## Selina Concise Mathematics Class 10 ICSE Solutions Probability

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Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability

### Probability Exercise 25(A) – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A coin is tossed once. Find the probability of:
(i) getting a tail
(ii) not getting a tail
Solution:

Question 2.
A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:
(i) a black ball.
(ii) a red ball.
(iii) a white ball.
(iv) not a red ball.
(v) not a black ball.
Solution:

Question 3.
In a single throw of a die, find the probability of getting a number:
(i) greater than 4.
(ii) less than or equal to 4.
(iii) not greater than 4.
Solution:

Question 4.
In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will not be an even number.
(iii) will be an odd number.
Solution:

Question 5.
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will:
(i) be a black card.
(ii) not be a red card.
(iii) be a red card.
(iv) be a face card.
(v) be a face card of red colour.
Solution:
Total number of cards = 52
Total number of outcomes = P(s) = 52
There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and diamond are black. So, there are 26 red cards and 26 black cards.
(i) Number of black cards in a deck = 26
P(E) = favourable outcomes for the event of drawing a black card = 26

(iv) There are 52 cards in a deck of cards, and 12 of these cards are face cards (4 kings, 4 queens, and 4 jacks).
P(E) = 12

Question 6.
(i) If A and B are two complementary events then what is the relation between P(A) and P(B)?
(ii) If the probability of happening an event A is 0.46. What will be the probability of not happening of the event A?
Solution:
(i) Two complementary events, taken together, include all the outcomes for an experiment and the sum of the probabilities of all outcomes is 1.
P(A) + P(B) = 1
(ii) P(A) = 0.46
Let P(B) be the probability of not happening of event A
We know,
P(A) + P(B) = 1
P(B) = 1 – P(A)
P(B) = 1 – 0.46
P(B) = 0.54
Hence the probability of not happening of event A is 0.54

Question 7.
In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of:
(i) winning of Geeta
(ii) not winning of Ritu
Solution:
(i) Winning of Geeta is a complementary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(winning of Geeta) = 1
P(winning of Geeta) = 1 – P(winning of Ritu)
P(winning of Geeta) = 1 – 0.73
P(winning of Geeta) = 0.27
(ii) Not winning of Ritu is a complementary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(not winning of Ritu) = 1
P(not winning of Ritu) = 1 – P(winning of Ritu)
P(not winning of Ritu) = 1 – 0.73
P(not winning of Ritu) = 0.27

Question 8.
In a race between Mahesh and John, the probability that John will lose the race is 0.54. Find the probability of:
(i) winning of Mahesh
(ii) winning of John
Solution:
(i) But if John looses, Mahesh wins
Hence, probability of John losing the race = Probability of Mahesh winning the race since it is a race between these two only
Therefore, P(winning of Mahesh) = 0.54
(ii) P(winning of Mahesh) + P(winning of John) = 1
0.54 + P(winning of John) = 1
P(winning of John) = 1 – 0.54
P(winning of John) = 0.46

Question 9.
(i) Write the probability of a sure event
(ii) Write the probability of an event when impossible
(iii) For an event E, write a relation representing the range of values of P(E)
Solution:

The number of elements in ‘E’ can’t be less than ‘0’ i.e. negative and greater than the number of elements in S.

Question 10.
In a single throw of die, find the probability of getting:
(i) 5
(ii) 8
(iii) a number less than 8
(iv) a prime number
Solution:

(ii) There are only six possible outcomes in a single throw of a die. If we want to find probability of 8 to come up, then in that case number of possible or favourable outcome is 0 (zero)

Question 11.
A die is thrown once. Find the probability of getting:
(i) an even number
(ii) a number between 3 and 8
(iii) an even number or a multiple of 3
Solution:

Question 12.
Which of the following cannot be the probability of an event?
(i) 3/5
(ii) 2.7
(iii) 43%
(iv) -0.6
(v) -3.2
(vi) 0.35
Solution:

Question 13.
A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:
(i) a white ball
(ii) a black ball
Solution:

Question 14.
A single letter is selected at random from the word ‘Probability’. Find the probability that it is a vowel.
Solution:

Question 15.

Solution:

### Probability Exercise 25(B) – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be:
(i) an even number
(ii) a multiple of 3
(iii) an even number and a multiple of 3
(iv) an even number or a multiple of 3
Solution:

(iv) From numbers 2 to 10, there are 7 numbers which are even numbers or a multiple of 3 i.e. 2, 3, 4, 6, 8, 9, 10
Favorable number of events = n(E) = 7

Question 2.
Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is:
(i) a multiple of 5
(ii) a multiple of 6
(iii) between 40 and 60
(iv) greater than 85
(v) less than 48
Solution:

Question 3.
From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 3
(ii) 5
(iii) 3 and 5
(iv) 3 or 5
Solution:

Question 4.
A die is thrown once. Find the probability of getting a number:
(i) less than 3
(ii) greater than or equal to 4
(iii) less than 8
(iv) greater than 6
Solution:

Question 5.
A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8?
Solution:

Question 6.
A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.
Solution:

Question 7.
If two coins are tossed once, what is the probability of getting:
(iii) both heads or both tails.
Solution:

Question 8.
Two dice are rolled together. Find the probability of getting:
(i) a total of at least 10.
(ii) a multiple of 2 on one die and an odd number on the other die.
Solution:

Question 9.
A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is:
(i) a spade(v) Jack or queen
(ii) a red card(vi) ace and king
(iii) a face card(vii) a red and a king
(iv) 5 of heart or diamond(viii) a red or a king
Solution:

Question 10.
A bag contains 16 colored balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is:
(i) red(v) green or red
(ii) not red(vi) white or green
(iii) white(vii) green or red or white
(iv) not white
Solution:

Question 11.
A ball is drawn at random from a box containing 12 white, 16 red and 20 green balls. Determine the probability that the ball drawn is:
(i) white(iii) not green
(ii) red(iv) red or white
Solution:

Question 12.
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is:
(i) a red card
(ii) a black card
(iv) an ace
(v) a black ace
(vi) ace of diamonds
(vii) not a club
(viii) a queen or a jack
Solution:

Question 13.
Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is:
(i) a multiple of 4 or 6
(ii) a multiple of 3 and 5
(iii) a multiple of 3 or 5
Solution:

Question 14.
In a single throw of two dice, find the probability of:
(i) a doublet
(ii) a number less than 3 on each dice
(iii) an odd number as a sum
(iv) a total of at most 10
(v) an odd number on one dice and a number less than or equal to 4 on the other dice.
Solution:

### Probability Exercise 25(C) – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A bag contains 3 red balls, 4 blue balls and 1 yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it; find the probability that the ball is:
(i) yellow
(ii) red
(iii) not yellow
(iv) neither yellow nor red
Solution:

Question 2.
A dice is thrown once. What is the probability of getting a number:
(i) greater than 2?
(ii) less than or equal to 2?
Solution:

Question 3.
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is:
(i) a face card
(ii) not a face card
(iii) a queen of black card
(iv) a card with number 5 or 6
(v) a card with number less than 8
(vi) a card with number between 2 and 9
Solution:

Question 4.
In a match between A and B:
(i) the probability of winning of A is 0.83. What is the probability of winning of B?
(ii) the probability of losing the match is 0.49 for B. What is the probability of winning of A?
Solution:

Question 5.
A and B are friends. Ignoring the leap year, find the probability that both friends will have:
(i) different birthdays?
(ii) the same birthday?
Solution:

Question 6.
A man tosses two different coins (one of Rs 2 and another of Rs 5) simultaneously. What is the probability that he gets:
Solution:

Question 7.
A box contains 7 red balls, 8 green balls and 5 white balls. A ball is drawn at random from the box. Find the probability that the ball is:
(i) white
(ii) neither red nor white.
Solution:

Question 8.
All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting:
(i) a black face card
(ii) a queen
(iii) a black card
Solution:

Question 9.
In a musical chairs game, a person has been advised to stop playing the music at any time within 40 seconds after its start. What is the probability that the music will stop within the first 15 seconds?
Solution:

Question 10.
In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that:
(i) it is acceptable to a trader who accepts only a good shirt?
(ii) it is acceptable to a trader who rejects only a shirt with major defects?
Solution:

Question 11.
Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 8
(ii) 13
(iii) less than or equal to 12
Solution:

Question 12.
Which of the following cannot be the probability of an event?
(i) 3/7
(ii) 0.82
(iii) 37%
(iv) -2.4
Solution:

Question 13.
If P(E) = 0.59; find P(not E)
Solution:
P(E) + P(not E) = 1
0.59 + P(not E) = 1
P(not E) = 1 – 0.59 = 0.41

Question 14.
A bag contains a certain number of red balls. A ball is drawn. Find the probability that the ball drawn is:
(i) black
(ii) red
Solution:

Question 15.
The probability that two boys do not have the same birthday is 0.897. What is the probability that the two boys have the same birthday?
Solution:
P(do not have the same birthday)+P(have same birthday) = 1
0.897 + P(have same birthday) = 1
P(have same birthday) = 1 – 0.897
P(have same birthday) = 0.103

Question 16.
A bag contains 10 red balls, 16 white balls and 8 green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be:
(i) not red?
(ii) neither red nor green?
(iii) white or green?
Solution:

Question 17.
A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin:
(i) will be a Re 1 coin?
(ii) will not be a Rs 2 coin?
(iii) will neither be a Rs 5 coin nor be a Re 1 coin?
Solution:

Question 18.
A game consists of spinning arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; as shown below.
If the outcomes are equally likely, find the probability that the pointer will point at:
(i) 6
(ii) an even number
(iii) a prime number
(iv) a number greater than 8
(v) a number less than or equal to 9
(vi) a number between 3 and 11

Solution:

Question 19.
One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:
(i) a queen of red color
(ii) a black face card
(iii) the jack or the queen of the hearts
(iv) a diamond
(v) a diamond or a spade
Solution:

