Algebra Class-6 ML Aggarwal ICSE Maths APC Solutions Chapter-9. We provide step by step Solutions of Exercise / lesson-9 Algebra ICSE Class-6th ML Aggarwal Mathematics .

Our Solutions contain all type Questions with Exe- 9.1, Exe-9.2, Exe-9.3, Exe-9.4, Exe- 9.5,  Objective Type Questions (includes : Mental Maths , MCQs ) and Check Your Progress to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-6 Maths.

## Algebra Class-6 ML Aggarwal ICSE Maths APC Solutions Chapter-9

–: Select Topic :–

Exercise 9.1 ,

Exercise-9.2,

Exercise-9.3,

Exercise 9.4 ,

Exercise 9.5 ,

Objective Type Questions,

Mental Maths,

Multiple Choice Questions ,(MCQ)

### Exercise – 9.1, Algebra Class-6 ML Aggarwal ICSE Maths APC Solutions

#### Question 1

Find the rule which gives the number of matchsticks required to make the following matchsticks patterns. Use a variable to write the rule.
(i) A pattern of letter T as
(ii) A pattern of letter V as
(iii) A pattern of letter Z as
(iv) A pattern of letter U as
(v) A pattern of letter F as
(vi) A pattern of letters S as

(i) Number of matchsticks requried = 2n
(ii) Number of matchsticks requried = 2n
(iii) Number of matchsticks requried = 3n
(iv) Number of matchsticks requried = 3n
(v) Number of matchsticks requried = An
(vi) Number of matchsticks requried = 5n

#### Question  2

If there are 24 mangoes in a box, how will you write the number of mangoes in terms of the number of boxes? (use b for the number of boxes.)

Total number of mangoes = 24b

#### Question 3.

Anuradha is drawing a dot Rangoli (a beautiful pattern of lines joining dots). She has 8 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 12 rows?

∵ Numbers of dots in 1 row = 8
∴ Number of dots in r rows = 8 × r = 8
Number of dots in 12 rows = 12 × 8 = 96

#### Question  4

Anu and Meenu are sisters. Anu is 5 years younger than Meenu. Can you write Anu’s age in terms of Meenu’s age? Take Meenu’s age as x years.

Yes! we can write Anu’s age in terms of Meenu’s age.
Age of Meenu = x
∵ Anu is 5 years younger than Meenu
∴ Age of Anu = (x – 5) years

#### Question  5

Oranges are to be transferred from larger boxes to smaller boxes. When a larger box is empited, the oranges from it fill 3 samller boxes and still 7 oranges are left. If the number of oranges in a small box are taken to be x, then what is the number of oranges in the larger box? .

Let the number of oranges in a smaller box be x.
∴ Number of oranges in three smaller boxes = 3x
Number of oranges remained outside = 7
∴ Number of oranges in the larger box = 3x + 7

#### Question  6

Harsha’s score in Mathematics is 15 more than three-fourth of her score in Science. If she scores x marks in Science, find her score in Mathemstics?

Let the score of science be x
Harsha score’s in Mathematics = $\frac{3}{4} \text { th of } x+15$
∴ Score of Harsha’s in Mathematics = $\frac{3}{4} x+15$

#### Question  7

Look at the following matchstick pattern of equilateral triangles. The triangles are not separate. Two neighbouring triangles have a common matchstick. Observe the pattern and find the rule that gives the number of matchsticks.

Number of matchsticks required = 2x + 1, where x is the number of triangles.

#### Question  8

Look at the following matchstick pattern of letter A. The A’s are not separate. Two neighbouring A’s have two common matchsticks. Observe the pattern and find the rule that gives the number of matchsticks.

Solution:

Number of matchsticks required = 4x + 2, where x is the number of letter ‘A’ formed.

### Algebra Class-6 ML Aggarwal ICSE Maths APC Solutions Exercise – 9.2

#### Question 1.

If the side of an equilateral triangle is l, then express the perimeter of the triangle in terms of l.
Solution:
Perimeter (P) of the equilateral triangle with side l = Sum of the lengths of sides of the equilateral triangle = l + l + l = 3l

#### Question 2.

The side of a regular hexagon is l. Express its perimeter in terms of l.

Perimeter (P) of the regular hexagon with side l = Sum of the lengths of all sides of the regular hexagon

= l + l + l + l + l + l = 6l

#### Question 3

The length of an edge of a cube isl. Find the formula for the sum of lengths of all the edges of the cube.

