Prime Mathematics 5

Multiplication of unlike terms with different variables

Multiplication of 2x and
3y is 2x × 3y = 6xy
Anima, what do you
understand about the
multiplication of 2x and 3y ?

When we multiply the terms with
different bases, we should multiply
only the coefficients of the bases.

In the case of multiplication of unlike terms having different variables, we
should only multiply their coefficients. In this multiplication, we can’t add
the power of the variables.

Multiplication of binomial expressions by monomial
expression

Multiply (2x+3) by 3x
The multiplication of (2x+3) and 3x is

3x × (2x+3)
= 2x × 3x + 3 ×3x
= 6x2 +9x

Each term of binomial
expression 2x+3 is multiplied

by monomial 3x.

Asha, can you multiply Let me try. -2x × (3x2 + 2x)
(3x2 + 2x ) by -2x ? = 3x2 × (-2x) + 2x × (-2x)
= -6x3 – 4x2

Asha, you have done Sir, Is my solution
correctly. Well done! correct or not?

When we multiply the binomial expressions by monomial expression then we
should multiply each terms of the binomial expressions separately by the
term of the monomial expression.

196 Prime Mathematics Book - 5

1. Multiply: Exercise: 11.5 c) 4x by 3y
a) a by b f) 2xy by 4z
d) 4a by 3b2 b) 2x by y i) -2ab by 3ac
g) 6x2 by 2y2 e) 2x2 by 5yz l) 9m2n by – 2pq2
j) 7x2z by -2y h) 5a2b by 2cd2
k) -4p by -5q2

2. Multiply:

a) (2x + 3) by x b) (3a - 1) by 2ab c) (x + 2) by 2x

d) (2y - 5) by 2y e) (4 + 5x) by 2x f) (2a2 + 1) by 3a2

g) (3a2 - 4) by 2a h) (h - 3m) by 2m i) (6p – 2q) by -3pq

j) (4x - y ) by 2x k) (3a2 - 2b2) by 5ab l) (x2 + y2) by 3xy

3. Simplify : b) 3x × 2y × z
a) a × b × 2c × 3d d) x × y × 3x ×2y
c) 2p – 4q × 3r2 f) ab × 3ab × 2cd
e) 2x × 3y × x × 2y h) 5m × (-2n) × (4p) × 3q
g) 2x2 × (-3xy) × 4y2
i) 2xy × 3pq × 5xq

4. Area of a rectangle is length (l) x breadth (b). Find the area of the
following rectangles.
a) b) c) d)

2y
2y
3x
2y

4x (2x+3) (5y+4) (2x+y)

5. a) If a= 2x + 3 and b= 4x2, find the value of a × b.
b) If w = 3y2 + 1 and z = 2x, find the value of wz.
c) If p= 2a + 3b and q = 4ab , find the value of p × q.

Prime Mathematics Book - 5 197

Division of algebraic expressions

Study and learn the following:

Let’s take the area of a rectangle of which length 2x unit be 4xy sq. units.
For finding the breadth of the rectangle, we know that

Area of the rectangle = length × breadth

or, 4xy = 2x × breadth

or, b42rxxeya=dtbhre=a2d4t2×h [dividing both sides by 2x]
or,
x × y
× x

∴ breadth = 2y units

From the above example, it is clear that the multiplication and division
process are inverse operations of each other.
From the above example, it is also clear that while dividing an algebraic
term by another term, we should divide the coefficient of dividend by the
coefficient of divisor and cancel the same base (variables) of the dividend by
the divisor.

Binu, can you divide
12x2 by 4x.

Binu, You are absolutely May I try , sir?
correct. Very good!

First we find the prime
factors of 12 and 4 and then
x2 can be written as x × x.
Divide 12x2 by 4x

= 12x2 ÷ 4x Then, we cancel the same
factors and variables of the
= 12x2 numerator and denominator.
4x Is my process correct, sir ?

= 2×2×3×x×x
2×2×x
=3x

198 Prime Mathematics Book - 5

Example 1: Divide 8x3 by 2x.

Solution : Here,

8x3 ÷ 2x
8x3
= 2x

= 2×2×2×x×x×x Factors of 8 = 2 × 2 × 2
2×x x3 = x × x × x

= 4x2

Example 2: Divide 6a2b2 by 3ab.

Solution: Here,

6a2b2 ÷ 3ab
6a2b2
= 3ab

= 2 × 3 × a × a × b × b = 2ab
3 × a × b

Division of a binomial expression by a monomial expression
Study and learn the following:

Divide (4x2 + 12x) by 2x.