Question 20.
From a deck of 52 cards, all the face cards are removed and then the remaining cards are shuffled. Now one card is drawn from the remaining deck. Find the probability that the card drawn is:
(i) a black card
(ii) 8 of red color
(iii) a king of black color
Solution:

Question 21.
Seven cards:- the eight, the nine, the ten, jack, queen, king and ace of diamonds are well shuffled. One card is then picked up at random.
(i) What is the probability that the card drawn is the eight or the king?
(ii) If the king is drawn and put aside, what is the probability that the second card picked up is:
a) an ace? b) a king?
Solution:

Question 22.
A box contains 150 bulbs out of which 15 are defective. It is not possible to just look at a bulb and tell whether or not it is defective. One bulb is taken out at random from this box. Calculate the probability that the bulb taken out is:
(i) a good one
(ii) a defective one
Solution:

Question 23.
(i) 4 defective pens are accidentally mixed with 16 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is drawn at random from the lot. What is the probability that the pen is defective?
(ii) Suppose the pen drawn in (i) is defective and is not replaced. Now one more pen is drawn at random from the rest. What is the probability that this pen is:
a) defective
b) not defective?
Solution:

Question 24.
A bag contains 100 identical marble stones which are numbered 1 to 100. If one stone is drawn at random from the bag, find the probability that it bears:
(i) a perfect square number
(ii) a number divisible by 4
(iii) a number divisible by 5
(iv) a number divisible by 4 or 5
(v) a number divisible by 4 and 5
Solution:

Question 25.
A circle with diameter 20 cm is drawn somewhere on a rectangular piece of paper with length 40 cm and width 30 cm. This paper is kept horizontal on table top and a die, very small in size, is dropped on the rectangular paper without seeing towards it. If the die falls and lands on paper only, find the probability that it will fall and land:
(i) inside the circle
(ii) outside the circle
Solution:

Question 26.
Two dice (each bearing numbers 1 to 6) are rolled together. Find the probability that the sum of the numbers on the upper-most faces of two dice is:
(i) 4 or 5
(ii) 7, 8 or 9
(iii) between 5 and 8
(iv) more than 10
(v) less than 6
Solution:

Question 27.
Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting:
(iv) all tails
(v) at least one tail
Solution:

Question 28.
Two dice are thrown simultaneously. What is the probability that:
(i) 4 will not come up either time?
(ii) 4 will come up at least once?
Solution:

Question 29.
Cards marked with numbers 1, 2, 3 ……… 20 are well shuffled and a card is drawn at random. What is the probability that the number on the card is:
(i) a prime number
(ii) divisible by 3
(iii) a perfect square
Solution:

Question 30.
Offices in Delhi are open for five days in a week (Monday to Friday). Two employees of an office remain absent for one day in the same particular week. Find the probability that they remain absent on:
(i) the same day
(ii) consecutive day
(iii) different days
Solution:

Question 31.
A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball; find the number of black balls in the box.
Solution:

Question 32.
From a pack of 52 playing cards, all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is
(i) A face card (King, Jack or Queen)
(ii) An even numbered red card?
Solution:

Question 33.
A die has 6 faces marked by the given numbers as shown below:

The die is thrown once. What is the probability of getting
(i) a positive integer?
(ii) an integer greater than -3?
(iii) the smallest integer?
Solution:
Given that the die has 6 faces marked by the given numbers as below:

When a die is rolled, total number of possible outcomes – 6
(i) For getting a positive integer, the favourable outoomes are: 1, 2, 3
⇒ Number of favourable outcomes – 3
⇒ Required probability = $$\frac { 3 }{ 6 } =\frac { 1 }{ 2 }$$

(ii) For getting an integer greater than -3, the favourable outcomes
are: -2,-1, 1, 2, 3
⇒ Number of favourable outcomes – 5
⇒ Required probability = $$\frac { 5 }{ 6 }$$

(iii) For getting a smallest integer, the favourable outoomes are: -3
⇒ Number of favourable outcomes = 1
⇒ Required probability = $$\frac { 1 }{ 6 }$$

Question 34.
A bag contains 5 white balls, 6 red balls and 9 green balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is:
(i) a green ball
(ii) a white or a red ball.
(iii)Neither a green ball nor a white ball
Solution:
Number of white balls = 5
Number of red balls = 6
Number of green balls = 9
∴ Total number of balls = 5 + 6 + 9 = 20

Question 35.
A game of numbers has cards marked with 11, 12, 13, ….., 40. A card is drawn at random. Find the probability that the number on the card drawn is:
(i) A perfect square
(ii) Divisible by 7.
Solution:
Total number of outcomes = 30
(i) The perfect squares from 11 to 40 are 16, 25 and 36. So, the number of possible outcomes = 3 Hence, the probability that the number on the card drawn is a perfect square
= $$=\quad \frac { Number\quad of\quad possible\quad outcomes }{ Total\quad number\quad of\quad outcomes } =\frac { 3 }{ 30 } =\frac { 1 }{ 10 }$$
(ii) Among the given numbers, 14, 21, 28 and 35 are divisible by 7. So, the number of possible outcomes = 4 Hence, the probability that the number on the card drawn is divisible by 7
= $$\frac{\text { Number of possible autoomes }}{\text { Total number of outoomes }}=\frac{4}{30}=\frac{2}{15}$$

Question 36.
Sixteen cards are labelled as a, b, c, … , m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is:
i. a vowel
ii. a consonant
iii. none of the letters of the word median?
Solution:
Here, Total number of all possible outcomes = 16
i. a, e, i and o are the vowels.
Number of favourable outcomes = 4
∴ Required Probability = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{4}{16}=\frac{1}{4}$$

ii. Number of consonants = 16 – 4 (vowels) = 12
∴ Number of favourable outcomes = 12
∴ Required Probability = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{12}{16}=\frac{3}{4}$$

iii. Median contains 6 letters.
∴ Number of favourable outcomes = 16 – 6 = 10
∴ Required Probability = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{10}{16}=\frac{5}{8}$$

Question 37.
A box contains a certain number of balls. On each of 60% balls, letter A is marked. On each of 30% balls, letter B is marked and on each of remaining balls, letter C is marked. A ball is drawn from the box at random. Find the probability that the ball drawn is:
i. marked C
ii. A or B
iii. neither B nor C
Solution:
A box contains,
60% balls, letter A is marked.
30% balls, letter B is marked.
10% balls, letter C is marked.
i. Total number of all possible outcomes = 100
Number of favourable outcomes = 10
∴ Required Probability = $$\frac{\text { Number of favou rable outcomes }}{\text { Total number of all possible outcomes }}=\frac{10}{100}=\frac{1}{10}$$

ii. The probability that the ball drawn is marked A = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{60}{100}=\frac{6}{10}$$ … (1)
The probability that the ball drawn is marked B = $$\frac{\text { Number of favou rable outcomes }}{\text { Total number of all possible outcomes }}=\frac{30}{100}=\frac{3}{10}$$ … (2)
∴ Required Probability = $$\frac{6}{10}+\frac{3}{10}=\frac{9}{10}$$
iii. The probability that the ball drawn is neither B nor C
= 1 – [P(B) + P(C)]
= 1 – $$\left[\frac{3}{10}+\frac{1}{10}\right]$$
= 1 – $$\frac{4}{10}$$
= $$\frac{6}{10}$$
= $$\frac{3}{5}$$

Question 38.
A box contains a certain number of balls. Some of these balls are marked A, some are marked B and the remaining are marked C. When a ball is drawn at random from the box P(A) = $$\frac{1}{3}$$ and P(B) = $$\frac{1}{4}$$. If there are 40 balls in the box which are marked C, find the number of balls in the box.
Solution:
P(C) = 1 – [P(A) + P(B)] = $$1-\left[\frac{1}{3}+\frac{1}{4}\right]=1-\frac{7}{12}=\frac{5}{12}$$
Probability = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}$$
Given that 40 balls in the box are marked C.
⇒ $$\frac{5}{12}=\frac{40}{\text { Total number of all possible outcomes }}$$
⇒ Total number of all possible outcomes = $$\frac{40 \times 12}{5}=96$$
∴ the number of balls in the box is 96.

More Resources for Selina Concise Class 10 ICSE Solutions

## Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere (Surface Area and Volume)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume)

### Cylinder, Cone and Sphere Surface Area and Volume Exercise 20A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The height of a circular cylinder is 20 cm and the radius of its base is 7 cm. Find :
(i) the volume
(ii) the total surface area.
Solution:

Question 2.
The inner radius of a pipe is 2.1 cm. How much water can 12 m of this pipe hold?
Solution:

Question 3.

Solution:

Question 4.
How many cubic meters of earth must be dug out to make a well 28 m deep and 2.8 m in diameter? Also, find the cost of plastering its inner surface at Rs 4.50 per sq meter.
Solution:

Question 5.
What length of solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of external diameter 20 cm, 0.25 cm thick and 15 cm long?
Solution:

Question 6.
A cylinder has a diameter of 20 cm. The area of curved surface is 100 sq cm. Find:
(i) the height of the cylinder correct to one decimal place.
(ii) the volume of the cylinder correct to one decimal place.
Solution:

Question 7.
A metal pipe has a bore (inner diameter) of 5 cm. The pipe is 5 mm thick all round. Find the weight, in kilogram, of 2 metres of the pipe if 1 cm3 of the metal weights 7.7 g.
Solution:
Inner radius of the pipe = r =$$\frac{5}{2}$$ = 2.5 cm
External radius of the pipe = R = Inner radius of the pipe + Thickness of the pipe
= 2.5 cm + 0.5 cm
= 3 cm
Length of the pipe = h = 2 m= 200 cm
Volume of the pipe = External Volume – Internal Volume

Since 1cm3 of the metal weights 7.7 9,
∴ Weight of the pipe = (1728.6 × 7.7)g = $$\left(\frac{1728.6 \times 7.7}{1000}\right)$$ kg = 13.31 kg

Question 8.
A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm x 14 cm 10.5 cm. Find the rise in level of the water when the solid is submerged.
Solution:

Question 9.
A cylindrical container with internal radius of its base 10 cm, contains water up to a height of 7 cm. Find the area of wetted surface of the cylinder.
Solution:

Question 10.
Find the total surface area of an open pipe of length 50 cm, external diameter 20 cm and internal diameter 6 cm.
Solution:

Question 11.