Total length (L) of the edges of a cube = Sum of the lengths of all (12) edges of the cube.
= l + l + l + l + l+ l + l + l + l + l + l + l = 12l

#### Question 4

If the radius of a circle is r units, then express the length of a diameter of the circle in terms of r.

### ML Aggarwal Solutions for ICSE Class-6 Maths Algebra Exercise – 9.3

#### Question 1

Form four expressions with numbers 7, 5 and 8 (no variables) using operations of addition, subtraction or multiplication with the condition that every number should be used but not more than once.

The possible expressions are:
5 × 7 + 8, 5 × 8 – 7
(5 + 8) – 7, 8 × (5 + 7)

#### Question 2

Which out of following are expressions with numbers only?
(i) 2y + 3
(ii) (7 × 20) – 82
(iii) 5 × (21 – 7) + 9 x 2
(iv) 5 -11 n
(v) (5 × 4) – 45 + p
(vi) 3 × (11 + 7) – 24 + 3

(iii) 5 × (21 – 7) + 9 × 2,
(vi) 3 × (11 + 7) – 24 ÷ 3 are expressions with numbers only.

#### Question 3

Identify the operations (addition, subtraction, multiplication, division) in forming the following expressions and tell how the expressions have been formed:
(i) x + 5
(ii) y – 7
(iii) 3z
(iv) $\frac{p}{5}$
(v) 2x + 17
(vi) 3y – 5
(vii) $-7 m+\frac{2}{3}$
(viii) $\frac{x}{3}-15$

(i) x + 5
(ii) y – 7
Subtraction → 7 subtracted from y.
(iii) 3z
Multiplication → z multiplied by 3.
(iv) $\frac{p}{5}$
Division → p divided by 5.
(v) 2x + 17
Multiplication and addition → First x mutliplied by 2 and then 17 added to the product.
(vi) 3y – 5
Multiplication and subtraction → First y multiplied by 3 and then 5 subtracted from the product.
(vii) $-7 m+\frac{2}{3}$
Multiplication and addition → First m multiplied by -7 and then $\frac{2}{3}$ added to the product.
(viii) $\frac{x}{3}-15$
Division and subtraction → First x divided by 3 and then 15 subtracted from the quotient.

#### Question 4

Write expression for the following:
(ii) p subtracted from 7
(iii) p multiplied by 7
(iv) p divided by 7
(v) 7 divided by p
(vi) 7 subtracted from -m
(vii) p multiplied by -5
(viii) -p divided by 5

(i) p + 7
(ii) 7 – p
(iii) 7p
(iv) $\frac{p}{7}$
(v) $\frac{7}{p}$
(vi) -m – 7
(vii) -5p
(viii) $\frac{-p}{5}$

#### Question 5

Write expression for the following:
(i) 11 added to 2 m
(ii) 11 subtracted from 2 m
(iii) 3 added to 5 times y
(iv) 3 subtracted from 5 times y
(v) y is multiplied by -8 and then 5 is added to the result
(vi) y is multiplied by 5 and then the result is subtracted from 16.

(i) 2m + 11
(ii) 2m – 11
(iii) 5y + 3
(iv) 5y – 3
(v) -8y + 5
(vi) 16 – 5y

#### Question 6

Write the following in mathematical form using signs and symbols:
(i) 6 more than thrice a number x.
(ii) 7 taken away from y.
(iii) 3 less than quotient of x by y.

(i) 3x + 6
(ii) $\frac{x}{y}-3$
(iii) y – 7

#### Question 7

Form six expressions using t and 4. Use not more than one number operation and every expression must have t in it.

t + 4, t – 4, 4 – t, 4t, $\frac{t}{4}, \frac{4}{t}$

#### Question 8

Form expressions using y, 2 and 7. Use only two different number operations and every expression must have y in it.
Solution:
2y + 7, 2y- 1, 7y + 2, 7y – 2, $\frac{y}{2}+7, \frac{y}{2}$ – 7, …..

#### Question 9

A student scored x marks in English but the teacher deducted 5 marks for bad handwriting. What was the student’s final score in English?

Marks in English = x
Deducted = 5
Final score = x – 5

#### Question 10

Raju’s father’s age is 2 years more than 3 times Raju’s age. If Raju’s present age is y years, then what is his father’s age?

