Probability of drawing black ball = 2
Probability of drawing blackball in first case
$$\frac{x+6}{18}=\frac{2 x}{12}$$

$$\frac{x+6}{3}=\frac{2 x}{2}$$

$$\frac{x+6}{3}$$ = x
x + 6 = 3x
6 = 3x – x
6 = 2x
x = 3
∴ Number of black balls = 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 marbles in the jar.
Solution:
Total number of marbles in jar =24
Let number of green marbles = x
∴ Number of blue marbles = 24 – x
P (Green marbles) = $$\frac{x}{24}$$
When a marble is drawn
Probability of drawing green marble = $$\frac{2}{3}$$ (Given)
$$\frac{x}{24}=\frac{2}{3}$$
x = $$\frac{24 \times 2}{3}$$
x = 16

∴ Number of green marbles = 16
∴ Number of blue marbles = 24 – x = 24 – 16 = 8.

## PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.2

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter Statistics Ex 14.2 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 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.

Solution:
For mode:
In the given data, Maximum frequency is 23 and it corresponds to the class interval 35 – 45
Modal class = 35 – 45
So, l = 35; f1 = 23; f0 = 21; f2 = 14 and h = 10
Using fonnula, Mode l = l + $$\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)$$ × h
Mode = 35 + $$\left[\frac{23-21}{2(23)-21-14}\right]$$ × 10
= 35 + $$\frac{2}{46-35}$$ × 10
= 35 + $$\frac{20}{11}$$ = 35 + 1.8 = 36.8.

For Mean:

From above data,
Assumed mean (a) = 30
Width of class (h) = 10
∴ $$\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{43}{80}$$ = 0.5375
Using formula, Mean $$(\overline{\mathrm{X}})=a+h \bar{u}$$
$$\bar{X}$$ = 30 + 10 (0.5375)
= 30 + 5.375 = 35.375 = 35.37
Hence, mode of given data is 36.8 years and mean of the given data is 35.37 years. Also, it is clear from above discussion that average age of a patient admitted in the hospital is 35.37 years and maximum number of patients admitted in the hospital are of age 36.8 years.

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

Determine the ¡nodal lifetimes of the components.
Solution:
In the given data.
Maximum frequency is 61 and it corresponds to the class interval 60 – 80.
∴ Model class = 60 – 80
So, l = 60; f1 = 61 ; f0 = 52; f2 = 38 and h = 20
Using formula, Mode = l + $$\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)$$ × h

Mode = 60 + $$\left(\frac{61-52}{2(61)-52-38}\right)$$ × 20

= 60 + $$\frac{9}{122-52-38}$$ × 20

= 60 + $$\frac{9}{32}$$ × 20

= 60 + $$\frac{180}{32}$$

= 60 + 5.625 = 65.625
Hence, modal Lifetimes of the components is 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:
For Mode: In the given data.
Maximum frequency is 40, and it corresponds to the class interval 1500 – 2000.
∴ Model class = 1500 – 2000
So, l = 1500; f1 = 40; f0 = 24; f2 = 33 and h = 500
Using formula, Mode = l + $$\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)$$ × h

= 1500 + $$\left\{\frac{40-24}{2(40)-24-33}\right\}$$ × 500

= 1500 + $$\left\{\frac{16}{80-24-33}\right\}$$ × 500

= 1500 + $$\frac{16 \times 500}{23}$$

= 1500 + $$\frac{8000}{23}$$ = 1500 + 347.83 = 1847.83

For Mean:

From above data,

Assumed Mean (a) = 2750
Length of width (h) = 500
$$\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=-\frac{35}{200}$$ = – 0.175
Using formula, Mean $$(\overline{\mathrm{X}})=a+h \bar{u}$$
$$\overline{\mathbf{X}}$$ = 2750 + 500 (- 0.175)
= 2750 – 87.50 = 2662.50
Hence, the modal monthly expenditure of family is 1847.83 and the mean monthly expenditure is 2662.50.

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

Solution:
For Mode:
In the given data,
Maximum frequency is 10 and it corresponds to the class interval is 30 – 35.
∴ Modal class = 30 – 35.
So, l = 30; f1 = 10; f0 = 9; f2 = 3 and h = 5
using formula, Mode = l + $$\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)$$ × h
Mode = 30 + $$\left(\frac{10-9}{2(10)-9-3}\right)$$ × 5
= 30 + $$\frac{1}{20-12}$$ × 5
= 30 + $$\frac{5}{8}$$ = 30 + 0.625 = 30.625 = 30.63 (approx.)

For mean:

From above data, Assumed Mean (a) = 32.5
Width of class (h) = 5
∴ $$\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=-\frac{23}{35}$$ = – 0.65

Using formula, Mean $$(\overline{\mathrm{X}})=a+h \bar{u}$$
$$\overline{\mathbf{X}}$$ = 32.5 + 5 (- 0.65)
= 32..5 – 3.25 = 29.25 (approx.)
Hence, mode and mean of given data is 30.63 and 29.25. Also, from above discussion, it clear that states/U.T. have student per teacher is 30.63 and on average, this ratio is 29.25.

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:
In the given data,
Maximum frequency is 18 and it corresponds to the class interval 4000 – 5000.
∴ Modal class = 4000 – 5000
So, l = 4000; f1 = 18; f0 = 4; f2 = 9 and h = 1000
Using formula, Mode = l + $$\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)$$ × h

Mode = 4000 + $$\left(\frac{18-4}{2(18)-4-9}\right)$$ × 1000

= 4000 + $$\frac{14}{36-13}$$ × 1000

= 4000 + $$\frac{14000}{23}$$ = 4000 + 608.6956

= 4000 + 608.7 = 4608.7 (approx.)
Hence, mode of the given data is 4608.7.

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:
In the given data,
Maximum frequency is 20 and it corresponds to the class interval 40 – 50
∴ Modal Class = 40 – 50
So, l = 40; f1 = 20; f0 = 12; f2= 11 and h = 10
Using formula, Mode = l + $$\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)$$ × h

Mode = 40 + $$\left(\frac{20-12}{2(20)-12-11}\right)$$ × 10

= 40 + $$\frac{8}{40-23}$$ × 10

= 40 + $$\frac{80}{17}$$ = 40 + 4.70588

= 40 + 4.7 = 44.7 (approx.)
Hence, mode of the given data is 44.7 cars.

## PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.1

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter Statistics Ex 14.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

Question 1.
A survey was conducted by a group of students as a part of their enviroment 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, number of plants and houses are small in their values; so we must use direct method

Mean X = $$\overline{\mathrm{X}}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}$$

$$\frac{162}{20}$$ = 8.1

Hence, mean number of plants per house is 8.1.

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:

From given data,
Assumed Mean (a) = 150
and width of the class (h) = 20
∴ $$\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}$$
= $$\frac{-12}{50}$$ = – 0.24

Using formula, Mean $$(\overline{\mathrm{X}})=a+h \bar{u}$$
= 150 + (20) (- 0.24)
= 150 – 4.8 = 145.2
Hence,mean daily wages of the workers of factory is ₹ 145.20.