Solution:

Question 12.
The radius of a solid right circular cylinder increases by 20% and its height decreases by 20%. Find the percentage change in its volume.
Solution:

Question 13.
The radius of a solid right circular cylinder decreases by 20% and its height increases by 10%. Find the percentage change in its :
(i) volume
(ii) curved surface area
Solution:

Question 14.
Find the minimum length in cm and correct to nearest whole number of the thin metal sheet required to make a hollow and closed cylindrical box of diameter 20 cm and height 35 cm. Given that the width of the metal sheet is 1 m. Also, find the cost of the sheet at the rate of Rs. 56 per m.
Find the area of metal sheet required, if 10% of it is wasted in cutting, overlapping, etc.
Solution:

Question 15.
3080 cm3 of water is required to fill a cylindrical vessel completely and 2310 cm3 of water is required to fill it upto 5 cm below the top. Find :
(ii) height of the vessel.
(iii) wetted surface area of the vessel when it is half-filled with water.
Solution:

Question 16.
Find the volume of the largest cylinder formed when a rectangular piece of paper 44 cm by 33 cm is rolled along it :
(i) shorter side.
(ii) longer side.
Solution:

Question 17.
A metal cube of side 11 cm is completely submerged in water contained in a cylindrical vessel with diameter 28 cm. Find the rise in the level of water.
Solution:

Question 18.
A circular tank of diameter 2 m is dug and the earth removed is spread uniformly all around the tank to form an embankment 2 m in width and 1.6 m in height. Find the depth of the circular tank.
Solution:

Question 19.
The sum of the inner and the outer curved surfaces of a hollow metallic cylinder is 1056 cm2 and the volume of material in it is 1056 cm3. Find its internal and external radii. Given that the height of the cylinder is 21 cm.
Solution:

Question 20.
The difference between the outer curved surface area and the inner curved surface area of a hollow cylinder is 352 cm2. If its height is 28 cm and the volume of material in it is 704 cm3;find its external curved surface area.
Solution:

Question 21.
The sum of the heights and the radius of a solid cylinder is 35 cm and its total surface area is 3080 cm2, find the volume of the cylinder.
Solution:

Question 22.
The total surface area of a solid cylinder is 616 cm2. If the ratio between its curved surface area and total surface area is 1 : 2; find the volume of the cylinder.
Solution:

Question 23.
A cylindrical vessel of height 24 cm and diameter 40 cm is full of water. Find the exact number of small cylindrical bottles, each of height 10 cm and diameter 8 cm, which can be filled with this water.
Solution:

Question 24.
Two solid cylinders, one with diameter 60 cm and height 30 cm and the other with radius 30 cm and height 60 cm, are metled and recasted into a third solid cylinder of height 10 cm. Find the diameter of the cylinder formed.
Solution:

Question 25.
Solution:

Question 26.

The given figure shows a solid formed of a solid cube of side 40cm and a solid cylinder of radius 20 cm and height 50 cm attached to the cube as shown.
Find the volume and the total surface area of the whole solid (Take π = 3.14)
Solution:

Question 27.
Two right circular solid cylinders have radii in the ratio 3 : 5 and heights in the ratio 2 : 3, Find the ratio between their :
(i) curved surface areas.
(ii) volumes.
Solution:

Question 28.
A dosed cylindrical tank, made of thin ironsheet, has diameter = 8.4 m and height 5.4 m. How much metal sheet, to the nearest m2, is used in making this tank, if $$\frac{1}{15}$$ of the sheet actually used was wasted in making the tank?
Solution:

### Cylinder, Cone and Sphere Surface Area and Volume Exercise 20B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the volume of a cone whose slant height is 17 cm and radius of base is 8 cm.
Solution:

Question 2.

Solution:

Question 3.
The circumference of the base of a 12 m high conical tent is 66 m. Find the volume of the air contained in it.
Solution:

Question 4.
The radius and height of a right circular cone are in the ratio 5:12 and its volume is 2512 cubic cm. Find the radius and slant height of the cone. (Take π = 3.14)
Solution:

Question 5.
Two right circular cones x and y are made, x having three times the radius of y and y having half the volume of x. Calculate the ratio between the heights of x and y.
Solution:

Question 6.
The diameters of two cones are equal. If their slant heights are in the ratio 5:4, find the ratio of their curved surface areas.
Solution:

Question 7.
There are two cones. The curved surface area of one is twice that of the other. The slant height of the latter is twice that of the former. Find the ratio of their radii.
Solution:

Question 8.
A heap of wheat is in the form of a cone of diameter 16.8 m and height 3.5 m. Find its volume. How much cloth is required to just cover the heap?
Solution:

Question 9.
Find what length of canvas, 1.5 m in width, is required to make a conical tent 48 m in diameter and 7 m in height. Given that 10% of the canvas is used in folds and stitching. Also, find the cost of the canvas at the rate of Rs. 24 per meter.
Solution:

Question 10.
A solid cone of height 8 cm and base radius 6 cm is melted and re-casted into identical cones, each of height 2 cm and diameter 1 cm. Find the number of cones formed.
Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.
A vessel, in the form of an inverted cone, is filled with water to the brim. Its height is 32 cm and diameter of the base is 25.2 cm. Six equal solid cones are dropped in it, so that they are fully submerged. As a result, one-fourth of water in the original cone overflows. What is the volume of each of the solid cones submerged?
Solution:

Question 14.

Solution:

### Cylinder, Cone and Sphere Surface Area and Volume Exercise 20C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.

Solution:

Question 2.

Solution:

Question 3.
A spherical ball of lead has been melted and made into identical smaller balls with radius equal to half the radius of the original one. How many such balls can be made?
Solution:

Question 4.
How many balls each of radius 1 cm can be made by melting a bigger ball whose diameter is 8 cm.
Solution:

Question 5.
8 metallic sphere; each of radius 2 mm, are melted and cast into a single sphere. Calculate the radius of the new sphere.
Solution:

Question 6.
The volume of one sphere is 27 times that of another sphere. Calculate the ratio of their:
(ii) surface areas
Solution:

Question 7.
If the number of square centimeters on the surface of a sphere is equal to the number of cubic centimeters in the volume, what is the diameter of the sphere?
Solution:

Question 8.
Solution:

Question 9.
The internal and external diameters of a hollow hemi-spherical vessel are 21 cm and 28 cm respectively. Find:
(i) internal curved surface area
(ii) external curved surface area
(iii) total surface area
(iv) volume of material of the vessel.
Solution:

Question 10.
A solid sphere and a solid hemi-sphere have the same total surface area. Find the ratio between their volumes.
Solution:

Question 11.
Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted and recasted into a single solid sphere. Taking ∏ = 3.1, find the surface area of solid sphere formed.
Solution:

Question 12.
The surface area of a solid sphere is increased by 21% without changing its shape. Find the percentage increase in its:
(ii) volume
Solution:

### Cylinder, Cone and Sphere Surface Area and Volume Exercise 20D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast.
Solution:

Question 2.
A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.
Solution:

Question 3.
The radii of the internal and external surfaces of a metallic spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid right circular cone of height 32 cm. find the diameter of the base of the cone.
Solution:

Question 4.
Total volume of three identical cones is the same as that of a bigger cone whose height is 9 cm and diameter 40 cm. find the radius of the base of each smaller cone, if height of each is 108 cm.
Solution:

Question 5.
A solid rectangular block of metal 49 cm by 44 cm by 18 cm is melted and formed into a solid sphere. Calculate the radius of the sphere.
Solution:

Question 6.
A hemi-spherical bowl of internal radius 9 cm is full of liquid. This liquid is to be filled into conical shaped small containers each of diameter 3 cm and height 4 cm. How many containers are necessary to empty the bowl?
Solution:
Radius of hemispherical bowl = 9 cm
Volume = $$\frac{1}{2} \times \frac{4}{3} \pi r^{3}=\frac{2}{3} \pi 9^{3}=\frac{2}{3} \pi \times 729=486 \pi \mathrm{cm}^{2}$$
Diameter each of cylindrical bottle = 3 cm
Radius = $$\frac{3}{2}$$cm, and height = 4 cm
∴ Volume of bottle = $$\frac{1}{3} \pi \pi^{2} n=\frac{1}{3} \pi \times\left(\frac{3}{2}\right)^{2} \times 4=3 \pi$$
∴ No. of bottles = $$\frac{486 \pi}{3 \pi}=162$$

Question 7.
A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone if it is completely filled.
Solution:

Question 8.
A solid cone of radius 5 cm and height 8 cm is melted and made into small spheres of radius 0.5 cm. Find the number of spheres formed.
Solution:

Question 9.

Solution:

Question 10.
A solid metallic cone, with radius 6 cm and height 10 cm, is made of some heavy metal A. In order to reduce weight, a conical hole is made in the cone as shown and it is completely filled with a lighter metal B. The conical hole has a diameter of 6 cm and depth 4 cm. Calculate the ratio of the volume of the metal A to the volume of metal B in the solid.

Solution:

Question 11.
A hollow sphere of internal and external radii 6 cm and 8 cm respectively is melted and recast into small cones of base radius 3 cm and height 8 cm. Find the number of cones.
Solution:

Question 12.
The surface area of a solid metallic sphere is 2464 cm2. It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate :
(i) the radius of the sphere.
(ii) the number of cones recast. (Take π = $$\frac{22}{7}$$)
Solution:

### Cylinder, Cone and Sphere Surface Area and Volume Exercise 20E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A cone of height 15 cm and diameter 7 cm is mounted on a hemisphere of same diameter. Determine the volume of the solid thus formed.
Solution:

Question 2.
A buoy is made in the form of a hemisphere surmounted by a right circular cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 m and its volume is two-third the volume of hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two decimal places.
Solution:

Question 3.

