# Free Linear Equations in One Variable 02 Practice Test - 8th Grade

### Question 1

28 is divided into two parts** **in such a way that 65 of one part is equal to 23 of the other. What would be the smaller part?

8

9

10

18

#### SOLUTION

Solution :C

Let one part be x.

Then, other part = 28−x

Given that

65 of one part = 23 of the other

⇒ 65x = 23(28−x)

⇒ (6×32)x = 5(28−x)

⇒ 9x=5(28−x)

⇒ 9x=140−5x

⇒ 14x=140

⇒ x=10The other part would be 28 - 10 = 18

Hence, the smaller part is 10.

### Question 2

The sum of the digits of a two-digit number is 7. If the number formed by interchanging the digits is less than the original number by 27, then find the original number.

43

61

52

25

#### SOLUTION

Solution :C

Given, the sum of the digits of a two-digit number is 7.

Let the units digit of the original number be x.

Then, the tens digit of the original number is 7−x.

The two-digit number =10(7−x)+(1×x)

=70−10x+x=70−9xWhen the digits are interchanged, the new number is 10(x)+1(7−x)=10x−x+7=9x+7

New number = Original number - 279x+7=70−9x−2718x=36x=2

The tens digit =7−x=7−2=5

∴The original number is 52.

### Question 3

The value of the variable after solving the equation 2x + 3 = 3x + 2 is

#### SOLUTION

Solution :In the given equation, x is the variable (whose value has to be found)

2x + 3 = 3x + 23x - 2x = 3 - 2

x = 1

### Question 4

The denominator of a rational number is greater than its numerator by 3. If the numerator is increased by 7 and the denominator is decreased by 1, then the new number becomes 32. Find the original number.

811

58

710

25

#### SOLUTION

Solution :A

Let the numerator of a rational number be x.

Then the denominator of a rational number would be (x+3).

When the numerator is increased by 7, it becomes (x+7)

When the denominator is decreased by 1, it becomes

⇒x+3−1=x+2

The new number formed =32

So, (x+7)(x+2)=32 2x+14=3x+6 3x−2x=14−6 x=8

∴ Numerator, x=8

Denominator, x+3=8+3=11

Hence, the original rational number is 811

### Question 5

If the sum of three consecutive multiples of 4 is 444, then these multiples are 144, 148, 152.

True

False

#### SOLUTION

Solution :A

Let us assume the first multiple of 4 as 4x. Hence, the next two multiples are 4x + 4 and 4x + 8

According to the question,

4x+(4x+4)+(4x+8)=444

⇒12x+12=444

⇒4(3x+3)=4(111)⇒3x+3=111

⇒3x=108

⇒ x = 36

⟹4x=4×36=144,

4x+4=144+4=148 and

4x+8=144+8=152Therefore, the multiples are 144, 148 and 152.

### Question 6

Solve for x:

x+1x+4 =23

3

4

5

6

#### SOLUTION

Solution :C

Given,

x+1x+4 =23

On cross multiplication, we get,

3×(x+1)=2×(x+4)⇒3x+3=2x+8

Transposing 2x from RHS to LHS and 3 from LHS to RHS, we get,

3x−2x=8−3x=5

### Question 7

Which of the following equation(s) is/are linear equation(s) in one variable?

2x + 3y = 10

4a + 3b = 3b + 6

2x = 3x + 4

2x + 4z = 5

#### SOLUTION

Solution :B and C

If the degree of an equation is one, then that equation is called a linear equation. In all the equations given, the degree of each equation is one as the highest exponent of variables in a term is one.

Equations which have only one variable or which can be reduced to equations having only one variable are known as linear equations in one variable.

Since, 4a + 3b = 3b + 6 can be reduced to a linear equation in variable "a" by subtracting "3b" from both sides, this is a linear equation in one variable.

Also, 2x = 3x + 4 is a linear equation in one variable as it has only one variable, x.

The other equations are also linear but they have two variables, so they are not linear equations in one variable.

### Question 8

The present age of Pratik and Arjun are in the ratio 2 : 3. After 4 years, the ratio of their ages will be 4 : 5. The present age of Pratik would be

#### SOLUTION

Solution :Let the present age of Pratik be 2x and Arjun be 3x.

4 years from now, Pratik 's age = 2x + 4 and Arjun 's age = 3x + 4

Given that, 2x+43x+4 = 45

⟹10x+20=12x+16

Thus, 2x=4⟹x=2

Pratik's present age is 2x=4.

### Question 9

A total of ₹10,000 is distributed among 150 persons as gifts. If a gift is either of ₹50 or ₹100 denominations, then the total number of gifts worth ₹50 is 50.

#### SOLUTION

Solution :B

Total number of gifts = 150

Let the number of gifts worth ₹50 be x

Then the number of gifts worth ₹100 would be (150 - x)

Amount spent on x gifts of ₹ 50 = ₹50x

Amount spent on (150 - x) gifts of ₹100 = ₹100(150 - x)

So, 50x + 100 (150 - x) = 10000

50x + 15000 - 100x = 10000

50x = 15000 - 10000 = 5000

x = 100

The total number of ₹50 gifts = x = 100 and hence the given statement is False.

### Question 10

The value of x will be

.

#### SOLUTION

Solution :The equation is, 7x - 2 = 2x + 8

or 7x-2x = 8+2

or 5x = 10

x = 2