NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 PDF Download

Linear Equations in one Variable

Exercise 2.2

Question 1:
 If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8. What is the number?

Answer 1:

Let the number be x.
According to the question,

  [Multiplying both sides by 2]

   [Adding both sides 1/2 ]

Hence, the required number is 3/4.

Question 2:
 The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and breadth?
Answer 2:
Let the breadth of the pool be x m.
Then, the length of the pool = (2x + 2) m
Perimeter = 2(l + b)
        154 = 2(2x + 2 + x)

      [Dividing both sides by 2] 

        77 =3x-2
       77 -2  =3x+2 -2      [Subtracting 2 from both sides]

        75 =3x

       75/3 = 3x/3  [Dividing both sides by 3]
        25 =x
      x =25 m

Length of the pool = 2x +2 =  2 x 25 + 2 = 50 + 2 = 52 m
Breadth of the pool = 25 m
Hence, the length of the pool is 52 m and breadth is 25 m.

Question 3:
The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is cm. What is the length of either of the remaining equal sides?
Answer 3:
Let each of equal sides of an isosceles triangle be x cm.
Perimeter of a triangle = Sum of all three sides

         [Subtracting 4/3 from both the sides]

                 [Dividing both sides by 2]

 Hence, each equal side of an isosceles triangle iscm.

Question 4:
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Answer 4:
Sum of two number = 95
Let the first number be x, then another number be x +15.
According to the question,
x + x+15 = 95

2x +15 = 95
 2x +15 - 15 = 95 - 15 [Subtracting 15 from both sides]
 2x = 80
 2x/2 = 80/2   [Dividing both sides by 2]
 x=40
So, the first number = 40 and another number = 40 + 15 = 55
Hence, the two numbers are 40 and 55.

Question 5:
Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Answer 5:
Let the two numbers be 5x and 3x.
According to question,
5x - 3x =18

   [Dividing both sides by 2]

Hence, first number = 5 x 9 = 45 and second number = 3 x 9 = 27.

Question 6:
Three consecutive integers add up to 51. What are these integers?
Answer 6:
Let the three consecutive integers be x, x +1 and x + 2.
According to the question,
x + x +1+ x + 2 = 51
 3x + 3 = 51
 3x + 3 - 3 = 51 -3    [Subtracting 3 from both sides]
 3x/3 = 48/3    [Dividing both sides by 3]
 x =16
Hence, first integer = 16, second integer = 16 + 1 = 17 and third integer = 16 + 2 = 18. 

Question 7:
The sum of three consecutive multiples of 8 is 888. Find the multiples.
Answer 7:
Let the three consecutive multiples of 8 be x, x + 8 and x + 16.
According to question, x + x + 8 + x +16 = 888
 3x +24 = 888
 3x + 24- 24 = 888 - 24    [Subtracting 24 from both sides]
 3x/3 = 864/3     [Dividing both sides by 3]
 x = 288
Hence, first multiple of 8 = 288, second multiple of 8 = 288 + 8 = 296 and third multiple of 8 = 288 + 16 = 304.

Question 8:
Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Answer 8:
Let the three consecutive integers be x, x + 1 and x + 2.
According to the question, 2x + 3( x +1) + 4( x + 2)= 74
 2x +3x +3+4x+8 = 74
 9x +11= 74
 9x +11-11= 74-11   [Subtracting 11 from both sides]
 9x/9 = 63/9    [Dividing both sides by 9]
 x = 7
Hence first integer = 7, second integer = 7 + 1 = 8 and third integer = 7 + 2 = 9.

Question 9:
 The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?
Answer 9:
Let the present ages of Rahul and Haroon be 5x years and 7x years respectively.
According to condition, (5x+4)(7x+4) =56
 12x + 8 = 56
 12x + 8 - 8 = 56-8    [Subtracting 8 from both sides]
 12x = 48    [Dividing both sides by 12]
 x = 4
Hence, present age of Rahul = 5 x 4 = 20 years and present age of Haroon = 7 x 4 = 28 years.