Solution:

Question 4.
The cubical block of side 7 cm is surmounted by a hemisphere of the largest size. Find the surface area of the resulting solid.
Solution:

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of the top which is open is 5 cm. It is filled with water. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:

Question 6.
A hemispherical bowl has negligible thickness and the length of its circumference is 198 cm. find the capacity of the bowl.
Solution:

Question 7.
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r cm.
Solution:

Question 8.

Solution:

Question 9.
A solid hemisphere of diameter 28 cm is melted and recast into a number of identical solid cones, each of diameter 14 cm and height 8 cm. Find the number of cones so formed.
Solution:

Question 10.
A cone and a hemisphere have the same base and same height. Find the ratio between their volumes.
Solution:

### Cylinder, Cone and Sphere Surface Area and Volume Exercise 20F – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base are removed. Find the volume of the remaining solid.
Solution:

Question 2.
From a solid cylinder whose height is 16 cm and radius is 12 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume and total surface area of the remaining solid.
Solution:

Question 3.
A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 80 m, calculate the total area of canvas required. Also, find the total cost of canvas used at Rs 15 per meter if the width is 1.5 m
Solution:

Question 4.
A circus tent is cylindrical to a height of 8 m surmounted by a conical part. If total height of the tent is 13 m and the diameter of its base is 24 m; calculate:
(i) total surface area of the tent
(ii) area of canvas, required to make this tent allowing 10% of the canvas used for folds and stitching.
Solution:

Height of the cylindrical part = H = 8 m
Height of the conical part = h = (13 – 8)m = 5 m
Diameter = 24 m → radius = r = 12 m
Slant height of the cone = l

Slant height of cone = 13 m
(i) Total surface area of the tent = 2πrh + πrl = πr(2h + l)

(ii)Area of canvas used in stitching = total area

Question 5.
A cylindrical boiler, 2 m high, is 3.5 m in diameter. It has a hemispherical lid. Find the volume of its interior, including the part covered by the lid.
Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.
A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm x 14 cm x 10.5 cm. Find the rise in level of the water when the solid is submerged.
Solution:

Question 9.
Spherical marbles of diameter 1.4 cm are dropped into beaker containing some water and are fully submerged. The diameter of the beaker is 7 cm. Find how many marbles have been dropped in it if the water rises by 5.6 cm.
Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.
A cylindrical can, whose base is horizontal and of radius 3.5 cm, contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate:
(i) the total surface area of the can in contact with water when the sphere is in it;
(ii) the depth of water in the can before the sphere was put into the can.
Solution:

Question 15.
A hollow cylinder has solid hemisphere inward at one end and on the other end it is closed with a flat circular plate. The height of water is 10 cm when flat circular surface is downward. Find the level of water, when it is inverted upside down, common diameter is 7 cm and height of the cylinder is 20 cm.
Solution:

### Cylinder, Cone and Sphere Surface Area and Volume Exercise 20G – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
What is the least number of solid metallic spheres, each of 6 cm diameter, that should be melted and recast to form a solid metal cone whose height is 45 cm and diameter is 12 cm?
Solution:

Question 2.
A largest sphere is to be carved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere. (Answer correct to the nearest integer)
Solution:

Question 3.
A right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled in identical cones of height 12 cm and diameter 6 cm having a semi-spherical shape on the top. Find the number of cones required.
Solution:

Question 4.
A solid is in the form of a cone standing on a hemisphere with both their radii being equal to 8 cm and the height of cone is equal to its radius. Find in terms of π, the volume of the solid.
Solution:

Question 5.
The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of wire.
Solution:

Question 6.
Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
A cylindrical water tank of diameter 2.8m and height 4.2m is being fed by a pipe of diameter 7 cm through which water flows at the rate of 4m/s. Calculate, in minutes, the time it takes to fill the tank.
Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.
A solid, consisting of a right circular cone, standing on a hemisphere, is placed upright, in a right circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, having given that the radius of the cylinder is 3 cm and its height is 6 cm; the radius of the hemisphere is 2 cm and the height of the cone is 4 cm. Give your answer to the nearest cubic centimeter.
Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.
Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.
Solution:

Question 19.
A certain number of metallic cones, each of radius 2 cm and height 3 cm, are melted and recast into a solid sphere of radius 6 cm. Find the number of cones used.
Solution:
Let the number of cones melted be n.
Let the radius of sphere be rs = 6 cm
Radius of cone be rc = 2 cm
And, height of the cone be h = 3 cm
Volume of sphere = n (Volume of a metallic cone)

Question 20.
A conical tent is to accommodate 77 persons. Each person must have 16m3 of air to breathe. Given the radius of the tent as 7m, find the height of the tent and also its curved surface area.
Solution:

More Resources for Selina Concise Class 10 ICSE Solutions

## Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles)

### Constructions Circles Exercise 19 – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Draw a circle of radius 3 cm. Mark a point P at a distance of 5 cm from the centre of the circle drawn. Draw two tangents PA and PB to the given circle and measure the length of each tangent.
Solution:
Steps of Construction:

1. Draw a circle with centre O and radius 3 cm.
2. From O, take a point P such that OP = 5 cm
3. Draw a bisector of OP which intersects OP at M.
4. With centre M, and radius OM, draw a circle which intersects the given circle at A and B.
5. Join AP and BP.
AP and BP are the required tangents.
On measuring AP = BP = 4 cm

Question 2.
Draw a circle of diameter of 9 cm. Mark a point at a distance of 7.5 cm from the centre of the circle. Draw tangents to the given circle from this exterior point. Measure the length of each tangent.
Solution:

1. Draw a circle of diameter 9 cm, taking O as the centre.
2. Mark a point P outside the circle, such that PO = 7.5 cm.
3. Taking OP as the diameter, draw a circle such that it cuts the earlier circle at A and B.
4. Join PA and PB.
Thus, PA and PB are required tangents. PA = PB = 6 cm

Question 3.
Draw a circle of radius 5 cm. Draw two tangents to this circle so that the angle between the tangents is 45º.
Solution:
Steps of Construction:

1. Draw a circle with centre O and radius BC = 5 cm
2. Draw arcs making an angle of 180º- 45º = 135º at O such that ∠AOB = 135º
3. AT A and B, draw two rays making an angle of 90º at each point which meet each other at point P, outside the circle.
4. AP and BP are the required tangents which make an angle of 45º with each other at P.

Question 4.
Draw a circle of radius 4.5 cm. Draw two tangents to this circle so that the angle between the tangents is 60º.
Solution:
Steps of Construction:

1. Draw a circle with centre O and radius BC = 4.5 cm
2. Draw arcs making an angle of 180º – 60º = 120º at O such that ∠AOB = 120º
3. AT A and B, draw two rays making an angle of 90º at each point which meet each other at point P, outside the circle.
4. AP and BP are the required tangents which make an angle of 60º with each other at P.

Question 5.
Using ruler and compasses only, draw an equilateral triangle of side 4.5 cm and draw its circumscribed circle. Measure the radius of the circle.
Solution:
Steps of construction:

1. Draw a line segment BC = 4.5 cm
2. With centers B and C, draw two arcs of radius 4.5 cm which intersect each other at A.
3. Join AC and AB.
4. Draw perpendicular bisectors of AC and BC intersecting each other at O.
5. With centre O, and radius OA or OB or OC draw a circle which will pass through A, B and C.
This is the required circumcircle of triangle ABC.
On measuring the radius OA = 2.6 cm

Question 6.
Using ruler and compasses only.
(i) Construct triangle ABC, having given BC = 7 cm, AB – AC = 1 cm and ∠ABC = 45°.
(ii) Inscribe a circle in the ∆ABC constructed in (i) above. Measure its radius.
Solution:
Steps of Construction:

i) Construction of triangle:

• Draw a line segment BC = 7 cm
• At B, draw a ray BX making an angle of 45º and cut off BE = AB – AC = 1 cm
• Join EC and draw the perpendicular bisector of EC intersecting BX at A.
• Join AC.
∆ABC is the required triangle.

ii) Construction of incircle:

• Draw angle bisectors of ∠ABC and ∠ACB intersecting each other at O.
• From O, draw perpendiculars OL to BC.
• O as centre and OL as radius draw circle which touches the sides of the ∆ABC. This is the required in-circle of ∆ABC.
On measuring, radius OL = 1.8 cm

Question 7.
Using ruler and compasses only, draw an equilateral triangle of side 5 cm. Draw its inscribed circle. Measure the radius of the circle.
Solution:
Steps of Construction:

1. Draw a line segment BC = 5 cm
2. With centers B and C, draw two arcs of 5 cm radius each which intersect each other at A.
3. Join AB and AC.
4. Draw angle bisectors of ∠B and ∠C intersecting each other at O.
5. From O, draw OL ⊥ BC.
6. Now with centre O and radius OL, draw a circle which will touch the sides of ∆ABC
On measuring, OL = 1.4 cm

Question 8.
Using ruler and compasses only,
(i) Construct a triangle ABC with the following data:
Base AB = 6 cm, BC = 6.2 cm and ∠CAB – 60°
(ii) In the same diagram, draw a circle which passes through the points A, B and C and mark its centre as O.
(iii) Draw a perpendicular from O to AB which meets AB in D.
(iv) Prove that AD = BD
Solution:
Steps of Construction:

1. Draw a line segment AB = 6 cm
2. At A, draw a ray making an angle of 60º with BC.
3. With B as centre and radius = 6.2 cm draw an arc which intersects AX ray at C.
4. Join BC.
∆ABC is the required triangle.
5. Draw the perpendicular bisectors of AB and AC intersecting each other at O.
6. With centre O, and radius as OA or OB or OC, draw a circle which will pass through A, B and C.
7. From O, draw OD ⊥ AB.
Proof: In right ∆OAD and ∆OBD
OA = OB (radii of same circle)
Side OD = OD (common)

Question 9.
Using ruler and compasses only construct a triangle ABC in which BC = 4 cm, ∠ACB = 45° and perpendicular from A on BC is 2.5 cm. Draw a circle circumscribing the triangle ABC.
Solution:
Steps of Construction:

1. Draw a line segment BC = 4 cm.
2. At C, draw a perpendicular line CX and from it, cut off CE = 2.5 cm.
3. From E, draw another perpendicular line EY.
4. From C, draw a ray making an angle of 45º with CB, which intersects EY at A.
5. Join AB.
∆ABC is the required triangle.
6. Draw perpendicular bisectors of sides AB and BC intersecting each other at O.
7. With centre O, and radius OB, draw a circle which will pass through A, B and C.
Measuring the radius OB = OC = OA = 2 cm

Question 10.
Perpendicular bisectors of the sides AB and AC of a triangle ABC meet at O.
(i) What do you call the point O?
(ii) What is the relation between the distances OA, OB and OC?
(iii) Does the perpendicular bisector of BC pass through O?
Solution:
Steps of Construction:

1. O is called the circumcentre of circumcircle of ∆ABC.
2. OA, OB and OC are the radii of the circumcircle.
3. Yes, the perpendicular bisector of BC will pass through O.