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;

From the above data,
Assumed mean (a) = 18
Using formula, Mean $$(\overline{\mathrm{X}})=a+\frac{\sum f_{i} d_{i}}{\Sigma f_{i}}$$

$$\bar{X}=18+\frac{2 f-40}{44+f}$$
But. Mean of data $$(\bar{x})$$ = 18 …(Given)
∴ 18 = 18 + $$\frac{2 f-40}{44+f}$$
or $$\frac{2 f-40}{44+f}$$ = 18 – 18 = 0
or 2f – 40 = 0
or 2f = 40
or f = $$\frac{40}{2}$$ = 20
Hence, missing frequency f is 20.

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:

From above data,
Assumed Mean (a) = 75.5
Width of Class (h) = 3
∴ $$\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}$$
= $$\frac{4}{30}$$ (approx.)
Using formula,
Mean $$(\overline{\mathrm{X}})=a+h \bar{u}$$
= 75.5 + 3 (0.13) = 75.5 + 0.39
$$\overline{\mathrm{X}}$$ = 75.89
Hence, mean heart beats per minute for women is 75.89.

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number 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:
Since, values of number of mamgoes and number of boxes are large numerically. So, we must use step-deviation method.

From above data,
Assumed Mean (a) = 57
Width of class (h) = 3
∴ $$\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}$$

$$\bar{u}=\frac{25}{400}$$ = 0.0625

Using formula, Mean $$(\overline{\mathrm{X}})=a+h \bar{u}$$
$$\bar{X}$$ = 57 + 3(0.0625)
= 57 + 0.1875 = 57.1875 = 57.19.
Hence, mean number of mangoes kept in a packing box is 57.19.

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:

From above data,
Assumed mean (a) = 225
Width of class (h) = 50
∴ $$\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}$$

$$\bar{u}=-\frac{7}{25}$$ = 0.28
Using formula,
Mean $$(\overline{\mathrm{X}})=a+h \bar{u}$$
$$\bar{X}$$ = 225 + 50 (- 0.28)
$$\bar{X}$$ = 225 – 14
$$\bar{X}$$ = 211
Hence, mean daily expenditure on food is ₹ 211.

Question 7.
To find out the concentration of SO2 in the air (in parts per million, 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:

From above data,
Assumed mean (a) = 0.10
Width of class (h) = 0.04
∴ $$\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}$$

$$\bar{u}=-\frac{1}{30}$$ = – 0.33(approx.)
Using formula, Mean $$(\overline{\mathrm{X}})=a+h \bar{u}$$
$$\bar{X}$$ = 0.10 + 0.04 (- 0.33)
= 0.10 – 0.0013 = 0.0987 (approx.)
Hence, mean concentration of SO2 in the air is 0.0987 ppm.

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

Solution:

From above data,
Assumed Mean (a) = 17
Using formula, Mean($$\bar{X}$$) = a + $$\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}$$
$$\bar{X}$$ = 17 + $$-\frac{181}{40}$$
= 17 – 4.52 = 12.48.
Hence, mean 12.48 number of days a student was absent.

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

Solution:

From the above data,
Assumed Mean (a) = 70
Width of class (h) = 10
∴ $$\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{-2}{35}$$ = – 0.057

Using formula, Mean $$(\overline{\mathrm{X}})=a+h \bar{u}$$
$$\bar{X}$$ = 70 + 10 (- 0.057)
= 70 – 0.57 = 69.43
Hence, mean literacy rate is 69.43%.

## PSEB 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.
A copper wire 3 mm in diameter is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to 8.88 g per cm3.
Solution:
Diametcr of wire (d) = 3 mm
∴ Radius of wire (r) = $$\frac{3}{2}$$ mm = $$\frac{3}{20}$$ cm

Diameter of cylinder = 10 cm
Radius of cylinder (R) = 5 cm
Height of cylinder (H) = 12 cm
Circumference of cylinder = length of wire used in one turn
2πR = length of wire used in one turn
2 × $$\frac{22}{7}$$ × 5 = length of wire used in one turn
$$\frac{220}{7}$$ = length of wire used in one turn
Number of turn used = $$\frac{\text { Height of cylinder }}{\text { Diameter of wire }}$$
= $$\frac{12 \mathrm{~cm}}{3 \mathrm{~mm}}=\frac{12}{3} \times 10$$ [1 mm = $$\frac{1}{10}$$ cm]
= $$\frac{120}{3}$$ = 40

∴ length of wire used = Number of turns × length of wire used in one turn
H = 40 × $$\frac{220}{7}$$ = 1257.14 cm
Volume of wire used = πr2H
= $$\frac{22}{7} \times \frac{3}{20} \times \frac{3}{20} \times 1257.14$$ = 88.89 cm3
Mass of 1 cm3 = 8.88 gm
Mass of 88.89 cm3 = 8.88 × 88.89 = 789.41 gm
Hence, length of wire is 1257.14 cm
and mass of wire is 789.41 gm.

Question 2.
A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area the double cone so formed. (Choose value of t as found
appropriate.)
Solution:
Let ∆ABC be the right triangle right angled at A whose sides AB and AC measure 3 cm and 4 cm respectively.
The length of the side BC (hypotenuse) = $$\sqrt{3^{2}+4^{2}}=\sqrt{9+16}$$ = 5 cm
Here, AO (or A’O) is the radius of the common base of the double cone formed revolving the right triangle about BC.
Height of the cone BAA’ is BO and slant height is 3 cm.
Height of the cone CAA’ is CO and slant height is 4 cm.
Now, ∆AOB ~ ∆CAB (AA similarity)

∴ $$\frac{\mathrm{AO}}{4}=\frac{3}{5}$$

⇒ AO = $$\frac{4 \times 3}{5}=\frac{12}{5}$$ cm

Also $$\frac{\mathrm{BO}}{3}=\frac{3}{5}$$

⇒ BO = $$\frac{\mathrm{BO}}{3}=\frac{3}{5}$$ cm
Thus CO = BC – OB
= 5 – $$\frac{9}{5}$$ = $$\frac{16}{5}$$ cm
Now, Volume of double cone = [Volume of Cone ABA’] + [Volume of cone ACA’]

∴ Volume of double cone = 30 cm3.
Also, surface area of double cone = [Surface area of Cone ABA’] + [surface area of Cone ACA’]
= π .AO.AB + π. AO.A’C
= $$\left(\frac{22}{7} \times \frac{12}{5} \times 3\right)+\left(\frac{22}{7} \times \frac{12}{5} \times 4\right)$$

= $$\left(\frac{792}{35} \times \frac{1056}{35}\right)$$
= $$\frac{1848}{35}$$ = 52.75 cm2

Question 3.
A cistern, Internally measuring 150 cm × 120 cm × 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 the water overflowing, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
Solution:
Volume of one brick = 22.5 × 7.5 × 6.5 cm3 = 1096.87 cm3
Volume of cistern = 150 × 120 × 110 cm3 = 1980000 cm3
Let number of bricks used = n
Total volume of n bricks = n (volume of one brick) = n [1096.871] cm3
Volume of water = 129600 cm3
Volume of water available for bricks = 1980000 – 129600 = 1850400 cm3

Each bricks absorb $$\frac{1}{17}$$th of its volume in water
Volume of water available for bricks = $$\frac{17}{16}$$ × volume of water available for bricks
= $$\frac{17}{10}$$ × 1850400
Volume of water available for bricks = 1966050 cm3
Total volume of n bricks = Volume of water available for bricks
n[1096.87] cm3 = 1966050 cm3

n = $$\frac{1966050}{1096.87}$$
n = 1792.42
Hence, Number of bricks used = 1792.