Here, (4x2 + 12x) ÷ 2x

= 4x2 + 12x
2x

= 4x2 + 12x
2x 2x

= 2 × 2× x × x + 2 × 2 × 3 × x
2× x 2 × x

= 2x + 6

While dividing a binomial expression by a
monomial expression, we should separately divide

each term of dividend (numerator) by divisor
(denominator).

Prime Mathematics Book - 5 199

Lakpa, can you divide
(12x3 - 15x4) by 3x2 ?

Yes sir! Let me try.

Lakpa, your The process Lakpa followed.
answer is absolutely

correct.

(12x3 – 15x4) ÷ 3x2

= 12x3 – 15x4
3x2

= 12x3 - 15x4
3x2 3x2

= 2 × 2 × 3 ×x × x × x - 3 × 5×x ×x × x × x
3 × x× x 3× x× x

= 4x - 5x2

Exercise: 11.6

1. Perform the following division:

a) 3x ÷ x b) 6x3 ÷ 3x c) 4x2 ÷ 2x
f) (-14m5) ÷ 2m2
d) 4a2 ÷ 2a2 e) 8y3 ÷ 2y i) 26b5 ÷ 13b3
l) ( -28z9) ÷ 4z5
g) 18p3 ÷ 9p h) 6a4 ÷ 2a2

j) 10x7 ÷ (-2x4) k) 35m6 ÷ 7m5

2. Simplify: b) 8x3y2 ÷ 4xy
a) 4a2b ÷ 2ab d) 20a3b2 ÷ 4a2b2
c) 12m3y4 ÷ 3my2 f) a6b4 ÷ (-a3b2)
e) 27p3q4 ÷ 9pq2 h) (-20p4q3) ÷ 10p2q
g) 45x6y3 ÷ 5x4y2
b) (x3y - xy2) ÷ xy
3. Simplify: d) (8a3 - 2a2) ÷ 2a
a) (a2 + ab) ÷ a f) (8x3y + 12x2y3) ÷ 4x2y
c) (2b2 + 6b) ÷ 2b h) (2p2qr2 - 6pq2r3) ÷ 2pqr2
e) (10p3 + 5p2) ÷ 5p
g) (16x2yz - 12x3y2z3) ÷ 4xyz

200 Prime Mathematics Book - 5

4. Find the length of unknown side of the given rectangle from the
following condition.
a) Area of the rectangle = 12x2 sq. units and length = 4x units.
b) Area of the rectangle = 6y2 sq. units and breadth = 2y units.
c) Area of the rectangle = 32x4y3 sq. units and length = 8x3y2 units.
d) Area of the rectangle = 27x3b2 sq. units and breadth = 3a2b units.
e) Area of the rectangle = (3a2b + 12ab) sq. units and length = 3ab
units.
f) Area of the rectangle = (12x3y4 + 15x4y5) sq. units and breadth =
3x2y3 units.

5. a) Divide the product of 3x2 and 8xy2 by 4xy.
b) Divide the sum of 2x3y and 6xy2 by 2xy.

Simplification of algebraic expression with brackets
Study and learn the following example:

2x2 + x (3x + 5)
= 2x2 + x × 3x + x × 5
= 2x2 + 3x2 + 5x
= 5x2 + 5x

While simplifying the algebraic expression
involved with brackets, we should remove the brackets.

To remove the brackets, the terms inside the bracket
is multiplied individually by the term out side the
bracket along with its sign.

Ramu, can you simplify Ma'am, let me try.
3x2 - 4x (2 - 3x)?

Ramu, your solution 3x2 - 4x (2 - 3x)
is 100% correct. Very = 3x2 - 4x × 2 - 4x × (-3x)
= 3x2 - 8x + 12x2
good ! = 15x2 - 8x

Prime Mathematics Book - 5 201

Exercise: 11.7

1. Simplify: b) 4x - 2(x - 1)
a) 2x +(4x + 5) d) 5p - 2(p - 2)
c) 5y2 - 2y(4 - 3y) f) 5b2 - 2b(b + 3)
e) 3x + 2(x - 5) h) 3(x - 2) - (2x - 5)
g) 9x2 + 3x(x - 2) j) x(x - y) + y(x + y)
i) -(3a + 4) - (5 - 7a) l) 12b2 - 4b(2b - a)
k) 2(xy - 3) + 3(xy - 2x)

2. Add the sum of (2x - 3) and 4x to (5x + 2).
3. Subtract the sum of 4x and 5 from 9x + 7.
4. Subtract the product of (2p + 3) and p from 3p2.
5. Add the product of (2x - 3) and 4x to (6x2 - 2).

