Math 8

Worked Out Examples



Example: 1

Add: 10112 and 11112
Solution:

Carry overs → 1 1 1 Steps

First number → 1 0 1 12 • 1 + 1 = 2 = 102 : set down 0, carry 1
Second number → + 1 1 1 12
• 1 + 1 + 1 = 3 = 112 : set down 1, carry 1
• 1 + 0 + 1 = 2 = 102 : set down 0, carry 1
• 1 + 1 + 1 = 3 = 112 : set down 1, carry1
1 1 0 1 02

Hence, 10112 + 11112 = 110102

Example: 2

Add: 1110112 and 1111112 Steps
Solution:
• 1 + 1 = 2 and 210 = 102. : set down 0 and carry 1.
Carry overs → 1 1 1 1 1 1 • (1) + 1 + 1 = 3 and 310 = 112. : set down 1 and carry 1.
• (1) + 0 + 1 = 2 and 210 = 102.: set down 0 and carry 1.
1 1 1 0 1 1 2 • (1) + 1 + 1 = 3 and 310 = 112. : set down 1 and carry 1.
• (1) + 1 + 1 = 3 and 310= 112. : set down 1 and carry 1.
+ 1 1 1 1 1 1 2 • (1) + 1 + 1= 3 and 310 = 112. : set down 1 and carry 1.
• We finally set down 1.
1 1 1 1 0 1 02

∴ 1110112+ 1111112 = 11110102

Example: 3

Add the binary numbers: 11011012, 10101012 and 10110102
Solution:


Carry overs → 1 1 1 1 1 1

1 1 0 1 1 0 1 2

1 0 1 0 1 0 1 2

+ 1 0 1 1 0 1 0 2
1 0 0 0 1 1 0 02

Hence, 1011012+10101012+10110102 = 1000111002

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Example: 4 Steps

Subtract: 111012 from 11100112. • 1 – 1 = 0, set 0 down.
Solution: • 1 – 0 = 1, set 1 down.
• As 0 < 1, borrow 1 from higher places then 10-1=1, set 1 down.
0 10 • 0 – 0=0, set 0 down.
• 1–1=0, set 0 down.
1 1 1 0 1 12 • set 1 down.
– 1 0 1 0 12
1 0 0 1 1 02

∴ 1110112 – 101012 = 1001102

Example: 5

Simplify: 110112 + 10012 – 11112
Solution:

10 11

1 1 0 1 12
+ 1 0 0 12
1 0 0 1 0 02

∴ 110112 + 10012 = 1001002
Again,

1 1 10 1 10

1 0 0 1 0 02
– 1 1 1 12
1 0 1 0 12
∴ 1001002 – 11112 = 101012
Hence, 11011+10012 – 11112 = 101012

Exercise 9.2

1. Carry out the addition of the following binary numbers.

(a) 112 + 112 (b) 1012 + 1112 (c) 11002 + 102

(d) 111112 + 110112 (e) 1000012 + 1011012

(f) 10101012 + 10000012

2. Carry out the addition of the following binary numbers.

(a) 111102 + 101112+ 100012 (b) 10101012 + 10110102 + 11011012

146 Oasis School Mathematics-8

3. Carry out the subtraction of the following binary numbers:

(a) 1112 – 102 (b) 11002 – 112 (c) 100002 – 12
(d) 1111112 – 1000012 (e) 1001012 - 110112
4. Simplify:

(a) 11102 + 10102 (b) 111012 – 11002 + 110012

(c) 1111012 – 10012 + 11002 (d) 1111112+101112 – 101012



Answer

1. (a) 1102 (b) 11002 (c) 11102 (d) 1110102 (e) 10011102 (f) 100101102
2. (a) 10001102 (b) 1000111002 3. (a) 1012 (b) 10012 (c) 11112 (d) 111102 (e) 10102
4. (a) 110002 (b) 1010102 (c) 10000002 (d) 10000012

9.5 Addition and Subtraction of Quinary Numbers

a. Addition of Quinary Numbers

In quinary system, when the sum of the digits becomes 5 or more, we begin to take
carry 1 five, 2 fives, 3 fives, etc. to the higher places.

As there are five numerals 0, 1, 2, 3 and 4 in quinary system, the following addition
table gives us the rules of addition of quinary numbers.

