## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 25 Probability Ex 25C.

Other Exercises

Question 1.
A bag contains 3 red balls, 4 blue balls and one 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 die ball is :
(i) yellow
(ii) red
(iii) not yellow
(iv) neither,yellow nor red
Solution:
In a bag. there are 8 balls in which there are Red balls = 3
blue balls = 4
and yellow ball = 1
Total possible out come = 8
(i) Yellow balls = 1
∴ Number of favourable outcome = 1

(ii) Red balls = 3

(iii) Not yellow balls = 3+4 = 7
∴ Number of favourable outcome = 8

(iv) Neither yellow nor red can be blue ball.
∴ Number of favourable outcome = 4

Question 2.
A dice is thrown once. What is the probability of getting a number :
(i) greater than 2 ?
(iii) less than or equal to 2 ?
Solution:
A die is thrown once
Numbers marked on the faces of a die are 1, 2,3,4, 5, 6
∴ Number of possible outcome = 6
(i) Greater than 2 number = 3, 4, 5, 6 = 4 numbers

(ii) Less than or equal to 2 are 2, 1 which are 2 in numbers.

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 colour.
(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:
A deck of playing cards has 52 cards
∴ Number of possible outcome = 52
(i) A face card ; face cards in the deck are = 3 x 4 = 12
∴ Number of favourable outcome = 12

(ii) Not a face card which are 52 – 12 = 40
∴ Number of favourable outcome = 40

(iii) A queen of black color which are 2 in numbers in the deck
∴ Number of favourable outcome = 2

(iv) A card with number 5 or 6 are 2 x 4 numbers
∴ Number of favourable outcome = 8

(v) A card with number less than 8 which can be 2, 3, 4, 5, 6, 7 = 6

(vi) A card with number between 2 and 9 can be 3, 4, 5, 6, 7, 8 = 6
∴ Number of favourable outcome = 6

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:
A match is played between two persons A and B
∴ Number of possible outcome = 1
(i) The probability of winning of A is 0.83
∴ Probability of winning of B = 1 – 0.83 = 0.17 [∵ P(E) + P($$\bar { E }$$ )=1]
(ii) The probability of losing the match is 0.49 by B.
∴ Probability of losing of B or winning of A = 0.49

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 birthdays ? (Ignore a leap year)
Solution:
Number of days in a year = 365
and birthday of a person can be on one day only.
(i) Different birthdays can be 365 – 1 = 364

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 :
(i) at least one head ?
Solution:
There are two coins : one of two rupees and
other is of 5-rupees
∴ Number of Heads =1 + 1=2
and number of tails are 2 i.e., 2 and 5
∴ Number of possible outcome = 2 x 2 = 4
Number of favourable outcome = 3

(ii) At the most one head = 3

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:
Number of balls in a box are 20 in which 7 are red, 8 are green and 5 are white
∴ Number of possible outcome = 20
(i) White = 5
∴ Number of favourable outcome = 5

(ii) Neither red not white
i.e., all the green which are 8

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:
A deck of playing cards = 52
Cards which are removed = 3 (3 face cards of spades)
Balance cards in the deck = 52 – 3 = 49
(i) A black face card which are 6 – 3 = 3 in number
∴ Number of favourable outcome = 3

(ii) A queen : In the deck there are 4 – 1 = 3 queen

(iii) A black cards : which are 26 – 3 = 23 cards in number

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:
Total time for the musical race = 0 to 40 seconds = 40 seconds.
Time taken by a player =15 seconds. (0 to 15 seconds)

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:
Total number of shirts in a bundle = 50
No. of good shirts = 44
Minor defected = 4
Major defected = 2
∴ Number of possible outcome = 50
(i) Acceptance of only for a good shirt = 44

(ii) Rejecting of totally defected shirts, number of remaining shirts = 44 + 4 = 48

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:
Two dice are thrown at the same time and each dice has 6 numbers 1, 2, 3, 4, 5, 6 on its faces
∴Number of possible outcome = 6 x 6 = 36
(i) Sum of two numbers on the top is 8 i.e..(2, 6), (3, 5), (4,4), (5, 3), (6, 2)
∴ Number of favourable outcome = 5

(ii) Sum of two number on the top is 13.
At the most, then sum can be (6, 6) = 12
∴ number of favourable outcome = 0
∴ P(E) = 0
(iii) Sum is less than or equal to 12 i.e.(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), (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) = 36
∴ Number of favourable outcome = 36

Question 12.
Which of the following cannot be the probability of an event ?
(i) $$\frac { 3 }{ 7 }$$
(ii) 0.82
(iii) 37%
(iv) -2.4
Solution:
(i) ∵ $$\frac { 3 }{ 7 }$$ is between 0 and 1
∴ It is a probability event
(ii) ∵ 0.82 is between 0 and 1
∴ It is a probability event.
(iii) 37% = $$\frac { 37 }{ 100 }$$
∵It is between 0 and 1
∴ It is a probability event.
(iv) -2.4
∵It is less than 0.
∴ It is not a probability event.

Question 13.
If P(E) = 0.59; find P(not E).
Solution:
P(E) = 0.59
But P(E) + P($$\bar { E }$$ )= 1
or 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:
In a bag, there are certain number red balls. Let it be x balls.
One ball is drawn out
(i) A black
∵ There is no black ball in the bag
∴ Probability of black ball = 0
(ii) A red ball

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:
Probability of two boys do not have the same birthday [P (E)] = 0.897
Let Probability of those boys having the same birthday = P(not E)
= P($$\bar { E }$$ )
But P(E) + P($$\bar { E }$$ )= 1
⇒ 0.897 +P($$\bar { E }$$ ) = 1
⇒P($$\bar { E }$$ )= 1 -0.897 = 0.103
Hence probability having the 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 tharthe ball drawn will be :
(i) not red ?
(ii) neither red nor green ?
(iii) white or green ?
Solution:
In a bag, there are 10 red balls, 16 white and 8 green balls
∴ Total balls =10 + 16 + 8= 34 ball
∴ Number of possible outcome = 34
(i) Not red ball
Number of favourable outcome = 16 + 8 = 24

(ii) Neither red nor green
∴ Number of outcome = 34 – (10 + 8)
= 34-18=16

(iii) White or green
∴ Number of outcome = 16 + 8 = 24

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:
In a bag, there are
5-rupee coins = 20
2-rupee coins = 50
1-rupee coin =30
Total number of coins in the bag = 20 + 50 + 30 = 100
∴ Number of possible outcome =100
(i) one-rupee coin = 30

(ii) When there is no 2-rupee coins
∴ Number of coins = 20 + 30 = 50

(iii) Neither be Rs. 5 coins nor be Re-one coin = 100 – (20 + 30) = 100 – 50 = 50

Question 18.
A game consists of spinning an 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 eve., 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:
There are 12 numbers on the spinning game.
∴ Number of possible outcome = 12
(i) 6 which is one

(ii) an even number which are 6 i.e. 2, 4, 6, 8, 10, 12

(iii) A prime number which are 2, 3, 5, 7, 11 and 5 in number

(iv) A number greater than 8 are 9, 10, 11, 12
which are 4 in number.

(v) A number less than or equal to 9 are
1, 2, 3, 4, 5, 6, 7, 8, 9 which are 9 in number

(vi) A number between 3 and 11 are 4, 5, 6, 7, 8, 9, 10 which are 7 in number

Question 19.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting :
(i) a queen of red colour.
(ii) a black face card.
(iii) the jack or the queen of hearts.
(iv) a diamond.
(v) a diamond or a spade.
Solution:
Number of cards in a deck of playing card = 52
∴ Number of possible outcome = 52
(i) A queen of red colour
Number of favourable outcome = 2
(As there are 2 red queens in the deck)

(ii) A black face card.
There are 3 + 3 = 6 black face cards in the deck.

(iii) The jack or the queen of hearts which are two in numbers

(iv) A diamond.
There are 13 cards of diamond in the deck.

(v) A diamond or a spade.
There are 13 cards of diamond and 13 cards of spade in the deck
∴ Number of favourable outcome = 13 + 13 = 26

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:
Number of cards is a deck of playing card = 52
All face cards are removed. Which are 3 x 4 = 12
(i) A black card which are 10 in numbers

(ii) 8 of red colors

(iii) A king of black color.
∵ In the deck of playing cards, face cards are removed
∴ There is no face cards.
∴ P(E) = 0.

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:
There are 7 cards which are the eight, the nine, the ten, the jack, the’ queen, the king and the ace of diamond.
∴ Number of possible outcome = 7
(i) card having the eight or a king which are 2.

(ii) If king is drawn, then number of remaining playing cards = 7-1=6
(a) An ace.

(b) A king
There is no card of king
∴ P(E) = 0

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:
Number of bulbs in a box = 150
No. of defective bulbs = 15
∴ No of good bulbs = 150 – 15 = 135
∴ Number of possible outcome = 150
(i) A good bulb and number of good bulbs =135

(ii) A defective bulb. Number of defective bulb = 15

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:
Number of defective pens = 4
(i) and number of good pens = 16
∴ Total pens = 4 + 16 = 20
∴ Number of possible outcome = 20
One defective pen, no. of defective pens = 4

(ii) One defective pen is drawn
∴ Remaining pens = 20 – 1 = 19
(a) Defective one
Number of remaining defective pens =4-1=3

(b) Not defective, number of good pens =16

Question 24.
A bag contains 100 identical marble stones which are numbered from 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:
Total number of stones = 100
On which numbers 1 to 100 are marked
∴ Number of possible outcome = 100
(i) A perfect square = which are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 which are 10 inmumbers

(ii) A number divisible by 4 are 4, 8, 12, 16, …, 96, 100 which one 25 in numbers

(iii) A number divisible by 5, are 5, 10, 15, 20, ….., 95, 100 which are 20 in numbers,

(iv) A number divisible by 4 or 5 are 4, 5, 8, 10,12, 15, 16, 20, 24, 25, 28, 30, 32, 35, 36, 40, 96,95, 100 which are 40 in numbers

(v) A number divisible by 4 and 5 are 20, 40, 60, 80 and 100 which are 5 in numbers

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 the paper only, find the probability that it will fall and land :
(i) inside the circle.
(ii) outside the circle.
Solution:
Diameter of the circle = 20 cm
Length of rectangular paper = 40 cm
and width = 30 cm
Area of rectangle = 40 x 30 = 1200 cm2
∴ Number of possible out come = 1200

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:
Two dice having numbers 1 to 6 are rolled together.
∴ Number of possible outcome = 6 x 6 = 36
(i) If sum is 4 or 5 of numbers on the upper most face,
Their number of favourable outcome = (1,3), (2, 2), (3, 1), (1, 4), (2, 3), (3, 2), (4, 1) which are 7 in numbers

(ii) If sum of number on the upper faces by 7, 8 or 9, then these can be (1, 6), (2, 6), (3, 6), (4, 3), (5. 2). (6, 1). (4, 3), (2, 5), (3, 4), (4, 4), (5, 3), (5, 4), (6, 2), (6, 3), (4, 5) which are total 15 in numbers.

(iii) Sum is between 5 and 8.
i.e. sum is 6 or 7.
These can be.
(1, 5), (2, 5), (1, 6), (5, 1), (5, 2), (6, 1), (3, 3), (3, 4), (4, 3), (2, 4), (4, 2) which are 11 in number.
∴ Number of favourable outcome = 11

(iv) If sum is more than 10, then these can be (5. 6), (6, 5), (6, 6) which are 3 in number

(v) If sum is less than 6, then there can be (1, 2), (2, 1), (1, 3), (3, 1), (4, 1), (1, 4), (2, 3), (3, 2), (2, 4), (4, 2)

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:
3 coins one tossed together.
∴ Number of possible outcome = 23 = 2 x 2 x 2 = 8
i.e. HHH, HHT, HTH, HTT, TTT, THH, THT, TTH
(i) exactly two heads these can be HHT, THH, HTH = 3 in numbers

(ii) At least two heads : These can be HHH,HHT, HTH, THH = 4 in numbers

(iii) Atmost two heads : These can be THH, HHT, HTH, HTT, THT, TTH, TTT which is 7 in numbers.