Question 4.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the 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:
Area of valley = 97280 km2
Rainfall in valley = 10 cm
∴ Volume of total rainfall = 97280 × $$\frac{10}{100}$$ × $$\frac{1}{1000}$$ km3
= 9.728 km3

Question 5.
An oil funnel made of tin sheet consists of a cylindrical portion 10 cm long attached to a frust.un of a cone. if the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

Solution:
Diameter of top of funnel = 18 cm
∴ Radius of top of funnel (R) = $$\frac{18}{2}$$ cm = 9 cm
Diameter of bottom of funnel = 8 cm
Radius of bottom of funnel (r) = 4 cm
Height of cylindrical portion (h) = 10 cm
Height of frustum (H) = (22 – 10) = 12 cm
Slant height of frustum (l)
= \sqrt{\mathrm{H}^{2}+(\mathrm{R}-r)^{2}}

= $$\sqrt{(12)^{2}+(9-4)^{2}}$$

= $$\sqrt{144+(5)^{2}}$$

= $$\sqrt{144+25}$$ = $$\sqrt{169}$$
Slant height of frustum (l) = 13 cm
Area of metal sheet required curved
surface area of cylindrical base + curved surface of frustum = 2πrh + πL [R + r]
= 2 × $$\frac{22}{7}$$ × 4 × 10 + $$\frac{212}{7}$$ × 13 [9 + 4] cm2
= 251.42 + 531.14 = 782.56 cm2

Question 6.
Derive the formula for the curved surface area and total surface area of a frustum of a cone, given to you in Section 13.5, using the symbols as explamed.
Solution:
A frustum of a right circular cone has two unequal flat circular bases and a curved surface. Let ACDB be the frustum of the cone obtained by removing the portion VCÐ. The line-segment OP joining the centres of two bases is called the height of the frustum. Each of the line-segments AC and BD of frustum ACDB is called its slant height.

Let R and r be the radii of the circular end faces (R > r) of the frustum ACDB of the cone (V, AB). We complete the conical part VCD. Let h and l be the vertical height and slant height respectively. Then OP = h and AC = BD = l.

The frustum of the right circular cone can be viewed as the difference of the two right circular VAB and VCD.
Let the height of the cone VAB be h1 and its slant height l1 i.e. VP = h1 and VA = VB = l1.

Now from right ∆DEB, we have
DB2 = DE2 + BE2

⇒ l2 = h2 + (R – r)2
⇒ l = $$\sqrt{h^{2}+(\mathrm{R}-r)^{2}}$$
Again ∆VOD ~ ∆VPB

⇒ $$\frac{V D}{V B}=\frac{O D}{P B}$$

⇒ $$\frac{l_{1}-l}{l_{1}}=\frac{r}{R}$$

⇒ 1 – $$\frac{l}{l_{1}}=\frac{r}{\mathrm{R}}$$

⇒ $$\frac{l}{l_{1}}=1-\frac{r}{\mathrm{R}}=\frac{\mathrm{R}-r}{\mathrm{R}}$$

⇒ l1 = $$\frac{l \mathrm{R}}{\mathrm{R}-r}$$

⇒ Now, l1 – l1 = $$\frac{l \mathrm{R}}{\mathrm{R}-r}-l=\frac{l r}{\mathrm{R}-r}$$

Curved surface area of the frustum of cone = πRl1 – πr(l1 – l)
[Curved surface area of a cone = π × r × 1]

= πR. $$\frac{l R}{R-r}$$ – πr. $$\frac{l r}{\mathrm{R}-r}$$

= πl ($$\frac{\mathrm{R}^{2}-r^{2}}{\mathrm{R}-r}$$) = $$\frac{\pi l(\mathrm{R}-r)(\mathrm{R}+r)}{(\mathrm{R}-r)}$$
= πl (R + r) sq. units.
∴ Curved snafaue area of the frustum of right circular cone = πl (R + r) sq. units,
where l = $$\sqrt{h^{2}+(\mathrm{R}-r)^{2}}$$
and total surface area of frustum of right circular cone = Curved surface area + Area of base + Area of top
= πl (R + r) + πR2 + πr2
∴ π [R2 + r2 + l (R + r)] sq. units.

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:
A frustum of a right circular cone has two unequal flat circular bases and a curved surface. Let ACDB be the frustum of the cone obtained by removing the portion VCD. The line-segment OP joining the centres of two bases is called the height of the frustum. Each of the line-segments AC and BD of frustum ACDB is called its slant height.

Ler R and r be the radii of the circular end faces (R > r) of the fnistum ACDB of the cone (V, AB). We complete the conical part VCD. Let h and l be the vertical height and slant height respectively. Then OP = h and AC = BD = l. The frustum of the right circular cone can be viewed as the difference of the two right circular VAB and VCD.
Let theheight of the cone VAB be h1 and its slant height l1 i.e.VP = h1 and VA = VB = l1.
∴ The height of the cone VCD = VP – OP = h1 – h
Since right ∆s VOD and VPB are similar
⇒ $$\frac{\mathrm{VO}}{\mathrm{VP}}=\frac{\mathrm{OD}}{\mathrm{PB}}=\frac{h_{1}-h}{h_{1}}=\frac{r}{\mathrm{R}}$$

⇒ 1 – $$\frac{h}{h_{1}}=\frac{r}{\mathrm{R}}$$

⇒ $$\frac{h}{h_{1}}=1-\frac{r}{\mathrm{R}}=\frac{\mathrm{R}-r}{\mathrm{R}}$$

⇒ h1 = $$\frac{h \mathrm{R}}{\mathrm{R}-r}$$

⇒ Height of the cone VCD = $$\frac{h \mathrm{R}}{\mathrm{R}-r}-h=\frac{h \mathrm{R}-h \mathrm{R}+h r}{\mathrm{R}-r}=\frac{h r}{\mathrm{R}-r}$$

Volume of the frustrum ACDB of the cone (V,AB) = Volume of the cone (V, AB) – Volume of the cone (V, CD)
= 

= $$\frac{\pi}{3}\left(\mathrm{R}^{2} \cdot \frac{h \mathrm{R}}{\mathrm{R}-r}-r^{2} \cdot \frac{h r}{\mathrm{R}-r}\right)$$

= $$\frac{\pi h}{3}\left(\frac{\mathrm{R}^{3}-r^{3}}{\mathrm{R}-r}\right)$$

= $$\frac{1}{3} \pi h \times \frac{(\mathrm{R}-r)\left(\mathrm{R}^{2}+\mathrm{R} r+r^{2}\right)}{(\mathrm{R}-r)}$$

= $$\frac{1}{3}$$ πh (R2 + Rr + r2)
Hence, volume of the frustum of cone is $$\frac{1}{3}$$ πh (R2 + Rr + r2).