(3y + 2) years

#### Question  11

Mohini is x years old. Express the following in algebraic form:
(i) three times Mohini’s age next year.
(ii) four times Mohini’s age 3 years ago.
(iii) the present age of Mohini’s uncle, if his uncle is 5 times as old as Mohini will be two years from now.
(iv) the present age of Mohini’s cousin, if her cousin is two years less than one-third of Mohini’s age five years ago

#### Question  12

A cuboidal box has height h cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.

Length of the box = 5h cm
Breadth of the box = (5h – 10) cm

#### Question  13

A bus travels at v km per hour. It is going from Delhi to Jaipur. After the bus has travelled 5 hours, Jaipur is still 20 km away. What is the distance from Delhi to Jaipur?

Speed of the bus = v km/hr
Distance travelled in 5 hours = 5v km
∴ Total distance = (5v + 20) km

#### Question  14

Change the following statements using expressions into statements in ordinary language:
(i) A notebook cost ₹ p. A book costs ₹ 3p.
(ii) The cost of rice per kg is ₹ p. The cost of oil per litre is ₹5p.
(iii) The speed of a truck is v km per hour. The speed of a bus is (v + 10) km per hour.
(iv) Tony’s box contains 8 times the marbles he puts on the table.
(v) The total number of students in the school is 20 times that of our class.
(vi) Raju is x years old. His uncle is 4x years old and his aunt is (4x – 3) years old.
(vii) In arrangement of dots there are r rows. Each row contains 5 dots.

(i) The cost of a book is 3 times the cost of a note book.
(ii) The cost of oil per litre is 5 times the cost of rice per kg.
(iii) The speed of a bus is 10 km per hour more than the speed of a truck.
(iv) Tony puts q marbles on the table. He has 8q marbles in his box.
(v) Our class has n students. The school has 20 n students.
(vi) Raju’s uncle is 4 times older than Raju and his aunt is 3 years younger than his uncle.
(vii) The total number of dots is 5 times the number of rows.

### Algebra Exercise – 9.4 for ICSE Class-6 ML Aggarwal Solutions

#### Question  1

Find the value of the following:
(i) 43
(ii) (-6)4
(iii) $\left(\frac{2}{3}\right)^{4}$
(iv) (-2)3 × 52

#### Question  2

Find the value of:
(i) 3x + 2y when x = 3 and y – 2
(ii) 5x – 3y when x = 2 and y = -5
(iii) a + 2b – 5c when a = 2, b = -3 and c = 1
(iv) 2p + 3q + 4r + pqr when p = -1, q = 2 and r = 3
(v) 3ab + 4bc – 5ca when a = 4, 6 = 5 and c = -2.

(i) 3x + 2y, x = 3,y = 2
(3 × 3) + (2 × 2) = 9 + 4 = 13
(ii) 5x – 3y, x = 2, y = -5
(5 × 2) – (3 × -5) = 10 + 15 = 25
(iii) a + 2b – 5c, a =2,b = -3, c = 1
2 + (2 × -3) -5 × (1)
= 2 – 6 – 5 = -9
(iv) 2p + 3q + 4r + pqr, p = -1, q = 2, r = 3
= (2 × -1) + (3 × 2) + (4 × 3) + (-1) × 2 × 3
= -2 + 6 + 12 – 6= 10
(v) 3ab + 4bc – 5ca, a = 4, b = 5, c = -2 (3 × 4 × 5) + (4 × 5 × -2) -5 × -2 × 4
= 60 – 40 + 40 = 60

#### Question  3

Find the value of:
(i) 2x2 – 3x + 4 when × = 2
(ii) 4x3 – 5x2 – 6x + 7 when x = 3
(iii) 3x3 + 9x2 – x + 8 when x = -2
(iv) 2x4 – 5x3 + 7x – 3 when x = -3

(i) 2x2 – 3x + 4, x = 2
= 2 × (2)2 -3x2 + 4
= 8 – 6 + 4 = 6
(ii) 4x3 – 5x2 – 6x + 7, x = 3
= 4(3)3 – 5(3)2 – 6(3) + 7
= 108 – 45 – 18 + 7 = 52
(iii) 3x3 + 9x2 – x + 8, x = -2
= 3(-2)3 + 9(-2)2 – (-2) + 8
= -24 + 36 + 2 + 8 = 22
(iv) 2x4 – 5x3 + 7x – 3, x = -3
= 2(-3)4 – 5(-3)3 + 7(-3) – 3
= 162 + 135 – 21 – 3 = 273

#### Question  4

If x = 5, find the value of:
(i) 6 – 7x2
(ii) 3x2 + 8x – 10
(iii) 2x3 – 4x2 – 6x + 25