(iv) All tails : There can be TTT i. e., only one

(v) At least one tail : There can be HHT, HTH, HTT, TTT, THH, THT, TTH = 7 in number

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:
Two dice are thrown simultaneous,
and each dice has 1-6 numbers on its faces
∴ Number of possible outcome = 6 x 6 = 36
(i) 4 will not come up either time
∴ Number of favourable outcomes = (6 – 1)² = (5)² = 25

(ii) 4 will come up at least once
∴ These can be (1, 4), (2, 4), (3, 4), (4, 4), (4, 5),(4, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)
∴ Number of favourable outcome =11

Question 29.
Cards marked with numbers 1, 2, 3, 4, …,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:
∵ Office are open for 5 days a week
∴ Number of possible outcomes for 2 employees = 5 x 5 = 25
Let the five days of working be denoted by M, T, W, Th, F for Mondays, Tuesday, Wednesday Thursday and Friday respectively
(i) On Same day.
Favourable outcome will be M;T;T;M;T;W;W;T;W;H; TH.W ; TH ; F and F, TH which are 8 in all

(iii) Absent on different days : Then = 1- P(absent on the same day)
= 1 – $$\frac { 1 }{ 5 }$$= $$\frac { 4 }{ 5 }$$ [from (i)]
∵ P(E) + P(not E) = 1

Question 31.
A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of white ball; find the number of black balls in the box.
Solution:
In a box, some balls are black while 30 balls are white
Let number of black balls = x
Then number of possible outcome = x + 30
Probability of drawing a black ball = $$\frac { 2 }{ 5 }$$ of a white balls.
Now, In case of black ball,

Either x + 30 = 0, then x = -30 which is not possible
or x – 12 = 0, then x = 12
Hence number of black balls = 12

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 ? (2011)
Solution:
No. of total cards = 52 cards removed of 4 colors = 3, 6, 9, 12 = 4 x 4 = 16
Remain using cards = 52 – 16 = 36
(i) No. of faces cards = 2 x 4 = 8 cards (excluding queen)
∴ Probebility P(E) = $$\frac { 8 }{ 36 }$$ = $$\frac { 2 }{ 9 }$$
(ii) An even number red cards = 2, 4, 8, 10 = 4×2 = 8 cards
∴ Probebility P(E) = $$\frac { 8 }{ 36 }$$ = $$\frac { 2 }{ 9 }$$

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:
Total outcomes n(S)= 6
(i) a positive integer = (1, 2, 3)
No. of favourables n(E) = 3

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. (2015)
Solution:
A bag contains 5 white balls, 6 red balls and 9 green balls.
∴ Total number of balls = 5 + 6 + 9 = 20 balls
One ball is drawn at random.
(i) Probability of a green ball = $$\frac { 9 }{ 20 }$$
(ii) Probability of a white or a red ball = $$\frac { 5 + 6 }{ 20 }$$ = $$\frac { 11 }{ 20 }$$
(iii) Probability of neither a green ball nor a white ball = $$\frac { 6 }{ 20 }$$ = $$\frac { 3 }{ 10 }$$ (Only red balls )

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:
(i) The perfect squares lying between 11 and 40 and 16, 25 and 36.
So the number of possible outcomes is 3.
Total number of cards from 11 to 40 is 40 – 11 + 1 = 30
Probability that the number on the card drawn is a perfect square

So, the probability that the number on the
card drawn is a perfect square is $$\frac { 1 }{ 10 }$$ .
(ii) The numbers lfom 11 to 40 that are divisible by 7 are 14, 21, 28 and 35.
So the number of possible outcomes is 4. Total number of cards from 11 to 40 is 30.
Probability that the number on the card drawn is divisible by 7

So, the probability that the number on the card drawn is divisible by 7 is $$\frac { 2 }{ 15 }$$.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5C.

Other Exercises

Question 1.
Solve each of the following equations, using the formula:
Solution:

Question 2.
Solve each of the following equations for x and give, in each case, your answer correct to one decimal place :
(i) x² – 8x + 5 = 0
(ii) 5x² + 10x – 3 = 0
Solution:

Question 3.
Solve each of the following equations for x and give, in each case, your answer correct to 2 decimal places :
(i) 2x² – 10x + 5 = 0
(ii) 4x² + $$\frac { 6 }{ x }$$ + 13 = 0
(iii) x² – 3x – 9 = 0 [2007]
(iv) x² – 5x – 10 = 0 [2013]
Solution:

Question 4.
Solve each of the following equations for x, giving your answer correct to 3 decimal places:
(i) 3x² – 12x – 1 = 0
(ii) x² – 16x + 6 = 0
(iii) 2x² + 11x + 4 = 0
Solution:

Question 5.
Solve:
(i) x4 – 2x² – 3 = 0
(ii) x4 – 10x² + 9 = 0
Solution:

Question 6.
Solve:
(i) (x² – x)² + 5 (x² – x) + 4 = 0
(ii) (x² – 3x)² – 16 (x² – 3x) – 36 = 0
Solution:

Question 7.
Solve :

Solution:

Question 8.
Solve the equation : 2x – $$\frac { 1 }{ x }$$ = 7. Write your answer correct to two decimal places.
Solution:

Question 9.
Solve the following equation and give your answer correct to 3 significant figures : 5x² – 3x – 4 = 0
Solution:

Question 10.
Solve for x using the quadratic formula. Write your answer correct to two significant figures, (x – 1)² – 3x + 4 = 0
Solution:

Question 11.
Solve the quadratic equation x² – 3 (x + 3) = 0; Give your answer correct to two significant figures.
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 16 Loci (Locus and Its Constructions) Ex 16B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 16 Loci (Locus and Its Constructions) Ex 16B

Other Exercises

Question 1.
Describe the locus for questions 1 to 13 given below: 1. The locus of a point at a distance 3 cm from a fixed point.

Solution:
The locus of a point which is 3 cm away from a fixed point is circumference of a circle whose radius is 3 cm and the fixed point is called the centre of the circle.

Question 2.
The locus of points at a distance 2 cm from a fixed line.
Solution:
A pair of straight lines 1 and m which are parallel to the given line at a distance of 2 cm, from it is the locus.

Question 3.
The locus of the centre of a wheel of a bicycle going straight along a level road.
Solution:

The locus of the centre of a wheel which is going straight along a level road will be a straight line parallel to the road at a distance equal to the radius of the wheel.

Question 4.
The locus of the moving end of the minute hand of a clock.
Solution:

The locus of the moving end of the minute hand of the clock will be a circle where radius will be of the length of the minute hand.

Question 5.
The locus of a stone dropped from the top of a tower.
Solution:

The locus of stone which is dropped from the top of the tower will be a vertical line through the point from which the stone is dropped.

Question 6.
The locus of a runner running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Solution:

The locus of the runner running round a circular track at a distance of 1.5 m from the inner edge will be the circum¬ference of a circle whose radius is equal to the radius of the inner circular track plus 1.5 m

Question 7.
The locus of the door-handle as the door opens.
Solution:

The locus of the door handle will be the circumfer-ence of a circle with centre at the axis of rotation of the door and radius equal to the distance between the door handle and the axis of rotation of the door.

Question 8.
The locus of points inside a circle and equidistant from two fixed points on the circumference of the circle.
Solution:

The locus of the points inside the circle which are equidistant from the fixed points on the circumference of the circle will be the diameter which is perpendicular bisector of the line joining the two fixed points on the circle.

Question 9.
The locus of the centres of all circles passing through two fixed points.
Solution:

The locus of the centre of all the circles which pass through two fixed points will be the perpendicular bisector of the line segment joining the two fixed points which are given.

Question 10.
The locus of vertices of all isosceles triangles having a common base.
Solution:

The locus of vertices of all isosceles triangles have a common base will be the perpendicular bisector of the common base of the triangles.

Question 11.
The locus of a point in space, which is always at a distance of 4 cm from a fixed point.
Solution:
The locus of a point in space is the surface of the sphere whose centre is the fixed point and radius equal to 4 cm.

Question 12.
The locus of a point P, so that:
AB2 = AP2+ BP2, where A and B are two fixed points.
Solution:

The locus of the point P is the circumference of a circle with AB as diameter and satisfies the condition AB2 = AP2+ BP2.

Question 13.
The locus of a point in rhombus ABCD, so that it is equidistant from
(i) AB and BC ; (ii) B and D.
Solution:
Locus of the point in a rhombus ABCD which is equidistant from.

Question 14.
The speed of sound is 332 metres per second. A gun is fired. Describe the locus of all the people on the earth’s surface, who hear the sound exactly after one second.
Solution:
The locus of all the people on earth’s surface is the circumference of a circle whose radius is 332 m and centre is the point where the gun is fired.

Question 15.
Describe:
(i) The locus of points at distances less than 3 cm from a given point.
(ii) The locus of points at distances greater than 4 cm from a given point.
(iii) The locus of points at distances less than or equal to 2.5 cm from a given point.
(iv) The locus of points at distances greater than or equal to 35 mm from a given point.
(v) The locus of the centre of a given circle which rolls around the outside of a second circle and is always touching it.
(vi) The locus of the centres of all circles that are tangent to both the arms of a given angle.
(vii) The locus of the mid-points of all chords par-allel to a given chord of a circle.
(viii) The locus of points within a circle that are equidistant from the end points of a given chord.
Solution:
(i) The space inside of the circle whose radius is 3 cm and centre is the fixed point which is given.
(ii) The space outside of the circle whose radius is 4 cm and centre is the fixed point which is given.
(iii) The space inside and circumference of the circle with a raduis of 2.5 cm and centre is the given fixed point.
(iv) The space outside and the circumference of a circle with a radius of 35 mm and centre is the given fixed point.
(v) Circumference of the circle concentric with

the second circle whose radius is equal to the sum of the radii of the two given circles.
(vi) The locus of the centre of all circle whose tan-gents are the arms of a given angle is the bisector of that angle.

(vii) The locus of the mid-points of the chords which are parallel to a given chords is the diameter perpendicular to the given circle.

The locus of points within a circle that are equidistant from the end points of a given chord is a diameter which is perpendicular bisector of the given chord.

Question 16.
Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
Solution:
Steps of Construction:

Draw a line XY parallel to the base BC from the vertex A.
This line is the locus of vertex A. All the tri-angles which have the base BC and altitude (length) equal to AD.

Question 17.
In the given figure, obtain all the points equidistant from lines m and n ; and 2.5 cm from O.

Solution:

Draw angle bisector PQ and XY of angles formed by the lines m and n. From O, draw arcs with radius 2.5 cm, which intersects the angle bisectors at A, B, C and D respectively.
Hence A, B, C and D are the required points.
P.Q.
By actual drawing obtain the points equidistant from lines m and n ; and 6 cm from a point P, where P is 2 cm above m, m is parallel to – n and m is 6cm above n
Solution:
Steps of construction:

(i) Draw a line n.
(ii) Take a point P on n and draw a perpendicular to n. ,
(iii) Cut off LM = 6 cm and draw a line q, the per pendicu lar bisector of LM
(iv) At M, draw a line n making an angle of 900 at.
(v) Produce LM and take a point P such that PM =2 cm.
(vi) From P, draw are an with 6 cm radius which intersects the line q, (he perpendicular bisector cf LM, atA and B.
A and B are the required points which are equidisant from rn and n and is at a distance of 6 cm from P

Question 18.
A straight line AB is 8 cm long. Draw and describe the locus of a point which is :
(i) always 4 cm from the line AB.
(ii) equidistant from A and B.
Mark the two points X and Y, which are 4 cm from AB and equidistant from A and B. Describe the figure AXBY.
Solution:
Steps of Construction:

(i) Draw a line segment AB equal to 8 cm.
(ii) Draw two parallel lines  ℓ and m to AB at a distance of 4cm

(iii) Draw the perpendicular bisector of AB which intersects the parallel lines  ℓ and m at X and Y respectively then X and Y are the required points.
(iv) Join AX.AY, BX and BY.
The figure so formed is a square as its diago¬nals are equal and intersect at 90°.

Question 19.
Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respec¬tively. Draw and describe the locus of a point which is :
(i) equidistant from BA and BC.
(ii) 4 cm from M.
(iii) 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
Solution:
(i) Draw an angle of 60° with AB = BC = 8 cm.
(ii) Draw the angle bisector BX of ∠ABC.

