# Free Linear Equations in One Variable 03 Practice Test - 8th Grade

### Question 1

The solution of the equation x+55x+8 = 1138 is

#### SOLUTION

Solution :x+55x+8 = 1138

38x + 190 = 55x + 88

55x -38x = 190 - 88

17x = 102

x = 10217 = 6

### Question 2

If 53x = 2527, then x lies between 1 and 2.

True

False

#### SOLUTION

Solution :A

53x = 2527

By cross multiplication, we get

5 × 27 = 3x×25

⇒3x=5×2725

⇒x=5×273×25⇒x = 95

= 1.8, which lies between 1 and 2.

### Question 3

The age of Sunil and his father are in the ratio 3:7. Four years later the sum of their ages will be 78 years. Which of the following represents the age of either Sunil or his father?

18 years

42 years

21 years

49 years

#### SOLUTION

Solution :C and D

Let the age of Sunil be 3x.

Then, the age of Sunil's father would be 7x.Four years later,

Sunil's age = 3x+4 and

Sunil's father's age = 7x+4.

Then, the sum of their ages four years later =3x+4+7x+4=10x+8.

Given that, 10x+8=78.

⇒10x=70 ⇒x=7∴ Present age of Sunil = 3 × 7

= 21 years

Present age of Sunil's father = 7 × 7

= 49 years

### Question 4

The value of x satisfying the equation x+94+2x+35=10 is 11.

True

False

#### SOLUTION

Solution :A

x+94+2x+35=10 is the given equation.

We can check if x=11 is a solution by substituting this value in the LHS to check whether it is equal to RHS or not(11+9)4+(22+3)5

= 5 + 5

= 10

Thus we see that x=11 makes LHS = RHS and is the solution of the given equation.

### Question 5

It is given that "b" has a linear dependency on "a" and "a" has a linear dependency on "c". The equation 3a + 2b + 4c = 5 can be reduced into a linear equation.

#### SOLUTION

Solution :A

Linear dependency signifies that "a","b"and "c" are linearly related, which means that if one of them can be found, the other two can be found.

Hence, the given equation can be reduced to a linear equation if the exact relation between "a","b"and "c" is given.

For a better understanding, let us express the dependencies as equations.

Since b is linearly dependent on a, we can write b = ka + p, where k and p are some constants. Note that in this equation, b and a are the only variables, whereas k and p are known.

Similarly, a can be written as a = gc + u, where g and u are constants (known values).

Substituting these values of a and b in the given equation, we get,

3a + 2b + 4c = 5

3(gc + u) + 2(ka + p) + 4c = 5

3gc + 3u + 2p + 2k(gc + u) + 4c = 5

(3g + 2kg + 4)c + (3u + 2p + 2ku - 5) = 0.

Note that in the final equation, the only c is unknown, the others are all constants and hence is a linear equation in one variable.

### Question 6

Solve for x:

x+56−x+19=x+34

−27

−17

7

−57

#### SOLUTION

Solution :B

Given,

x+56−x+19=x+34Multiplying both sides by 36 (LCM of 4, 6 and 9), we get,

(x+56×36)−(x+19×36)=(x+34×36)

⇒6(x+5)–4(x+1)=9(x+3)

⇒6x+30–4x−4=9x+27

⇒2x+26=9x+27

Transposing 9x from RHS to LHS, and 26 from LHS to RHS, we get,

2x−9x=27−26

⇒−7x=1

Dividing both sides by -7, we get,

x=−17

### Question 7

Pawan is a cashier in a bank. He has currency notes of denominations 1000, 500 and 100 respectively. The ratio of the number of these notes is 2:3:5. If the total cash with Pawan is ** **₹5,00,000, then how many notes of ** **₹1000 denomination does he have?

250

375

625

125

#### SOLUTION

Solution :A

Given that the denomination notes of 1000, 500, 100 are in the ratio 2:3:5.Number of 1000 denomination notes with Pawan = 2x

Therefore, amount = ₹ 2000xNumber of 500 denomination notes with Pawan = 3x

Therefore, amount = ₹ 1500xNumber of 100 denomination notes with Pawan = 5x

Therefore, amount = ₹ 500xTotal amount = ₹ 5,00,000 (Given)

2000x+1500x+500x=₹5,00,0004000x=5,00,000

x=5,00,0004,000=125

Therefore, number of 1000 denomination notes = 2x=2×125

= 250 notes.

### Question 8

The total number of variables in the equation 3a + b + 6ab = 0 is 3. State True or False.

True

#### SOLUTION

Solution :B

A variable is an unknown quantity whose value is not constant.

In the given equation 3a + b + 6ab = 0, "a" and "b" are the variables. "ab" is not considered as a separate variable because, if "a" and "b" are individually found, "ab" can be found. Hence, there are 2 variables in the equation and not 3.

### Question 9

Solve the equation 3x+1=65x−1.

1

2

3

9

#### SOLUTION

Solution :A

3x+1 = 65x−1

On cross multiplication, we get

⇒3(5x−1)=6(x+1)⇒ 15x−3=6x+6⇒ 9x=9⇒ x=1

### Question 10

If you subtract 14 from a number and multiply the result by 12, you get 14. What is the number?

14

24

34

1

#### SOLUTION

Solution :C

Let the number be x.

'Subract ¼ from the number' is mathematically written as 'x - ¼'.

Multiplying the above obtained result by ½ is mathematically shown as:

'(x - ¼) × ½'.

According to the question,

(x−14)×12=14

⇒ x−14=2×14

(by multiplying both the sides by 2)

⇒ x−14=12

⇒ x = 12+14

(by transposing ¼ from LHS to the RHS)

⇒ x = 34

So, the required number is 34.