(i) 6 – 7x2 = 6 – 7(5)2 = 6 – 7(25)
= 6 – 175 = -169
(ii) 3(5)2 + 8(5) – 10
= 3(25) + 40 – 10
= 75 + 40 – 10 = 75 + 30 = 105
(iii) 2(5)3 – 4(5)2 – 6(5) + 25
= 2(125) – 4(25) – 30 + 25
= 250 – 100 – 30 + 25= 145

#### Question  5

If x = 2, y = 3 and z = -1, find the values of:
(i) x + y
(ii) $\frac{x y}{z}$
(iii) $\frac{2 x+3 y-4 z}{3 x-z}$

#### Question  6

If a = 2, b = 3 and c = -2, find the value of a2 + b2 + c2 – 2ab – 2bc – 2ca + 3abc.

a = 2,b = 3,c = -2
a2 + b2 + c2 – 2ab – 2bc – 2ca + 3abc
= (2)2 + (3)2 + (-2)2 – 2 × 2 × 3 – 2 × 3 × – 2 – 2 × -2 × 2 + 3 × 2 × 3 × -2
= 4 + 9 + 4 – 12 + 12 + 8 – 36
= 25 – 36 = -11

#### Question  7

If p = 4, q = -3 and r = 2, find the value of: p3 + q3 – r3 – 3pqr.

p = 4, q = -3, r = 2
p3 + q3 – r3 – 3pqr
= (4)3 + (-3)3 – (2)3 – 3 × 4 × -3 × 2
= 64 – 27 – 8 + 72
= 136 – 35 = 101

#### Question  8

If m = 1, n = 2 and p = -3, find the value of 2mn4 – 15m2n + p.

m = 1, n = 2, p = -3
2mn4 – 15m2n + p
= 2(1 )(2)4 – 15(1 )2(2) + (-3)
= 32 – 30 – 3 = —1

#### Question  9

State true or false:
(i) The value of 3x – 2 is 1 when x = 0.
(ii) The value of 2x2 – x – 3 is 0 when x = -1.
(iii) p2 + q2 – r2 when p = 5, q = 12 and r = 13.
(iv) 16 – 3x = 5x when x = 2.

(i) The value of 3x – 2 is 1 when x = 0. False
Correct :
∵ 3 × 0 – 2 = -2
(ii) The value of 2x2 – x – 3 is 0 when x = -1. True
2(-1 )2 – (-1) – 3
= 2 + 1 – 3 = 0
(iii) p2 + q2 = r2 when p = 5, q = 12 and r = 13. True
(5)2 + (12)2 = (13)2
= 25 + 144 = 169
⇒ 169= 169
(iv) 16 – 3x = 5x when x = 2. True
16 – 3x2 = 5x2
16 – 6 = 10
⇒ 10 = 10

#### Question  10

For x = 2 and y = -3, verify the following:
(i) (x + y)2 = x2 + 2xy + y2
(ii) (x – y)2 = x2 – 2xy + y2
(iii) x2 – y2 = (x + y) (x – y)
(iv) (x + y)2 = (x – y)2 + 4xy
(v) (x + y)3 = x3 + y3 + 3x2y + 3xy2

x = 2, y = -3
(i) (x + y)2 = x2 + 2xy + y2
L.H.S. = (x + y)2 = (2 – 3)2 = (-1)2 = 1
R.H.S. = x2 + 2xy + y2
= (2)2 + 2 × 2 (-3) + (-3)2
= 4 – 12 + 9 = 13 – 12 = 1
L.H.S. = R.H.S.

(ii) (x – y)2 = x2 – 2xy + y2
L.H.S. = (x – y)2 = [2 – (-3)]2
= (2 + 3)2 = (5)2 = 25
R.H.S. = x2 – 2xy + y2
= (2)2 – 2 × 2 × (-3) + (-3)2
= 4 + 12 + 9 = 25
∴ L.H.S. = R.H.S.

(iii) x2 – y2 = (x + y)(x – y)
L.H.S. = (x)2 – (y)2 = (2)2 – (-3)2
= 4 – 9 = -5
R.H.S. = (x + y)(x – y)
= (2 – 3) [2 – (-3)] = (2 – 3) (2 + 3) = -1 × 5 = -5
∴ L.H.S. = R.H.S.