(iii) With centre M and N, draw circles of radius equal to 4 cm, which intersect each other at P. P is the required point.
(iv) Join MP, NP.
BMPN is a rhombus.
∵ MP = PM = BN = PN = 4 cm.

Question 20.
Draw a triangle ABC in which AB = 6 cm, BC = 4.5 cm and AC = 5 cm. Draw and label :
(i) the locus of the centres of all circles which touch AB and AC ;
(ii) the locus of the centres of all the circles of radius 2 cm which touch AB.
Hence, construct the circle of radius 2 cm which touches AB and AC.
Solution:

(i) Draw a line segment BC = 4.5 cm
(ii) With centre B and radius 6 cm and with centre C and radius 5 cm, draw arcs which intersects each other at point A
(iii) Join AB and AC.
ABC is a required triangle.
(iv) Draw the angle bisector of ∠BAC.
(v) Draw a lines parallel to AC and AB at a distance of 2 cm. which intersects each other AD at O.
(vi) With centre O and radius 2 cm, draw a circle which touches AB and AC.

Question 21.
Construct a triangle ABC, having given AB = 4.8 cm. AC = 4 cm and ∠A = 75°. Find a point P.
(i) inside the triangle ABC.
(ii) outside the triangle ABC.
equidistant from B and C; and at a distance of 1.2 cm from BC.
Solution:

(i) Draw a line segment AB = 4.8 cm.
(ii) At, A draw a ray AX making an angle of 75°.
(iii) Cut off AC = 4 cm from AX.
(iv) Join BC.
(v) Draw two lines  ℓ and m parallel to BC at a distance of 1.2 cm.
(vi) Draw the perpendicular bisector of BC which intersect  ℓ and m at P and P .
P and P1 are the required points which are in¬side and outside the given triangle ABC.

P.Q.
O is a fixed point. Point P moves along a fixed line AB. Q is a point on OP produced such that OP = PQ. Prove that the locus of point Q is a line parallel to AB.
Solution:
O is fixed point. P moves along AB; a fixed line. OP is joined and produced it to Q such that OP = PQ, Now we have to prove that locus of P is a line parallel to AB.

Proof:
∵ P moves along AB, and Q moves in such a way that PQ is always equal to OP.
P is the mid-point of OQ.
Now is A OQQ’
P and P’ are the mid-point of OQ and OQ’
AB || OQ’
Locus of Q is a line CD, which is parallel to AB.

Question 22.
Draw an angle ABC = 75°. Find a point P such that P is at a distance of 2 cm from AB and 1.5 cm from BC.
Solution:

Steps of Construction:
(i) Draw a ray BC.
(ii) At B, draw a ray BA making an angle of 75° with BC.
(iii) Draw a line  ℓ parallel to AB at a distance of 2 cm.
(iv) Draw another line m parallel to BC at a distance of 1.5 cm. which intersects m at P.
∴ P is the required point.

Question 23.
Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained.
Solution:

(i) Draw a line segment AB = 5.6 cm.
(ii) From A and B, as centres and radius 9.2 cm, draw the arcs which intersect each other at C.
(iii) Join CA and CB.
(iv) Draw two lines m and n parallel to BC at a distance of 2 cm each.
(v) Draw the angle bisector of ∠CAB which inter-sects the parallel lines m and n at P and Q respectively.
P and Q are the required points which are equi-distant from AB and AC.
On measuring the distance between P and Q is 4.3 cm.

Question 24.
Construct a triangle ABC, with AB = 6 cm, AC = BC = 9 cm. Find a point 4 cm from A and equidistant from B and C.
Solution:

(iii) Join CA and CB.
(iv) Draw the perpendicular bisector of BC.
(v) A as centre and radius 4 cm, draw an arc which intersect the perpendicular bisector of BC, at P.
P is the required point which is equidistant from B and C and at a distance of 4 cm from A.

Question 25.
thythtyhRuler and compasses may be used in this question. All construction lines and arcs must be clearly shown and be of sufficient length and clarity to permit assessment.
(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
(ii) Construct the locus of all points inside triangle ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
(iv) Mark the point Q, in your construction, which would make A QBC equal in area to A ABC, and isosceles.
(v) Measure and record the length of CQ. [1998]
Solution:
(i) Draw a line segument BC = 6 cm.
(ii) At B, draw a ray BX making an angle of 60H and cut off BA = 9 cm.
(iii) Join AC, then A ABC is the given triangle. ..(i)
(iv) Draw perpendicular bisector of BC which
intersects BA in M, then any point on LM, is the equidistant from B and C. ….(ii)

(v) Though A, draw a line m II BC.
(vi) The perpendicular bisector of BC and the par-allel line intersect each other at Q.
(vii) Join QB and QC.
Then A QBC is equal in the area of A ABC and through any point on line m, and bace BC, every triangle is equal in area to the given triangle ABC. Length of CQ, on measuring.

Question 26.
State the locus of a point in a rhombus ABCD, wi.r’ch is equidistant
(i) from AB and AD; (ii) from the vertices A and C.
[1998]
Solution:
In rhombus ABCD, draw the angle bisector of ∠A which meets in C
∴ Join BD, which intersects AC at O.
O is the required locus.
From O, draw OL ⊥ AB and OM ⊥ AD.
In Δ AOL and Δ AOM

∴ O is equidistant from AB and AD.
∵ Diagonal AC and BD bisect each other at O at right angles.
∴ AO = OC
O is equidistant from A and C.

Question 27.
Use graph paper for this question. Take 2 cm = 1 unit on both the axes.
(i) Plot the points A (1,1), B (5,3) and C (2.7).
(ii) Construct the locus of points equidistant from A and B.
(iii) Construct the locus of points equidistant from ABandAC.
(iv) Locate the point P such that PA = PB and P is equidistant from AB and AC.
(v) Measure and record the length PA in cm.
[1999]
Solution:

Plot the points A (1, 1), B (5, 3) and C (2, 7) on the graph and join AB, BC and CA.
Draw the perpendicular bisector of AB and angle bisector of ∠A which intersect each other at P. P is the required point,
∵ P lies on the perpendicular bisector of AB.
∴P is equidistant from A and B.
Again,
∵ P lies on the angle bisector of ∠A
∴ P is equidistant’from AB and AC
Now, on measuring the length of PA, it is 5.2 cm
(Approx.)

Question 28.
Construct an isosceles triangle ABC such that AB = 6 cm, BC = AC = 4 cm. Biscet ∠C internally and mark a point P on this bisector such that CP = 5 cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB. [2001]
Solution:
(i) Draw a line segment AB = 6 cm.
(ii) With centres A and B and radius 4 cm, draw two arcs which intersect each other at C.

(iii) Join CA and CB.
(iv) Draw the angle bisector of ∠C and cut off CP – 5 cm.
(v) A line m is drawn parallel to AB at a distance of 5 cm.
(vi) P as centre and radius 5 cm, draw arcs which
intersect the line m at Q and R.
(vii) Join PQ, PR and AQ.
Q and R are the required points.
P.Q
Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of lengths 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
(ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC. [1995]
Solution:

(i) Draw a circle with radius = 4 cm and O is the centre.
(ii) Take a point A on it.
(iii) A as centre and radius 6 cm draw an arc which intersects the circle at B.
(iv) Again A as centre and radius 5 cm, draw another arc which intersects the circle at C.
(v) Join AB and AC.
(vi) Draw the perpendicular bisector of AC, which intersects AC at M and meets the circle at E and F. EF is the locus of the points inside the circle which are equidistant from A and C.
(vii) Join AE, AF, CE and CF.
Proof:

Similarly, we can prove that CF = AF Hence EF is the locus of points which are equidistant from A and C.
(ii) Again draw the bisector of ∠A which meets the circle at N.
∴ Locus of points inside the circle which are equidistant from AB and AC is the perpendicular bisector of ∠A.

Question 29.
Plot the points A (2,9), B (-1,3) and C (6,3) on a graph paper. On the same graph paper, draw the locus of point A so that the area of ΔABC remains the same as A moves.
Solution:

Draw axis XOX’ and YOY’ on the graph paper. Plot the points A (2, 9), B (-1,3) and C (6, 3) Join AB, BC and CA which form a ΔABC. From A, draw a line l parallel to BC on x-axis The locus of point A is the line l.
∵ l || BC and triangles on the same base BC and between the same parallel are equal in area.
∴ l is the required locus of point A.

Question 30.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
(i) Complete the rectangle ABCD such that:
(a) P is equidistant from A B and BC.
(b) P is equidistant from C and D.
(ii) Measure and record the length of AB.
(2007)
Solution:
(i) Draw a line segment BC = 5 cm.
(ii) At B, draw a ray making an angle of 45° and cut off BP = 4 cm.
(iii) Join PC.
(iv) v P is equidistant from AB and BC.
∴ P lies on the bisector of ∠ABC.
Now draw a ray BY making an angle of 90°.
P is equidistant from C and D P lies on the perpendicular bisector of CD.

(v) From C, draw CZ ⊥ BC which intersect the perpendicular bisector at Q.
(vi) Cut off QD = CQ and from BP, cut off BA = CD.
Then ABCD is the required rectangle. Measuring the length of AB, it is 5.7 cm approximately.

Question 31.
Use ruler and compasses only for the following questions. All constructions lines and arcs must be clearly shown.
(i) Construct a ΔABC in which BC = 6.5 cm, ∠ABC = 60°, AB = 5 cm.
(ii) Construct the locus of points at a distance of 3.5 cm from A.
(iii) Construct the locus of points equidistant from AC and BC.
(iv) Mark 2 points X and Y which are at
distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY. (2016)
Solution:
(i) Steps of construction :
(1) Draw BC = 6.5 cm using a ruler.
(2) At B, draw ∠CBP = 60°
From BP, cut off BA = 5 cm.
(3) Join AC to get the required triangle.
(4) With A as a centre and radius equal to 3.5 cm, draw a circle. This circle is the required
locus of points at a distance of 3.5 cm from A.
(5) Draw the bisector of ∠ACB. This bisector is the locus of points equidistant from AC and BC.
(6) The angle bisector drawn above cuts the circle at X and Y. These are the points which are at a distance of 3.5 cm from A and also equidistant from AC and BC. On measuring, the length of XY comes out to be 5.2 cm.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 16 Loci (Locus and Its Constructions) Ex 16B Heights and Distances Ex 22C are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5B.

