Chapter 11 Algebra

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 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 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)  = 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)  = 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

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 13.
Complete the table and find the solution of the equation 19 – x = 13
Algebra Class 6 Extra Questions Maths Chapter 11
Solution:
By inspection, we have
Algebra Class 6 Extra Questions Maths Chapter 11
Thus, the required solution is 6.


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

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.

Chapter 11 Algebra