Algebraic equation

Introduction

Ram has some marbles (suppose x). His friend gave him 4 marbles. Now, he
has (x + 4) marbles altogether. So, x + 4 is a mathematical statement. In x +
4, x can have any value because x + 4 is satisfied by any value of x.

After getting 4 marbles from his friend, Ram has 10 marbles altogether. In
this case, the above statement can be written in the form of x + 4 = 10. It
is also an open mathematical statement but containing equal (=) sign. This
statement is true for only one value (fixed value) of x. Such type of open
mathematical statement is called equation.

What value of x substituted in x + 4, gives 10 ? When x is replaced by 6, then
x + 4 is 6 + 4 = 10. So, x = 6 is the solution of x + 4 = 10.

The other examples of equation are y - 3 = 9, 2x + 3 = 17, x = 4, 3x = 27,
etc. 2

Let's take an equation x + 5 = 9.

In the equation x + 5 = 9, x + 5 is the expression on left side of equal (=) sign.
So, it is called the expression of left hand side (L.H.S.). Similarly, 9 is the
expression on right side of equal (=) sign. So, it is called the expression of
right hand side (R.H.S.).

202 Prime Mathematics Book - 5

Solution of an equation:
Let's take a group of numbers i.e. {1, 2, 3, ......} and an equation x + 3 = 7.
Now, substitute the place of x by each element of the given set.

When, x = 1, then x + 3 = 7 becomes
1+3=7
Or, 4 ≠ 7, So, x can't be replaced by 1.

When, x = 2, then x + 3 = 7 becomes
2+3=7
Or, 5 ≠ 7, So, x can't be replaced by 2

When, x = 3, then x + 3 = 7 becomes
3+3=7
Or, 6 ≠ 7, So, x can't be replaced by 3
When, x = 4, then x + 3 = 7 becomes
4+3=7
Or, 7 = 7, which is true. So, x can be replaced by 4.
∴ 4 is called the solution of x + 3 = 7.
From the above example, it is clear that the variable contained in the
equation has only one value which satisfies the given equation.

The Facts for Solving Equations:
Solving an equation means to find the value of the variables that satisfies
the given equation. To solve the given equation, we use generally four facts
which are:
1. When we add an equal number on the both sides of two equal quantities,

the sum will remain equal. For example, If y = 2, then y + 3 = 2 + 3.

2. When we subtract an equal number from two equal quantities, the
difference will remain equal. For example,
If x = 5, then x - 2 = 5 - 2 .

3. When we multiply two equal quantities by equal numbers, the product
will remain equal. For example,
If y = 5, then y × 3 = 5 × 3.

Prime Mathematics Book - 5 203

4. When we divide two equal quantities by equal numbers, the quotient
will remain equal. For example,
x 8
If x = 8, then 2 = 2 .

Exercise: 11.8

1. A = { 1,2,3,4,5,6} is a set. Substitute the variables by each element
of set A and find the solution of the following equations.
a) x + 4 = 6 b) x + 3 = 7 c) x - 2 = 4
d) y + 1 = 4 e) y - 3 = 2 f) z + 1 = 6

2. What should be added to the both sides of the following equations to
find the value of the variables?
a) x - 3 = 7 b) y - 3 = 5 c) z - 7 = 15
d) x - 5 = 13

3. What should be subtracted from the both sides of the following
equations to find the value of the variables?
a) x + 3 = 8 b) y + 7 = 15 c) a + 5 = 19 d) z + 4 = 9

4. By what should both sides of the following equations be multiplied
to find the value of the variables?
x y a y
a) 4 = 2 b) 3 = 7 c) 5 =3 d) 2 = 5

5. By what should both sides of the following equations be divided to
find the value of the variables?
a) 3x = 21 b) 2y = 10 c) 4b = 28 d) 5x = 15

Working rule for solving equations

Study and learn the following examples:

1. Solve : x + 5 = 14

Here, x + 5 = 14 To remove 5 from the L.H.S., subtract
or, x + 5 - 5 = 14 - 5 5 on the both sides.
∴x=9