Other Exercises

Solve equations, number 1 to number 20, given below, using factorisation method :
Question 1.
x² – 10x – 24 = 0
Solution:
x² – 12x + 2x – 24 = 0
⇒ x (x – 12) + 2 (x – 12) = 0
⇒ (x – 12) (x + 2) = 0
Either x – 12 = 0, then x = 12
or x + 2 = 0, then x = – 2
x = 12, – 2

Question 2.
x² – 16 = 0
Solution:
⇒ x² – (4)² = 0
⇒ (x + 4) (x – 4) = 0
Either x + 4 = 0, then x = – 4
or x – 4 = 0, then x = 4
x = 4, – 4

Question 3.
2x² – $$\frac { 1 }{ 2 }$$ x = 0
Solution:
⇒ 4x² – x = 0
⇒ x (4x – 1) = 0
Either x = 0,
or 4x – 1 = 0, then 4x = 1 ⇒ x = $$\frac { 1 }{ 4 }$$
x = 0, $$\frac { 1 }{ 4 }$$

Question 4.
x (x – 5) = 24
Solution:
⇒ x² – 5x – 24 = 0
⇒ x² – 8x + 3x – 24 = 0
⇒ x (x – 8) + 3 (x – 8) = 0
⇒ (x – 8) (x + 3) = 0
Either x – 8 = 0, then x = 8
or x + 3 = 0, then x = – 3
x = 8, – 3

Question 5.
$$\frac { 9 }{ 2 }$$ x = 5 + x²
Solution:
⇒ 9x = 10 + 2x²
⇒ 2x² – 9x + 10 = 0
⇒ 2x² – 4x – 5x + 10 = 0
⇒ 2x (x – 2) – 5 (x – 2) = 0
⇒ (x – 2) (2x – 5) = 0
Either x – 2 = 0, then x = 2
or 2x – 5 = 0, then 2x = 5 ⇒ x = $$\frac { 5 }{ 2 }$$
x = 2, $$\frac { 5 }{ 2 }$$

Question 6.
$$\frac { 6 }{ x }$$ = 1 + x
Solution:
⇒ 6 = x + x²
⇒ x² + x – 6 = 0
⇒ x² + 3x – 2x – 6 = 0
⇒ x (x + 3) – 2 (x + 3) = 0
⇒ (x + 3) (x – 2) = 0
Either x + 3 = 0, then x = – 3
or x – 2 = 0, then x = 2
x = 2, – 3

Question 7.
x = $$\frac { 3x + 1 }{ 4x }$$
Solution:
⇒ 4x² = 3x + 1
⇒ 4x² – 3x – 1 = 0
⇒ 4x² – 4x + x – 1 = 0
⇒ 4x (x – 1) + 1 (x – 1) = 0
⇒ (x – 1) (4x + 1) = 0
Either x – 1 = 0, then x = 1
or 4x + 1 = 0, then 4x = -1 ⇒ x = $$\frac { -1 }{ 4 }$$
x = 1, $$\frac { -1 }{ 4 }$$

Question 8.
x + $$\frac { 1 }{ x }$$ = 2.5
Solution:

Question 9.
(2x – 3)² = 49
Solution:
⇒ 4x² – 12x + 9 = 49
⇒ 4x² – 12x + 9 – 49 = 0
⇒ 4x² – 12x – 40 = 0
⇒ x² – 3x – 10 = 0 (Dividing by 4)
⇒ x² – 5x + 2x – 10 = 0
⇒ x (x – 5) + 2 (x – 5) = 0
⇒ (x – 5) (x + 2) = 0
Either x – 5 = 0, then x = 5
or x + 2 = 0, then x = – 2
x = 5, – 2

Question 10.
2 (x² – 6) = 3 (x – 4)
Solution:
⇒ 2x² – 12 = 3x- 12
⇒ 2x² – 3x – 12 + 12 = 0
⇒ 2x² – 3x = 0
⇒ x (2x – 3) = 0
Either A = 0,
or 2x – 3 = 0, then 2x = 3 ⇒ x = $$\frac { 3 }{ 2 }$$
x = 0, $$\frac { 3 }{ 2 }$$

Question 11.
(x + 1) (2x + 8) = (x + 7) (x + 3)
Solution:

Question 12.
x² – (a + b) x + ab = 0
Solution:
⇒ x² – ax – bx + ab = 0
⇒ x (x – a) – b (x – a) = 0
⇒ (x – a) (x – b) = 0
Either x – a = 0, then x = a
or x – b = 0, then x = b
x = a, b

Question 13.
(x + 3)² – 4 (x + 3) – 5 = 0
Solution:
Let x + 3 = y, then
⇒ y² – 4y – 5 = 0
⇒y² – 5y + y – 5 = 0
⇒ y (y – 5) + 1 (y – 5) = 0
⇒ (y – 5) (y + 1) = 0
Substituting the value of y,
⇒ (x + 3 – 5) (x + 3 + 1) = 0
⇒ (x – 2) (x + 4) = 0
Either x – 2 = 0, then x = 2
or x + 4 = 0, then x = – 4
x = 2, -4

Question 14.
4 (2x – 3)² – (2x – 3) – 14 = 0
Solution:
Let 2x – 3 = y, then
⇒ 4y² – y – 14 = 0
⇒ 4y² – 8y + 7y – 14 = 0
⇒ 4y (y – 2) + 7 (y – 2) = 0
⇒ (y – 2) (4y + 7) = 0
Substituting the value of y,
⇒ (2x – 3 – 2) (8x – 12 + 7) = 0
⇒ (2x – 5) (8x – 5) = 0
Either 2x – 5 = 0, then 2x = 5 ⇒ x = $$\frac { 5 }{ 2 }$$
or 8x – 5 = 0, then 8x = 5 ⇒ x = $$\frac { 5 }{ 8 }$$
x = $$\frac { 5 }{ 2 }$$ , $$\frac { 5 }{ 8 }$$

Question 15.

Solution:
⇒ (3x – 2) (x + 4) = (3x – 8) (2x – 3)
⇒ 3x² + 12x – 2x – 8 = 6x² – 9x – 16x + 24
⇒ 3x² + 12x – 2x – 8 – 6x² + 9x + 16x – 24 = 0
⇒ – 3x² + 35x – 32 = 0
⇒ 3x² – 35x + 32 = 0
⇒ 3x² – 3x – 32x + 32 = 0
⇒ 3x (x – 1) – 32 (x – 1) = 0
⇒ (x – 1) (3x – 32) = 0
⇒ 3x (x – 1) – 32 (x – 1) = 0
⇒ (x – 1) (3x – 32) – 0
Either x – 1 = 0, then x = 1
or 3x – 32 = 0, then 3x = 32 ⇒ x = $$\frac { 32 }{ 3 }$$
x = 1, $$\frac { 32 }{ 3 }$$ or 1, 10$$\frac { 2 }{ 3 }$$

Question 16.
2x² – 9x + 10 = 0, when :
(i) x ∈ N
(ii) x ∈ Q.
Solution:
2x² – 9x + 10 = 0
⇒ 2x² – 4x – 5x + 10 = 0
⇒ 2x (x – 2) – 5 (x – 2) = 0
⇒ (x – 2) (2 – 5) = 0
Either x – 2 = 0, then x = 2
or 2x – 5 = 0, then 2x = 5 ⇒ x = $$\frac { 5 }{ 2 }$$
(i) When x ∈ N, then x = 2
(ii) When x ∈ Q, then x = 2 , $$\frac { 5 }{ 2 }$$

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.
Find the quadratic equation, whose solution set is :
(i) {3, 5}
(ii) {-2, 3}
Solution:
(i) Solution set is {3, 5} or x = 3 and x = 5
Equation will be
(x – 3) (x – 5) = 0
⇒ x² – 5x – 3x + 15 = 0
⇒ x² – 8x + 15 = 0
(ii) Solution set is {-2, 3} or x = -2, x = 3
Equation will be
(x + 2) (x – 3) = 0
⇒ x² – 3x + 2x – 6 = 0
⇒ x² – x – 6 = 0

Question 22.

Solution:

Roots are not real.
Hence there is no possible real value of x.

Question 23.
Find the value of x, if a + 1 = 0 and x² + ax – 6 = 0.
Solution:
a + 1 = 0 ⇒ a = -1
Now the equation x² + ax – 6 = 0 will be x² + (-1) x – 6 = 0
⇒ x² – x – 6 = 0
⇒ x² – 3x + 2x – 6 = 0
⇒ x (x – 3) + 2 (x – 3) = 0
Either x – 3 = 0, then x = 3
or x + 2 = 0, then x = – 2
x = 3, – 2

Question 24.
Find the value of x, if a + 7 = 0; b + 10 = 0 and 12x² = ax – b.
Solution:
a + 7 = 0, then a = -7
and b + 10 = 0, then b = -10
Now, substituting the value of a and b in
12x² = ax – b
⇒ 12x² = – 7x – (-10)
⇒ 12x² = – 7x + 10
⇒ 12x² + 7x – 10 = 0
⇒ 12x² + 15x – 8x – 10 = 0
⇒ 3x (4x + 5) – 2 (4x + 5) = 0
⇒ (4x + 5) (3x – 2) = 0
Either 4x + 5 = 0, then 4x = -5 ⇒ x = $$\frac { 5 }{ 4 }$$
or 3x – 2 = 0, then 3x = 2 ⇒ x = $$\frac { 2 }{ 3 }$$
x = $$\frac { 5 }{ 4 }$$, $$\frac { 2 }{ 3 }$$

Question 25.
Use the substitution y = 2x + 3 to solve for x, if 4 (2x + 3)² – (2x + 3) – 14 = 0.
Solution:
y = 2x + 3, then equation
4 (2x + 3)² – (2x + 3) – 14 = 0 will be 4y² – y – 14 = 0
⇒ 4y² – 8y + 7y – 14 = 0
⇒ 4y (y – 2) + 7 (y – 2) = 0
⇒ (y – 2) (4y + 7) = 0
Either y – 2 = 0, then y = 2

Question 26.
Without solving the quadratic equation 6x² – x – 2 = 0, find whether x = $$\frac { 2 }{ 3 }$$ is a solution of this equation or not.
Solution:

Question 27.
Determine whether x = -1 is a root of the equation x² – 3x + 2 = 0 or not.
Solution:
x² – 3x + 2 = 0
x = -1
Substituting the value of x = -1, in the quadratic equation
L.H.S. = x² – 3x + 2 = (-1)² – 3(-1) + 2 = 1 + 3 + 2 = 6 ≠ 0
Remainder is not equal to zero
x = -1 is not its root.

Question 28.
If x = $$\frac { 2 }{ 3 }$$ is a solution of the quadratic equation 7x² + mx – 3 = 0; find the value of m.
Solution:

Question 29.
If x = -3 and x = $$\frac { 2 }{ 3 }$$ are solutions of quadratic equation mx² + 7x + n = 0, find the values of m and n.
Solution:
x = -3, x = $$\frac { 2 }{ 3 }$$ are the solution of the quadratic equation, mx² + 7x + n = 0
Then these values of x will satisfy it
(i) If x = -3, then mx² + 7x + n = 0
⇒ m(-3)² + 7(-3) + n = 0
⇒ 9m – 21 + n = 0
⇒ n = 21 – 9m ……(i)

Question 30.
If quadratic equation x² – (m + 1) x + 6 = 0 has one root as x = 3; find the value of m and the other root of the equation.
Solution:
In equation, x² – (m + 1) x + 6 = 0
x = 3 is its root, then it will satisfy it
⇒ (3)² – (m + 1) x 3 + 6 = 0
⇒ 9 – 3m – 3 + 6 = 0
⇒ -3m + 12 = 0

Question 31.
Give that 2 is a root of the equation 3x² – p (x + 1) = 0 and that the equation px² – qx + 9 = 0 has equal roots, find the values of p and q.
Solution:
3x² – p (x + 1) = 0
⇒ 3x² – px – p = 0
2 is a root of the equal It will satisfy it
3(2)² – p(2) – p = 0
⇒ 3 x 4 – 2p – p = 0
⇒ 12 – 3p = 0
⇒ 3p = 12
⇒ p = 4
px² – qx + 9 = 0
Here, a = p, b = -q, c = 9
D = b² – 4ac = (-q)² – 4 x p x 9 = q² – 36p
Roots are equal.
D = 0
⇒ q² – 36p = 0
⇒ q² – 36 x 4 = 0
⇒ q² = 144
⇒ q² = (±12)²
⇒ q = ± 12
Hence, p = 4 and q = ±12

Question 32.

Solution:

Question 33.
Solve: ( $$\frac { 1200 }{ x }$$ + 2 ) (x – 10) – 1200 = 60
Solution:

By cross multiplication,
⇒ 1200x – 12000 + 2x² – 20x – 1260x = 0
⇒ 2x² + 1200x – 20x – 1260x – 12000 = 0
⇒ 2x² – 80x – 12000 = 0
⇒ x² – 40x – 6000 = 0
⇒ x² – 100x + 60x – 6000 = 0
⇒ x (x – 100) + 60 (x – 100) = 0
⇒ (x – 100) (x + 60) = 0
Either x – 100 = 0, then x = 100
or x + 60 = 0, then x = -60
x = 100, -60

Question 34.
If -1 and 3 are the roots of x² + px + q = 0, find the values of p and q.
Solution:
-1 and 3 are the roots of the equation
x² + px + q = 0
Substituting the value of x = -1 and also x = 3, then
(-1 )² + p(-1) + q = 0
⇒ 1 – p + q = 0
⇒ p – q = 1
⇒ p = 1 + q …(i)
and (3)² + p x 3 + q = 0
⇒ 9 + 3p + q = 0
⇒ 9 + 3 (1 + q) + q = 0 [From(i)]
⇒ 9 + 3 + 3q + q = 0
⇒ 12 + 4q = 0
⇒ 4q = -12
⇒ q = -3
p = 1 + q = 1 – 3 = -2
Hence, p = -2, q = -3

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5B are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5A.