Question 11.
The bisectors of angles A and B of a scalene triangle ABC meet at O.
i) What is the point O called?
ii) OR and OQ are drawn perpendiculars to AB and CA respectively. What is the relation between OR and OQ?
iii) What is the relation between angle ACO and angle BCO?
Solution:
Steps of Construction:

1. O is called the incentre of the incircle of ∆ABC.
2. OR and OQ are the radii of the incircle and OR = OQ.
3. OC is the bisector of angle C
∴ ∠ACO = ∠BCO

Question 12.
i) Using ruler and compasses only, construct a triangle ABC in which AB = 8 cm, BC = 6 cm and CA = 5 cm.
ii) Find its incentre and mark it I.
iii) With I as centre, draw a circle which will cut off 2 cm chords from each side of the triangle.
Solution:
Steps of Construction:

1. Draw a line segment BC = 6 cm.
2. With centre B and radius 8 cm draw an arc.
3. With centre C and radius 5 cm draw another arc which intersects the first arc at A.
4. Join AB and AC.
∆ABC is the required triangle.
5. Draw the angle bisectors of ∠B and ∠A intersecting each other at I. Then I is the incentre of the triangle ABC
6. Through I, draw ID ⊥ AB
7. Now from D, cut off $$D P=D Q=\frac{2}{2}=1 \mathrm{cm}$$
8. With centre I, and radius IP or IQ, draw a circle which will intersect each side of triangle ABC cutting chords of 2 cm each.

Question 13.
Construct an equilateral triangle ABC with side 6 cm. Draw a circle circumscribing the triangle ABC.
Solution:
Steps of Construction:

1. Draw a line segment BC = 6 cm
2. With centers B and C, draw two arcs of radius 6 cm which intersect each other at A.
3. Join AC and AB.
4. Draw perpendicular bisectors of AC, AB and BC intersecting each other at O.
5. With centre O, and radius OA or OB or OC draw a circle which will pass through A, B and C.
This is the required circumcircle of triangle ABC.

Question 14.
Construct a circle, inscribing an equilateral triangle with side 5.6 cm.
Solution:
Steps of Construction:

1. Draw a line segment BC = 5.6 cm
2. With centers B and C, draw two arcs of 5.6 cm radius each which intersect each other at A.
3. Join AB and AC.
4. Draw angle bisectors of ∠B and ∠Cintersecting each other at O.
5. From O, draw OL ⊥ BC.
6. Now with centre O and radius OL, draw a circle which will touch the sides of ∆ABC
This is the required circle.

Question 15.
Draw a circle circumscribing a regular hexagon of side 5 cm.
Solution:
Steps of Construction:

1. Draw a regular hexagon ABCDEF with each side equal to 5 cm and each interior angle 120º.
2. Join its diagonals AD, BE and CF intersecting each other at O.
3. With centre as O and radius OA, draw a circle which will pass through the vertices A, B, C, D, E and F.
This is the required circumcircle.

Question 16.
Draw an inscribing circle of a regular hexagon of side 5.8 cm.
Solution:
Steps of Construction:

1. Draw a line segment AB = 5.8 cm
2. At A and B, draw rays making an angle of 120° each and cut off AF = BC = 5.8 cm
3. Again F and C, draw rays making an angle of 120° each and cut off FE = CD = 5.8 cm.
4. Join DE. Then ABCDEF is the regular hexagon.
5. Draw the bisectors of ∠A and ∠B intersecting each other at O.
6. From O, draw OL ⊥ AB
7. With centre O and radius OL, draw a circle which touches the sides of the hexagon.
This is the required in circle of the hexagon.

Question 17.
Construct a regular hexagon of side 4 cm. Construct a circle circumscribing the hexagon.
Solution:
Steps of Construction:

1. Draw a circle of radius 4 cm with centre O
2. Since the interior angle of regular hexagon is 60o, draw radii OA and OB such that ∠AOB = 60°
3. Cut off arcs BC, CD, EF and each equal to arc AB on given circle
4. Join AB, BC, CD, DE, EF, FA to get required regular hexagon ABCDEF in a given circle.
The circle is the required circum circle, circumscribing the hexagon.

Question 18.
Draw a circle of radius 3.5 cm. Mark a point P outside the circle at a distance of 6 cm from the centre. Construct two tangents from P to the given circle. Measure and write down the length of one tangent.
Solution:
Steps of Construction:

1. Draw a line segment OP = 6 cm
2. With centre O and radius 3.5 cm, draw a circle
3. Draw the midpoint of OP
4. With centre M and diameter OP, draw a circle which intersect the circle at T and S
5. Join PT and PS.
PT and PS are the required tangents. On measuring the length of PT = PS = 4.8 cm

Question 19.
Construct a triangle ABC in which base BC = 5.5 cm, AB = 6 cm and m∠ABC =120˚.
i. Construct a circle circumscribing the triangle ABC.
ii. Draw a cyclic quadrilateral ABCD so that D is equidistant from B and C.
Solution:
Steps of Construction:
i.

a. Draw a line BC = 5.4 cm.
b. Draw AB = 6 cm, such that m ∠ABC = 120°.
c. Construct the perpendicular bisectors of AB and BC, such that they intersect at O.
d. Draw a circle with O as the radius.
ii.
(e) Extend the perpendicular bisector of BC, such that
it intersects the circle at D.
(f) Join BD and CD.
(g) Here BD = DC.

Question 20.
Using a ruler and compasses only :
(i) Construct a triangle ABC with the following data: AB = 3.5 cm, BC = 6 cm and ∠ABC = 120°.
(ii) In the same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC.
(iii) Measure ∠BCP.
Solution:
Steps of constructions:

1. Draw a line segment BC = 6 cm.
At B, draw a ray BX making an angle of 120° with BC.
With B as centre and radius 3.5 cm, cut-off AB = 3.5 cm.
Join AC
Thus, ABC is the required triangle.

2. Draw perpendicular bisector MN of BC which cuts BC at point o.
With O as centre and radius = OB, draw a circle.
Draw angle bisector of ∠ABC which meets the cirde at point P.
Thus, point P is equidistant from AB and BC

3. On measuring, ∠BCP = 30°

Question 21.
Construct a ∆ABC with BC = 6.5 cm, AB = 5.5 cm, AC = 5 cm. Construct the incircle of the triangle. Measure and record the radius of the incircle.
Solution:
Steps of construction:

1. Draw BC = 6.5 cm.
2. With B as centre, draw an arc of radius 5.5 cm.
3. With C as oentre, draw an arc of radius 5 cm.
Let this arc meets the previous arc at A.
4. Join AB and AC to get ∆ABC.
5. Draw the bi sectors of ∠ABC and ∠ACB.
Let these bisectors meet each other at O.
6. Draw ON ⊥ BC.
7. With O as centre and radius ON, draw a inarcle that touches all the sides of ∆ABC
8. By measurement, radius ON = 1.5 cm

Question 22.
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence :
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Solution:
Steps of construction:

1. Draw AB = 5.5 cm
2. Construct ∠BAR = 1050
3. With centre A and radius 6 cm, aut off arc on AR at C.
4. Join BC. ABC is the required triangle.
5. Draw angle bisector BD of ∠ABC, which is the loss of points equidistant from BA and BC.
6. Draw perpendicular bisector EF of BC, which is the locus of points equidistant from B and C.
7. BD and EF intersect each other at point P.
Thus, P satisfies the above two lod.
By measurement, PC = 4.8 cm

Question 23.
Construct a regular hexagon of side 5 cm. Hence construct all its lines of symmetry and name them. (2016)
Solution:
Steps of construction:

1. Draw AF measuring 5 cm using a ruler.
2. With A as the centre and radius equal to AF, draw an arc above AF.
3. With F as the centre, and same radius cut the previous arc at Z
4. With Z as the centre, and same radius draw a circle passing through A and F.
5. With A as the centre and same radius, draw an arc to cut the circle above AF at B.
6. With B as the centre and same radius, draw an arc to cut the circle at C.
7. Repeat this process to get remaining vertices of the hexagon at D and E.
8. Join consecutive arcs on the circle to form the hexagon.
9. Draw the perpendicular bisectors of AF, FE and DE.
10. Extend the bisectors of AF, FE and DE to meet CD, BC and AB at X, L and O respectively.
11. Join AD, CF and EB.

These are the 6 lines of symmetry of the regular hexagon.