+0 1 2 3 4 Remember !

1 1 2 3 4 105 1 + 0 = 1 3+0=3
1 + 1 = 2 3+1=4
22 3 4 105 115 1 + 2 = 3 3 + 2 = 105
1 + 3 = 4 3 + 3 = 115
3 3 4 105 115 125 1 + 4 = 105 3 + 4 = 125

44 105 115 125 135 2 + 0 = 2 4+0=4
2 + 1 = 3 4 + 1 = 105
2 + 2 = 4 4 + 2 = 115
2 + 3 = 105 4 + 3 = 125
For example, 2 + 4 = 115 4 + 4 = 135

Add: 344105 + 323435

Carry over → 1 1 1 Steps:

3 4 4 1 0 5 • 0 + 3 = 3. : set 3 down.
+ 3 2 3 4 3 5 • 1 + 4 = 5 and 510 = 105. : set 0 down and carry 1.
• 4 + 3 +( 1) = 8 and 810 = 135. : set 3 down and carry 1.
1 2 2 3 0 3 5 • 4 + 2 + (1) = 7 and 710 = 125. : set 2 down and carry 1.
• 3 + 3 + (1) = 7 and 710 = 125. : set 2 down and carry 1.
⇒ 3 4 4105 + 323435 = 1223035

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b. Subtraction of Quinary Numbers

In quinary system, when the greater quinary numeral is subtracted from the smaller

one, we borrow 5 from the previous digit at higher place value to the digit at lower

place value.

Let's see the following example, Remember !

For example, 105-1 = 4 115-1=105 125-1=115 135-1=125
105-2=3 115-2=4 125-2=105 135-2=115
105-3=2 115-3=3 125-3=4 135-3=105
Subtract: 32105 – 21145 105-4=1 115-4=2 125-4=3 135-4=4

Steps

1 10 10 • As 0 < 4, we borrow 1 from higher place then 105– 4=1, set 1 down.
• As 0 < 1, borrow 1 from higher place105–1=4, set 4 down.
3 2 1 05 • 1 – 1 = 0. So, set 0 down
– 2 1 1 45

1 0 4 15 • 3 – 2 = 1. So, set 1 down.

∴ Required difference = 10415

Worked Out Examples



Example: 1

Simplify: 2 3 4 1 2 5 + 1 0 2 1 4 5
Solution:

11 Steps

23 412 • 2 + 4 = 115, set 1 down, carry 1.
5 • 1 + 1 + 1 = 3, set 3 down.
• 4 + 2 = 115, set 1 down, carry 1.
+1 0 2 1 4 • 1 + 3 = 4, set 4 down.
5 • 2 + 1 = 3, set 3 down.

3 413 1 5



∴ 234125 + 102145 = 341315

Example: 2

Carry out the subtraction of the following quinary numbers: 231435 – 204145.
Solution:

2 11 3 13 Steps

23143 • 3 < 4, borrow 1 from higher place then 135 – 4 = 4, set 4 down.
5 • 3 – 1 = 2, set 2 down.
• 1 < 4, borrow, 1 from higher place 115 – 4 = 2, set 2 down
–2 0 4 1 4 • 2 – 0 = 2, set 2 down.
5

2 2 2 45

∴ Required difference = 22245

148 Oasis School Mathematics-8

Example: 3

Carry out the addition of the following quinary numbers (i.e. base-5 numbers).

324105 + 123445 + 310015 Steps

Solution: (i) 0 + 4 + 1 = 105. So, set 0 down and carry 1.
(ii) 1 + 4 + 0 + (1)=115. So, set 0 down and carry 1.
Carry overs → 1 1 1 1 (iii) 4 + 3 + 0 + (1)=135. So, set 3 down and carry 1.
3 2 4 1 05 (iv) 2 + 2 + 1 + (1)=115. So, set 1 down and carry 1.
+ 1 2 3 4 45 (v) 3 + 1 + 3 + (1)=135. So, set 3 down and carry 1.
+ 3 1 0 0 15 (vi) Set down 1.
1 3 1 3 1 05

∴ Required sum = 1313105

Exercise 9.3

1. Carry out the addition of the following quinary numbers (base 5 numbers).

(a) 13425 + 4135 (b) 100045 + 4325 (c) 444445 + 111115
(d) 123445 + 343335 (e) 33335 + 10045