204 Prime Mathematics Book - 5

2. Solve : 2x-3 = 11

Here, 2x-3 = 11 To find the value of x, first of all,
we need to remove 3, add 3 on
or, 2x - 3 + 3 = 11 + 3 the both sides. Then, the result
is 2x = 14. Now, to find the value
or, 2x = 14
of x, divide both sides by 2.
or, 2x = 14 7
x 2 [Multiplying both sides by 5]
[Dividing both sides by 3]
∴ x=7
[Multiplying both sides by 3]
3. SHoore,lrvee,3 5:3y5y35×y=5=2=12211 × 5 [Adding 3 on both sides]
or, 3y = 105 [Dividing both sides by 2]

or, 3y = 103 35
3 3
∴ y=5

4. Solve : 2x-3 = 1

Here, 2x-3 = 1
3
2x-3
or, 3 × 3 = 1 × 3

or, 2x - 3 = 3

or, 2x - 3 + 3 = 3 + 3

or, 2x = 6

or, 2x = 63
2 2
∴ x=3

From the above examples, it is clear that
1. If there is a number with plus (+) sign in the side of the

variable, subtract that number from the both sides.
2. If there is a number with minus (-) sign in the side of the

variable, add that number from the both sides.
3. If there is a number as coefficient of a variable, divide

both sides by that number.
4. If there is a number as denominator of a variable, multiply

both sides by that number.

Prime Mathematics Book - 5 205

Exercise: 11.9

1. Solve: b) y + 2 = 7 c) x + 9 = 12
a) x + 3 = 9 e) x - 2 = 5 f) y - 5 = 4
d) y + 6 = 15 h) m + 5 = 13 i) 2y + 1 = 9
g) a - 3 = 7 k) 3 + 5y = 28 l) 4m - 2 = 14
j) 3y + 3 = 12 n) 2x + 5 = 19 o) 5x - 1 = 14
m) 3x - 4 = 11
p) 3p + 2 = 14

2. Solve:

a) x = 4 b) x = 2 c) y = 3
3 5 4
y 2x 3a
d) 3 = 7 e) 5 = 4 f) 4 = 9

g) 5x = 35 h) 4m = 12 i) p - 2 = 4
7 5 3
+3 3x - 1 2x
j) x 2 = 5 k) 4 = 5 l) 3 + 5 = 7

m) 5x + 2 = 4 n) 8 + 2y = 3 o) 3a - 2 = 10
3 4 4
3y + 2
p) 2 = 7

3. Solve :

a) 2x + 3 = x + 5 b) 2x - 3 = 7 - 3x c) 7y - 8 = 5y + 4

d) 4x - 2 = 3 e) 5y + 2 = y f) 3m + 4 = m
x 3 2
g) 9 - (2 - 3x) = 10 h) 3(x - 2) + 1 = 10 i) 2(y + 3) - y = 7

4. Solve: b) c)
a)

2x+5kg 9 kg 2x+x+3 x+9 x+x+x+2kg 11kg

206 Prime Mathematics Book - 5

Word Problems

We use an equation to solve the word problems. Firstly, we should write
the word problems into the mathematical form with consideration of the
unknown quantity as variable, like x, y, z etc. The mathematical form is the
equations. Then, we solve the equation and find the value of the variable
which is the required solution of the given word problems.

Study and learn the following examples:

Example 1. If 5 is added to a number, the sum becomes 19. Find the
number.

Solution: Let the number be x
Then, according to question,
x + 5 = 19
or, x + 5 - 5 = 19 - 5
or, x = 14
∴ The required number is 14.

Example 2: When 2 is subtracted from one third of a number, the result
is 20. Find the number.

Solution: Let the number be x.

Then, according to question

x – 2 = 20
3
x
or, 3 - 2 + 2 = 20 + 2

or, x = 22
3
x
or 3 × 3 = 22 × 3

∴ x = 66

∴ The required number is 66.

Prime Mathematics Book - 5 207

Example 3. A boy thought of a number. He doubled it and added 7. If he
got 23, find the number which he thought.

Solution: Let the number which the boy thought be x

Then,

2x + 7 = 23 x 2x 2+x7 + 7

or, 2x + 7 - 7 = 23 - 7 Reverse operation -7

or, 2x = 16 and then ÷ 2.

or, 2x = 16 8
2 2
∴ x=8

∴ The required number which he thought is 8.

Example 4. The perimeter of a rectangle is 34cm. If the length of the
rectangle is 10cm, find its breadth.

Solution: Let the breadth be x cm.

We know that,

Perimeter = 2 (length + breadth )

= 2(10 + x)

But according to question,

Perimeter = 34cm The perimeter of a rectangle
is two times the sum of length
∴ 2(10 + x) = 34
and breadth.
or, 2(10 + x) = 3417 Perimeter = 2(length +breadth)
2 2

or, 10 + x = 17

or, 10 + x - 10 = 17 – 10

∴ x=7

∴ The breadth of the rectangle is 7cm.

