Linear Equations in One Variable Class 8 MCQ with Answers

Answer: (a) 25

Explanation: Let the number be x

Now, Three decreased from a number means x – 3

So, x – 3 = 22

x = 22 + 3

x = 25

Hence, the required number is 25

Answer: (a) 16

Explanation: Let the number be x

Number subtracted from 32 means 32 – x

Twice the number = 2x

According to the given condition,

2x – ( 32 – x ) = 16

2x – 32 + x = 16

3x = 16 + 32

Or, 3x = 48

x = 48/3

x = 16

Hence, the required number is 16

Answer: (a) 4

Explanation: Let the number be x

Then, 15 times the number is 15x

Now 24 is subtracted from 15 times the number, i.e., 24 is subtracted form 15x , which mean 15x – 24

The result of such equation is 36

So, we can write the equation as

15x – 24 = 36

15x = 36 + 24

x = 60

x = 60/15

x = 4

Hence, the number is 4

Answer: (c) 4

Explanation: Let the number be x

Then, 11 times the number is 11x

Now 30 is subtracted from 11 times the number, i.e., 30 is subtracted form 11x , which mean 11x – 30

The result of such equation is 14

So, we can write the equation as

11x – 30 = 14

11x = 14 + 30

x = 44

x = 44/11

x = 4

Hence, the number is 4

Answer: (b) 67 , 68 , 69

Explanation: Let the first number be x

Then, the second number is x+1

Third number is x+2

According to the given condition:

x + ( x+1 ) + ( x+2 ) = 204

3x + 3 = 204

3x = 204 – 3

Or, 3x = 201

x = 201/3

x = 67

First number = x = 67

Second number = x+1 = 67 + 1 = 68

Third number = x+2 = 67 + 2 = 69

Hence, the required numbers are: 67 , 68 , 69

Answer: (c) 62000

Explanation: Let the income on which tax was paid be ₹ a

Tax on ₹ 1 = ₹ 0.05

Tax on ₹ a = ₹ 0.05 x a

According to the given condition:

Tax on income of Rs. a = 3100

0.05 x a = 3100

a = ₹ 3100/0.05

a = ₹ 62000

Hence, the income earned by the man on which he paid tax of ₹ 3100 = ₹ 62000

Answer: (a) 27, 29, 31

Explanation: We know that, consecutive odd numbers differ by 2

Let the smallest, odd natural number be a

Therefore, The other two odd natural numbers are a , a+2 and a+4

Given, the sum of there three consicutive odd numbers is equal to 87

So, a + ( a + 2 ) + ( a + 4 )

3a + 6 = 87

3a = 87 – 6

3a = 81

a = 81/3

a = 27

a+2 = 27 + 2 = 29

a+4 = 27 + 4 = 31

Hence, the required numbers are 27 , 29 , 31

Answer: (b) 58 and 60

Explanation: Let the first number be 2a

Then, the second number is 2a + 2 , As even number differ by 2

According to the given condition:

2a + ( 2a + 2 ) = 118

4a + 2 = 118

4a = 118 – 2

4a = 116

a = 116/4

a = 29

Hence, The first number is 2a

= 2 x 29

= 58

The second number is 2a + 2

= 2 x 29 + 2

= 60

Answer: (a) 59 and 61

Explanation: Let the first number be 2x + 1

The second number be 2x + 3 , As odd number differ by 2

According to the given problem,

( 2x + 1 ) + ( 2x + 3 ) = 120

4x + 4 = 120

4x = 120 – 4

4x = 116

x = 1164

x = 29

Hence, the first number = 2x + 1

= 2 x 29 + 1

= 59

Second number = 2x + 3

= 2 x 29 + 3

= 61

Answer: (a) 20

Explanation: Let the required number be = x

when 10 is added to twice of it, we get,

= 2 x x + 10

Since, the sum is equal to 50

therefore, 2 x x + 10 = 50

2x = 50 – 10

2x = 40

x = 20

Hence, the required number is 20

Answer: (b) 6

Explanation: Let the number be a

Eight times the number diminished by 4 means (8a−4)

We are given that the result of this equation ( 8a – 4 ) is equal to 44

Therefore, the equation will be

( 8a – 4 ) = 44

Moving all constants to the other side of the equation, we get

8a = 44 + 4

a = 48/8

a = 6

Hence, the required number is = 6

Answer: (c) 15

Explanation: Let the number be a

Then, Twice a number means 2 x a or 2a

If Twice a number is increased by 15 , it would mean 2a is increased by 15

Or,

2a + 15

We know it is equal to 45 . So we can write the equation as

2a + 15 = 45

Moving all constants to other side of the equation, we get

2a = 45 – 15

2a = 30

a = 30/2

a = 15

Therefore, the number is 15

Answer: (b) 35, 121

Explanation: Let one of the numbers be y

then the Second number = 80 + y (Since the difference of both the numbers is 80 , if one number is y the other would be 80 + y )

