Table of Contents

## Extra Questions Very Short Answer Type

Question 1.

Identify the algebraic linear equations from the given expressions.

(a) x^{2} + x = 2

(b) 3x + 5 = 11

(c) 5 + 7 = 12

(d) x + y^{2} = 3

Solution:

(a) x^{2} + x = 2 is not a linear equation.

(b) 3x + 5 = 11 is a linear equation.

(c) 5 + 7 = 12 is not a linear equation as it does not contain variable.

(d) x + y^{2} = 3 is not a linear equation.

Question 2.

Check whether the linear equation 3x + 5 = 11 is true for x = 2.

Solution:

Given that 3x + 5 = 11

For x = 2, we get

LHS = 3 × 2 + 5 = 6 + 5 = 11

LHS = RHS = 11

Hence, the given equation is true for x = 2

Question 3.

Form a linear equation from the given statement: ‘When 5 is added to twice a number, it gives 11.’

Solution:

As per the given statement we have

2x + 5 = 11 which is the required linear equation.

Question 4.

If x = a, then which of the following is not always true for an integer k. (NCERT Exemplar)

(a) kx = ak

(b) \(\frac { x }{ k }\) = \(\frac { a }{ k }\)

(c) x – k = a – k

(d) x + k = a + k

Solution:

Correct answer is (b).

Question 5.

Solve the following linear equations:

(a) 4x + 5 = 9

(b) x + \(\frac { 3 }{ 2 }\) = 2x

Solution:

(a) We have 4x + 5 = 9

⇒ 4x = 9 – 5 (Transposing 5 to RHS)

⇒ 4x = 4

⇒ x = 1 (Transposing 4 to RHS)

(b) We have x + \(\frac { 3 }{ 2 }\) = 2x

⇒ \(\frac { 3 }{ 2 }\) = 2x – x

⇒ x = \(\frac { 3 }{ 2 }\)

Question 6.

Solve the given equation 3\(\frac { 1 }{ x }\) × 5\(\frac { 1 }{ 4 }\) = 17\(\frac { 1 }{ 2 }\)

Solution:

We have 3\(\frac { 1 }{ x }\) × 5\(\frac { 1 }{ 4 }\) = 17\(\frac { 1 }{ 2 }\)

Question 7.

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

Solution:

We have 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 verified.

Question 8.

Solution:

⇒ 3x × 3 – (2x + 5) × 4 = 5 × 6

⇒ 9x – 8x – 20 = 30 (Solving the bracket)

⇒ x – 20 = 30

⇒ x = 30 + 20 (Transposing 20 to RHS)

⇒ x = 50

Hence x = 50 is the required solution.

Question 9.

The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles of the triangle.

Solution:

Let the angles of a given triangle be 2x°, 3x° and 4x°.

2x + 3x + 4x = 180 (∵ Sum of the angles of a triangle is 180°)

⇒ 9x = 180

⇒ x = 20 (Transposing 9 to RHS)

Angles of the given triangles are

2 × 20 = 40°

3 × 20 = 60°

4 × 20 = 80°

Question 10.

The sum of two numbers is 11 and their difference is 5. Find the numbers.

Solution:

Let one of the two numbers be x.

Other number = 11 – x.

As per the conditions, we have

x – (11 – x) = 5

⇒ x – 11 + x = 5 (Solving the bracket)

⇒ 2x – 11 = 5

⇒ 2x = 5 + 11 (Transposing 11 to RHS)

⇒ 2x = 16

⇒ x = 8

Hence the required numbers are 8 and 11 – 8 = 3.

Question 11.

If the sum of two consecutive numbers is 11, find the numbers.

Solution:

Let the two consecutive numbers be x and x + 1.

As per the conditions, we have

x + x + 1 = 11

⇒ 2x + 1 = 11

⇒ 2x = 11 – 1 (Transposing 1 to RHS)

⇒ 2x = 10

x = 5

Hence, the required numbers are 5 and 5 + 1 = 6.

### Linear Equations in One Variable Class 8 Extra Questions Short Answer Type

Question 12.

The breadth of a rectangular garden is \(\frac { 2 }{ 3 }\) of its length. If its perimeter is 40 m, find its dimensions.

Solution:

Let the length of the garden be x m

its breadth = \(\frac { 2 }{ 3 }\) × m.

Perimeter = 2 [length + breadth]

Question 13.

The difference between two positive numbers is 40 and the ratio of these integers is 1 : 3. Find the integers.

Solution:

Let one integer be x.

Other integer = x – 40

As per the conditions, we have

\(\frac { x-40 }{ x }\) = \(\frac { 1 }{ 3 }\)

⇒ 3(x – 40) = x

⇒ 3x – 120 = x

⇒ 3x – x = 120

⇒ 2x = 120

⇒ x = 2

Hence the integers are 60 and 60 – 40 = 20.

Question 14.

Solve for x:

Solution:

Question 15.

The sum of a two-digit number and the number obtained by reversing its digits is 121. Find the number if it’s unit place digit is 5.

Solution:

Unit place digit is given as 5

Let x be the tens place digit

Number formed = 5 + 10x

Number obtained by reversing the digits = 5 × 10 + x = 50 + x

As per the conditions, we have

5 + 10x + 50 + x = 121

⇒ 11x + 55 = 121

⇒ 11x = 121 – 55 (Transposing 55 to RHS)

⇒ 11x = 66

⇒ x = 6

Thus, the tens place digit = 6

Hence the required number = 5 + 6 × 10 = 5 + 60 = 65

### Linear Equations in One Variable Class 8 Extra Questions Higher Order Thinking Skills (HOTS)

Question 16.

If the length of the rectangle is increased by 40% and its breadth is decreased by 40%, what will be the percentage change in its perimeter?

Solution:

Let the length of the rectangle be x m and its breadth be y m

Perimeter = 2(x + y)

Now the length of the rectangle becomes after a 40% increase

Question 17.

A fruit seller buys some oranges at the rate of ₹ 5 per orange. He also buys an equal number of bananas at the rate of ₹ 2 per banana. He makes a profit of 20% on oranges and a profit of 15% on bananas. In the end, he sold all the fruits. If he earned a profit of ₹ 390, find the number of oranges.

Solution:

Let the number of oranges bought by him be x and also the number of bananas be x.

Cost of x oranges at the rate of ₹ 5 per orange = ₹ 5x

Cost of x bananas at the rate of ₹ 2 per banana = ₹ 2x

Question 18.

A steamer goes downstream from one point to another in 7 hours. It covers the same distance upstream in 8 hours. If the speed of stream be 2 km/h, find the speed of the steamer in still water and the distance between the ports. (NCERT Exemplar)

Solution:

Let speed of steamer in still water = x km/h

Speed of stream = 2 km/h

Speed downstream = (x + 2) km/h

Speed upstream = (x – 2) km/h

Distance covered in 7 hours while downstream = 7(x + 2)

Distance covered in 8 hours while upstream = 8(x – 2)

According to the condition,

7(x + 2) = 8(x – 2)

⇒ 7x + 14 = 8x – 16

⇒ x = 30 km/h

Total Distance = 7(x + 2) km = 7(30 + 2) km = 7 × 32 km = 224 km.