208 Prime Mathematics Book - 5

Exercise: 11.10
1. Express the following problems in mathematical equation and solve

them to find the unknown quantities:
a. The sum of two numbers is 17. If one of them is 8, find the another

number.
b. The difference of two numbers is 6. If the greater number is 17,

find the another number.
c. If 19 is added to y, the sum becomes 41. Find the value of y.
d. What should be added to 26 to make the sum 45?
e. What should be subtracted from 27 to get 12?
2. Make the equations from the following problems and solve them:
a. When 3 is added to one fourth of a number, the result is 11. Find

the number.
b. When 4 is subtracted from one third of a number, the result is 7.

Find the number.
c. If 4 times a number is 28, find the number.
d. If two third of a number is 14, find the number.
e. Find the number whose three fourth is 27.
3. Make the equation and solve them:
a. If one number is greater than other by 12 and their sum is 50, find

the numbers.
b. If one number is smaller than other by 9 and their sum is 31, find

the numbers.
c. Adarsh thought of a number. He doubled it and added 11. If he got

45, find the number which he had thought.
d. Anima thought of a number. She added 5 to it and the sum is

multiplied by 3. If she got 66, what number did she think? Find it.
4. a. The perimeter of a rectangle is 38cm. If the length of the rectangle

is 12cm, find its breadth.
b. The perimeter of a rectangle is 42cm. If the length of the rectangle

is 3cm, more than its breadth, find the length and breadth of the
rectangle.
c. The three sides of a triangle are (x + 2)cm, xcm and (x - 1) cm
respectively. If the perimeter of the triangle is 22cm, find the
sides of the triangle.

Prime Mathematics Book - 5 209

Unit Revision Test

1. Write the coefficient, base and power of the following terms:

a. 5x3 b. -2y7 c. - 5 (xy)-2
7

2. Write the term having :-
a. Base = x, coefficient = 3 and power = 4
b. Base = a, coefficient = -2 and power = 3

3. If x = 2, y = 3 and z = -2, find the values of :
a. 3x2 + y - 2z b. xy + 3y2 - 3z c. 2x2 - xy2

4. Add the following:
a. 2a + 5b, 3a - 2b + 3, a + 7 + 2b
b. 2x + 3y - 3, 2x - 5y + z, 4x + 7z + 4y + 5

5. Subtract:
a. 2x + 3 from 5x - 2
b. 5x + 8y - 4z from 9x + 12y + 2z
c. 3a2 + 2ab + 5b2 from 5a2 + 5ab - 2b2

6. Simplify: b. 2x × 3xy × 2y2
a. x + 4y - 3x + 2y

7. Multiply: b. (4x - 1) by 2xy c. (a2 + b2) by 3ab
a. 4a2 × 3a × -2a3

8. If a = 2x - 5 and b = 3x3, find the value of a × b.

9. Divide: b. (12x2y + 20xy2) ÷ 4xy
a. 18x3 ÷ 6x

10. Solve: b. 4x - 3 = 17 + 2x
a. 3x - 2 = 19

11. When 6 is added to two third of a number, the result is 22. Find the
number.

12. The perimeter of a rectangle is 40cm. If the length of the rectangle
is 12cm, find its breadth.

210 Prime Mathematics Book - 5

Answers:

Exercise: 11.1

1. a) constant b) variable c) variable d) variable

2. a) 3xy, 1 b) 2x, 3yz ; 2 c) 2a2b, - 4bc, 2b2d; 3 d) 4xy2, - 6yz; 2

3. a) binomial b) trinomial c) monomial d) binomial

e) monomial f) multinomial g) monomial h) multinomial

4. a) coefficient = 2 b) coefficient = - 3 c) coefficient = 7a

base = x base = p base = x

power = 5 power = 1 power = 3

d) coefficient = -9b e) coefficient = 23
base = y
base = x

power = 7 power = 2

5. a) 3x2 b) 7y5 c) 5az3 d) -2bz7

6. a) like b) like c) unlike d) unlike e) unlike f) unlike

7. a) -1 b) 9 c) 13 d) 18 e) 21

8. a) 8 b) 26 c) 2 d) 24 e) 2

9. a) 10 b) -3 c) 26 d) 14 e) 18

10. a) 11 b) 24 c) 34 d) 2 e) 4

11. a) x+y+z, 10 b) x+y+2z, 12 c) 3x+y+2z, 18

Exercise: 11.2

1. a) 2x b) 3x c) 4y d) 6y e) 9x f) 5a2 g) 12a2
l) 9abc m) 3x n) 2y
h) 9a5 i) 5ab j) 22ab k) 22yz s) 2xy t) -4m3m2
e) x2y f) 17ab2 g) 18x3
o) 4z2 p) 4x3 q) 2ab r) 5a2b
e) 8ab
2. a) 6x b) 5a c) 10x d) 5ab
3. a) 8 b) -3 c) 12 d) 18