We are also given

(80+y)/y = 25/9

9( 80 + y ) = 25y

720 + 9y = 25y

720 = 25y – 9y

720 = 16y

or 16y = 720

y = 720/16

y = 45

Hence, First number = y

= 45

Second number = 80 + y

= 80 + 45

= 125

Answer: (c) 90

Explanation: Let the number be a

Then, One-ninth of a number is = a/9

One-fifteenth of a number is = a/15

According to the given condition,

(a/9) – (a/15) = 4

Solving LHS of the given equation,

L.C.M of 9 & 15 is 45

(5a−3a)/45 = 4

2a = 4 x 45

a = 180 / 2

Hence, the number is 90

Answer: (b) 330

Explanation: Let the number be x

according to the given condition:

435 – x = x – 225

Or, 435 + 225 = 2x

660 = 2x

Or, x = 660/2

x = 330

Answer: (c) 15

Explanation: 40 – 5( a – 12 ) = a + 10

40 – 5a + 60 = a + 10

100 – 5a = a + 10

Taking like terms on the same side

100 – 10 = a + 5a

90 = 6a

or, 6a = 90

a = 90/6

a = 15

Answer: (c) y = 30

Explanation: (y+4)/9 – (y−8)/6 = 1/3

LCM of 9 & 6 is 18

(6(y+4)−9(y−8))/18 = 1/3

(6y+24−9y+72)/18 = 1/3

(−3y+96)/18 = 1/3

3( -3y + 96 ) = 1 x 18

-9y + 288 = 18

-9y = 18 – 288

y = −270/−9

y = 30

Answer: (c) 150 , 151 , 152 , 153

Explanation: Let the required positive consecutive integers be a , a+1 , a+2 and a+3

Now, smallest number is a

Greatest number is a + 3

According to the given problem:

1/8 of Smallest Number – (1/12) of ( Largest Number ) = 6

1/8 of a – (1/12) of ( a+3 ) = 6

1/8 x a – (1/12) x ( a+3 ) = 6

L.C.M of 8 and 12 is 24

(3×a−2×(a+3))/24 = 6

(3a−2a−6)/24 = 6

a – 6 = 6 x 24

a – 6 = 144

a = 144 + 6

a = 150

Therefore, First integer is = a = 150

Second integer is a+1 = 150 + 1

= 151

Third integer is a+2 = 150 + 2

= 152

Fourth integer is a+3 = 150 + 3

= 153

Hence, the four positive consecutive integers are : 150 , 151 , 152 , 153

Answer: (b) 69˚, 21˚, 90˚

Explanation: Let the two angles of a triangle be 23y˚ and 7y˚ (respectively)

Third angle = 23y˚ + 7y˚

= 30y˚

Since the sum of all the angles of a triangle = 180˚

23y + 7y + 30y = 180

60y = 180

y = 180/60

y = 3 ˚

Ist angle = 23y

= 23 x 3

= 69 ˚

Iind angle = 7y

= 7 x 3

= 21 ˚

IIIrd angle = 30y

= 30 x 3

= 90 ˚

Answer: (c) 54 , 26

Explanation: Let first number be y

Then the Second number would be = 80 – y (Since the sum of both the numbers is 80 , if one number is y the other would be 80 – y )

We are also given

y/(80−y) =7/3

3y = 7 x ( 80 – y )

3y = 560 – 7y

3y + 7y = 560

10 y = 560

y = 560/10

y = 56

Hence, the First number = y

= 56

Second number = 80 – y

= 80 – 56

= 24

Answer: (b) Lalit’s age = 16 years , Anuj’s age = 20 years

Explanation: Let Lalit’s age be = x years

Then, Anuj’s age = ( Lalit’s age be + 4 ) years

Or

= ( x + 4 ) years

We are given that Anuj’s age + Lalit’s age = 36 years.

i.e.,

x + ( x + 4 ) = 36

2x + 4 = 36

2x = 36 – 4

2x = 32

x = 32/2

x = 16

Therefore, Lalit’s age = x years = 16 years

Anuj’s age = ( x + 4 ) years

= ( 16 + 4 ) years

= 20 years

Answer: (c) 20 and 80

Explanation: Let one part be a

Then, another part would be 100 – a (Since the sum of both the parts is 100 , if one part is a the other would be 100 – a )

According to the given conditions

a = 1/4 ( 100 – a )

a = (1/4) x 100 – (1/4) x a

(a/1) + (1a/4) = 25

(4a+1a)/4 = 25

5a/4 = 25

a = 25 x (4/5)

a = 20

Hence, one of the part a = 20

Other part is 100 – a

= 100 – 20

= 80