(iv) (x + y)2 = (x – y)2 + 4xy
L.H.S. = (x + y)2 = [2 + (-3)]2
= (2 – 3)2 = (-1)2 = 1
R.H.S. = (x – y)2 + 4xy
= [2 – (-3)]2 + 4 × 2 × (-3)
= (2 + 3)2 + 4 × 2 × (-3)
= (5)2 – 24 = 25 – 24 = 1
∴ L.H.S. = R.H.S.

(v) (x + y)3 = x3 + y2 + 3x2y + 3xy2
L.H.S. = (x + y)3 = [2 + (-3)]3 = (2 – 3)3
= (-1)3 = (-1) × (-1) × (-1) = -1
R.H.S. = x3 + y3 + 3x2y + 3xy2
= (2)3 + (-3)3 + 3 (2)2 (-3) + 3 × 2 (-3)2
=8 – 27 + 3 × 4 × (-3) + 6 (9)
= 8 – 27 – 36 + 54 = 62 – 63 = -1
∴ L.H.S. = R.H.S.

### ML Aggarwal Solutions of Algebra Exercise – 9.5 for ICSE Class-6

#### Question  1

State which of the following are equations with a variable. In case of an equation with a variable, identify the variable.

(i) 17 + x = 5
(ii) 2b – 3 = 7
(iii) (y – 7) > 5
(iv) $\frac{9}{3}=3$
(v) 7 × 3 – 19 = 2
(vi) 5 × 4 – 8 = 31
(vii) 2p < 15
(viii) 7 = 11 × 5 – 12 × 4
(ix) $\frac{3}{2} q=5$

(i) 17 + x= 5 Is an equation → L.H.S. = R.H.S. → Related variable x.
(ii) 2b – 3 = 7 Is an equation → L.H.S. = R.H.S. → Related variable b.
(iii) (y- 7) >5
Is not an equation → L.H.S. ≠ R.H.S.
It has no sign of equality (=).
(iv) $\frac{9}{3}=3$
Is an equation = L.H.S. = R.H.S.
It has no variable.
(v) 7 × 3 – 19 = 2
Is an numerical equation = L.H.S. = R.H.S. It has no variable.
(vi) 5 × 4 – 8 = 31
Is an equation = L.H.S. = R.H.S. → Related variable t.
(vii) 2p < 15
Is not an equation = L.H.S. ≠ R.H.S.
It has no sign of equality.
(viii) 7 = 11 × 5 – 12 × 4
Is an numerical equation = L.H.S. = R.H.S. It has no variable.
(ix) $\frac{3}{2} q=5$
Is an equation → L.H.S. = R.H.S. → Related variable q.

#### Question  2

Solve each of the following equations :
(i) x + 6 = 8
(ii) 2 – x = 5
(iii) 4x = -6
(iv) $\frac{x}{2}=5$
(v) 2y – 3 = 2
(vi) 4 – 5y = 2

(i) x + 6 = 8
= x = 8 – 6 ⇒ x = 2
(ii) 2 – x = 5
= -x = 5 – 2 ⇒ -x = 3 ⇒ x = – 3
(iii) 4x = -6
$=x=\frac{-6}{4}=\frac{-3}{2}$
(iv) $\frac{x}{2}=5$
= x = 5 × 2 ⇒ x = 10
(v) 2y – 3 = 2
= 2y = 2 + 3 ⇒ 2y = 5 ⇒ y = $\frac{5}{2}$
(vi) 4 – 5y = 2
= 4 – 2 = 5y ⇒ 5y = 2
⇒ y = $\frac{2}{5}$

#### Question  3

Solve the following linear equations:
(i) 5(x + 1) = 25
(ii)2(3x – 1) = 10
(iii) $\frac{3 x-1}{4}=11$

(i) Givrn 5(x + 1) = 25
⇒ $\frac{5(x+1)}{5}=\frac{25}{5}$ (dividing both sides by 5)
⇒ x + 1 = 5
⇒ x + 1 – 1 = 5 – 1 (Subtracting 1 from both sides)
⇒ x = 4

(ii) 2(3x – 1) = 10
⇒ $\left(\frac{2(3 x-1)}{2}\right)=\frac{10}{2}$ (dividing both sides by 2)
⇒ 3x – 1 = 5
⇒ 3x – 1 + 1 = 5 + 1 (adding 1 to both sides)
⇒ 3x = 6
⇒ $\frac{3 x}{3}=\frac{6}{3}$ (dividing both sides by 3)
⇒ x = 2