Other Exercises

Question 1.
Without solving, comment upon the nature of roots of each of the following equations:
(i) 7x² – 9x + 2 = 0
(ii) 6x² – 13x + 4 = 0
(iii) 25x² – 10x + 1 = 0
(iv) x² + 2√3 x – 9 = 0
(v) x² – ax – b² = 0
(vi) 2x² + 8x + 9 = 0
Solution:

Question 2.
Find the value of ‘p’, if the following quadratic equations have equal roots :
(i) 4x² – (p – 2) x + 1 = 0
(ii) x² + (p – 3) x + p = 0
Solution:

Question 3.
The equation 3x² – 12x + (n – 5) = 0 has equal roots. Find the value of n.
Solution:
3x² – 12x + (n – 5) = 0
Here a = 3, b = -12, c = (n – 5)
D = b² – 4ac

Question 4.
Find the value of ‘m’, if the following equation has equal roots :
(m – 2) x² – (5 + m) x + 16 = 0
Solution:

Question 5.
Find the value of k for the which the equation 3x² – 6x + k = 0 has distinct and real root.
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5A are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 25 Probability Ex 25B.

Other Exercises

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:
No. of cards = 9
Having numbers marked on it = 2 to 10
∴ Number of possible outcomes = 9
(i) An even number i.e. 2, 4, 6, 8, 10 = 5
∴ Number of even numbers = 5

(ii) A multiple of 3 are 3, 6, 9
Number of multiple of 3 = 3

(iii) An even number and a multiple of 3 are 6
which is one in number

(iv) An even number or a multiple of 3 are 2, 3, 4, 6, 8, 9, 10
which are 7 in numbers

Question 2.
Hundred identical cards are numbered from 1 to 100. The cards are well shuffled and then a card is drawn. Find the probability that the number on the 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:
Number of cards = 100
Marked with numbers from 1 to 100
∴ Number of possible outcome = 100
(i) A multiple of 5 are 5, 10, 15 95, 100
which are 20 in numbers.
∴ Number of favourable outcome = 20

(ii) A multiple of 6 are 6, 12, 18, 24, 90, 96
which are 16 in numbers.
∴ Number of favourable outcome =16

(iii) Between 40 and 60 are 41, 42, ……. , 58, 59,
which are 19

(iv) Greater than 85 are 86 to 100 which are 15 in numbers.

(v) Less than 48 are 1 to 47 which are 47 in numbers

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:
Number of identical cards = 25
Numbers marked on their are 1 to 25 i.e.
1, 2, 3, 4, 5, …. 21, 22, 23, 24, 25
∴ Number of possible outcome = 25
(i) Multiple of 3 are 3, 6, 9, 12, 15, 18, 21, 24
Which are 8 in numbers.

(ii) Multiple of 5 are 5, 10, 15, 20, 25
which are 5 in numbers

(iii) Multiple of 3 and 5 are = 15
which is 1 in number

(iv) Multiple of 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25
which are 12 in numbers

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:
A die has 6 numbers i.e., 1, 2, 3, 4, 5, 6
∴ Number of possible outcome = 6
(i) Less than 3 are 1, 2 which are 2 in numbers

(ii) Greater than or equal to 4 are 4, 5, 6
which are 3 in number.

(iii) Less than 8 are 1, 2, 3, 4, 5, 6
which are 6 in numbers

(iv) Greater than 6 is nothing on the die

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:
Number of pages of the book = 85
which are from 1 to 85
Number of possible outcome = 85
∴ Number of pages whose sum of its page is 8 are : 17, 26, 35, 44, 53, 62, 71, 80 and 8

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:
Numbers marked on each die = 6
∴ Total number of cases = 6 x 6 = 36
∵ Favourable come are (5, 5), (5, 6) as 5 appears on the first, which are 2 in number

Question 7.
If two coins are tossed once, what is the probability of getting :
(iii) both heads or both tails.
Solution:
∵ A coins has two faces Head and Jail or H, T
∴ Two coins are tossed
∴ Number of coins = 2 x 2 = 4
which are HH, HT, TH, TT
(i) When both are Head, then
∴ Number of outcome = 1

(ii) At least one head, then
Number of outcomes = 3

(iii) When both head or both tails, then 1
Number of outcomes = 2

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:
∵ A die has 6 faces which are 1, 2, 3, 4, 5,6
∴ On rolling two dice at a time, number of comes = 6 x 6 = 36
∴ Number of possible outcome = 36
(i) a total of atleast 10, the favourable can be (4,6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6) which are 6 in number

(ii) A multiple of 2 on one die and an odd number on the other
∴ Outcome can be (2, 1), (2, 3), (g, 5), (4, 1), (4,3) , (4, 5), (6, 1), (6, 3), (6, 5), (1, 2), (3, 2), (5, 2), (1, 4), (3, 4), (5, 4), (1, 6), (3, 6), 5, 6) which are 18 in numbers.
Number of favourable outcome

Question 9.
A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is :
(ii) a red card.
(iii) a face card.
(iv) 5 of heart or diamond.
(v) Jack or queen.
(vi) ace and king.
(vii) a red and a king.
(viii) a red or a king.
Solution:
A pack of playing card has 52 cards
∴ Number of possible outcome = 52
∵ there are 13 cards of spade
∴ Number of favourable outcome = 13

(ii) A red card.
∵ There are 13 + 13 = 26 red card

(iii) A face card.
∵ There are 3 x 4 = 12 faces which are red.

(iv) 5 of heart or diamond.
∴ Number of cards =1 + 1=2

(v) Jack or queen
There are 4 + 4 = 8 such cards

(vi) ace and king.
There is no such card which is ace and king both
∴ P(E) = 0.
(vii) a red and a king
There are 2 such cards which are red kings

(viii) a red or a king
There are 26 cards which are red in which 2 kings are red and 2 more kings which are black = 26 + 2 = 28

Question 10.
A bag contains 16 coloured 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
(ii) not red
(iii) white
(iv) not white
(v) green or red
(vi) white or green
(vii) green or red or white.
Solution:
Number of balls in a bag = 16
Green balls = 6
White balls = 3
Red balls = 7
∴ Total possible outcome =16
(i) Red balls = 7

(ii) Not red balls = 16-7 = 9

(iii) White balls = 3

(iv) Not white balls = 16 – 3= 13

(v) Green or red balls = 6 + 7 = 13

(vi) White or green balls = 3 + 6 = 9

(vii) Green or red or white balls =16

Question 11.
A ball is drawn at random from a box .ontammg 12 white. 16 red and 20 green balls. Determine the probability that the ball drawn is:
(i) white
(ii) red
(iii) not green
(iv) red or white.
Solution:
Number of balls in a box
White = 12
Red = 16
Green = 20
Total balls = 12 + 16 +20 = 48
∴ Total possible outcome = 48
(i) White balls = 12
∴ Number of favourable outcome =12

(ii) Red balls = 16
∴ Number of favourable outcome =16

(iii) Not green
Number of balls =12 + 16 = 28
∴ Number of favourable outcome = 28

(iv) Red or white balls =12 + 16 = 28
∴ Number of favourable outcome = 28

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:
Number of cards in playing card deck = 52
∴ Number of possible outcome = 52
(i) A red card
There are 13 + 13 = 26 red cards in the deck
∴ Number of favourable outcome = 26

(ii) A black card
There are 13 + 13 = 26 black cards
∴ Number of favourable outcome = 26

There are 13 spade cards in the deck
∴ Number of favourable outcome = 13

(iv) An Ace
There use 4 aces in the deck
Number of favourable outcome = 4

(v) A black ace
There are two black aces in a deck
∴ Number of favourable outcome = 2
Number of cards in playing card deck = 52

(vi) Ace of diamonds
∴ There are only one ace of diamonds
∴ Number of favourable of outcome = 1

(vii) Not a club
There are 13 x 3 = 39 cards which are not a club
∴ Number of favourable outcome = 39

(viii) A queen or a Jack
There are 4 queen cards and 4 Jack cards
∴ Number of favourable outcome = 4 + 4 = 8

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:
There are 30 cards which are marks with numbers 1 to 30 and one card is drawn
(i) A multiple of 4 or 6.
∴ There are multiple of 4 or 6 = 4,6, 8, 12, 16, 18, 20, 24, 28, 30 which are 10
∴ Number of favourable outcome = 10

(ii) A multiple of 3 and 5 are 15, 30 which are 2
∴ Number of favourable outcome = 2

(iii) A multiple of 3 or 5
which are 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30 which are 14
∴ Number of favourable outcome = 14

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:
Number of dice thrown = 2
Each die has 1 to 6 numbers on its faces Number of possible outcomes = 6 X 6 = 36
(i) A doublet : These can be (1, 1), (2, 2),(3, 3), (4, 4), (5, 5) and (6, 6) which are 6
∴ Number of favourable outcome = 6

(ii) A number less than 3 on each die which can be (1, 1), (1, 2), (2, 1), (2, 2) which are 4 in numbers
∴ Number of favourable outcome = 4

(iii) An odd number as a sum which can be (1, 1), (1,3), (1,5), (2, 1), (2,3), (2,5), (3, 1), (3,3), (3,5), (4, 1), (4,3), (4, 5), (5,1),(5,3), (5,5), (6, 1), (6, 3), (6, 5) which are 18 in numbers.
∴ Number of favourable outcome = 18

(iv) Total of at most 10
Which can be = 36 – 3 (which can be (5, 6), (6, 6), (6, 5) = 33
∴ Number of favourable outcome = 33

(v) An odd number on one die and a number less than or equal to 4 on the other die
(1, 1), (1,2), (1,3), (1,4), (3,1), (3,2), (3, 3), (3,4) , (5, 1), (5, 2), (5, 3), (5, 4), (2, 1), (3, 1), (4, 1),(1, 3), (2, 3), (2, 5), (3, 1), (3, 5), (4, 1), (4, 3), (4,5) which are 23 in numbers
∴ Number of favourable outcome = 23

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability Ex 25B  are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems (Based on Quadratic Equations) Ex 6A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A.

Other Exercises

Question 1.
The product of two consecutive integers is 56. Find the integers.
Solution:
Let the first integer = x
Then second integer = x + 1
Now according to the condition given
x (x + 1) = 56
⇒ x² + x – 56 = 0
⇒ x² + 8x – 7x – 56 = 0
⇒ x (x + 8) – 7 (x + 8) = 0
⇒ (x + 8) (x – 7) = 0
Either x + 8 = 0, then x = – 8
or x – 7 = 0, then x = 7
(i) If x = -8, then
first integer = -8
and second integer = – 8 + 1 = – 7
(ii) If x = 7, then
first integer = 7
and second integer = 7 + 1 = 8
Integers are 7, 8 or -8, -7

Question 2.
The sum of the squares of two consecutive natural numbers is 41. Find the numbers.
Solution:
Let the first natural number = x
Then second natural number = x + 1
Now according to the condition given,
(x)² + (x + 1)² = 41
⇒ x² + x² + 2x + 1 = 41
⇒ 2x² + 2x + 1 – 41 = 0
⇒ 2x² + 2x – 40 = 0
⇒ x² + x – 20 = 0 (Dividing by 2)
⇒ x² + 5x – 4x – 20 = 0
⇒ x (x + 5) – 4 (x + 5) = 0
⇒ (x + 5) (x – 4) = 0
Either x + 5 = 0 then x = – 5 But it is not a natural number
or x – 4 = 0, Then x = 4
Numbers are 4 and 5

Question 3.
Find the two natural numbers which differ by 5 and the sum of whose squares is 97.
Solution:
Let the first natural number = x
Then second natural number = x + 5
Now according to the given condition,
(x)² + (x + 5)² = 97
⇒ x² + x² + 10x + 25 – 97 = 0
⇒ 2x² + 10x – 72 = 0
⇒ x² + 5x – 36 = 0 (Dividing by 2)
⇒ x² + 9x – 4x – 36 = 0
⇒ x (x + 9) – 4 (x + 9) = 0
⇒ (x + 9) (x – 4) = 0
Either x + 9 = 0, then x = – 9 But it is not a natural number
or x – 4 = 0, then x = 4
First number = 4
and second number = 4 + 5 = 9