4. a) 12cm b) 20cm c) 23cm d) 20cm

5. a) 13x b) 6x c) 8a d) 5xy (f) 8ab

Exercise: 11.3

1. a) 3a + 7 b) 5x + 9 c) 11x - 3 d) 5a + 8b
e) 11m - 5n h) 5x + 8y + 3z
i) 8a - 5b + 7c f) 7x2 + 5y g) 7ab + 7bc l) 9a2 + 7ab + 7b2

j) 9a2 + 8b2 + 16 k) 11x2 + 9xy + 11y2

Prime Mathematics Book - 5 211

m) 5x3 + 9x2 + 5x - 2 n) 11x + 8y o) 11a + 11b p) 6x2 + 3x + 12

q) 10ab + 12bc + 14ab

2. a) 4x + 5 b) 5x + 6 c) 3a + 6b

d) 5m + 5n e) 2x2 + 9y2 f) 5x2 + 2b2

g) 3x + 2y + 4 h) 2x + 5y + 3z i) -2a - 11b + 6c

j) 3a2 - 5ab + 4b2 k) 4ab - 10bc + 6ca l) 3x3 + 3x2 - 3x + 8

3. a) 4a + b b) 9x + 7y c) 5x2 + 2x + 8

d) 3m + 9n + 2p e) a2 + 2a + 7 f) -5pq + 2qr + rp

4. a) 5x + 12 b) 2a + 7b c) 5x + 6y + 8 d) 6a + 3b + 8c

5. a) 3x + 4 b) 6a + 2b c) 4a + 4b + 2c

6. a) 6a + 5b, 2a - b b) 5x - 10, 13x + 4 c) 12x + 11y + 11, x + 12y + 5

7. 24cm 8. 34cm

Exercise: 11.4

1. a) 12a b) 14x2 c) 12x2 d) a5 e) x3 f) 10b3

g) 18m5 h) 48p10 i) -12x3 j) -20x6 k) 35q4 l) -60z8

2. a) a3 b) 6x3 c) -12y3 d) 8y6 e) -6b6 f) 10x4

g) -14y6 h) 6p6 i) 24a6 j) 216z5 k) 252x4

3. a) x2cm2; 9cm2 b) 4x2cm2; 36cm2 c) 16x2cm2; 144cm2 d) 9x2cm2; 81cm2

4. a) 2y2cm2; 8cm2 b) 6y2cm2; 24cm2 c) 15y2cm2; 60cm2

5. a) 8a3cm3; 64cm3 b) 18a3cm3; 144cm3 c) 24a3cm3; 192cm3

6. a) 48 b) 108

1. a) ab b) 2xy Exercise: 11.5 d) 12ab2 e) 10x2yz
c) 12xy i) -6a2bc j) -14x2yz

f) 8xyz g) 12x2y2 h) 10a2bcd2 d) 4y2 - 10y e) 12x2y2
h) 2mn - 6m2
k) 20pq2 l) -18m2npq2 l) 3x3y + 3xy3
d) 6x2y2
2. a) 2x2 + 3x b) 6a2b - 2ab c) 2x2+ 4x i) 30x2ypq2
d) 4xy + 2y2
e) 8x + 10x2 f) 6a4 + 3a2 g) a3 - 8a

i) -18p2q + 6pq2 j) 8x2 - 2xy k) 15a3b - 10ab3

3. a) 6abcd b) 6xyz c) 24pqr2

f) 6a2b2cd g) -24x3y3 h) -120mnpq

4. a) 8xy b) 4xy + 6y c) 15xy + 12x

5. a) 8x3 + 12x2 b) 6xy2 + 2x c) 8a2b + 12ab2

212 Prime Mathematics Book - 5

Exercise: 11.6

1. a) 3 b) 2x2 c) 2x d) 2 e) 4y2 f) -7m3
k) 5m l) -7z4
g) 2p2 h) 3a2 i) 2b2 j) -5a3 e) 3p2q2 f) -a3b2
2. a) 2a b) 2x2y c) 4m2y2 d) 5a
f) 2x + 3y2
g) 9x2y h) -2p2q2
f) 4xy + 5x2y2
3. a) a + b b) x2 - y c) b + 3 d) 4a2 - a e) 2p2 + p
h) p - 3qr
g) 4x - 3x2yz2 c) 4xy