(iii) Given $\frac{3 x-1}{4}=11$
⇒ $4 \times \frac{3 x-1}{4}=4 \times 11$ (multiplying both sides by 4)
⇒ 3x – 1 = 44
⇒ 3x – 1 + 1 = 44 + 1 (adding 1 to both sides)
⇒ 3x = 45
⇒ $\frac{3 x}{3}=\frac{45}{3}$ (dividing both sides by 3)
⇒ x = 15

#### Question  4

Solve the following linear equations:

(i) 5x – 6 = 12 – x
⇒ 5x + x = 12 + 6

#### Question  5

(i) 3(x + 7) = 18
(ii) 2(x- 1) = x + 2
(iii) $3 x-\frac{1}{3}=2\left(x-\frac{1}{2}\right)+5$
(iv) 4(2x – 1) -2(x – 5) = 5(x + 1) + 3

⇒ 4(4 – 1) -2 (-3) = 5(3) + 3
⇒ 4 × 3 + 6 = 15 + 3
⇒12 + 6 = 18
⇒ 18 = 18

### Objective Type Questions, Algebra Class-6 ML Aggarwal

#### Question  1

Fill in the blanks:
(i) In algebra, we use …………… to represent variables (generalized numbers).
(ii) A symbol or letter which can be given various numerical values is called a ……………
(iii) If Jaggu’s present age is x years, then his age 7 years from now is ……………
(iv) If one pen costs ₹X x, then the cost of 9 pens is ……………
(v) An equation is a statement that the two expressions are ……………
(vi) Trial an error is one of methods to obtain …………… of an equation.
(vii) 7 less than thrice a number y is ……………
(viii) If 3x + 4 = 19, then the value of x is ……………
(ix) The number of pencils bought for ₹ x at the rate of ₹2 per pencil is ……………
(x) In the expression (-7)5, base = …………… and exponent = ……………
(xi) If base = 6 and exponent = 5, then the exponential form = …………… .

(i) In algebra, we use letters to represent variables (generalized numbers).
(ii) A symbol or letter which can be given various numerical values is called a variable.
(iii) If Jaggu’s present age is x years, then his age 7 years from now is (x + 7) years.
(iv) If one pen costs ₹ x, then the cost of 9 pens is ₹9x.
(v) An equation is a statement that the two expressions are equal.
(vi) Trial an error is one of methods to obtain the solution of an equation.
(vii) 7 less than thrice a num bery is 3y – 7.
(viii) If 3x + 4 = 19, then the value of x is 5.
(ix) The number of pencils bought for ₹ x at the rate of ₹2 per pencil is $\frac{x}{2}$.
(x) In the expression (-7)5, base = -7 and exponent = 5.
(xi) If base = 6 and exponent = 5, then the exponential form = 65.

#### Question  2

State whether the following statements are ture (T) or false (F):
(i) If x is variable then 5x is also variable.
(ii) If y is variable then y – 5 is also variable.
(iii) The number of angles in a triangle is a variable.
(iv) The value of an algebraic expression changes with the change in the value of the variable.
(v) If the length of a rectangle is twice its breadth, then its area is a constant.
(vi) An equation is satisfied only for a definite value of the variable.
(vii) If x toffees are distributed equally among 5 children, then each child gets 5x toffees.
(viii) t minutes are equal to 60 t seconds.
(ix) If x is a negative integer, then -x is a positive integer.
(x) x = 5 is a solution of the equation 3x + 2 = 13.
(xi) 2y- 7 > 13 is an equation.
(xii) ‘One third of a number x added to itself gives 8’ can be expressed as $\frac{x}{3}$ + 8 = x.
(xiii)The difference between the ages of two sisters Lata and Asha is a variable.

(i) If x is variable then 5x is also variable. True
(ii) If y is variable then y – 5 is also variable. True
(iii) The number of angles in a triangle is a variable. False
(iv) The value of an algebraic expression changes with the change in the value of the variable. True
(v) If the length of a rectangle is twice its breadth, then its area is a constant. False
(vi) An equation is satisfied only for a definite value of the variable. True
(vii) If x toffees are distributed equally among 5 children, then each child gets 5x toffees. False
(viii) t minutes are equal to 60 t seconds. True
(ix) If x is a negative integer, then -x is a positive integer. True
(x) x = 5 is a solution of the equation 3x + 2 = 13. False
(xi) 2y – 1 > 13 is an equation. False
(xii) ‘One third of a number x added to itself gives 8’ can be expressed as $\frac{x}{3}$ + 8 = x. False
(xiii)The difference between the ages of two sisters Lata and Asha is a variable. False