Question 4.
The sum of a number and its reciprocal is 4.25. Find the number.
Solution:
Let the number = x
Now according to the condition,

⇒ 4x (x – 4) – 1 (x – 4) = 0
⇒ (x – 4) (4x – 1) = 0
Either x – 4 = 0, then x = 4
or 4x – 1 = 0, 4x = 1 then x = $$\frac { 1 }{ 4 }$$
Number is 4 or $$\frac { 1 }{ 4 }$$

Question 5.
Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is $$\frac { 7 }{ 10 }$$
Solution:
Let the first natural number = x
Then second natural number = x + 3
Now according to the given condition,

But it is not a natural number
or x – 2 = 0, then x = 2
First number = 2
and second number = 2 + 3 = 5

Question 6.
Divide 15 into two parts such that the sum of their reciprocals is $$\frac { 3 }{ 10 }$$.
Solution:
Let first part = x
Then second part = 15 – x (sum = 15)
Now according to the given condition,

⇒ x (x – 5) – 10 (x – 5) = 0
⇒ (x – 5) (x – 10) = 0
Either x – 5 = 0, then x = 5
or x – 10 = 0, then x = 10
If x = 5, then first part = 15 – 5 = 10
If x = 10, then second part = 15 – 10 = 5
Parts are 5, 10

Question 7.
The sum of the squares of two positive integers is 208. If the square of the larger number is 18 times the smaller number, find the numbers.
Solution:
Let x be the larger number and y be the smaller number, then
According to the conditions x2 = 18 ….(i)
and x² + y² = 208 ….(ii)
⇒ 18y + y² = 208 [From (i)]
⇒ y² + 18y – 208 = 0
⇒ y² + 26y – 8y – 208 = 0
⇒ y (y + 26) – 8 (y + 26) = 0
⇒ (y + 26) (y – 8) = 0
Either y + 26 = 0, then y = -26
But it is not possible as it is not positive
or y – 8 = 0, then y = 8
Then x² = 18y ⇒ y² = 18 x 8 ⇒x² = 144 = (12)² ⇒ x = 12
Number are 12, 8

Question 8.
The sum of the squares of two consecutive positive even numbers is 52. Find the numbers.
Solution:
Let first even number = 2x
and second even number = 2x + 2
Now according to the given condition,
(2x)² + (2x + 2)² = 52
⇒ 4x² + 4x² + 8x + 4 = 52
⇒ 4x² + 4x² + 8x + 4 – 52 = 0
⇒ 8x² + 8x – 48 = 0
⇒ x² + x – 6 = 0 (Dividing by 8)
⇒ x² + 3x – 2x – 6 = 0
⇒ x (x + 3) – 2 (x + 3) = 0
⇒ (x + 3) (x – 2) = 0
Either x + 3 = 0, then x = – 3 But it is not possible because it is not positive.
or x – 2 = 0, then x = 2
First even number = 2 x 2 = 4
Second number = 4 + 2 = 6
Hence numbers are 4, 6

Question 9.
Find two consecutive positive odd numbers, the sum of whose squares is 74.
Solution:
Let first odd number = 2x -1
Second odd number = 2x – 1 + 2 = 2x + 1
Now according to the given condition,
(2x – 1)² + (2x + 1)² = 74
⇒ 4x² – 4x + 1 + 4x² + 4x + 1 = 74
⇒ 8x² + 2 – 74 = 0
⇒ 8x² – 72 = 0
⇒ x² – 9 = 0 (Dividing by 8)
⇒ (x + 3) (x – 3) = 0
Either x + 3 = 0, then x = -3 But it is not possible because it is not positive.
or x – 3 = 0, then x = 3
First odd number = 2x – 1 = 2 x 3 – 1 = 6 – 1 = 5
and second odd number = 5 + 2 = 7
Number are 5, 7

Question 10.
The denominator of a positive fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2.9; find the fraction.
Solution:
Let numerator of a fraction = x
Then denominator = 2x + 1

Question 11.
Three positive numbers are in the ratio $$\frac { 1 }{ 2 }$$ : $$\frac { 1 }{ 3 }$$ : $$\frac { 1 }{ 4 }$$ Find the numbers; if the sum of their squares is 244.
Solution:
Multiply each ratio by L.C.M. of de-nominators i.e. by 12.
We get, 6 : 4 : 3
Let first positive number = 6x
Then second number = 4x
and third number = 3x
According to the given condition,
(6x)² + (4x)² + (3x)² = 244
⇒ 36x² + 16x² + 9x² = 244
⇒ 61x² = 244
⇒ x² – 4 = 0
⇒ (x + 2) (x – 2) = 0
Either x + 2 = 0, then x = -2 But it is not possible because it is not positive
or x – 2 = 0, then x = 2
First number = 6x = 6 x 2 = 12
Second number = 4x = 4 x 2 = 8
and third number = 3x = 3 x 2 = 6
Numbers are 12, 8, 6

Question 12.
Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
Solution:
Let first part = x
Then second part = 20 – x (Sum = 20)
Now, according to the given condition,
3(x)² – (20 – x) = 10
⇒ 3x² – 20 + x – 10 = 0
⇒ 3x² + x – 30 = 0
⇒ 3x² + 10x – 9x – 30 = 0
⇒ x (3x + 10) – 3 (3x + 10) = 0
⇒ (3x + 10) (x – 3) = 0
Either 3x + 10 = 0, then 3x = -10 ⇒ x = $$\frac { -10 }{ 3 }$$
But it is not possible.
or x – 3 = 0 then x = 3
Then first part = 3
and second part = 20 – 3 = 17

Question 13.
Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Assume the middle number to be x and form a quadratic equation satisfying the above statement Hence ; find the three numbers.
Solution:
Let middle number = x
Then, first number = x – 1
and third number = x + 1
Now according to the condition,
(x)²= [(x + 1)² – (x – 1)²] + 60
⇒ x² = [x² + 2x + 1 – x² + 2x – 1 ] + 60
⇒ x² = 4x + 60
⇒ x² – 4x – 60 = 0
⇒ x² – 10x + 6x – 60 = 0
⇒ x (x – 10) + 6 (x – 10) = 0
⇒ (x – 10) (x + 6) = 0
Either x – 10 = 0 then x = 10
or x + 6 = 0 then x = -6. But it is not a natural number.
Middle number = 10
First number = 10 – 1 = 9
and third number = 10 + 1 = 11
Hence numbers are 9, 10, 11

Question 14.
Out of three consecutive positive integers, the middle number is p. If three times the square of the largest is greater than the sum of the squares of the other two numbers by 67; calculate the value of p.
Solution:
Middle number = p
then smallest number = p – 1
and greatest number = p + 1
Now, according to the condition.
3 (p + 1)² – (p – 1)² – p² = 67
⇒ 3 (p² + 2p + 1) – (p² – 2p + 1) – p² = 67
⇒ 3p² + 6p + 3 – p² + 2p – 1 – p² – 67 = 0
⇒ p² + 8p + 2 – 67 = 0
⇒ p² + 8p – 65 = 0
⇒ p² + 13p – 5p – 65 = 0
⇒ p (p + 13) – 5 (p + 13) = 0
⇒ (p + 13) (p – 5) = 0
Either p + 13 = 0, then p = -13 But it is not possible.
or p – 5 = 0, then p = 5
p = 5

Question 15.
A can do a piece of work in ‘x’ days and B can do the same work in (x + 16) days. If both working together can do it in 15 days. Calculate ‘x’.
Solution:

Question 16.
One pipe can fill a cistern in 3 hours less than the other. The two pipes together can fill the cistern in 6 hours 40 minutes. Find the time that each pipe will take to fill the cistern.
Solution:
Let first pipe can fill the cistern in = x hrs.
Second pipe will fill the cistern in = x – 3 hrs.

But it is not possible.
x = 15
First pipe can fill in 15 hrs.
and second pipe in 15 – 3 = 12 hrs.

Question 17.
A positive number is divided into two parts such that the sum of the squares of the two parts is 20. The square of the larger part is 8 times the smaller part. Taking A as the smaller part of the two parts. Find the number. (2010)
Solution:
Let larger part = y
and smaller part = x
x² + y² = 20 ….(i)
and y² = 8x ….(ii)
Substituting the value of y² in „
x² + 8x = 20
⇒ x² + 8x – 20 = 0
⇒ x² + 10x – 1x – 20 = 0 {-20 = 10 x (-2), 8 = 10 – 2}
⇒ x (x + 10) – 2(x + 10) = 0
⇒ (x + 10) (x – 2) = 0
Either x + 10 = 0, then x = – 10 which is not possible because it is negative
or x – 2 = 0, then x = 2
Smaller number = 2
and larger number = y² = 8x = 8 x 2 = 16
⇒ y² = 16 = (4)²
⇒ y = 4
Number = x + y = 2 + 4 = 6

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 6 Solving Problems Ex 6A are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode) Ex 24A.

Other Exercises

Question 1.
Find the mean of 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 a students 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:
(a) Hence x = 9

(b) If marks of each students be increased by 4 then new mean will be = 59 + 4 = 63

Question 3.
Find the mean of natural numbers from 3 to 12.
Solution:
Numbers betweeen 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
Here n = 10

Question 4.
(a) Find the means 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:
(a) The mean of 7, 11, 6, 5 and 6.

(b) If we subtract 2 from each number, then the mean will be 7 – 2 = 5

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 number (Σxi) = 5 x 8 = 40 …(i)
But Σxi = 6 + 4 + 7 + a+10 = 27 + a ….(ii)
From (i) and (ii)
27 + a = 40 ⇒ a = 40 – 27
∴ 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 (Σxi) = 5 x 8 = 40 ….(i)
But sum of numbers given = 6 + y + 7 + x + 14
= 21 + y + x + ….(ii)
From (i) and (ii)
27 + y + x = 40
⇒ y = 40 – 27 – x
⇒ y= 13 – x

Question 7.
The ages of 40 students are given in the following table :

Find the arithmetic mean.
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 numbers = 69,5 x 10 = 695 ….(i)
But sum of given number = 72 + 70 + x + 62 + 50 + 71 + 90 + 64 + 58 + 82 = 619+x ….(ii)
From (i) and (ii)
619 + x = 695
⇒ x = 695 – 619 = 76

Question 9.
The following table gives the heights of plants in centimeter. If the mean height of plants is 60.95 cm; find the value of ‘f’.

Solution:

Question 10.
From the data given below, calculate the mean wage, correct to the nearest rupee.

(i) If the number of workers in each category is doubled, what would be the new mean wage? [1995]
(ii) If the wages per day in each category are increased by 60%; what is the new mean wage?
(iii) If the number of workers in each category is doubled and the wages per day per worker are reduced by 40%. What would be the new mean wage?
Solution:

(i) Mean remains the same if the number of workers in each catagory is doubled.
∴ Mean = 80.
(ii) Mean will be increased by 60% if the wages per day per worker is increased by 60%.
∴ New mean = 80 x $$\frac { 160 }{ 100 }$$= 128
(iii) No change in the mean if the number of worker is doubled but if wages per worker is reduced by 40%, then
New mean = 80 x $$\frac { 60 }{ 100 }$$= 48

Question 11.
The contents of 100 match boxes were checked to determine the number of matches they contained.

(i) Calculate, correct to one decimal place, the mean number of matches per box.
(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to bring the mean up to exactly 39 matches. [1997]
Solution:

(ii) In the second case,
New mean = 39 matches
∴ Total contents = 39 x 100 = 3900
But total no of matches already given = 3813
∴ Number of new matches to be added = 3900 – 3813 = 87

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

Solution:
Mean = 3

Question 13.
In the following table, Σf= 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:
Let the Assumed mean = 30

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

Solution:
Let the Assumed mean A = 45

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency Ex 24A are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D.

Other Exercises

Solve each of the following equations :
Question 1.