Question 24.
Draw a line AB = 5 cm. Mark a point C on AB such that AC = 3 cm. Using a ruler and a compass only, construct:
(i) A circle of radius 2.5 cm, passing through A and C.
(ii) Construct two tangents to the circle from the external point B. Measure and record the length of the tangents.   (2016)
Solution:
Steps for construction:

1. Draw AB = 5 cm using a ruler.
2. With A as the centre cut an arc of 3 cm on AB to obtain C.
3. With A as the centre and radius 2.5 cm, draw an arc above AB.
4. With same radius, and C as the centre draw an arc to cut the previous arc and mark the intersection as O.
5. With O as the centre and radius 2.5 cm, draw a circle so that points A and C lie on the circle formed.
6. Join OB.
7. Draw the perpendicular bisector of OB to obtain the mid-point of OB, M.
8. With the M as the centre and radius equal to OM, draw a circle to cut the previous circle at points P and Q.
9. Join PB and QB. PB and QB are the required tangents to the given circle from exterior point B.

QB = PB = 3 cm
That is, length of each tangent is 3 cm.

Solution 25.
Steps of construction :

1. Draw a line AB = 7 cm
2. Taking P as centre and same radius, draw an arc of a circle which intersects AB at M.
3. Taking M as centre and with the same radius as before drawn an arc intersecting previously drawn arc, at point N.
4. Draw the ray AX passing through N, then ∠XAB = 60°
5. Taking A as centre and radius equal to 5 cm, draw an arc cutting AX at C.
6. Join BC
7. The required triangle ABC is obtained.
8. Draw angle bisector of ∠CAB and ∠ABC
9. Mark their intersection as O
10. With O as center, draw a circle with radius OD

Solution 26.
Steps for construction :

1. Draw BC = 6.8 cm.
2. Mark point D where BD = DC = 3.4 cm which is mid-point of BC.
3. Mark a point A which is intersection of arcs AD = 4.4 cm and AB = 5 cm from a point D and B respectively.
4. Join AB, AD and AC. ABC is the required triangle.
5. Draw bisectors of angle B and angle C which are ray BX and CY where I is the incentre of a circle.
6. Draw incircle of a triangle ABC.

Solution 27.
Steps for construction :

1. Draw concentric circles of radius 4 cm and 6 cm with centre of O.
2. Take point P on the outer circle.
3. Join OP.
4. Draw perpendicular bisectors of OP where M is the midpoint of OP.
5. Take a distance of a point O from the point M and mark arcs from M on the inner circle it cuts at point A and B respectively.
6. Join PA and PB.

We observe that PA and PB are tangents from outer circle to inner circle are equal of a length 4.5 cm each.

Solution 28.
Steps for construction :

1. Draw BC = 7.2 cm.
2. Draw an angle ABC = 90°using compass.
3. Draw BD perpendicular to AC using compass.
4. Join BD.
5. Draw perpendicular bisectors of AB and BC which intersect at I, where I is the circumcentre of a circle.
6. Draw circumcircle using circumcentre I. we get radius of a circle is 4.7 cm.

More Resources for Selina Concise Class 10 ICSE Solutions

## Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency (Mean, Median, Quartiles and Mode)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency

### Measures of Central Tendency Exercise 24A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the mean of the following set of numbers:
(i) 6, 9, 11, 12 and 7
(ii) 11, 14, 23, 26, 10, 12, 18 and 6
Solution:

Question 2.
Marks obtained (in mathematics) by 9 student are given below:
60, 67, 52, 76, 50, 51, 74, 45 and 56
(a) find the arithmetic mean
(b) if marks of each student be increased by 4; what will be the new value of arithmetic mean.
Solution:

Question 3.
Find the mean of the natural numbers from 3 to 12.
Solution:

Question 4.
(a) Find the mean of 7, 11, 6, 5, and 6
(b) If each number given in (a) is diminished by 2, find the new value of mean.
Solution:

Question 5.
If the mean of 6, 4, 7, ‘a’ and 10 is 8. Find the value of ‘a’
Solution:
No. of terms = 5
Mean = 8
Sum of numbers = 8 x 5 = 40 .(i)
But, sum of numbers = 6+4+7+a+10 = 27+a ..(ii)
From (i) and (ii)
27+a = 40
a = 13

Question 6.
The mean of the number 6, ‘y’, 7, ‘x’ and 14 is 8. Express ‘y’ in terms of ‘x’.
Solution:
No. of terms = 5 and mean = 8
Sum of numbers = 5 x 8 = 40 ..(i)
but sum of numbers = 6+y+7+x+14 = 27+y+x .(ii)
from (i) and (ii)
27 + y + x = 40
x + y = 13
y = 13 – x

Question 7.

Solution:

Question 8.
If 69.5 is the mean of 72, 70, ‘x’, 62, 50, 71, 90, 64, 58 and 82, find the value of ‘x’.
Solution:
No. of terms = 10
Mean = 69.5
Sum of the numbers = 69.5 x 10 = 695 ……….(i)
But sum of numbers = 72+70+x+62+ 50+71+90+64+58+82
= 619 + x ……(ii)
from (i) and (ii)
619 + x = 695
x = 76

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.
If the mean of the following distribution is 3, find the value of p.

Solution:

Question 13.
In the following table, ∑f = 200 and mean = 73. Find the missing frequencies f1, and f2.

Solution:

Question 14.
Find the arithmetic mean (correct to the nearest whole-number) by using step-deviation method.

Solution:

Question 15.
Find the mean (correct to one place of decimal) by using short-cut method.

Solution:

### Measures of Central Tendency Exercise 24B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
Calculate the mean of the distribution, given below, using the short cut method :

Solution:

Question 11.
Calculate the mean of the following distribution:

Solution:

### Measures of Central Tendency Exercise 24C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A student got the following marks in 9 questions of a question paper.
3, 5, 7, 3, 8, 0, 1, 4 and 6.
Find the median of these marks.
Solution:
Arranging the given data in descending order:
8, 7, 6, 5, 4, 3, 3, 1, 0
The middle term is 4 which is the 5th term.
Median = 4

Question 2.
The weights (in kg) of 10 students of a class are given below:
21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24.
Find the median of their weights.
Solution:

Question 3.
The marks obtained by 19 students of a class are given below:
27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28. Find:
(i) median
(ii) lower quartile
(iii) upper quartile
(iv) interquartile range
Solution:

Question 4.
From the following data, find:
(i) Median
(ii) Upper quartile
(iii) Inter-quartile range
25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83
Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

### Measures of Central Tendency Exercise 24D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the mode of the following data:
(i) 7, 9, 8, 7, 7, 6, 8, 10, 7 and 6
(ii) 9, 11, 8, 11, 16, 9, 11, 5, 3, 11, 17 and 8
Solution:
(i) Mode = 7
Since 7 occurs 4 times
(ii) Mode = 11
Since it occurs 4 times

Question 2.

Solution:
Mode is 122 cm because it occur maximum number of times. i.e. frequency is 18.

Question 3.

Solution:

Question 4.

Solution:

Question 5.
Find the median and mode for the set of numbers:
2, 2, 3, 5, 5, 5, 6, 8 and 9
Solution:

Question 6.
A boy scored following marks in various class tests during a term; each test being marked out of 20.
15, 17, 16, 7, 10, 12, 14, 16, 19, 12 and 16
(i) What are his modal marks?
(ii) What are his median marks?
(iii) What are his total marks?
(iv) What are his mean marks?
Solution:

Question 7.
Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks.
0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7 and 8.
Solution:

Question 8.

Solution:

### Measures of Central Tendency Exercise 24E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.
The mean of 1, 7, 5, 3, 4 and 4 is m. The numbers 3, 2, 4, 2, 3, 3 and p have mean m-1 and median q. Find p and q.
Solution:

Question 5.

Solution:

Question 6.
The marks of 20 students in a test were as follows:
2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19 and 20.
Calculate:
(i) the mean (ii) the median (iii) the mode
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.
The median of the observations 11, 12, 14, (x – 2) (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
Solution:
Data in ascending order:
11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47
Total number of observations = n = 9 (odd)
⇒ Median – $$\left(\frac{n+1}{2}\right)^{t h}$$ term = $$\left(\frac{9+1}{2}\right)^{t h}$$ term =5th term
Given, median = 24
⇒ 5th term = 24
⇒ x + 4 = 24
⇒ x = 20
Thus, the observation are as follows:
11, 12, 14, 18, 24, 29, 32, 38, 47
∴ Mean = $$\frac{\sum x}{n}=\frac{11+12+14+18+24+29+32+38+47}{9}=\frac{225}{9}=25$$

Question 17.
The number 6, 8, 10, 12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x.
Solution:

Question 18.
(Use a graph paper for this question). The daily pocket expenses of 200 students in a school are given below :

Draw a histogram representing the above distribution and estimate the mode from the graph.
Solution:
Histogram is as follows:

In the highest rectangle which represents modal class draw two lines AC and BD intersecting at E.
From E, draw a perpendicular to x-axis meeting at L.
Value of L is the mode. Hence, mode = 21.5

Question 19.
The marks obtained by 100 students in a mathematics test are given below :

Draw an ogive for the given distribution on a graph sheet.
Use a scale of 2 cm = 10 units on both the axes.
Use the ogive to estimate :
(i) Median
(ii) Lower quartile
(iii) Number of students who obtained more than 85% marks in the test.
(iv) Number of students failed, if the pass percentage was 35.
Solution:

The ogive is as follows:

Question 20.
The mean of following numbers is 68. Find the value of ‘x’.
45, 52, 60, x, 69, 70, 26, 81 and 94.
Hence, estimate the median.
Solution:

Question 21.
The marks of 10 students of a class in an examination arranged in ascending order is as follows:
13, 35, 43, 46, x, x + 4, 55, 61, 71, 80
If the median marks is 48, find the value of x. Hence, find the mode of the given data.
Solution:

Question 22.
The daily wages of 80 workers in a project are given below.

Use a graph paper to draw an ogive for the above distribution. (Use a scale of 2 cm = Rs. 50 on x – axis and 2 cm = 10 workers on y – axis). Use your ogive to estimate.
i. the median wages of the workers.
ii. thelower quartile wage of workers.
iii. the number of workers who earn more than Rs. 625 daily.
Solution:

Question 23.
The histogram below represents the scores obtained by 25 students in a Mathematics mental test. Use the data to:
i. Frame a frequency distribution table.
ii. To calculate mean.
iii. To determine the Modal class.