2. Carry out the subtraction of the following quinary numbers (base 5 numbers).

(a) 3315 – 1315 (b) 43205 – 32235
(c) 20325 – 3435 (d) 40005 – 235

3. Simplify:

(a) 203145 + 112035 – 21135
(b) 32015 – 21105 + 314215
(c) 421305 + 220335 – 100345
(d) 1102145 – 20125 – 312035



Answer

1. (a) 23105 (b) 104415 (c) 1111105 (d) 1022325 (e) 43425
2. (a) 2005 (b) 10425 (c) 11345 (d) 34225
3. (a) 244045 (b) 330125 (c) 1041245 (d) 214445

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Unit

10 IInntetgeegresrs

10.1 Integers

Introduction:

We have already learnt about natural numbers. Let us consider the set of natural
numbers.

N = {1, 2, 3, 4, 5, ......... }

Natural numbers are closed under addition and multiplication. i.e. sum and product
of two natural numbers is also a natural number.

Let us be clear with an example.

3 and 5 are natural numbers.
3 + 5 = 8 which is a natural number.
3 × 5 = 15 which is also a natural number.

Let us see the case of subtraction.
(3 – 5) = – 2 which is not a natural number.

i.e. natural numbers may not always be closed under subtraction. So, in order to
make the operation of subtraction meaningful, negative numbers and zero are
introduced.

The set of natural numbers together with their negatives including zero are called
integers.

The set of integers is denoted by Z.

∴ Z = { ..........., –4, –3, –2, –1, 0, 1, 2, 3, 4, ......... }

Z+ = {1, 2, 3, 4, ........... }

= a set of positive integers.
Z - = {–1, –2, –3, –4,........... }

= a set of negative integers.


Integers are closed under addition, subtraction and multiplication.

150 Oasis School Mathematics-8

Laws of addition on integers

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

Take any three integers – 3, 4 and 1

–3 + 4 = 1, which is an integer.

4 + 1 = 5, which is an integer.

(–3 + 1) + 4 = 2, which is an integer.

Hence, we conclude that the sum of integers is again an integer.

If a, b and c be any three integers then a + b is also an integer and a + b + c is also
an integer.

This property is called closure property.

Again, – 3 + 4 = 1, 4 + (–3) = 1
i.e, – 3 + 4 = 4 + (–3)

∴ a+b = b+a

This property is called commutative property.

Again, (–3 + 4) + 1 = 2

– 3 + (4 + 1) = 2

i.e. (–3 + 4) + 1 = – 3 + (4 +1)

∴ (a + b) + c = a + (b + c)

This property is called associative property.

Again,

– 3 + 3 = 0

3 + (–3) = 0

i.e. a + (–a) = 0 Remember !

This property is the inverse property. + ve + ve = + ve
+ ve + – ve = + ve [If greater number is+ve]
Here, –a is the additive inverse of a. + ve + – ve = – ve [If greater number is–ve]
– ve + – ve = – ve
Remember !

If a, b and c be any three integers, then

i. a + b and a + b + c are also integers. [Closure property]

ii. a + b = b + a [Commutative property]

iii. (a + b) + c = a + (b + c) [Associative property]

iv. a+ 0 = 0 + a = a [Identity property]

v. a + (–a) = (–a) + a = 0 [Inverse property]

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Laws of multiplication on integers

Let's take any three integers 2, –3, 5.

Here, 2 × (–3) = – 6 is an integer.

2 × 5 = 10 is an integer.

∴ If a and b be any two integers then a × b is also an integer.

It is called closure property.

Again, 2 × (–3) = – 6

–3 × 2 = – 6

∴ 2 × (–3) = –3 × 2

i.e. a×b = b×a

It is called commutative property.

Again, (2×(–3)) × 5 = – 6 × 5 = – 30
2 × (–3 × 5) = 2 × (–15) = – 30
(2 × (–3)) × 5 = 2 × (–3 × 5)
∴

i.e. a × (b × c) = (a × b) × c

This property is called associative property.