Question 10:
The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength?
Answer 10:
Let the number of girls be x.
Then, the number of boys = x +8.
According to the question,  

 5(x+8) = 7x
 5x + 40 = 7x
 5x -7x = - 40     [Transposing 7x to L.H.S. and 40 to R.H.S.]
 -2x = -40
 -2x/2= - 40/-2   [Dividing both sides by -2]
 x = 20
Hence the number of girls = 20 and number of boys = 20 + 8 = 28.

Question 11:
Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Answer 11:
Let Baichung’s age be x years, then Baichung’s father’s age = (x + 29) years and
Baichung’s granddaughter’s age = (x+29+26) = (x+55) years.
According to condition, x + x + 29 + x +55 =135
 3x + 84 =135
 3x +84 - 84 =135-84     [Subtracting 84 from both sides]
 3x/3 = 51/3          [Dividing both sides by 3]
x =17 years
Hence, Baichung’s age = 17 years, Baichung’s father’s age = 17 + 29 = 46 years and
Baichung’s granddaughter’s age = 17 + 29 + 26 = 72 years.

Question 12:
Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Answer 12:
Let Ravi’s present age be x years.
After fifteen years, Ravi’s age = 4x years.
Fifteen years from now, Ravi’s age = (x +15) years.
According to question, 4x = x +15

 4x -x =15  [Transposing x to L.H.S.]
 3x =15
 3x/3=15/3    [Dividing both sides by 3]
 x=5

Hence, Ravi’s present age be 5 years.

Question 13:
A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get -7/12. What is the number?
Answer 13:
Let the rational number be x.
According to the question,

 [Subtracting 2/3 from both sides]

 [Dividing both sides by 60]

         x=-1/2

Hence, the rational number is -1/2.

Question 14:
Lakshmi is a cashier in a bank. She has currency notes of denominations Rs.100, Rs.50 and Rs.10 respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is Rs.4,00,000. How many notes of each denomination does she have?
Answer 14:
Let number of notes be 2x, 3x and 5x.
According to question, 100 * 2x + 50* 3x +10*5x =  4,00,000
 200x +150x + 50x = 4,00,000
 400x = 4,00,000

 [Dividing both sides by 400]

 x= 1000

Hence, number of denominations of Rs.100 notes = 2 x 1000 = 2000
Number of denominations of Rs.50 notes = 3 x 1000 = 3000
Number of denominations of Rs.10 notes = 5 x 1000 = 5000
Therefore, required denominations of notes of Rs.100, Rs.50 and Rs.10 are 2000, 3000 and 5000 respectively.            

Question 15:
 I have a total of Rs.300 in coins of denomination Rs.1, Rs.2 and Rs.5. The number of Rs.2 coins is 3 times the number of Rs.5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Answer 15:
Total sum of money = Rs.300
Let the number of Rs.5 coins be x, number of Rs.2 coins be 3x and number of Rs.1 coins
be 160-(x+3x) =160-4x.

[Subtracting 160 from both sides]

[Dividing both sides by 7]

         x=20
Hence, the number of coins of Rs.5 denomination = 20
Number of coins of Rs. 2 denomination = 3 x 20 = 60
Number of coins of Rs.1 denomination = 160 – 4 x 20 = 160 – 80 = 80

Question 16:
 The organizers of an essay competition decide that a winner in the competition gets a prize of Rs.100 and a participant who does not win, gets a prize of Rs.25. The total prize money distributed is Rs.3,000. Find the number of participants is 63.
Answer 16:
Total sum of money = Rs.3000
Let the number of winners of Rs.100 be x.
And those who are not winners = 63- x
According to the question, 100*x+25*(63-x) =3000

   [Subtracting 1575 from both sides]

    [Dividing both sides by 7]

Hence the number of winner is 19.