Solution:

More Resources for Selina Concise Class 10 ICSE Solutions

## Selina Concise Mathematics Class 10 ICSE Solutions Similarity (With Applications to Maps and Models)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 15 Similarity (With Applications to Maps and Models)

### Similarity Exercise 15A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In the figure, given below, straight lines AB and CD intersect at P; and AC // BD. Prove that:
(i) ∆APC and ∆BPD are similar.
(ii) If BD = 2.4 cm AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.

Solution:
(i)

(ii)

Question 2.
In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that:
(i) ∆APB is similar to ∆CPD
(ii) PA × PD = PB × PC
Solution:
(i)

(ii)

Question 3.
P is a point on side BC of a parallelogram ABCD. IfDPproduced meets AB produced at point L, prove that:
(i) DP: PL = DC: BL.
(ii) DL: DP=AL: DC.
Solution:
(i)

(ii)

Question 4.
In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO=2DO; show that:
(i) ∆AOB is similar to ∆COD.
(ii) OA × OD – OB × OC.
Solution:
(i)

(ii)

Question 5.
In ∆ABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that:
(i) CB: BA=CP: PA
(ii) AB × BC = BP × CA
Solution:
(i)

(ii)

Question 6.
In ∆ABC; BM ⊥ AC and CN ⊥ AB; show that:
$$\frac{\mathbf{A B}}{A C}=\frac{B M}{C N}=\frac{A M}{A N}$$
Solution:

Question 7.
In the given figure, DE//BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.
(i) Write all possible pairs of similar triangles.
(ii) Find lengths of ME and DM.

Solution:
(i)

(ii)

Question 8.
In the given figure, AD =AE and AD2 = BD × EC
Prove that: triangles ABD and CAE are similar.

Solution:

Question 9.
In the given figure, AB // DC, BO = 6 cm and DQ = 8 cm; find: BP × DO.

Solution:

Question 10.
Angle BAC of triangle ABC is obtuse and AB =AC. P is a point in BC such that PC = 12 cm. PQ and PR are perpendiculars to sides AB and AC respectively. If PQ = 15 cm and PR=9 cm; find the length of PB
Solution:

Question 11.
State, true or false:
(i) Two similar polygons are necessarily congruent.
(ii) Two congruent polygons are necessarily similar.
(iii) All equiangular triangles are similar.
(iv) All isosceles triangles are similar.
(v) Two isosceles-right triangles are similar.
(vi) Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.
(vii) The diagonals of a trapezium, divide each other into proportional segments.
Solution:
(i) False
(ii) True
(iii) True
(iv) False
(v) True
(vi) True
(vii) True

Question 12.
Given = ∠GHE = ∠ DFE = 90°, DH = 8, DF = 12, DG = 3x + 1 and DE = 4x + 2.

Find; the lengths of segments DG and DE.
Solution:

Question 13.
D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that CA2 = CB × CD.
Solution:

Question 14.
In the given figure, ∆ABC and ∆AMP are right angled at B and M respectively. Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) Proe that ∆ABC ~ ∆AMP
(ii) Find AB and BC.
Solution:

Question 15.
Given : RS and PT are altitudes of A PQR prove that:
(i)∆PQT ~ ∆QRS,
(ii) PQ × QS = RQ × QT.
Solution:

Question 16.
Given : ABCD is a rhombus, DPR and CBR are straight lines

Prove that: DP × CR = DC × PR.
Solution:

Question 17.
Given: FB = FD, AE ⊥ FD and FC ⊥ AD. Prove : $$\frac{\mathbf{F B}}{\mathbf{A D}}=\frac{\mathbf{B C}}{\mathbf{E} \mathbf{D}}$$
Solution:

Question 18.
In ∆ PQR, ∠ Q = 90° and QM is perpendicu¬lar to PR, Prove that:
(i) PQ2 = PM × PR
(ii) QR2 = PR × MR
(iff) PQ2 + QR2 = PR2
Solution:

Question 19.
In ∆ ABC, ∠ B = 90° and BD × AC.
(i) If CD = 10 cm and BD = 8 cm; find AD.
(ii) If AC = 18 cm and AD = 6 cm; find BD.
(iii) If AC = 9 cm, AB = 7 cm; find AD.
Solution:

Question 20.
In the figure, PQRS is a parallelogram with PQ = 16 cm and QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N.

Find the lengths of PN and RM.
Solution:

Question 21.
In quadrilateral ABCD, diagonals AC and BD intersect at point E. Such that
AE : EC = BE :’ED.
Show that ABCD is a parallelogram
Solution:

Question 22.
In ∆ ABC, AD is perpendicular to side BC and AD2 = BD × DC.
Show that angle BAC = 90°

Solution:

Question 23.
In the given figure AB // EF // DC; AB ~ 67.5 cm. DC = 40.5 cm and AE = 52.5 cm.

(i) Name the three pairs of similar triangles.
(ii) Find the lengths of EC and EF.
Solution:

Question 24.
In the given figure, QR is parallel to AB and DR is parallel to QB.

Prove that— PQ2 = PD × PA.
Solution:

Question 25.
Through the mid-point M of the side CD o£. a parallelogram ABCD, the line BM is drawn ‘ intersecting diagonal AC in L and AD produced in E.
Prove that : EL = 2 BL.
Solution:

Question 26.
In the figure given below P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC.

(i) Calculate the ratio PQ : AC, giving reason for your answer.
(ii) In triangle ARC, ∠ ARC = 90° and in triangle PQS, ∠ PSQ = 90°. Given QS = 6 cm, calculate the length of AR. [1999]
Solution:

Question 27.
In the right angled triangle QPR, PM is an altitude.

Given that QR = 8 cm and MQ = 3.5 cm. Calculate, the value of PR., [2000]
Given— In right angled ∆ QPR, ∠ P = 90° PM ⊥ QR, QR = 8 cm, MQ = 3.5 cm
Calculate— PR
Solution:

Question 28.
In the figure given below, the medians BD and CE of a triangle ABC meet at G.
Prove that—
(i) ∆ EGD ~ ∆ CGB
(ii) BG = 2 GD from (i) above.
Solution:

### Similarity Exercise 15B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In the following figure, point D divides AB in the ratio 3:5. Find:

Also, if:
(iv) DE = 2.4 cm, find the length of BC.
(v) BC = 4.8 cm, find the length of DE.

Solution:
(i).

(ii)

(iii)

(iv)

(v)

Question 2.
In the given figure, PQ//AB;
CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find:
(i) $$\frac{\mathbf{C P}}{\mathbf{P A}}$$
(ii) PQ
(iii) If AP=x, then the value of AC in terms of x.

Solution:
(i)

(ii)

(iii)

Question 3.
A line PQ is drawn parallel tp the side BC of AABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA=6.0 cm and AQ = 4.2 cm, find the length of AP.
Solution:

Question 4.
In ∆ABC, D and E are the points on sides AB and AC respectively.
Find whether DE // BC, if:
(i) AB=9 cm, AD=4 cm, AE=6 cm and EC = 7.5 cm.
(ii) AB=63 cm, EC=11.0 cm, AD=0.8 cm and AE = 1.6 cm.
Solution
(i).

(ii).

Question 5.
In the given figure, ∆ABC ~ ∆ADE. If AE: EC = 4 :7 and DE = 6.6 cm, find BC. If ‘x’ be the length of the perpendicular fromA to DE, find the length of perpendicular from

A to DE find the length of perpendicular from A to BC in terms of ‘x’.
Solution:

Question 6.
A line segment DE is drawn parallel to base BC of AABC which cuts AB at point D and AC at point E. If AB = 5 BD and EC=3.2 cm, find the length of AE.
Solution:

Question 7.
In the figure, given below, AB, Cd and EFare parallel lines. Given AB = 7.5 cm, DC =y cm, EF=4.5 cm, BC=x cm and CE=3 cm, calculate the values of x and y.

Solution:

Question 8.
In the figure, given below, PQR is a right- angle triangle right angled at Q. XY is parallel to QR, PQ = 6 cm, P Y=4 cm and PX : XQ = 1:2. Calculate the lengths of PR and QR.

Solution:

Question 9.
In the following figure, M is mid-point of BC of a parallelogram ABCD. DM intersects the diagonal AC at P and AB produced at E. Prove that PE = 2PD.

Solution:

Question 10.
The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. IfAE=4 cm, AF = 8 cm and AB = 12 cm, find the perimeter of the parallelogram ABCD.

Solution:

### Similarity Exercise 15C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
(i) The ratio between the corresponding sides of two similar triangles is 2 is to 5. Find the ratio between the areas of these triangles.
(ii) Areas of two similar triangles are 98 sq. cm and 128 sq. cm. Find the ratio between the lengths of their corresponding sides.
Solution:

Question 2.
A line PQ is drawn parallel to the base BC, of ∆ ABC which meets sides AB and AC at points P and Q respectively. If AP = $$\frac{1}{3}$$ PB; find the value of:

Solution:

Question 3.
The perimeters of two similar triangles are 30 cm and 24cm. If one side of first triangle is 12cm, determine the corresponding side of the second triangle.
Solution:

Question 4.
In the given figure AX : XB = 3 : 5

Find :
(i) the length of BC, if length of XY is 18 cm.
(ii) ratio between the areas of trapezium XBCY and triangle ABC.
Solution:

Question 5.
ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP : AB.
Given— In ∆ ABC, PQ || BC in such away that area APQ = area PQCB
To Find— The ratio ol’ BP : AB

Solution:

Question 6.
In the given triangle PQR, LM is parallel to QR and PM : MR = 3 : 4

Solution:

Question 7.
The given diagram shows two isosceles triangles which are similar also. In (he given dia¬gram, PQ and BC are not parallel:
PC = 4, AQ = 3, QB = 12, BC = 15 and AP = PQ.