Again if A1 and A2 (A1 > A2) are the surface areas of the two circular bases, then A1 = πR2 andA2 = πr2

Now volume of the frustum of cone,

= $$\frac{1}{3}$$ πh (R2 + r2 + Rr)

= $$\frac{h}{3}$$ (πR2 + πr2 + $$\sqrt{\pi \mathrm{R}^{2}} \sqrt{\pi r^{2}}$$)

= $$\frac{h}{3}$$ (A1 + A2 + $$\sqrt{A_{1} A_{2}}$$)

## PSEB 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:

Radius of upper end (R) = 2 cm
Radius of lower end (r) = 1 cm
Height of glass (H) = 14 cm
Glass is in the shape of frustum
Volume of frustum = $$\frac{1}{3}$$ π [R2 + r2 +Rr]H

= $$\frac{1}{3}$$ × $$\frac{22}{7}$$ [(2)2 + (1)2 + 2 × 1] 14

= $$\frac{1}{3}$$ × $$\frac{22}{7}$$ [4 + 1 + 2] 14

= $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 7 × 14

= $$\frac{22 \times 14}{3}$$
Hence, Volume of glass = 102.67 cm3.

Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
Slant height of frustum = 4 cm

Let radius of upper end and lower ends are R and r
Circumference of upper end = 18 cm
2πR = 18
R = $$\frac{18}{2 \pi}=\frac{9}{\pi}$$ cm
Circumference of lower end = 6 cm
2πr = 6 cm
r = $$\frac{6}{2 \pi}=\frac{3}{\pi}$$ cm
Curved surface area of frustum = π [R + r] l
= π $$\left[\frac{9}{\pi}+\frac{3}{\pi}\right]$$ 4
= π $$\left[\frac{9+3}{\pi}\right]$$ 4
= 12 × 4 = 48 cm2
Hence, Curved surface area of frustum = 48 cm2.

Question 3.
A fez, the cap used by the Turks, is shaded like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution:

Radius of lower end of frustum (R) = 10 cm
Radius of upper end of frustum (r) = 4 cm
Slant height of frustum (l) = 15 cm
Curved surface area of frustum = πL [R+ r]
= $$\frac{22}{7}$$ × 15 [10 + 4]
= $$\frac{22}{7}$$ × 15 × 14
= 22 × 15 × 2 = 660 cm2
Area of the closed side = πr2 = $$\frac{22}{7}$$ × (4)2
= $$\frac{22}{7}$$ × 4 × 4 = $$\frac{352}{7}$$ cm2
Total area of the material used = Curved surface area of frustum + Area of the closed side
= 660 + 50.28 = 710.28 cm2
Hence, Total material used = 710.28 cm2.

Question 4.
A container opened from the top is made up of a metal sheet ¡s in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate, of ₹ 20 per litre. Also fmd the cost of metal sheet used to make the container, If it costs ₹ 8 per 100 cm2. (Take π = 3:14.)
Solution:

Radius of upper end of container (R) = 20 cm
Radius of lower end of container (r) = 8 cm
Height of container (H) = 16 cm
Slant height (l) = $$\sqrt{\mathrm{H}^{2}+(\mathrm{R}-r)^{2}}$$
= $$\sqrt{(16)^{2}+(20-8)^{2}}$$ = $$\sqrt{256+144}$$
Slant height (l) = $$\sqrt{400}=\sqrt{20 \times 20}$$ = 20 cm

Capacity of the container = $$\frac{1}{3}$$ πH[R2 + r2 + Rr]
= $$\frac{1}{3}$$ × 3.14 × 16 [(20)2 + (8)2 + 20 × 8]
= $$\frac{3.14 \times 16}{3}$$ [400 + 64 + 100]
= 3.14 × 16 × 624 = 10449.92 cm3
Milk in the container = 10449.92 cm3
= $$\frac{10449.92}{1000}$$ litres [∵ 1 cm3 = $$\frac{1}{1000}$$ litres]
∴ Milk in the container = 10.45 litres
Cost of 1 it. milk = ₹ 20
∴ Cost of 10.45 litre = ₹ 20 × 10.45
Total cost of milk = ₹ 209
Curved surface area of frustum = πL [R + r]
= 3.14 × 20[20 + 8]
= 3.14 × 20 × 28 cm2 = 1758.4 cm2
Area of base of container = πr2
= 3.14 × (8)π = 3.14 × 64 = 200.96 cm2

Total metal used to make coniainer = curved surface area of frustum + area of base
= (1758. 4 + 200.96) cm2 = 1959.36 cm2
Cost of 100 cm2 metal sheet used = ₹ 8
Cost of 1 cm2 metal sheet used = ₹ $$\frac{8}{100}$$
Cost of 1959.36 cm2 metal sheet used = ₹ $$\frac{8}{100}$$ × 1959.36
= ₹ 156.748 = ₹ 156.75
Hence, Total cost of sheet used = ₹ 156.75
and Total cost of milk is ₹ 209.

Question 5.
A metallic right circular cone 20 cm high and whose vertical angle Ls 600 is cut into two parts at the middle of its height by a plane parallel to Its base. If the frustum so obtained be drawn into a wire of diameter $$\frac{4}{4}$$ cm, find the length of the wire.
Solution:
Vertical angle of cone = 60°
Altitude of cone divide vertical angle ∠EOF = 30°

In ∆ODB,
$$\frac{\mathrm{BD}}{\mathrm{OD}}$$ = tan 30°

$$\frac{r}{10}=\frac{1}{\sqrt{3}}$$

r = $$\frac{10}{\sqrt{3}}$$ cm

In ∆OEF,

$$\frac{E F}{O E}$$ = tan 30°

$$\frac{\mathrm{R}}{20}=\frac{1}{\sqrt{3}}$$

R = $$\frac{20}{\sqrt{3}}$$ cm

Volume of frustum = $$\frac{\pi}{3}$$h[R2 + r2 + Rr]
= $$\frac{22}{73} \times \frac{10}{3}\left[\left(\frac{20}{\sqrt{3}}\right)^{2}+\left(\frac{10}{\sqrt{3}}\right)^{2}+\frac{20}{\sqrt{3}} \times \frac{10}{\sqrt{3}}\right]$$

= $$\frac{22}{7} \times \frac{10}{3}\left[\frac{400}{3}+\frac{100}{3}+\frac{200}{3}\right]$$