A gain, 2 × (– 3 + 5) = 2 × 2 = 4

2 × (–3) + 2 × 5 = – 6 + 10 = 4

∴ 2 × (–3 + 5) = 2 × (–3) + 2 × 5

i.e. a × (b + c) = a × b + a × c

This property is called distributive property. Remember !
Again,
2 × 1 = 2, 5 × 1 =5 , –3 × 1 = 3 + ve × + ve = + ve, + ve × – ve = – ve
– ve × – ve = +ve
i.e a × 1 = a +ve ÷ + ve = + ve, + ve ÷ – ve = – ve
This property is called identity property. – ve ÷ + ve = –ve, – ve, ÷ – ve = + ve

Hence, if a, b and c be any three integers then
i. a × b and a × b × c are also integers. (Closure property)
ii. a × b = b × a (Commutative property)
iii. (a × b) × c = a × (b × c) (Associative property)
iv. a × (b + c) = a × b + a × c (Distributive property)
v. a × 1 = 1 × a = a (Identity property)
vi. a × 0 = 0 × a = 0
Note : Division of any integers by zero is not defined.

152 Oasis School Mathematics-8

Exercise 10.1

1. Write the additive inverse of the following integers.
(a) +6 (b) –9 (c) +7 (d) –2

2. Evaluate: (b) + 8 +(–3) (c) +7 + (–11)
(a) –5 + 4 (e) (–6) + (–10) (f) (–7) – (–8)
(d) (–8) + (–3) (h) +18 – (–12)
(g) (–12) + (–4)
3. Evaluate: (b) (–5)×(–3) (c) (–7) × (–4)
(a) (–3) × (+4) (e) (+5) × (+6) (f) (–8) × (+4)
(d) (+2) × (–8)
4. Evaluate : (b) –24 ÷ (–3) (c) (–15) ÷ (+5)
(a) 12 ÷ (–3) (e) (+48) ÷ (+8) (f) (–16) ÷ (–4)
(d) (+30) ÷ (–6)
5. Simplify: (b) (+15) × (–3) × (–5)
(a) (+5) × (–3) × (+2) (d) (+18) × (–2) × (–3)
(c) (+12) × (–3) × (–1)

Answer

1. (a) – 6 (b) + 9 (c) – 7 (d) + 2 2. (a) – 1 (b) + 5 (c) – 4 (d) – 11 (e) – 16 (f) + 1
(g) – 16 (h) + 30 3. (a) – 12 (b) + 15 (c) + 28 (d) – 16 (e) + 30 (f) – 32
4. (a) – 4 (b) + 8 (c) – 3 (d) – 5 (e) + 6 (f) + 4 5. (a) – 30 (b) + 225 (c) + 36 (d) + 108

10.2 Simplification of Integers

Order of Operations for Simplification

Some mathematical problems contain mixed operations: addition, subtraction,
multiplication, division and different brackets. When more than one operation is involved,
we have to use the following rules for the simplification of given expression of integers.

This rule is known as 'BODMAS'

Steps

Remove bar — , brackets ( ), { }, [ ] in order. By simplifying all the operations within it or by
simplifying, all the operations within it. (B)
Perform the operation involving 'of' (O)

Perform the operation involving division (D)

Perform the operation involving multiplication (M)
Perform the operation involving addition (A)

Perform the operation involving subtraction (S)

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Worked Out Examples



Example: 1

Simplify: 115 + 51 ÷ 17 × 5 – 80

Solution:
115 + 51 ÷ 17 × 5 – 80

= 115 + 3 × 5 – 80 [Operation of division]

= 115 + 15 – 80 [Operation of multiplication]

= 1 30 – 80 [Operation of addition]

= 50 [Operation of subtraction]

Example: 2

Simplify: 50 ÷ {18 – (4 ×10 ÷ 2)}

Solution:
50 ÷ {18 – (4 ×10 ÷ 2)}
= 50 ÷ {18 – (4 × 5)}
= 50 ÷ {18 – 20}
= 50 ÷ – 2 = – 25

Example: 3

Simplify: 64 ÷ 8 – 2 [3 + {7 – 3(3 + 4 – 2 )}]
Solution:
64 ÷ 8 – 2 [3 + {7 – 3(3 + 4 – 2 )}]
= 64 ÷ 8 – 2 [3 + {7 – 3 (3 + 2)}]
= 64÷8 – 2[3 + {7 – 3(5)]
= 64÷8 – 2 [3 + {7 – 15}]
= 64÷8 – 2 [3 + (–8)]
= 64÷8 – 2[–5]
= 64÷8 + 10 = 8 + 10 = 18

