Important Questions for Class 8 Maths - Linear Equations in One Variable

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Q1: Verify that x = 2 is the solution of the equation 4.4x – 3.8 = 5.

Sol: 4.4x – 3.8 = 5
Putting x = 2, we have
4.4 × 2 – 3.8 = 5
⇒ 8.8 – 3.8 = 5
⇒ 5 = 5
L.H.S. = R.H.S.
Hence, it is verified.

Q2: Linear equation in one variable has ________.
(a) only one variable with any power
(b) only one term with a variable
(c) only one variable with power 1
(d) only constant term
Ans: (c) Only one variable with power 1

Q3: The sum of two numbers is 11, and their difference is 5. Find the numbers.
Sol: Let one of the numbers from the two numbers be x.
Let the other number = 11 – x.
As per the given conditions, we have
x – (11 – x) = 5
⇒ x – 11 + x =5
⇒ 2x – 11 = 5
⇒ 2x = 5 + 11……………… (Transposing 11 to R.H.S.)
⇒ 2x = 16
⇒ x = 8
Hence, the required numbers for the given question are 8 and 11 – 8 = 3, respectively.

Q4: Arpita’s present age is thrice of Shilpa. If Shilpa’s age three years ago was x. Then Arpita’s present age is 
(a) 3(x – 3)
(b) 3x + 3
(c) 3x – 9
(d) 3(x + 3)
Ans: (d)
Sol:
Given:- Shilpa’s age three years ago was x
Then, Shilpa’s present age is= x + 3
Arpita’s present age is thrice of Shilpa’s age =3(x + 3)

Q5: The number of boys and girls in a class is in the ratio of 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?
Sol: Let the number of boys in a class be 7x
Let the number of girls in a class be 5x 
According to the question,
7x = 5x + 8
7x -5x = 8
2x = 8
x = 8/2
x =  4
Therefore, Number of boys = 7 x 4 =28
Number of girls = 5 x 4 =20
Total number of students = 20 + 28= 48

Q6: Find the number whose fifth part increased by 5 is equal to its fourth part diminished by 5.
Sol: Let the number be x.
According to the question, we get
(1/5)x + 5 = (1/4) x – 5
On rearranging the given equation,
(1/5) x – (1/4) x = -5-5
(1/5) x – (1/4)x =-10
By taking L.C.M., we will get,
(4x-5x)/20=-10
Again by transposing
-x = -200
 x = 200

Q7: The sum of three consecutive multiples of 8 is 888. Find the multiple.
Sol: Let the three consecutive multiples be x, x + 8, x + 16
According to the given question
The sum of three consecutive multiples of 8 is 888
x + x + 8 + x + 16 = 888
3x + 24 = 888
3x = 888 – 24
3x = 864
x = 864/3
x = 288
Therefore the three consecutive multiples are:
x =288
x + 8 = 296
x + 16 = 304, respectively. 

Q8: If (5x/3) – 4 = (2x/5), then the numerical value of 2x – 7 is
(a) 19/13
(b) -13/19
(c) 0
(d) 13/19
Ans: (b)
Sol: (B) -13/19
Given :- (5x/3) – 4 = (2x/5)
(5x/3) – (2x/5) = 4
L.C.M. of 3 and 5 is 15
(25x  –  6x)/15 = 4
19x = 4 × 15
19x = 60
X = 60/19
Substituting x=60/19 in the given equation,
= (2 × (60/19)) – 7
= (120/19) – 7
= (120  –  133)/19
= – 13/19

Q9: The angles of a triangle are in the ratio 2 : 3: 4. Find the angles of the triangle.
Sol: Let the angles of the triangle be 2x°, 3x° and 4x°.
From the given question, we get,
2x + 3x + 4x = 180 
∵ The sum of all the angles of a triangle is 180°)
⇒ 9x = 180
⇒ x = 20……….. (Transposing 9 to R.H.S.)
Hence, The angles of the given triangle are 
2× 20 = 40°,
3 × 20 = 60°,
4 × 20 = 80°.

Q10: If 8x – 3 +17x, then x ________.
(a) is a fraction
(b) is an integer
(c) is a rational number
(d) cannot be solved 
Ans: (c)
Sol: Given:- 8x - 3 = 25 + 17x
Moving -3 to R.H.S. and becomes 3 and 17x to L.H.S. 
We obtain,
8x – 17x = 25 +3 
-9x = 28
x = -28/9
Thus, x is a rational number.

Q11: Find three consecutive odd numbers whose sum is 147.
Sol: Let the first, second, and third consecutive odd numbers be (2x +1),(2x + 3) and (2x +  5), respectively.
Hence the sum of the consecutive odd numbers is
 (2x + 1) + (2x + 3) + (2x + 5) = 147.
On further simplifying, we get
2x + 2x + 2x + 1 + 3 + 5 = 147
 6x + 9 = 147.
On rearranging, we obtain
6x = 147 – 9
6x = 138
X =  138/6 = 23,
So the three consecutive odd numbers are (2x + 1)= 47
(2x + 3) = 49
(2x + 5) = 51.  

Q12: The perimeter of a rectangular swimming pool is 154m. Its length is 2m, more than twice its breadth. What is the length and the breadth of the pool? 
Sol: Let the breadth of the swimming pool be x m. 
The length of the swimming pool will be = (2x + 2) m. 
Perimeter of swimming pool:- 2 (l + b) =154
2 (2x + 2 + x)=154
2 (3x + 2)=154
∴ Dividing both sides by 2, we obtain
(3x + 2)=77
On transporting two on the R.H.S., we get
3x = 77 – 2
3x = 75
x = 75/3
x = 25 m
Hence, the breadth of the swimming pool is x = 25m 
The length of the swimming pool will be= (2x + 2) m. 
= (2 х 25 + 2) m
= (50 + 2) m
= 52 m
Thus, the length of the swimming pool is 52m, and the breadth of the swimming pool is 25m. 

Q13: What is the share of A when Rs 25 are divided between A and B so that A gets Rs 8 more than B is 16.5?
Sol: Let the share of B be x.
Let the share of A be (x + 8).
From this, we get,
x + x + 8 = 25
2x = 25 – 8
2x= 17
x = 17/2
x = 8.5
Therefore, A’s share will be 8.5.

Q14: Ram’s father is 26 years younger than Ram’s grandfather and 29 years older than Ram. The sum of the ages of all three is 135 years. What is the age of each one of them?
Sol: Let Ram’s present age be x years
Rams father’s present age is = (x + 29) years
Rams grandfather’s present age =(x + 29 + 26) years
The sum of all three ages adds up to 135 years
Hence,
x + (x + 29) + (x + 29 + 26) = 135
x + x + x + 29 + 29 + 26 = 135
3x + 84= 135
3x = 135-84
3x = 51
x = 51/3
x = 17
Hence, Ram’s present age is x=17 years
Ram’s father’s present age =(x + 29)
= (17 + 29)
= 46 years
Ram’s grandfather’s age =(x + 29 + 26)
= (17 + 29 + 26)
= 72 years 

Q15: 3x+  2/3 = 2x + 1
Sol: 3x + 2/3 = 2x + 1
By transposing the above equation, we get
3x + 2 = 3(2x + 1)
3x + 2 = 6x + 3
By moving all the variables on the L.H.S., we get,
3x - 6x = 3-2
-3x = 1
x = -1/3