= $$\frac{22}{7} \times \frac{10}{3}\left[\frac{400+100+200}{3}\right]$$

Volume of frustum = $$\frac{22}{7} \times 10 \times \frac{700}{9}$$ cm3

= $$\frac{22}{7} \times \frac{7000}{9}$$ cm3
Frustum is made into wiie, which is in shape of cylinder having diameter $$\frac{1}{16}$$ cm
∴ Radius of cylinderical wire (r1) = $$\frac{1}{2} \times \frac{1}{16} \mathrm{~cm}=\frac{1}{32} \mathrm{~cm}$$

Let height of cylinder so formed be H cm
On recasting volume remain same
Volume of frustum = Volume of cylindrical wire

$$\frac{22}{7} \times \frac{7000}{9}$$ = πr12H

$$\frac{22}{7} \times \frac{7000}{9}=\frac{22}{7} \times\left(\frac{1}{32}\right)^{2} \times \mathrm{H}$$

H = $$\frac{\frac{22}{7} \times \frac{7000}{9}}{\frac{22}{7} \times \frac{1}{32} \times \frac{1}{32}}$$

= $$\frac{7000}{9}$$ × 32 × 32

H = 796444.44 cm

H =  = 7964.44 m
Hence, Length of cylindrical wire (H) = 7964.44 m

## PSEB 6th Class Punjabi Solutions Chapter 1 ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ

Punjab State Board PSEB 6th Class Punjabi Book Solutions Chapter 1 ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 6 Punjabi Chapter 1 ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ

ਪਾਠ-ਅਭਿਆਸ ਪ੍ਰਸ਼ਨ-ਉੱਤਰ

1. ਛੋਟੇ ਉੱਤਰ ਵਾਲੇ ਪ੍ਰਸ਼ਨ

ਪ੍ਰਸ਼ਨ 1.
‘ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਕਵਿਤਾ ਵਿਚ ਕਿਸ ਪੁੱਤ ਦਾ ਜ਼ਿਕਰ ਹੈ ?
ਉੱਤਰ :
ਪੰਜਾਬ ਦਾ ।

ਪ੍ਰਸ਼ਨ 2.
“ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਕਵਿਤਾ ਦਾ ਕਵੀ ਕੌਣ ਹੈ ?
ਉੱਤਰ :
ਸਰਦਾਰ ਅੰਜੁਮ ।

2. ਸੰਖੇਪ ਉੱਤਰ ਵਾਲੇ ਪ੍ਰਸ਼ਨ

ਪ੍ਰਸ਼ਨ 1.
ਹੇਠ ਲਿਖੀ ਕਾਵਿ-ਸਤਰ ਦਾ ਭਾਵ-ਅਰਥ ਲਿਖੋ :
‘ਇਹ ਮਿੱਟੀ ’ਤੇ ਘਰ-ਘਰ ਜਾ ਕੇ ਪਿਆਰ ਦਾ ਬੂਟਾ ਲਾਵੇ ।
ਉੱਤਰ :
ਪੰਜਾਬੀ ਲੋਕਾਂ ਦਾ ਮੁਢਲਾ ਸੁਭਾ ਸਾਰੀ ਮਨੁੱਖਤਾ ਵਿਚ ਪਿਆਰ ਦਾ ਪਸਾਰ ਕਰਨ ਵਾਲਾ ਹੈ ।

ਪ੍ਰਸ਼ਨ 2.
‘ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਕਵਿਤਾ ਦੀਆਂ ਚਾਰ ਸਤਰਾਂ ਜ਼ਬਾਨੀ ਲਿਖੋ ।
ਉੱਤਰ :
ਇਹਦਾ ਰੰਗ ਸੰਧੂਰੀ ਹੈ, ਇਹ ਗੋਰੀ ਚਿੱਟੀ ਹੈ ।
ਇਹਨੂੰ ਮੈਲੀ ਨਾ ਕਰਨਾ, ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਹੈ ।
ਇਸ ਮਿੱਟੀ ਵਿਚ ਕੋਈ ਕੁੜੱਤਣ, ਕਿੱਦਾਂ ਕੋਈ ਉਗਾਵੇ ।
ਇਹ ਮਿੱਟੀ ‘ਤੇ ਘਰ-ਘਰ ਜਾ ਕੇ ਪਿਆਰ ਦਾ ਬੂਟਾ ਲਾਵੇ ।

ਪ੍ਰਸ਼ਨ 3.
“ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਕਵਿਤਾ ਨੂੰ ਟੋਲੀ ਬਣਾ ਕੇ ਆਪਣੀ ਜਮਾਤ ਵਿਚ ਗਾਓ ।
ਉੱਤਰ:
(ਨੋਟ :-ਵਿਦਿਆਰਥੀ ਆਪ ਕਰਨ ।)

