# Free Linear Equations in One Variable 01 Practice Test - 8th Grade

### Question 1

Which of the following are equations?

4b = 6

24x - 36 = 0

a = b

#### SOLUTION

Solution :B, C, and D

An algebraic expression consists of operands (variables and numbers) and operators (+, −, ×, ÷).

When an algebraic expression is equated to another algebraic expression (which may be even 0), we get an algebraic equation.

So, 3x - 5y + 6z is not an equation, but an expression.

### Question 2

If a sequence follows the pattern 2.5n−10, then find the 10th term of the sequence.

5

10

15

20

#### SOLUTION

Solution :C

By substituting n = 10 in the given equation, we can find out the 10th term i.e.,

2.5x−10 = (2.5 × 10) - 10

= 25 - 10

= 15

### Question 3

The perimeter of a rectangular swimming pool is 64 m. If its length and breadth are in the ratio of 5 : 3, then find the length of the pool (in metre).

10

20

30

40

#### SOLUTION

Solution :B

Perimeter of swimming pool =64 (Given)

Perimeter of rectangle =2(length+breadth)

Since length and breadth are given to be in the ratio 5 : 3, we can assume them to be 5x and 3x respectively.

⇒ Length = 5x and breadth = 3x

Substituting these in the formula of perimeter, we get

64=2(5x+3x)⇒ 8x=32⇒ x=4∴ Length of the swimming pool

=5x=5×4=20 m

### Question 4

The solution of the equation 37 + x = 177 is 2.

True

False

#### SOLUTION

Solution :A

37 + x = 177

x = (177)–(37) = 147 = 2

Hence, the solution of the equation 37 + x = 177 is 2.

Hence the given statement is true.

### Question 5

Anita thinks of a number and subtracts 32 from it. She multiplies the result by 8. The result now obtained is 5 times the same number she thought of. What is the number she thought of ?

4

6

8

10

#### SOLUTION

Solution :A

Let, the number which Anita initially thought of be x.

She subtracts 32 from x and then multiplies it by 8.

⇒(x –32)×8

According to the question, (x –32)×8=5×x8x –12=5x8x –5x=123x=12x=123x=4

Thus, the number that Anita initially thought of is 4.

### Question 6

The digit at tens place of a two digit number is two times the digit at its units place. If the digits are interchanged and added to the original number, then the result will be 66. Find the original number.

42

63

84

21

#### SOLUTION

Solution :A

Let the digit in units place be x.

Then, the digit in tens place will be 2x.So, the number will be

(10×2x)+(1×x)=20x+x=21xWhen we interchange the digits, the number will be

(10×x)+(1×2x)=10x+2x=12xIt is given that,

21x+12x=66

⇒ 33x=66

⇒ x=6633

⇒ x=2

∴The digit in units place =2

The digit in tens place =4

The original number is 42.

Verification:

Original number =42

Interchanging the digits of 42, we get 24.

Sum of the original number and the interchanged number is 42+24=66.

### Question 7

The present age of Vinay’s mother is three times the present age of Vinay. After 5 years, sum of their ages will be 70 years. Find their present ages.

10 years, 30 years

15 years, 45 years

20 years, 60 years

18 years, 54 years

#### SOLUTION

Solution :B

Let Vinay’s present age be xThen, his mother’s present age =3x

After 5 years,Vinay’s age = x+5Vinay's mother’s age = 3x+5

Sum of Vinay and Vinay's Mother's age after 5 years = 70

⇒x+5+3x+5=70⇒4x+10=70⇒4x=60⇒x=15⇒3x=45∴Vinay’s age is 15 years and his mother’s age is 45 years.

### Question 8

Twenty years from now, Arjun will become three times as old as he is now. His present age is

#### SOLUTION

Solution :Let Arjun be x years old now

After 20 years, his age will be x + 20

According to question,

x + 20 = 3x or 3x = x + 20

3x - x = 20

2x = 20

x = 10

Therefore, Arjun's present age is 10 yrs.

### Question 9

Half the students of a class are dancing and three-fourths of the remaining are playing cricket. If the remaining 9 students are studying in the library, then find the total number of students in the class.

72

80

84

96

#### SOLUTION

Solution :A

Let the total number of students be x

According to the question,

Number of students dancing =x2

Number of students remaining =x−x2

=x2

Students playing cricket

=(34)(x2)

=3x8Remaining students =9

⇒ (x−x2)−3x8=9⇒ x8=9⇒ x=8×9⇒ x=72

Hence, the total number of students in the class = 72

### Question 10

The altitude of a triangle is three-fifth the length of its corresponding base. If the altitude is increased by 4 cm and the base is decreased by 2 cm, then the area of the triangle remains the same. Find the base of the triangle in cm.

357

257

267

367

#### SOLUTION

Solution :C

Let the base of the triangle be x cm.

Then, the corresponding altitude of the triangle would be 3x5 cm.

If altitude is increased by 4 cm, the new altitude =3x5+4 cm

New base =x−2 cm

Area of triangle =12×base×height

Since the area remains the same, 12×x×3x5=12×(3x5+4)×(x−2)

⇒ 3x25 = 3x25+4x−6x5−8

⇒ 4x−6x5=8⇒20x−6x5=8

⇒14x=40

⇒ x=4014=207⇒ x=267 cm

Hence, base of the triangle is267 cm long.