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

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8  [Multiplying both sides by 2]

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8   [Adding both sides 1/2 ]

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

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)
  NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8      154 = 2(2x + 2 + x)

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8      [Dividing both sides by 2] 

 NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8       77 =3x-2
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8       77 -2  =3x+2 -2      [Subtracting 2 from both sides]

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8        75 =3x

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8       75/3 = 3x/3  [Dividing both sides by 3]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8        25 =x
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8      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 
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8cm. 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

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8         [Subtracting 4/3 from both the sides]

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8                 [Dividing both sides by 2]

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8
 Hence, each equal side of an isosceles triangle isNCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8cm.

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
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 2x +15 - 15 = 95 - 15 [Subtracting 15 from both sides]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 2x = 80
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 2x/2 = 80/2   [Dividing both sides by 2]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 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

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8   [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
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 3x + 3 = 51
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 3x + 3 - 3 = 51 -3    [Subtracting 3 from both sides]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 3x/3 = 48/3    [Dividing both sides by 3]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 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
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 3x +24 = 888
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 3x + 24- 24 = 888 - 24    [Subtracting 24 from both sides]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 3x/3 = 864/3     [Dividing both sides by 3]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 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
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 2x +3x +3+4x+8 = 74
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 9x +11= 74
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 9x +11-11= 74-11   [Subtracting 11 from both sides]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 9x/9 = 63/9    [Dividing both sides by 9]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 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
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 12x + 8 = 56
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 12x + 8 - 8 = 56-8    [Subtracting 8 from both sides]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 12x = 48    [Dividing both sides by 12]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 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,  NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 5(x+8) = 7x
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 5x + 40 = 7x
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 5x -7x = - 40     [Transposing 7x to L.H.S. and 40 to R.H.S.]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 -2x = -40
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 -2x/2= - 40/-2   [Dividing both sides by -2]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 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
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 3x + 84 =135
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 3x +84 - 84 =135-84     [Subtracting 84 from both sides]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 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

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 4x -x =15  [Transposing x to L.H.S.]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 3x =15
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 3x/3=15/3    [Dividing both sides by 3]
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 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, NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 [Subtracting 2/3 from both sides]

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

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 [Dividing both sides by 60]

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8         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
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 200x +150x + 50x = 4,00,000
NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 400x = 4,00,000

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8 [Dividing both sides by 400]

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

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8[Subtracting 160 from both sides]

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8[Dividing both sides by 7]

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

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8   [Subtracting 1575 from both sides]

NCERT Solution (Ex-2.2) - Chapter 2: Linear Equations in one variable, Maths, Class 8    [Dividing both sides by 7]

Hence the number of winner is 19.

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

1. What is a linear equation in one variable?
Ans. A linear equation in one variable is an equation that can be written in the form ax + b = 0, where a and b are constants and x is the variable. It represents a straight line on a graph and has only one variable with degree 1.
2. What is the solution of a linear equation in one variable?
Ans. The solution of a linear equation in one variable is the value of the variable that makes the equation true. In other words, it is the value of x for which ax + b = 0 is satisfied. The solution is usually denoted by the symbol 'x =' followed by the value of x.
3. How do you solve a linear equation in one variable?
Ans. To solve a linear equation in one variable, we need to isolate the variable on one side of the equation. We can do this by performing the same operation on both sides of the equation. The operations that we can use are addition, subtraction, multiplication, and division. We continue this process until we have the variable on one side and the constant on the other side.
4. What is the significance of linear equations in one variable in real-life situations?
Ans. Linear equations in one variable have a wide range of applications in real-life situations. For example, they can be used to calculate the cost of goods or services, determine the speed of an object, or calculate the distance between two points. They can also be used to solve problems related to finance, such as calculating interest rates or determining the value of investments.
5. What are the different methods to solve a linear equation in one variable?
Ans. There are different methods to solve a linear equation in one variable, such as the method of substitution, the method of elimination, and the graphical method. In the method of substitution, we substitute the value of one variable from one equation into the other equation to find the value of the other variable. In the method of elimination, we eliminate one variable by adding or subtracting the two equations. In the graphical method, we plot the two equations on a graph and find the point of intersection, which gives us the solution of the equations.
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