Example: 4

Convert the given statement into mathematical expression and simplify.
4 times 20 is divided by 5 and is added to 3 times 5.
Solution:
4 × 20 ÷ 5 + 3 × 5
= 4 × 4 + 3 × 5 = 16 + 15 = 31

154 Oasis School Mathematics-8

Exercise 10.2

Simplify (b) 6 × 8 ÷ 4 – 2 × 5 + 7
1. (a) 5 × 6 + 10 ÷ 2 (d) 16 ÷ 8 – 4 × 3 + 12 ÷ 3
(c) 40 ÷ 8 + 5 × 4 – 10 (f) (–3) × 6 – (–18) ÷ 6 + 28 ÷ 7
(e) 115 + 51 ÷ 17 × 5 – 80

2. (a) 12 + [12 – {4 + (15 ÷ 5 × 3)}] (b) 10 – [20 ÷ {20 – 5(4 – 1)}]
(d) 150 ÷ {30 – (4 × 10 ÷ 2}
(c) 28 – {24 – 3(4 + 2)} (f) 3 [{84 – (6 × 8)} ÷ 12] – 8
(h) 5[125 – {27 – 3(6 – 2)}]
(e) 42 ÷ [3 + 16 ÷ {2 + 8 ÷ (6 – 2)}] (j) 39 – 7 {28 ÷ (17 – 10) + 1}
(l) {(4 × 5) – (12 ÷ 3)} × 7
(g) 5{ 24 + 8 ÷ (6 – 4)} ÷ (16 × 5 – 10)
(i) 102 ÷ [15 ×7 – 15 ÷ 5]

(k) 20 + [122 ÷ {6 + (16 × 3) + 7)]
(m) (15 + 60 – 42) ÷ [(–33) ÷ 3]

3. (a) 25 + (35 – 9 + 6 ) (b) 70 + [18 × 3 – 16 – 6 + 9 × 4]
(d) 60 ÷ [28 – {24 – (20 – 8 – 4 )}]
(c) 20 + [16 – {4 + (3 + 6 – 4 )}]

(e) 5 [40 ÷ {8 × 4 – 5 × 4 – (8 – 5 + 1 )}]

(f) 70 [460 ÷ {12 × 5 – 30 ÷ 6 – (15 – 8+2 )}]

4. Convert the following statements into mathematical expression and simplify.
(a) 25 is added to the product of 6 and 3.
(b) 18 is subtracted from 3 times the difference of 13 and 5.
(c) 6 is multiplied to one-fourth of 20 and 12 is added to the result.
(d) 15 is added to the difference of 12 and 7 and the result is divided by 4.
(e) 6 times the sum of 7 and 13 is divided by the sum of 5 and 7.
(f) One third of 30 is added to the one tenth of 70 and the result is divided by 17.



Answer
1. (a) 35 (b) 9 (c) 15 (d) –6 (e) 50 (f) –11
2. (a) 11 (b) 6 (c) 22 (d) 15 (e) 6 (f) 1 (g) 2 (h) 550 (i) 1 (j) 4 (k) 22

(l) 112 (m) -3 3. (a) 45 (b) 48 (c) 27 (d) 3 (e) 20 (f) 700
4. (a) 43 (b) 6 (c) 42 (d) 5 (e) 10 (f) 1

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Unit Rational and
11 Irrational Numbers

11.1 Rational Numbers

As we know that,

N = The set of natural numbers
= {1, 2, 3, 4, .......................}

W = The set of whole numbers
= {0, 1, 2, 3, ....................}

Z = The set of integers
= {............, -3, -2, -1, 0, 1, 2, 3, ............}

Let's take any two integers, 2 and 5

2 + 5 = 7 is an integer.

2 – 5 = –3 is an integer.

2 × 5 = 10 is an integer.

2÷ 5= 2 is not an integer.
5

To make the division of integers meaningful rational number is introduced.

2 is a rational number who 2 and 5 both are integers.
5
p
Hence, any number in the form of q where p ∈ Z, q ∈ Z and q ≠ 0, is a rational

number.