Solution:

Question 2.
(2x + 3)² = 81
Solution:
(2x + 3)2 = 81
⇒ 4x² + 12x + 9 = 81
⇒ 4x² + 12x + 9 – 81 = 0
⇒ 4x² + 12x – 72 = 0
⇒ x² + 3x – 18 = 0 (Dividing by 4)
⇒ x² + 6x – 3x – 18 = 0
⇒ x (x + 6) – 3 (x + 6) = 0
⇒ (x + 6) (x – 3) = 0
Either x + 6 = 0, then x = -6
or x – 3 = 0, then x = 3
x = 3, – 6

Question 3.
a² x² – b² = 0
Solution:
a² x² – b² = 0
⇒ (ax)² – (b)² =0
⇒ (ax + b) (ax – b) =
Either ax + b = 0, then x = $$\frac { -b }{ a }$$
or ax – b = 0. then x = $$\frac { b }{ a }$$
x = $$\frac { b }{ a }$$ , $$\frac { -b }{ a }$$

Question 4.

Solution:

Question 5.
x + $$\frac { 4 }{ x }$$ = – 4; x ≠ 0
Solution:
x + $$\frac { 4 }{ x }$$ = -4
⇒ x² + 4 = -4x
⇒ x² + 4x + 4 = 0
⇒ (x + 2)² = 0
⇒ x + 2 = 0
⇒ x = – 2

Question 6.
2x4 – 5x² + 3 = 0
Solution:
2x4 – 5x² + 3 = 0
⇒ 2(x²)² – 5x² + 3 = 0
⇒ 2(x²)² – 3x² – 2x² + 3 = 0
⇒ 2x4 – 3x² – 2x² + 3 = 0
⇒ x² (2x² – 3) – 1 (2x² – 3) = 0
⇒ (2x² – 3) (x² – 1) = 0

Question 7.
x4 – 2x² – 3 = 0
Solution:
x4 – 2x² – 3 = 0
⇒ (x²)² – 2x² – 3 = 0
⇒ (x²)² – 3x² + x² – 3 = 0
⇒ x² (x² – 3) + 1 (x² -3) = 0
⇒ (x² – 3) (x² + 1) = 0
Either x² – 3 = 0, then x² = 3 ⇒ x = √3
or x² + 1 = 0, then x² = – 1 In this case roots are not real
x = ±√3 or √3 , – √3

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.
(x² + 5x + 4)(x² + 5x + 6) = 120
Solution:
Let x² + 5x + 4 = y then x² + 5x + 6 = y + 2
Now (x² + 5x + 4) (x² + 5x + 6) = 120
⇒ y (y + 2) – 120 = 0
⇒ y² + 2y – 120 = 0
⇒ y² + 12y – 10y – 120 = 0
⇒ y (y + 12) – 10 (y + 12) = 0
⇒ (y + 12) (y – 10) = 0
Either y + 12 = 0, then y = – 12
or y – 10 = 0, then y = 10
(i) when y = -12, then x² + 5x + 4 = -12
⇒ x² + 5x + 4 + 12 = 0
⇒ x² + 5x + 16 = 0
Here a = 1, b = 5, c = 16
D = b² – 4ac = (5)² – 4 x 1 x 16 = 25 – 64 = -39
D < 0, then roots are not real
(ii) When y = 10, then x² + 5x + 4 = 10
⇒ x² + 5x + 4 – 10 = 0
⇒ x² + 5x – 6 = 0
⇒ x² + 6x – x – 6 = 0
⇒ x (x + 6) – 1 (x + 6) = 0
⇒ (x + 6) (x – 1) = 0
Either x + 6 = 0, then x = – 6
or x – 1 = 0, then x = 1
x = 1, -6

Question 12.
Solve each of the following equations, giving answer upto two decimal places:
(i) x² – 5x – 10 = 0 [2005]
(ii) 3x² – x – 7 = 0 [2004]
Solution:
(i) Given Equation is : x² – 5x – 10 = 0
On comparing with, ax² + bx + c = 0
a = 1, b = -5 , c = -10

Question 13.

Solution:

Question 14.
Solve:
(i) x² – 11x – 12 = 0; when x ∈ N
(ii) x² – 4x – 12 = 0; when x ∈ I
(iii) 2x² – 9x + 10 = 0; when x ∈ Q.
Solution:
(i) x² – 11x – 12 = 0
⇒ x² – 12x + x – 12 = 0
⇒ x (x – 12) + 1 (x – 12) = 0
⇒ (x – 12) (x + 1) = 0
Either x – 12 = 0, then x = 12
or x + 1 = 0, then x = -1
x ∈ N
x = 12
(ii) x² – 4x – 12 = 0
⇒ x² – 6x + 2x – 12 = 0
⇒ x (x – 6) + 2 (x – 6)=0
⇒ (x – 6) (x + 2) = 0
Either x – 6 = 0, then x = 6
or x + 2 = 0, then x = -2
x ∈ I
x = 6, -2
(iii) 2x² – 9x + 10 = 0
⇒ 2x² – 4x – 5x + 10 = 0
⇒ 2x (x – 2) – 5 (x – 2) = 0
⇒ (x – 2) (2x – 5) = 0
Either x – 2 = 0, then x = 2
or 2x – 5 = 0, then 2x = 5 ⇒ x = $$\frac { 1 }{ 2 }$$
x ∈ Q
x = 2, $$\frac { 5 }{ 2 }$$ or 2, 2.5

Question 15.
Solve: (a + b)² x² – (a + b) x – 6 = 0, a + b ≠ 0.
Solution:
(a + b)² x² – (a + b) x – 6 = 0
Let (a + b) x = y, then y² – y – 6 = 0
⇒ y² – 3y + 2y – 6 = 0
⇒ y (y – 3) + 2 (y – 3) = 0
⇒ (y – 3) (y + 2) = 0
Either y – 3 = 0, then y = 3
or y + 2 = 0, then y = – 2
(i) If y = 3, then

Question 16.

Solution:

Either x + p = 0, then x = -p
or x + q = 0, then x = -q
Hence x = -p, -q

Question 17.

Solution:
(i) x (x + 1) + (x + 2) (x + 3) = 42
⇒ x² + x + x² + 3x + 2x + 6 – 42 = 0
⇒ 2x² + 6x – 36 = 0
⇒ x² + 3x – 18 = 0
⇒ x² + 6x – 3x – 18 = 0
⇒ x (x + 6) – 3(x + 6) = 0
⇒ (x + 6) (x – 3) = 0
Either x + 6 = 0, then x = -6
or x – 3 = 0, then x = 3
Hence x = 3, -6

Question 18.
For each equation, given below, find the value of ‘m’ so that the equation has equal roots. Also, find the solution of each equation:
(i) (m – 3) x² – 4x + 1 = 0
(ii) 3x² + 12x + (m + 7) = 0
(iii) x² – (m + 2) x + (m + 5) = 0
Solution:

Question 19.
Without solving the following quadratic equation, find the value of ‘p’ for which the roots are equal. px² – 4x + 3 = 0.
Solution:
px² – 4x + 3 = 0 …..(i)
Compare (i) with ax² + bx + c = 0
Here a = p, b = -4, c = 3
D = b² – 4ac = (-4)² – 4.p.(3) = 16 – 12p
As roots are equal, D = 0
16 – 12p = 0
⇒ $$\frac { 16 }{ 12 }$$ = p
⇒ p = $$\frac { 4 }{ 3 }$$

Question 20.
Without solving the following quadratic equation, find the value of m for which the given equation has real and equal roots : x² + 2 (m – 1) x + (m + 5) = 0.
Solution:
x² + 2 (m – 1) x + (m + 5) = 0.
Here, a = 1, b = 2 (m – 1), c = m + 5
So, discriminant, D = b² – 4ac
= 4(m – 1)² – 4 x 1 (m + 5)
= 4m² + 4 – 8m – 4m – 20
= 4m² – 12m – 16
For real and equal roots D = 0
So, 4m² – 12m – 16 = 0
⇒ m² – 3m – 4 = 0 (Dividingby4)
⇒ m² – 4m + m – 4 = 0
⇒ m (m – 4) + 1 (m – 4) = 0
⇒ (m – 4) (m + 1) = 0
⇒ m = 4 or m = -1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations Ex 5D are helpful to complete your math homework.

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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 16 Loci (Locus and Its Constructions) Ex 16A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 16 Loci (Locus and Its Constructions) Ex 16A

Other Exercises

Question 1.
Given— PQ is perpendicular bisector of side AB of the triangle ABC

Solution:

Question 2.
Given— CP is bisector of angle C of A ABC.

Prove: p is equidistant from AC and BC
Solution:

Question 3.
Given— AX bisects angle BAG and PQ is perpendicular bisector of AC which meets AX at point Y.

Prove:
(i) X is equidistant from AB and AC.
(ii) Y is equidistant from A and C.
Solution:
Construction: From X, draw XL ⊥ AC and XM ⊥ AB and join YC.

Question 4.
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
Solution:
Given: In Δ ABC, AB, = 4.2 cm, BC = 6.3 cm and AC = 5cm

Steps of Construction:
(i) Draw a line segment BC = 6.3 cm.
(ii) With centre B-and radius 4.2 cirr draw mi are.
(iii) With centre C mid radius 5 cm, draw another arc which intersect the first arc at A.
(iv) Join AB mid AC.
A ABC is the required triangle.
(v) Again with centre B mid C mid radius greater
than $$\frac { -1 }{ 2 }$$ BC, draw arcs which intersects each other at L mid M.
(vi) Join LM intersecting AC at D mid BC at E.
(vii) Join DB.

Question 5.
In each of the given figures: PA = PH and QA = QB.

Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of points equidistant from two given fixed points.
Solution:
(i) Construction: Join PQ which meets AB in D.

Proof:
P is equidistant from A mid B
∴ P lies on die perpendicular bisector of AB similarly Q is equidistant from A mid B.
∴ Q lies on perpendicular bisector of AB P mid Q both lies on the perpendicular bisector of AB.
∴ PQ is Hie perpendicular bisector of AB

Hence locus of die points which are equidistant from two fixed points, is a perpendicular, bisector of die line joining die fixed points.         Q.E.D.

Question 6.
Construct a right angled triangle PQR, in which ∠Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR.
Solution:

Steps of Construction
(i) Draw a line segment QR = 4.5 cm.
(ii) At Q, draw a ray QX making an angle of 90°.
(iii) With centre P mid radius 8 cm, draw mi arc which intersects QX at P.
(iv) Join RP.
A-PQR is the required triangle.
(v) Draw the bisector of ∠PQR which meets PR in T.
(vi) From T, draw perpendicular PL and PM respec- lively on PQ and QR.

Question 7.
Construct a triangle ABC in which angle ABC = 75°. AB = 5 cm and BC = 6.4 cm.
Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisectors intersect each other at point P ; prove that P is equidistant from B and C ; and also from AC and BC.
Solution:

Steps of Construction:
(i) Draw a line segment BC = 6.4 cm.
(ii) At B, draw a ray BX making an angle of 75° with BC and cut off BA = 5 cm.
(iii) Join AC.
Δ ABC is the required triangle.
(iv) Draw the perpendicular bisector of BC.
(v) Draw the angle bisector of ∠ACB which intersects the perpendicular bisector of ,BC at P
(vi) Join PB and draw PL ⊥ AC.

Question 8.
In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B.
Prove that P is equidistant from AB and BC.

Solution:
Given:  In || gm ABCD. AB > BC and bisector of ∠B meets diagonal AC at P.
To Prove:  P is equidistant from AB and BC.
Construction: From P, draw PL ⊥ AB and PM ⊥ BC.

Question 9.
In triangle LMN, bisectors of interior angles at L and N intersect each other at point A.
Prove that:
(i) point A is equidistant from all the three sides of the triangle.
(ii) AM bisects angle LMN.
Solution:

Given: In A LMN, angle bisectors of ∠L and ∠N
meet at A, AM is joined.
To Prove:
(i) A is equidistant from all the sides of A LMN.
(ii) AM is the bisector of ∠M.
Proof: ∴ A lies on the bisector of ∠N
∴ A is equidistant from MN and LN Again
∴ A lies on the bisector of ∠L A is equidistant from LN and LM Hence
∴ A is equidistant from all sides of the triangle LMN.
∴ A lies on the bisector of ∠M                               Q.E.D.