#### Question  3

Choose the correct answer from the given four options (3 to 19):
Question 3.
I think of a number x, add 5 to it. The result is then multiplied by 2 and the final result is 24. The correct algebraic statement is
(a) x + 5 × 2 = 24
(b) (x + 5) × 2 = 24
(c) 2 × x + 5 = 24
(d) x + 5 = 2 × 24

Let number = x
⇒ i.e. x + 5
Now multiply result with 2
i.e. (x + 5) × 2
Now, final result is 24
i.e. (x + 5) × 2 = 24 (b)

#### Question  4

Which of the following is an equation?
(a) x + 5
(b) 7x
(c) 2y + 3 = 11
(d) 2p < 1

2y + 3 = 11 (c)

#### Question  5

If each matchbox contains 48 matchsticks, then the number of matchsticks required to fill n such boxes is
(i) 48 + n
(b) 48 – n
(c) 48 ÷ n
(d) 48n

Matchstick required to fill 1 matchbox
= 48 × 1 = 48
Matchstick required to fill 2 matchbox
= 48 × 2 = 96
Matchstick required to fill 3 matchbox
= 48 × 3 – 144
∴ Matchstick required to fill n matchbox
= 48 n (d)

#### Question  6

If the perimeter of a regular hexagon is x metres, then the length of each of its sides is
(a) (x + 6) metres
(b) (x – 6) metres
(c) (x ÷ 6) metres
(d) (6 ÷ x) metres

Perimeter of hexagon = x metres
6(side) = x metres
Side = (x ÷ 6) metres
∴ Side = (x ÷ 6) metres (c)

#### Question  7

x = 3 is the solution of the equation
(a) x + 7 = 4
(b) x + 10 = 7
(c) x + 7 = 10
(d) x + 3 = 7

When put the value of x = 3
3 + 7=10 (c)

#### Question  8

The solution of the equation 3x – 2 = 10 is
(a) x = 1
(b) x = 2
(c) x = 3
(d) x = 4

3x – 2 = 10
3x = 10 + 2
$x=\frac{12}{3}=x=4$ (d)

#### Question  9

The operation not involved in forming the expression 5x + $\frac{5}{x}$ from the variable x and number 5 is
(b) subtraction
(c) multiplication
(d) division

Subtraction (b)

#### Question  10

The quotient of x by 3 added to 7 is written as

$\frac{x}{3}+7$ (a)

#### Question  11

If there are x chairs in a row, then the number of persons that can be seated in 8 rows are
(a) 64
(b) x + 8
(c) 8x
(d) none of these

Let the no. of chairs in a row = x
⇒ Number of persons that can be seated in a row = x
Hence, number of persons that can be seoted in 8 row = 8x (c)

#### Question  12

If Arshad earns ₹ x per day and spends ₹ y per day, then his saving for the month of March is
(a) ₹(31x – y)
(b) ₹31(x – y)
(c) ₹31 (x + y)
(d) ₹31 (y – x)

Earning of Arshad for 1 day = ₹ x
Spending of Arshad for 1 day = ₹ y
Saving for 1 day = ₹(x – y)
Saving for 31 days = ₹31 (x – y) (b)

#### Question  13

If the length of a rectangle is 3 times its breadth and the breadth is x units, then its perimeter is
(a) 4x units
(b) 6x units
(c) 8x units
(d) 10x units

Breadth of rectangle = x units
Length of rectangle = 3(Breadth) = 3x
Perimeter of rectangle = 2(l + b)
= 2(3x + x)
= 2(4x) = 8x units (c)

#### Question  14

Rashmi has a sum of ₹ x. She spend ₹800 on grocery, ₹600 on cloths and ₹500 on education and received as ₹200 as a gift. How much money (in ₹) is left with her?
(a) x – 1700
(b) x – 1900
(c) x + 200
(d) x – 2100

Total money = ₹ x
Money spent = ₹800 on grocery
Money spent = ₹600 on cloths .
Money spent = ₹500 on education
Money left with Rashmi
= x – ₹800 + ₹600 + ₹500
= x – 1900
∴ Money left = x – 1900 + 200
= x – 1700 (a)

#### Question  15

For any two integers a and b, which of the following suggests that the operation of addition is commutative?
(a) a × b = b × a
(b) a + b = b + a
(c) a – b = b – a
(d) a + b > a

a + b = b + a

#### Question  16

In $\left(\frac{3}{4}\right)^{5}$, the base is
(a) 3
(b) 4
(c) 5
(d) $\frac{3}{4}$