Calculate—
(i) the length of AP
(ii) the ratio of the areas of triangle APQ and triangle ABC.
Solution:

Question 8.
In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP : PC =1:2. DP produced meets AB produced at Q. Given the area of triangle CPQ = 20 cm2.
Calculate—
(i) area of triangle CDP
(ii) area of parallelogram ABCD [1996]

Solution:

Question 9.
In the given figure. BC is parallel to DE. Area of triangle ABC = 25 cm2.
Area of trapezium BCED = 24 cm2 and DE = 14 cm. Calculate the length of BC.
Also. Find the area of triangle BCD.

Solution:

Question 10.
The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point P. If AP : CP = 3 : 5.

Find:
(i) ∆ APB : ∆ CPB
(ii) ∆ DPC : ∆ APB
(iii) ∆ ADP : ∆ APB
(iv) ∆ APB : ∆ ADB
Solution:

Question 11.
In the given figure, ARC is a triangle. DE is parallel to BC and $$\frac{A D}{D B}=\frac{3}{2}$$.
(i) Determine the ratios $$\frac{A D}{A B}, \frac{D E}{B C}$$.
(ii) Prove that ∆DEF is similar to ∆CBF. Hence, find $$\frac{E F}{F B}$$.
(iii) What is the ratio of the areas of ∆DEF and ∆BFC?

Solution:

Question 12.
In the given figure, ∠B = ∠E, ∠ACD = ∠BCE, AB=10.4 cm and DE=7.8 cm. Find the ratio between areas of the ∆ABC and ∆DEC.

Solution:

Question 13.
Triangle ABC is an isosceles triangle in which AB = AC = 13 cm and BC = 10 cm. AD is perpendicular to BC. If CE = 8 cm and EF ⊥ AB, find:

Solution:

### Similarity Exercise 15D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A’ B’ C’. Calculate:
(i) the length of AB, if A’ B’ = 6 cm.
(ii) the length of C’ A’ if CA = 4 cm.
Solution:
(i)

(ii)

Question 2.
A triangle LMN has been reduced by scale factor 0.8 to the triangle L’ M’ N’. Calculate:
(i) the length of M’ N’, if MN = 8 cm.
(ii) the length of LM, if L’ M’ = 5.4 cm.
Solution:
(i)

(ii)

Question 3.
A triangle ABC is enlarged, about the point O as centre of enlargement, and the scale factor is 3. Find:
(i) A’ B’, if AB = 4 cm.
(ii) BC, if B’ C’ = 15 cm.
(iii) OA, if OA’= 6 cm.
(iv) OC’, if OC = 21 cm.
Also, state the value of:

Solution:
(i)

(ii)

(iii)

(iv)

Question 4.
A model of an aeroplane is made to a scale of 1:400. Calculate:
(i) the length, in cm, of the model; if the length of the aeroplane is 40 m.
(ii) the length, in m, of the aeroplane, if length of its model is 16 cm.
Solution:

(ii)

Question 5.
The dimensions of the model of a multistorey building are 1.2 m × 75 cm × 2 m. If the scale factor is 1:30; find the actual dimensions of the building.
Solution:

Question 6.
On a map drawn to a scale of 1: 2,50,000; a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and angle ABC = 90°.
Calculate:
(i) the actual lengths of AB and BC in km.
(ii) the area of the plot in sq. km.
Solution:

(ii)

Question 7.
A model of a ship of made to a scale 1 : 300
(i) The length of the model of ship is 2 m. Calculate the lengths of the ship.
(ii) The area of the deck ship is 180,000 m2. Calculate the area of the deck of the model.
(iii) The volume of the model in 6.5 m3. Calculate the volume of the ship. (2016)
Solution:

Question 7(old).

Solution:

Question 8.
An aeroplane is 30 in long and its model is 15 cm long. If the total outer surface area of the model is 150 cm2, find the cost of painting the outer surface of the aeroplane at the rate of ₹ 120 per sq. m. Given that 50 sq. m of the surface of the aeroplane is left for windows.
Solution:

### Similarity Exercise 15E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5cm and BC = 18 cm.

Find:

Solution:
(i)

(ii)

Question 2.
In the following figure, ABCD to a trapezium with AB//DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP=6 cm and BE = 15 cm.
Calculate:
(i) EC
(ii) AF
(iii) PE

Solution:
(i)

(ii)

(iii)

Question 3.
In the following figure, AB, CD and EF are perpendicular to the straight line BDF.

If AB = x and CD = z unit and EF = y unit, prove that : $$\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$$.
Solution:

Question 4.
Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that:
$$\frac{A B}{P Q}=\frac{A D}{P M}$$.
Solution:

Question 5.
Triangle ABC is similar to triangle PQR. If AD and PM are altitudes of the two triangles,
prove that: $$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}$$
Solution:

Question 6.
Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that: $$\frac{A B}{P Q}=\frac{A D}{P M}$$.
Solution:

Question 7.
In the following figure, ∠AXY = ∠AYX. If $$\frac{\mathbf{B X}}{\mathbf{A X}}=\frac{\mathbf{C} \mathbf{Y}}{\mathbf{A} \mathbf{Y}}$$, show that triangle ABC is isosceles.

Solution:

Question 8.
In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown.

Prove that: $$\frac{A B}{B C}=\frac{P Q}{Q R}$$
Solution:

Question 9.
In the following figure, DE //AC and DC //AP. Prove that: $$\frac{B E}{E C}=\frac{B C}{C P}$$

Solution:

Question 10.
In the figure given below, AB//EF// CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm.
Calculate:
(i) EF
(ii) AC

Solution:
(i)

(ii)

Question 11.
In ∆ABC, ∠ABC = ∠DAC. AB = 8 cm, AC = 4 cm, AD = 5 cm.
(i) Prove that ∆ACD is similar to ∆BCA.
(ii) Find BC and CD.
(iii) Find area of ∆ACD: area of ∆ABC. (2014)
Solution:
(i)

(ii)

(iii)

Question 12.
In the given triangle P, Q and R are the midpoints of sides AB, BC and AC respectively. Prove that triangle PQR is similar to triangle ABC.

Solution:

Question 13.
In the following figure, AD and CE are medians of ∆ ABC. DF is drawn parallel to CE. Prove that:
(i) EF = FB;
(ii) AG : GD = 2 : 1

Solution:

Question 14.
The two similar triangles are equal in area. Prove that the triangles are congruent.
Solution:

Question 15.
The ratio between the altitudes of two similar triangles is 3 : 5; write the ratio between their:
(i) medians
(ii) perimeters
(iii) areas
Solution:

Question 16.
The ratio between the areas of two similar triangles is 16 : 25. Find the ratio between their:
(i) perimeters
(ii) altitudes
(iii) medians.
Solution:

Question 17.
The following figure shows a triangle PQR in which XY is parallel to QR. If PX: XQ = 1:3 and QR = 9 cm, find the length of XY.
Further, if the area of ∆ PXY = x cm2; find in terms of x, the area of :
(i) triangle PQR.
(ii) trapezium XQRY.

Solution:

Question 18.
On a map, drawn to a scale of 1 : 20000, a rectangular plot of land ABCD has AB = 24 cm, and BC = 32 cm. Calculate :
(i) The diagonal distance of the plot in kilometre
(ii) The area of the plot in sq. km.
Solution:

Question 19.
The dimensions of the model of a multistoreyed building are lm by 60 cm by 1.20 m. If the scale factor is 1 : 50,. find the actual
dimensions of the building. Also, find :
(i) the floor area of a room of the building, if the floor area of the corresponding room in the model is 50 sq cm.
(ii) the space (volume) inside a room of the model, if the space inside the corresponding room of the building is 90m3.
Solution:

Question 20.
In ∆ABC, ∠ACB = 90° and CD ⊥ AB. Prove that : $$\frac{\mathrm{BC}^{2}}{\mathrm{AC}^{2}}=\frac{\mathrm{BD}}{\mathrm{AD}}$$.
Solution:

Question 21.
A triangle ABC with AB = 3 cm, BC = 6 cm and AC = 4 cm is enlarged to ∆DEF such that the longest side of ∆DEF = 9 cm. Find the scale factor and hence, the lengths of the other sides of ∆DEF.
Solution:

Question 22.
Two isosceles triangles have equal vertical angles. Show that the triangles are similar.
If the ratio between the areas of these two triangles is 16 : 25, find the ratio between their corresponding altitudes.
Solution:

Question 23.
In ∆ABC, AP: PB = 2 :3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA.

Find:
(i) area ∆APO: area ∆ABC.
(ii) area ∆APO: area ∆CQO.
Solution:

Question 24.
The following figure shows a triangle ABC in which AD and BE are perpendiculars to BC and AC respectively.

Show that:
(ii) CA × CE = CB × CD
(iii) ∆ ABC – ∆DEC
(iv) CD × AB = CA × DE
Solution:

Question 25.
In the given figure, ABC is a triangle-with ∠EDB = ∠ACB.
Prove that ∆ABC ~ ∆EBD.
If BE=6 cm, EC = 4 cm,
BD = 5 cm and area of
∆BED = 9 cm2. Calculate the
(i) length of AB
(ii) area of ∆ABC

Solution:

Question 26.
In the given figure, ABC is a right-angled triangle with ZBAC = 90°.
(ii) If BD = 18 cm, CD = 8 cm, find AD.
(iii) Find the ratio of the area of ∆ADB is to area of ∆CDA.

Solution:

Question 27.
In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ∆ABC ~ ∆DEC
(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm. Calculate CD.
(iii) Find the ratio of the area of ∆ABC : area of ∆DEC.

Solution:
(i)

(ii)

(iii)

Question 28.
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:

(ii) If AC = 13 cm, BC = 5 cm and AE=4 cm. Find DEandAD.
Solution:
(i)

(ii)

(iii)

Question 29.
Given: AB // DE and BC // EF. Prove that:
(i) $$\frac{\mathrm{AD}}{\mathrm{DG}}=\frac{\mathrm{CF}}{\mathrm{FG}}$$
(ii) ∆DFG ~ ∆ACG.

Solution:

Question 30.

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