3. ਅਧਿਆਪਕ ਲਈ

ਪ੍ਰਸ਼ਨ 1.
ਪੰਜਾਬ ਦੀ ਮਹਾਨਤਾ ਬਾਰੇ ਬੱਚਿਆਂ ਨੂੰ ਹੋਰ ਜਾਣਕਾਰੀ ਦਿੱਤੀ ਜਾਵੇ ।
ਉੱਤਰ :
ਪੰਜਾਬ ਪੰਜਾਂ ਦਰਿਆਵਾਂ ਦੀ ਧਰਤੀ ਹੈ । 1947 ਵਿਚ ਇਹ ਭਾਰਤੀ ਪੰਜਾਬ ਤੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਪਾਕਿਸਤਾਨੀ ਪੰਜਾਬ ਵਿਚ ਵੰਡੀ ਗਈ ਹੈ , ਜਿਸ ਕਰਕੇ ਹੁਣ ਢਾਈ ਦਰਿਆ, ਸਤਲੁਜ, ਬਿਆਸ ਤੇ ਅੱਧਾ ਰਾਵੀ ਇਧਰ ਰਹਿ ਗਏ ਹਨ ਤੇ ਢਾਈ ਦਰਿਆ ਚਨਾਬ, ਜਿਹਲਮ ਤੇ ਅੱਧਾ ਰਾਵੀ ਉਧਰ । ਇਹ ਦਸ ਗੁਰੂ ਸਾਹਿਬਾਨ ਤੇ ਸੂਫ਼ੀ ਫ਼ਕੀਰਾਂ ਦੀ ਧਰਤੀ ਹੈ । ਇਸੇ ਧਰਤੀ ਉੱਤੇ ਸਿੱਖ ਧਰਮ ਦਾ ਜਨਮ ਹੋਇਆ ਤੇ ਸ੍ਰੀ ਗੁਰੂ ਗ੍ਰੰਥ ਸਾਹਿਬ ਦੀ ਰਚਨਾ ਹੋਈ । ਸੰਸਾਰ ਦੇ ਸਭ ਤੋਂ ਪੁਰਾਤਨ ਧਰਮ ਗੰਥ ਰਿਗਵੇਦ ਦੀ ਰਚਨਾ ਇੱਥੇ ਹੀ ਹੋਈ, । ਮਹਾਂਭਾਰਤ ਦਾ ਯੁੱਧ ਤੇ ਸੀ । ਮਦ ਭਗਵਦ ਗੀਤਾ ਦੀ ਰਚਨਾ ਵੀ ਪੁਰਾਤਨ ਪੰਜਾਬ ਦੀ ਧਰਤੀ ‘ਤੇ ਹੀ ਹੋਈ ।ਇਸਦੀ ਬੋਲੀ ਪੰਜਾਬੀ ਹੈ । ਅਜੋਕੇ ਪੰਜਾਬ ਦੇ ਪੱਛਮੀ ਪਾਸੇ ਪਾਕਿਸਤਾਨ ਲਗਦਾ ਹੈ । ਉੱਤਰੀ ਪਾਸੇ ਕਸ਼ਮੀਰ, ਪੂਰਬੀ ਪਾਸੇ ਹਿਮਾਚਲ, ਦੱਖਣ-ਪੂਰਬੀ ਪਾਸੇ ਹਰਿਆਣਾ ਤੇ ਦੱਖਣ-ਪੱਛਮੀ ਪਾਸੇ ਰਾਜਸਥਾਨ ਲਗਦੇ ਹਨ । ਇਸਦਾ ਖੇਤਰਫਲ 50,362 ਵਰਗ ਕਿਲੋਮੀਟਰ ਹੈ । ਕਣਕ ਤੇ ਝੋਨਾ ਇਸਦੀਆਂ ਮੁੱਖ ਫ਼ਸਲਾਂ ਹਨ । ਇਸਦੇ 22 ਜ਼ਿਲ੍ਹੇ ਹਨ । ਇਸਦੀ ਆਬਾਦੀ 2 ਕਰੋੜ 80 ਲੱਖ ਹੈ । ਪੰਜਾਬੀ ਲੋਕ ਆਪਣੇ ਖੁੱਲ੍ਹੇ-ਡੁੱਲੇ, ਮਿਹਨਤੀ ਤੇ ਅਣਖੀਲੇ ਸੁਭਾ ਕਰਕੇ ਸੰਸਾਰ ਭਰ ਵਿਚ ਪ੍ਰਸਿੱਧ ਹਨ । ਉਹ ਕਿਰਤ ਕਰਨ, ਨਾਮ ਜਪਣ ਤੇ ਵੰਡ ਛਕਣ ਵਿਚ ਵਿਸ਼ਵਾਸ ਰੱਖਦੇ ਹਨ । ਅਫ਼ਸੋਸ ਕਿ ਅੱਜ ਦੀ ਨੌਜਵਾਨ ਪੀੜੀ ਖੇਡਾਂ ਵਿਚ ਨਾਮਣਾ ਕਮਾਉਣ ਤੇ ਸਰੀਰ ਪਾਲਣ ਦੇ ਸ਼ੌਕ ਛੱਡ ਕੇ ਨਸ਼ਿਆਂ ਵਿਚ ਗ਼ਰਕ ਹੋ ਚੁੱਕੀ ਹੈ ।

ਪਸ਼ਨ 1.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਉ) ਇਹਦਾ ਰੰਗ ਸੰਧੂਰੀ ਹੈ ਇਹ ਗੋਰੀ ਚਿੱਟੀ ਹੈ ।
ਇਹਨੂੰ ਮੈਲੀ ਨਾ ਕਰਨਾ, ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਹੈ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਕਹਿੰਦਾ ਹੈ ਕਿ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਦਾ ਰੰਗ ਦਿਲ-ਖਿਚਵਾਂ ਸੰਧੂਰੀ ਹੈ । ਇਹ ਮਿੱਟੀ ਗੋਰੀ-ਚਿੱਟੀ ਅਰਥਾਤ ਸਾਫ਼-ਸੁਥਰੀ ਹੈ । ਸਾਡਾ ਸਭ ਦਾ ਫ਼ਰਜ਼ ਹੈ ਕਿ ਪੰਜਾਬ ਦੀ ਇਸ ਸਾਫ਼-ਸੁਥਰੀ ਮਿੱਟੀ ਨੂੰ ਕਿਸੇ ਤਰ੍ਹਾਂ ਵੀ ਮੈਲੀ ਨਾ ਕਰੀਏ ।

ਔਖੇ ਸ਼ਬਦਾਂ ਦੇ ਅਰਥ-ਸੰਧੂਰੀ = ਸੰਧੂਰ ਦੇ ਰੰਗ ਵਰਗੀ ।

ਪ੍ਰਸ਼ਨ 2.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਅ) ਇਸ ਮਿੱਟੀ ਵਿਚ ਕੋਈ ਕੁੜੱਤਣ, ਕਿੱਦਾਂ ਕੋਈ ਉਗਾਵੇ ।
ਇਹ ਮਿੱਟੀ ’ਤੇ ਘਰ-ਘਰ ਜਾ ਕੇ, ਪਿਆਰ ਦਾ ਬੂਟਾ ਲਾਵੇ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਕਹਿੰਦਾ ਹੈ ਕਿ ਇਹ ਕਿਵੇਂ ਹੋ ਸਕਦਾ ਹੈ ਕਿ ਕੋਈ ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਵਿਚ ਫ਼ਿਰਕੂਪੁਣੇ ਦੀ ਕੁੜੱਤਣ ਪੈਦਾ ਕਰ ਦੇਵੇ, ਜਦਕਿ ਇਹ ਮਿੱਟੀ ਤਾਂ ਘਰ-ਘਰ ਜਾ ਕੇ ਆਪਸੀ ਪਿਆਰ ਦਾ ਬੂਟਾ ਲਾਉਣ ਦਾ ਕੰਮ ਕਰਦੀ ਹੈ, ਅਰਥਾਤ ਪੰਜਾਬੀ ਲੋਕਾਂ ਦਾ ਮੁਢਲਾ ਸੁਭਾ ਸਾਰੀ ਮਨੁੱਖਤਾ ਵਿਚ ਆਪਸੀ ਪਿਆਰ ਦਾ ਪਸਾਰਾ ਕਰਨ ਵਾਲਾ ਹੈ ।