4. a) 3x b) 3y d) 9ab e) a + 4

5. a) 6x2y b) x2 + 3y

1. a) 6x + 5 b) 2x + 2 Exercise: 11.7 e) 5x - 10 f) 3b2 - 6b
c) 11y2 - 8y d) 3p + 4

g) 12x2 - 6x h) x - 1 i) 4a - 9 j) x2 + y2 k) 5xy - 6x - 6 l) 4b2 + 4ab

2. 11x - 1 3. 5x + 2 4. p2 - 3p 5. 14x2 - 12x - 2

1. a) 2 b) 4 c) 6 Exercise: 11.8 e) 5 f) 5
2. a) 10 b) 8 c) 22 d) 3
3. a) 5 b) 8 c) 14 d) 18
4. a) 8 b) 21 c) 15 d) 5
5. a) 7 b) 5 c) 7 d) 10
d) 3

1. a) 6 b) 5 c) 3 Exercise: 11.9 g) 10 h) 8 i) 4
j) 3 k) 5 l) 4 d) 9 e) 7 f) 9 p) 4 h) 15 i) 14
b) 10 c) 12 g) 49 h) 5 i) 1
2. a) 12 k) 7 l) 3 m) 5 n) 7 o) 3 p) 4
j) 7 b) 2 c) 6 g) 1
b) 3 c) 3 d) 21 e) 10 f) 12
3. a) 2
4. a) 2 m) 2 n) 2 o) 14

d) 2 e) -1 f) -4

Exercise: 11.10

1. a) 9 b) 11 c) 22 d) 19 e) 15
2. a) 32 b) 33 e) 36
3. a) 31, 19 b) 20, 11 c) 7 d) 21
4. a) 7cm b) 9cm, 12cm
c) 17 d) 17

c) 9cm, 7cm, 6cm

Prime Mathematics Book - 5 213

Some additional question for practice

1) Find the value of x form the following figure.
DC

80°

A x° 4x° B
O

2) From the following figure, find the value of x.
P

x

65° R
Q

3) If 2x, 3x, 4x, 5x, 6x are the interior angle of a pentagon. Find the value
of x.

4) A group of 729 students are arranged in rows and columns. If equal
number of students are kept in rows and columns. What is the number
of students in a row?

5) Find the HCF of 275, 325 and 400.
6) Find the LCM of 48, 72 and 96.
7) The number of vehicles in three towns are 45673, 54065 and 63068. Find

the total number of vehicles in the town.

214 Prime Mathematics Book - 5

8) The population of a town in 526729. If the male population and female
population of the town are 200583 and 225308 respectively excluding
children find the number of children in the town.

9) Find the smallest number that should be added to 22141. So that the
result is exactly divisible by 35.

10) Simplify:

81 + [51 - {48 ÷ 3 - 8 - 7 + 4} + 5]

11) A square has its area and perimeter are equal find its length.

12) Find the volume of a cupboard in which l = 6.2m, b = 4.2m and j=1.25m.

13) Simplify

4 + 8 + 3
5 20 10

14) Ram got 460 marks out of 800 full marks in the final examination of
grade 4 what percent makes did he get?

15) Divide 7kg 250gm by 7.

16) Find the volume of given figure.

4 cm

3 cm 5 cm

17) Simplify

1 3 + 2 7 + 3 4
4 10 15

18) Evaluate :

a) 40% of 300

b) 33% of 900

19) If the simple interest of Rs. 100 per year is Rs. 6, find the interest of Rs.
25000 for 5 year.

Prime Mathematics Book - 5 215

20) Find the selling price when CP = Rs 560 and profit = 25%
21) The marks scored by Prinsha in first term in different subject are given.

Show the information in a bar graph.

Subjects Nepali English Maths Science Social
Marks Scored 75 80 95 85 55

22) Find the perimeter of given figure. If x = 3, y = 5 and z = 2.
A

z

B x+y C
23) Subtract : 5x - zy + z form 7x + 3y - 4z
24) Multiply : (3xy + 2y2) by 5xy
25) Divide : (10p3 + 5p2) ÷ 5p

216 Prime Mathematics Book - 5

Subject: Mathematics Model Question Full Marks: 100
Grade: V Pass Marks: 40
Final Examination Time: 3 hours

Attempt all the questions :

Group ‘A’ (10 × 1 = 10)
iv. 1800
1. Select the correct answer :

a. What is the complement of 600 ?

