## Algebra Class 6 Extra Questions Maths Chapter 11

**Extra Questions for Class 6 Maths Chapter 11 Algebra**

### Algebra Class 6 Extra Questions Very Short Answer Type

Question 1.

Six less than a number equals to two. What is the number?

Solution:

Let the number be ‘x’.

According to condition, we have x – 6 = 2

By inspections, we have 8 – 6 = 2

∴ x = 8

Thus, the required number is 8.

Question 2.

Write an algebraic expression for each of the following:

(a) 3 subtracted from a number y.

(b) 5 is added to three times a number x.

Solution:

(a) The required expression is y – 3

(b) The required expression is 5 + 3x

Question 3.

Write an algebraic expression for the following expressions:

(a) The sum of a number x and 4 is doubled.

(b) One fourth of a number x is added to one third of the same number.

Solution:

(a) The required expression is 2 x (x + 4)

(b) The required expression is \(\frac { 1 }{ 4 }\)x + \(\frac { 1 }{ 3 }\)x

Question 4.

Think of a number x. Multiply it by 3 and add 5 to the product and subtract y subsequently. Find the resulting number.

Solution:

Required number is (3x + 5)

Now we have to subtract y from the result i.e., 3x + 5 – y

Question 5.

Here is a pattern of houses with matchsticks:

Write the general rule for this pattern.

Solution:

One house is made of 6 matchsticks i.e. 6 x 1

Two houses are made of 12 matchsticks i.e. 6 x 2

Three houses are made of 18 matchsticks i.e. 6 x 3

∴ Rule is 6n where n represents the number of houses.

Question 6.

If the side of an equilateral triangle is x, find its perimeter.

Solution:

We know that the three sides of an equilateral triangle are equal.

∴ x + x + x = 3x.

Thus, the required perimeter = 3x units.

Question 7.

If x = 3, find the value of the following:

(i) x + 5

(ii) 2x – 3

(iii) x – 7

(iv) \(\frac { x }{ 3 }\) – 1

Solution:

Given that x = 3

(i)x + 5 = 3 + 5 = 8

(ii) 2x – 3 = 2 x 3 – 3 = 6 – 3 = 3

(iii) x – 7 = 3- 7 = -4

(iv) \(\frac { x }{ 3 }\) – 1 = \(\frac { 3 }{ 3 }\) -1 = 1 – 1 = 0

Question 8.

If x = 2, y = 3 and 2 = 5, find the value of;

(a) 2x + y + z

(b) 4x -y + z

(c) x – y + z

Solution:

(a) Given that: x – 2, y = 3 and z = 5

∴ 2x + y + 2 = 2 x 2 + 3 + 5

= 4 + 3 + 5 = 12

(b) 4x – y + z = 4 x 2 – 3 + 5

= 8 – 3 + 5 = 5 + 5 = 10

(c) x – y + z = 2 – 3 + 5 = -1 + 5 = 4

Question 9.

State which of the following are equations with a variable?

(a) 12 = x – 5

(b) 2x > 7

(c) \(\frac { x }{ 2 }\) = 5

(d) 5 + 7 = 3 + 9

(e) 7 = (11 x 5) – (12 x 4)

Solution:

(a) 12 = x – 5 is an equation with a variable x.

(b) 2x > 7 is not an equation because it does not have ‘=’ sign.

(c) \(\frac { x }{ 2 }\) = 5 is an equation with a variable x.

(d) 5 + 7 = 3 + 9 is not an equation because it has no variable.

(e) 7 = (11 x 5) – (12 x 4) is not an equation because it has no variable.

Question 10.

Think of a number, add 2 to it and then multiply the sum by 6, the result is 42.

Solution:

Let the number be x.

∴ Sum of x and 2 = x + 2

Now by multiplying the sum by 6, we get

6 × (x + 2) = 42

⇒ 6 × x + 6 × 2 = 42

⇒ 6x + 12 = 42

By inspection, we get

6 × 5 + 12 = 42

⇒ 30 + 12 = 42

∴ 42 = 42

So, the required number = 5

### Algebra Class 6 Extra Questions Short Answer Type

Question 11.

The side of a regular hexagon is s cm. Find its perimeter.

Solution:

Each side of a regular hexagon = s

∴ its perimeter = s + s + s + s + s + s = 6s cm

Question 12.

If a = 3, b = \(\frac { 1 }{ 2 }\) and c = \(\frac { 1 }{ 4 }\), find the value of

Solution:

Question 13.

Complete the table and find the solution of the equation 19 – x = 13

Solution:

By inspection, we have

Thus, the required solution is 6.

Question 14.

If x = –\(\frac { 1 }{ 2 }\), y = \(\frac { 1 }{ 4 }\) and z = 0, find the value of the given expressions

(a) 8z + 2x -y

(b) z – y + 3x

Solution:

Question 15.

Fill in the blanks:

(a) 5 added to -5 = ……….

(b) If x = 3, then 3x – 5 = ……….

(c) If x = 1 and y = 2, then 2x + 3y = ……….

(d) If 10x – 6 = 14, then x = ……….

(e) 4 less than a number x = ……….

Solution:

(a) 0

(b) 4

(c) 8

(d) 2

(e) x – 4

### Algebra Class 6 Extra Questions Higher Order Thinking Skills (HOTS)

Question 16.

A starts his car from Delhi at 6.00 am to Amritsar. The uniform speed of his car is x km/h. At 12.00 noon, he finds that he is still 50 km away from Amritsar. Find the distance between Delhi and Amritsar.

Solution:

Time taken by A to reach Amritsar = 12.00 noon – 6.00 am = 6 hour.

The uniform speed of the car = x km/ hr

∴ Total distance covered by A = Time x speed = 6x km.

∴ Distance between Delhi and Amritsar = (6x + 50) km.

Question 17.

Anshika’s Score in Science is 15 more than the two-third of her score in Sanskrit. If she scores x marks in Sanskrit, find her score in Science.

Solution:

Anshika’s score in Sanskrit = x

∴ Her marks in Science = \(\frac { 2 }{ 3 }\)x +15

∴ Thus, Anshika’s score in Science = \(\frac { 2 }{ 3 }\)x + 15