ਪ੍ਰਸ਼ਨ 3.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਈ) ਪੁੱਛ ਲਓ ਫ਼ਕੀਰਾਂ ਤੋਂ, ਇਹਦੀ ਬਾਣੀ ਮਿੱਠੀ ਹੈ ।
ਇਹਨੂੰ ਮੈਲੀ ਨਾ ਕਰਨਾ, ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਹੈ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਕਹਿੰਦਾ ਹੈ ਕਿ ਇਹ ਗੱਲ ਸੂਫ਼ੀ ਫ਼ਕੀਰਾਂ ਤੇ ਗੁਰੂ ਸਾਹਿਬਾਂ ਤੋਂ ਪੁੱਛ ਕੇ ਸਮਝੀ ਜਾ ਸਕਦੀ ਹੈ ਕਿ ਪੰਜਾਬ ਦੀ ਬੋਲੀ ਬਹੁਤ ਮਿੱਠੀ ਹੈ, ਇਸੇ ਕਰਕੇ ਹੀ ਉਨ੍ਹਾਂ ਨੇ ਆਪਣੇ ਭਾਵਾਂ ਤੇ ਵਿਚਾਰਾਂ ਦਾ ਪ੍ਰਗਟਾਵਾ ਇਸ ਬੋਲੀ ਵਿਚ ਕੀਤਾ ਹੈ । ਜਿਸ ਪੰਜਾਬ ਦੀ ਧਰਤੀ ਉੱਤੇ ਇਹ ਬੋਲੀ ਬੋਲੀ ਜਾਂਦੀ ਹੈ, ਉਸਦੀ ਮਿੱਟੀ ਨੂੰ ਕਿਸੇ ਤਰ੍ਹਾਂ ਵੀ ਮੈਲੀ ਨਹੀਂ ਕਰਨਾ ਚਾਹੀਦਾ

ਔਖੇ ਸ਼ਬਦਾਂ ਦੇ ਅਰਥ-ਬਾਣੀ = ਬੋਲੀ ।

ਪ੍ਰਸ਼ਨ 4.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਸ) ਕਰੋ ਦੁਆ ਕਦੇ ਇਸ ਮਿੱਟੀ ’ਤੇ ਉਹ ਮੌਸਮ ਨਾ ਆਵੇ।
ਇਸ ਤੇ ਉੱਗਿਆ ਹਰ ਕੋਈ ਸੂਰਜ, ਕਾਲਖ਼ ਹੀ ਬਣ ਜਾਵੇ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਸਾਨੂੰ ਸਭ ਨੂੰ ਰੱਬ ਅੱਗੇ ਇਹ ਅਰਦਾਸ ਕਰਨ ਲਈ ਕਹਿੰਦਾ ਹੈ ਕਿ ਪੰਜਾਬ ਦੀ ਧਰਤੀ ਦੀ ਮਿੱਟੀ ਉੱਤੇ ਕਦੇ ਵੀ ਫ਼ਿਰਕੂ ਜ਼ਹਿਰ ਨਾਲ ਭਰਿਆ ਉਹ ਮੌਸਮ ਨਾ ਆਵੇ ਕਿ ਇਸ ਉੱਤੇ ਉੱਗਿਆ ਹਰ ਇਕ ਸੁਰਜ ਭਾਵ ਇਸ ਉੱਤੇ ਚੜ੍ਹਨ ਵਾਲਾ ਹਰ ਦਿਨ ਚਾਨਣ ਦੀ ਥਾਂ ਨਫ਼ਰਤ ਦੀ ਕਾਲਖ਼ ਦਾ ਪਸਾਰ ਕਰ ਦੇਵੇ ।

ਔਖੇ ਸ਼ਬਦਾਂ ਦੇ ਅਰਥ-ਦੁਆ = ਬੇਨਤੀ, ਅਰਦਾਸ ।

ਪ੍ਰਸ਼ਨ 5.
ਹੇਠ ਲਿਖੇ ਕਾਵਿ-ਟੋਟੇ ਦੇ ਸਰਲ ਅਰਥ ਕਰੋ :
(ਹ) ਰੱਬ ਦੇ ਘਰ ਤੋਂ ਅੰਜੁਮ, ਆਈ ਇਹ ਚਿੱਠੀ ਹੈ ।
ਇਹਨੂੰ ਮੈਲੀ ਨਾ ਕਰਨਾ, ਮੇਰੇ ਪੰਜਾਬ ਦੀ ਮਿੱਟੀ ਹੈ ।
ਉੱਤਰ :
ਕਵੀ ਅੰਜੁਮ ਕਹਿੰਦਾ ਹੈ ਕਿ ਰੱਬ ਦੇ ਘਰ ਤੋਂ ਵੀ ਸਾਨੂੰ ਚਿੱਠੀ ਰਾਹੀਂ ਇਹ ਸੰਦੇਸ਼ ਪੁੱਜਾ ਹੈ ਕਿ ਅਸੀਂ ਆਪਣੇ ਪੰਜਾਬ ਦੀ ਸਾਫ਼-ਸੁਥਰੀ ਮਿੱਟੀ ਨੂੰ ਨਫ਼ਰਤ ਦੀ ਜ਼ਹਿਰ ਭਰ ਕੇ ਇਸਨੂੰ ਮੈਲੀ ਨਾ ਕਰੀਏ, ਸਗੋਂ ਸਾਫ਼-ਸੁਥਰੀ ਹੀ ਰਹਿਣ ਦੇਈਏ ।. ਔਖੇ ਸ਼ਬਦਾਂ ਦੇ ਅਰਥ-ਅੰਜੁਮ = ਕਵੀ ਦਾ ਨਾਂ, ਸਰਦਾਰ ਅੰਜੁਮ ॥

## PSEB 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of cylinder of radius 6 cm. Find the height of the cylinder.
Solution:

Radius of sphere (r) = 4.2 cm
Radius of cylinder (R) = 6 cm
Let height of cylinder be H cm
On recasting volume remain same
Volume of sphere = Volume of cylinder
$$\frac{4}{3}$$ πR3 = πR2H

$$\frac{4}{3}$$ × $$\frac{22}{7}$$ × 4.2 × 4.2 × 4.2 = $$\frac{22}{7}$$ × 6 × 6 × H
∴ H = $$\frac{\frac{4}{3} \times \frac{22}{7} \times 4.2 \times 4.2 \times 4.2}{\frac{22}{7} \times 6 \times 6}$$

= $$\frac{4}{3} \times \frac{42}{10} \times \frac{42}{10} \times \frac{42}{10} \times \frac{1}{6 \times 6}$$

= $$\frac{2744}{1000}$$ = 2.744 cm

Hence, Height of cylinder (H) = 2.744 cm.

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:

Radius of first sphere (r1) = 6 cm
Radius of second sphere (r2) = 8 cm
Radius of third sphere (r3) = 10 cm
Let radius of new sphere formed be R cm
On recasting volume remain same
Volume of all three spheres = Volume of big sphere

$$\frac{4}{3}$$ πr13 + $$\frac{4}{3}$$ πr23 + $$\frac{4}{3}$$ πr33 = $$\frac{4}{3}$$ πR3

$$\frac{4}{3}$$ π[(6)3 + (8)3 + (10)3] = $$\frac{4}{3}$$ πR3

R3 = $$\frac{\frac{4}{3} \pi[216+512+1000]}{\frac{4}{3} \pi}$$

R3 = 1728
R = $$\sqrt[3]{1728}$$ = $$\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3}$$

= 2 × 2 × 3
R= 12 cm
Hence, Radius of sphere = 12 cm.