i. 1200 ii. 300 iii. 900

b. What is the sum of angles of a trangle?

i. 600 ii. 900 iii. 1800 iv. 3600

c. Which is the greatest number of 6 digits?

i. 6 ii. 100000 iii. 666666 iv. 999999

d. How many lakhs are there in 1 million?

i. 1 ii. 10 iii. 100 iv. 1000

e. Cube root of 125 is…….. iii. 10 iv. 100
i. 5 ii. 25

f. A set of even prime numbers between 10 and 20 is a
i. Finite set ii. Infinite set iii. Null set iv. Singleton set

g. 1 litre is equal to

i. 1 m3 ii. 1000cm3 iii. 100cm3 iv. 100m3

h. Value of (-11) × (-6) is iii. -17
i. -66 ii. 66 iv. 5

i. 4 ÷ 1 is equivalent to
5

i. 4 ii. 5 iii. 1 iv. 20
5 4 20
Prime Mathematics Book - 5 217

j. 25 % of Rs 100 is

i. Rs 25 ii. Rs 2.5 iii. Rs 0.25 iv. Rs 50

Group ‘B’ (17 × 2 = 34)

2. a. Find the value of x from the given figure without actual
measurement.
C

A x 50O B
O

b. Construct an angle of 1200 using protractor.

3 a. Write the number name of

i. 7,27,608 ii. 371,456,072

b. Write the place value of the following underlined digits:

i. 4,32,765,892 ii. 76,53,24,089

4. a. Write all the prime numbers between 30 and 50.
b. Find the H.C.F. of 60 and 75 using prime factorization method

5. a. Compare: 2 and 67.
3

b. Convert 15% into fraction in simple form.

6. a. Write the given statement in mathematical form and simplify.
i. Add 6 with the product 4, 9 and 4

ii. 6 times the difference of 19 and 7 is divided by 18.

b. Write the following sets in roster form.
i. The set of even numbers less than 11.
ii. Set of the days of a week starting from the letter ‘s’.

218 Prime Mathematics Book - 5

7. a. Simplify.
16bc + 3cd – 6bc – cd

b. Solve.
5x - 1 = 14

8. a. Subtract Rs. 64 and 85p from Rs. 212 and 28p.
b. Convert 2 kg 4 hg 25gm into gm.

9. a. Find area of the rectangle having length = 12cm and breadth =
8cm.

b. Calculate the volume of the cuboid given along side.
3cm

6.5cm 4.2cm

c. The cost of 7 pencils is Rs. 63. What is the cost of 12 such pencils?

Group ‘C’ (14 × 4 = 56)

10. From the figure given alongside, find the value of a and b.

100o

a 25o

b 75o

11. What should be added to 52,437 to make it the greatest 6 digit
number ?

12. Put the number 321564908 in local place value table and write the
number name.

13. Find the cube root of 216.
14. Simplify:

6 ÷ [5 - 14 ÷ {5 + 8 ÷ (6 - 2)}]

Prime Mathematics Book - 5 219

15. Himani distributed 144 chocolates among some of her friends on her
birthday. If each got as many chocolates as there were her friends,
for how many friends did she distribute the chocolates to?

16. Simplify: 8
2 5 9
1 3 + 6 -

17. The product of two fractions is 2185. If one of the fraction is 1 4 , find
the other fraction. 5

18. If the interest of Rs. 100 for a year is Rs. 8, what is the interest of
Rs. 400 in 2 years?

19. An aquarium is 1.5 m long, 90cm broad and 72cm high. How many
litres of water is required to fill the aquarium?

20. A fruit seller brought 200 oranges. If 10% of them are rotten, how
many of them are good?

21. The perimeter of a rectangle is 42cm. If length of the rectangle is
3cm more than it’s breadth, find the length of the rectangle.

22. Isha from Chabahil bought two shirts for Rs. 1050 each, a pant for Rs.
1450 and three T-shirts for Rs. 450 each from U.F.O. Kathmandu on
16/05/2068. Prepare a bill for Isha of her purchase.

23. The table given below shows number of students enrolled in different
classes of XYZ primary school in a year.

Classes I II III IV V

No. of students 15 25 40 30 50

Represent the information in a bar diagram.

220 Prime Mathematics Book - 5

Prime Mathematics Series

Prime Mathematics - 1
Prime Mathematics - 2
Prime Mathematics - 3
Prime Mathematics - 4
Prime Mathematics - 5
Prime Mathematics - 6
Prime Mathematics - 7
Prime Mathematics - 8
Prime Mathematics - 9
Prime Mathematics - 10

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