$\frac{3}{4}$

#### Question  17

a × a × b × b × b can be written as
(a) a2b3
(b) a3b2
(c) a3b3
(d) a5b5

a × a × b × b × b
= a2 × b3 = a2b3 (a)

#### Question  18

(-5)2 × (-1)3 is equal to
(a) 25
(b) -25
(c) 10
(d) -10

(-5)2 × (-1)3
⇒ (-5) × (-5) x (-1) × (-1) × (-1)
⇒ 25 × (-1) = -25 (b)

#### Question  19

(-2)3 × (-3)2 is equal to
(a) 65
(b) (-6)5
(c) 72
(d) -72

(-2)3 × (-3)2
⇒ (-2) × (-2) × (-2) × (-3) × (-3)
⇒ -8 × 9 = -72 (d)

#### Question  1

Look at the following matchstick pattern of polygons. Complete the table. Also write the general rule that gives the number of matchsticks.

#### Question  2

Write an algebraic expression for each of the following:
(i) If 1 metre cloth costs ₹ x, then what is cost of 6 metre cloth?
(ii) If the cost of a notebook is ₹ x and the cost of a book is ₹ y, then what is the cost of 5 notebooks and 2 books?
(iii) The score of Ragni in Mathematics is 23 more than two-third of her score in English. If she scores x marks in English, what is her score in Mathematics?
(iv) If the length of a side of a regular pentagon is x cm, then what is the perimeter of the pentagon?

(i) Cost of 1 metre cloth = ₹ x
Cost of 6 metre cloth = ₹ x × 6 = ₹6x
(ii) Cost of 1 notebook = ₹ x
Cost of 1 book = ₹y
Cost of 5 notebooks = 5 × (₹x) = ₹5x
Cost of 2 books = (₹y) × 2 = ₹2y
Total cost of 5 notebooks and 2 books = ₹(5x + 2y)
(iii) Score of Ragni in English = x marks
Score of Ragni in Mathematics = 23 more than two-third of her score in English
i.e. = $23+\frac{2}{3}(x)$
$23+\frac{2}{3} x$
(iv) Side of regular pentagon = x cm
Perimeter of pentagon = 5 × Side = 5x cm

#### Question  3

When x = 4 and y = 2, find the value of:
(i) x + y
(ii) x – y
(iii) x2 + 2
(iv) x2 – 2xy + y2

(i) x + y
Put x = 4 and y = 2
We get,
⇒ 4 + 2 = 6

(ii) x – y
Put x = 4, y = 2
We get,
⇒ 4 – 2 = 2

(iii) x2 + 2
Put x = 4 We get,
⇒ (4)2 + 2
= 16 + 2= 18

(iv) x2 – 2xy + y2
Put x = 4 and y = 2

#### Question  4

When a = 3 and b = -1, find the value of 2a3 – b4 + 3a2b3.

2a3 – b4 + 3a2b3
Put the value of a = 3, b = -1
= 2(3)3 – (-1)4 + 3(3)2 (-1)3
= 2 × 3 × 3 × 3 – (-1) × (-1) × (-1) × (-1) + 3 × 3 × 3 × (-1) × (-1) × (-1)
= 54 – 1 – 27 = 26

#### Question  5

When a = 3, b = 0, c = -2, find the values of:
(i) ab + 2bc + 3ca + 4abc
(ii) a3 + b3 + c3 – 3abc

(i) ab + 2bc + 3ca + 4abc
Put the values of a = 3, b = 0, c = -2
3 × 0 + 2 × 0 × -2 + 3 × -2 × 3 + 4 × 3 × 0 × -2
= 0 + 0 – 18 + 0 = -18

(ii) a3 + b3 + c3 – 3abc
Put the values of a = 3, b = 0, c
= -2 (3)3 + (0)3 + (-2)3 – 3 × 3 × 0 × -2
= 27 + 0 – 8 – 0 = 19

#### Question  6

Solve the following linear equations:
(i) 2x – $1 \frac{1}{2}=4 \frac{1}{2}$
(ii) 3(y – 1) = 2(y + 1)
(ii) n – 3 = 5n + 21
(iv) $\frac{1}{3}(7 x-1)=\frac{1}{4}$

-: End of Algebra Class-6 ML Aggarwal Solutions  :–