Question 3.
A 20 m deep well with-diameter 7 m is dug and the earth from digging is evenly spread out to form of platform 22 m by 14 m. Find the height of the platform.
Solution:
Diameter of well = 7 m
Radius of well (cylinder) R = $$\frac{7}{2}$$ m
Height of well (H) = 20 m
Length of Platform (L) =22 m
Width of Platform (B) = 14 m
Let height of Platform be H m

Volume of earth dug out from well = Volume of platform formed

πR2H = L × B × H
× × × 20 = 22 × 14 × H
∴ H = $$\frac{\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20}{22 \times 14}$$
H = 2.5 cm
Hence, Height of Platform H = 2.5 cm.

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of ¡t has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Depth of well (h) = 14 m
Radius of well (r) = $$\frac{3}{2}$$ m

Embankment is in the shape of hollow cylinder whose inner radius is same as radius of well and width of embañkment 4 m
Timer radius of embankment = Radius of well(r) = $$\frac{3}{2}$$ m
Outer radius of embankment (R) = ($$\frac{3}{2}$$ + 4) m
R = $$\frac{11}{2}$$ = 5.5 m
Volume of earth dug out = Volume of embankment (so formed)
πR2h = volume of outer cylinder – volume of inner cylinder
πr2h = πR2H – πr2H
= πH[R2 – r2]

$$\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times 14$$ = $$\frac{22}{7}$$ × H[(5.5)2 – (1.5)2]

H = $$\frac{\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times 14}{\frac{22}{7} \times(5.5-1.5)(5.5+1.5)}$$

= $$\frac{1.5 \times 1.5 \times 14}{4 \times 7}$$ = 1.125 m.

Hence, Height of embankment H = 1.125 m.

Question 5.
A container shaped like a right circular cylinder having 4iameter 12 cm and height 15 cm Is full of ice-cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be tilled with ice-cream.

Dinieter of cvlrnder (D) = 12 cm
.. Radius of cylinder (R) = 6 cm
Height of cylinder (H) = 15 cm
Diameter of cone = 6 cm
Radius of cone (r) = 3 cm
Radius of hemisphere (r) = 3 cm
Height of cone (h) = 12 cm
Let us suppose number of cones used to fill the ice-cream = n
Volume of ice cream in container = n [Volume of ice cream in one cone]

n = 10
Hence, Number of cones formed = 10.

Question 6.
How many silver coins 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions
5.5 cm × 10 cm × 3.5 cm.
Solution:
Silver coin is in the form of cylinder
Diameter of silver coin = 1.75 cm
∴ Radius of silver coin (r) = $$\frac{1.75}{2}$$ cm
Thickness of silver coin = Height of cylinder (H) = 2 mm

i.e., h = $$\frac{2}{10}$$ cm.
Length of cuboid (L) = 5.5 cm
Width of cuboid (B) = 10 cm
Height of cuboid (H) = 3.5 cm
Let number of coins melted to form cuboid = n
∴ Volume of cuboid = n [volume of one silver coin]
= n[πr2h]
5.5 × 10 × 3.5 = n × $$\frac{22}{7} \times \frac{1.75}{2} \times \frac{1.75}{2} \times \frac{2}{10}$$

$$\frac{\frac{55}{10} \times 10 \times \frac{35}{10}}{\frac{22}{7} \times \frac{175}{200} \times \frac{175}{200} \times \frac{2}{10}}$$ = n
n = 400
Hence, Number of corns so formed = 400.

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:

Radius of cylindrical bucket (R) = 18 cm
Height of cylindrical bucket (H) = 32 cm
Height of conical heap (h) = 24 cm
Let ‘r’ cm and ‘l’ cm be the radius and slant height of cone
Volume of sand in bucket = Volume of sand in cone
πR2H = $$\frac{1}{3}$$ πr2h

$$\frac{22}{7}$$ × 18 × 18 × 32 = $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × r2 × 24
$$\frac{\frac{22}{7} \times 18 \times 18 \times 32}{\frac{1}{3} \times \frac{22}{7} \times 24}$$ = r2
r2 = $$\frac{18 \times 18 \times 32}{8}$$
r2 = 1296
r = $$\sqrt{1296}$$
r = 36
∴ Radius of cone (r) = 36 cm
As we know,
(Slant height)2 = (Radius)2 + (Height)2
l = r2 + h2
l = $$\sqrt{(36)^{2}+(24)^{2}}$$

= $$\sqrt{1296+576}$$ = $$\sqrt{1872}$$

= $$\sqrt{12 \times 12 \times 13}$$

l = 12$$\sqrt{13}$$ cm

∴ Slant height of cone (1) = 12$$\sqrt{13}$$ cm.

Question 8.
Water in a canal 6m wide and 1.5m deep is flowing with a speed of 10 km/k How much area will it irrigate in 30 minutes, If 8 cm of standing water is needed?
Solution:
Width of canal = 6 m
Depth of water in canal = 1.5 m
Velocity at which water is flowing = 10km/hr
Volume of water discharge in one hour = (Area of cross section) velocity
= (6 × 1.5m2) × 10 km
= 6 × 1.5 × 10 × 6 × 1.5 × 10 × 1000 × 1000 m3
∴ Volume of water discharge in $$\frac{1}{2}$$ hour = $$\frac{1}{2}$$ × $$\frac{6 \times 15}{10}$$ × 1000 = 45000 m3
Let us suppose area to be irrigate = (x) m2

According to question, 8 cm standing water is required in field
∴ Volume of water discharge by canal in $$\frac{1}{2}$$ hours = Volume of water in field 45000 m3 = (Area of field) × Height of water
45000 m3 = x × ($$\frac{8}{100}$$ m)
$$\frac{4500}{8}$$ × 100 = x
x = 562500 m2
x = $$\frac{562500}{10000}$$ hectares
[1 m2 = $$\frac{1}{10000}$$ hectares]
x = 56.25 hectares
Hence, Area of field = 56.25 hectares.

Question 9.
A farmer connects a pipe of Internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate 0f 3km/h, in how much time will the tank be filled?
Solution:
Velocity of water = 3 km/hr
Diameter of pipe = 20 cm
Radius of pipe (r) = 10 cm = $$\frac{10}{100}$$ m = $$\frac{1}{10}$$ m
Diameter of tank = 10 m
Radius of tank (R) = 5 m
Depth of tank (H) = 2 m
Let us suppose pipe filled a tank in n minutes
Volume of water tank = Volume of water through the pipe in n minutes
πR2H = n[Area of cross section × Velocity of water]

πR2H = n[(πr2) × 3 km/h]

$$\frac{22}{7}$$ × (5)2 × 2 = n$\frac{22}{7} \times \frac{1}{10} \times \frac{1}{10} \times \frac{3 \times 1000}{60}$

25 × 2 × $$\frac{22}{7}$$ = $$\frac{11}{7}$$ n

n = $$\frac{25 \times 2 \times 22 \times 7}{11 \times 7}$$
n = 100 minutes
Hence, Time taken to fill the tank = 100 minutes.