Question 10.
Use ruler and compasses only for this question:
(i) construct ΔABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB. (2010)
Solution:
Steps of construction:
1. Draw a line BC = 6 cm and an angle CBX = 60°. Cut off AB = 3.5 cm. Join AC, ΔABC is the required triangle.
2. Draw ⊥ bisector of BC and bisector of ∠B.
3. Bisector of ∠B meets bisector of BC at P
∴ BP is the required length, where PB = 3.5 cm
4. P is the point which is equidistant from BA and BC, also equidistant fromB and C

Question 11.
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point E Prove that:

(i) F is equidistant from A and B.
(ii) F is equidistant from AB and AC.
Solution:
Given : In the figure,
In ΔABC, AD is the bisector of ∠BAC Which meets BC at D EG is the perpendicular bisector of AB which intersects AD at F
To prove :
(i) F is equidistant from A and B.
(ii) F is equidistant from AB and AC.
Proof:
(i) ∴ F lies on the perpendicular bisector of AB F is equidistant from A and B
(ii) Again,
∴ F lies onthe bisector of ∠BAC
∴ F is equidistant from AB and AC.
(10 cm theorem)
Hence proved.

Question 12.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Solution:
In quadrilateral ABCD, the bisectors of ∠B and ∠C meet each other at P.

To prove : D is equidistant from the sides AB and CD.
Proof:
∴ P lies on the bisector of ∠B P is equidistant from AB and BC ….(i)
Similarly, P lies on the bisector of ∠C P is equidistant from BC and CD ….(ii)
From (i) and (ii),
∴ P is equidistant from AB and CD
Hence proved.

Question 13.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Solution:
Steps of Construction:
(i) Draw a line segment AB = 6 cm
(ii) Draw perpendicular bisector LM of AB. LM is the required locus.
(iii) Take any point on LM say P.
(iv) Join PA and PB
∵ P lies on the right bisector of line AB

∴ P is equidistant from A and B.
∴ PA = PB
∴ Perpendicular bisector of AB is the locus of all points which are equidistant from A and B.

Question 14.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
Solution:

Steps of Construction:
(I) Draw a ray BC.
(ii) Construct a ray RA making an angle of 750 with BC.
(iii) ∴ ∠ABC = 75°
(iv) Draw the angle bisector BP of ∠ABC. BP is the required locus.
(v) Take any point D on BP.
(vi) From D, draw DE ⊥ AB and DF ⊥ BC.
∵ D lies on the angle bisector ∠ABC.
∴ D is equidistant from AB and BC.
∴ DE = DF
Similarly any point on BP, is equidistant from AB and BC.
∴BP is the locus of all points which are equidistant from AB and BC.

Question 15.
Draw an angle ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC ; and also equidistant from A and B
Solution:

Steps of Construction:
(i) Draw a line segment BC = 5 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BA = 4.6 cm.
(iii) Draw the angle bisector of ∠ABC.
(iv) Draw the perpendicular bisector of AB which intersects the angle bisector at P.
P is the required point which is equidistant from AB and BC as well as from points A and B.

Question 16.
In the figure given below, find a point P on CD equidistant from points A and B.

Solution:

Steps of Construction:
(i) In the figure AB and CD are two line segments.
(ii) Draw the perpendicular bisector of AB which intersects CD in P.
P is the required point which is equidistant from A and B
∵ P lies on the perpendicular bisector of AB.
∴ PA = PB.

Question 17.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.

Solution:

Steps of Construction:
(i) In the given triangle, draw the angle bisector of ∠BAC.
(ii) Draw the perpendicular bisector of BC which intersects the angle bisector of ∠A at P.
P is the required point which is equidistant from AB and AC as well as from B and C.
∵ P lies on the angle bisector of ∠BAC.
∴ It is equidistant from AB and AC. Again
∵ P lies on the perpendicular bisector of BC.
∴ P is equidistant from B and C.

Question 18.
Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that :
(i) P is equidistant from B and C.
(ii) P is equidistant from AB and BC.
(iii) Measure and record the length of PB.
(2000)
Solution:
Steps of Construction :
1. Draw a line segment AB = 7 cm.
2. Draw angle ∠ABC = 60° with the help of compass.
3. Cut off BC = 8 cm.
4. Join A and C.
5. The triangle ABC so formed is required triangle.
(i) Draw perpendicular bisector of line BC. The point situated on this line will be equidistant from B and C.
(ii) Draw angular bisector of ∠ABC. Any
point situated on this angular bisector is equidistant from lines AB and BC.

The point which fulfills the condition required in (i) and (ii) is the intersection point of bisector of line BC and angular bisector of ∠ABC.
(iii) Length of PB is 4.5 cm.

Question 19.
On a graph paper, draw the lines x = 3 and y = -5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
Solution:
On the graph paper, draw axis XOX’ and YOY’ Draw a line l, x = 3 which is parallel to y-axis and another line m, y = -5, which is parallel to x-axis

These two lines intersect eachother at P.
Now draw the angle bisector p of ∠P
∵ p is the bisector of ∠P
∴ Any point on P, is equidistant from l and m
∴ This line p is equidistant from l and m.

Question 20.
On a graph paper, draw the line x = 6. Now, on the same graph paper, draw the locus of the point which moves in such a way that its distance from the given line is always equal to 3 units.
Solution:

On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 6
Which is parallel to -axis
Take point P and Q which are at a distance of 3units from the line l
Draw line rn and n from P and Q parallel to P respectively
The line m and n are the required locus of the points P and Q
Which arc always 3 units from the line l.
Hence proved.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 16 Loci (Locus and Its Constructions) Ex 16B  are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations (In one variable) Ex 4B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations Ex 4B.

Other Exercises

Question 1.
Represent the following inequalities on real number lines:
(i) 2x – 1 < 5
(ii) 3x + 1 ≥ – 5
(iii) 2 (2x – 3) ≤ 6
(iv) -4 < x < 4
(v) -2 ≤ x < 5 (vi) 8 ≥ x > -3
(vii) -5 < x ≤ -1
Solution:

Question 2.
For each graph given below, write an inequation taking x as the variable :

Solution:

Question 3.
For the following inequations, graph the solution set on the real number line :
(i) – 4 ≤ 3x – 1 < 8
(ii) x – 1 < 3 – x ≤ 5
Solution:

Question 4.
Represent the solution of each of the following inequalities on the real number line :
(i) 4x – 1 > x + 11
(ii) 7 – x ≤ 2 – 6x
(iii) x + 3 ≤ 2x + 9
(iv) 2 – 3x > 1 – 5x
(v) 1 + x ≥ 5x – 11
(vi) $$\frac { 2x + 5 }{ 2 }$$ > 3x – 3
Solution:

Question 5.
x ∈ {real numbers} and -1 < 3 – 2x ≤ 7, evaluate x and represent it on a number line.
Solution:

Question 6.
List the elements of the solution set of the inequation – 3 < x – 2 ≤ 9 -2x ; x ∈ N.
Solution:

Question 7.
Find the range of values of x which satisfies

Graph these values of x on the number line.
Solution:

Question 8.
Find the values of x, which satisfy the inequation:

Graph the solution on the number line. (2007)
Solution:

Question 9.
Given x ∈ {real numbers}, find the range of values of x for which – 5 ≤ 2x – 3 < x + 2 and represent it on a real number line.
Solution:

Question 10.
If 5x – 3 ≤ 5 + 3x ≤ 4x + 2, express it as a ≤ x ≤ b and then state the values of a and b.
Solution:
Here in, 5x – 3 ≤ 5 + 3x ≤ 4x + 2
⇒ 5x – 3 ≤ 5 + 3x and 5 + 3x ≤ 4x + 2
⇒ 5x – 3x ≤ 5 + 3 and 3x – 4x ≤ 2 – 5
⇒ 2x ≤ 8 and – x ≤ – 3
⇒ x ≤ 4 and x ≥ 3
Solution is 3 ≤ x ≤ 4
a = 3 and b = 4

Question 11.
Solve the following inequation and graph the solution set on the number line :
2x – 3 < x + 2 ≤ 3x + 5; x ∈ R.
Solution:

Question 12.
Solve and graph the solution set of:
(i) 2x – 9 < 7 and 3x + 9 ≤ 25; x ∈ R.
(ii) 2x – 9 ≤ 7 and 3x + 9 > 25; x ∈ I.

(iii) x + 5 ≥ 4 (x – 1) and 3 – 2x < -7; x ∈ R.
Solution:
(i) 2x – 9 < 7 and 3x + 9 ≤ 25; x ∈ R.

Question 13.
Solve and graph the solution set of:
(i) 3x – 2 > 19 or 3 – 2x ≥ – 7; x ∈ R.
(ii) 5 > p – 1 > 2 or 7 ≤ 2p – 1 ≤ 17; p ∈ R.
Solution:

Question 14.
The diagram represents two inequations A and B on real number lines :

(i) Write down A and B in set builder notation.
(ii) Represent A ∩ B and A ∩ B’ on two different number lines.
Solution:

Question 15.
Use real number line to find the range of values of x for which :
(i) x > 3 and 0 < x < 6
(ii) x < 0 and -3 ≤ x < 1
(iii) -1 < x ≤ 6 and -2 ≤ x ≤ 3
Solution:

Question 16.
Illustrate the set {x : -3 ≤ x < 0 or x > 2 ; x ∈ R} on a real number line.
Solution:

Question 17.
Given A = {x : -1 < x < 5, x ∈ R} and B = {x : – 4 < x < 3, x ∈ R}
Represent on different number lines:
(i) A ∩ B
(ii) A’ ∩ B
(iii) A – B
Solution:

Question 18.
P is the solution set of 7x – 2 > 4x + 1 and Q is the solution set of 9x – 45 ≥ 5 (x – 5); where x ∈ R. Represent:
(i) P ∩ Q
(ii) P – Q
(iii) P ∩ Q’ on different number lines.
Solution:

Question 19.
If P = {x : 7x – 4 > 5x + 2, x ∈ R} and Q = {x : x – 19 ≥ 1 – 3x , x ∈ R}: find the range of set P ∩ Q and represent it on a number line.
Solution:

Question 20.
Find the range of values of x, which satisfy:

Graph, in each of the following cases, the values of x on the different real number lines:
(i) x ∈ W
(ii) x ∈ Z
(iii) x ∈ R.
Solution:

Question 21.
Given A = {x : – 8 < 5x + 2 ≤ 17, x ∈ I}, B = {x : -2 ≤ 7 + 3x < 17, x ∈ R}
Where R = {real numbers} and I = {integers}
Represent A an B is on two different number lines. Write down the elements of A ∩ B.
Solution:

Question 22.
Solve the following inequation and represent the solution set on the number line 2x – 5 ≤ 5x + 4< 11, where x ∈ I.
Solution:

Question 23.
Given that x ∈ I, solve the inequation and graph the solution on the number line:

Solution:

Question 24.
Given:
A = {x : 11x – 5 > 7x + 3, x ∈ R} and B = {x : 18x – 9 ≥ 15 + 12x, x ∈ R}.
Find the range of set A ∩ B and represent it on a number line. (2005)
Solution:

Question 25.
Find the set of values of x, satisfying:

Solution:

Question 26.

Solution:

Question 27.
Solve the inequation:

Graph the solution set on the number line.
Solution:

Question 28.
Find three consecutive largest positive integers such that the sum of one-third of first, one-fourth of second and one-fifth of third is atmost 20.
Solution:
Let first positive integer = x
Then, second integer = x + 1
and third integer = x + 2
According to the condition,

Question 29.
Solve the given inequation and graph the solution on the number line.
2y – 3 < y + 1 < 4y + 7; y ∈ R (2008)
Solution:

Question 30.
Solve the inequation:
3z – 5 ≤ z + 3 < 5z – 9; z ∈ R.
Graph the solution set on the number line.
Solution:

Question 31.
Solve the following inequation and represent the solution set on the number line.

Solution:

Question 32.
Solve the following inequation and represent the solution set on the number line:

Solution:

Question 33.
Solve the following inequation, write the solution set and represent it on the number

Solution:

Question 34.

Solution:

Question 35.
Solve the following inequation and write the solution set:
13x – 5 < 15x + 4 < 7x + 12, x ∈ R
Represent the solution on a real number line. (2015)
Solution:

Question 36.
Solve the following inequation, write the solution set and represent it on the number line.
-3 (x – 7) ≥ 15 – 7x > $$\frac { x + 1 }{ 3 }$$, x ∈ R. (2016)
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 4 Linear Inequations Ex 4B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.