Note: • Every integer is a rational number.
• Rational number is either a terminating or non terminating recuring decimals.

Terminating and non-terminating recurring decimals

Let's convert some rational numbers into decimal,

21 = 0.5 (terminating decimals)
41 = 0.25 (terminating decimals)
31 = 0.3333 ............ (non-terminating recurring decimals)

3 = 0.272727 ........ (non terminating recurring block of digits)
11

∴ Therefore, a rational number is either a terminating or non-terminating

recurring decimals.

Thus, all integers and fractions, whether positive or negative, including zero are

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called rational numbers.

11.2 Irrational Number

A set of numbers which have non-terminating and non-recurring decimals, i.e. they
p
cannot be expressed in the form q where p and q both are integers and q ≠ 0, such

numbers are irrational numbers.

2 = 1.4142135 ........ (non-terminating, non recurring decimals)

3 = 1.7320507… (non-terminating, non recurring decimals)

∴ π = 3.14159 … (non-terminating, non recurring decimals)

Hence, an irrational number can be expressed as a non-terminating, non-recurring
(non-repeating) decimal.

Irrational number in number line

Draw a number line. Take a point P on the number line, where OP = 1 unit. Draw
OPQ = 900.

Cut OP = OQ = 1 unit. Join OQ.

Now, OQ2 = OP2 + PQ2 = 1 + 1 = 2. Q

Taking 'O' as the centre and OQ as radius draw an

arc which meets number line at R, which represents P2
-3 -2 -1 0 1 R 2 3
2 as OQ = OR.

11.3 Real Numbers

The set of all rational and irrational numbers taken together is called the set of real
numbers. The set of real numbers is denoted by R and is written as

R = Q ∪ Q' = {x : x ∈ Q or x ∈ Q' } .…(i) R
Z
where Q = the set of rational numbers
OW
Q' = the set of irrational numbers F N=Z

This shows that every rational number is a real number and Z–

every irrational number is a real number. Thus, natural numbers Q Q'

constitute a proper subset of integers and the integers constitute

a proper subset of rational numbers and then latter constitute a proper subset of real

numbers.

i.e., N ⊂ W ⊂ Z ⊂ Q ⊂ R.

The relation between above set of numbers is shown in the above diagram.

Here, Z+ = N = the set of positive integers or natural numbers

Z– = the set of negative integers

W = the set of whole numbers

F = the set of fractional numbers.

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The classification of the set of real numbers is shown in the following diagram.

Worked Out Examples



Example: 1

Examine whether the following numbers are rational or irrational.

(i) 2 (ii) 0.5 (iii) 2 (iv) 49
3

Solution:

(i) Here, 2 = 1.4142134….

This is a non-terminating, non-recurring decimal.

So, 2 is an irrational number.

(ii) This is a terminating decimal. So, 0.5 is a rational number.
(iii)
2 = 0.6666.
3

This is a non-terminating recurring decimal. So, it is a rational number.

(iv) 49 = 7.
As the square root of 49 gives perfect whole number, it is a rational number.

Example: 2

Find rational number pq, if, (i) p = 4, q = 10 (ii) p = 0.3, q = 0.6
Solution:

(i) Rational number (ii) Rational number

= p = 4 = 2 = pq = 0.3 = 1
q 10 5 0.6 2

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Exercise 11.1

1. State whether the following numbers are rational or irrational?

(a) 2 (b) 3 (c) 1.317432...
5

(d) 16 (e) 1 (f) 1.3333...
4

(g) 272 (h) – 25 (i) 3.272727...

2. Convert the following numbers into decimal and state whether they are

terminating or non-terminating decimal.

(a) 15 (b) 54 (c) 53 (d) 32 p (e) 313
3. Write the rational number in the form q
if,

(a) p = 3, q = 5 (b) p = 0.5, q = 0.6

(c) p = 1, q = 3 (d) p = 1.5, q = 2.5

4. State whether the following statements are true or false.

(a) Every integer is a rational number.

(b) Every rational number is an integer.

(c) 1 is the smallest natural number.

(d) An rational number is a non-terminating and recurring decimal.

(e) A rational number is a non-terminating and non-recurring decimal.

5. From the adjoining figure, write the elements of the following sets.
(a) the set of natural numbers (N)
(b) the set of whole numbers (W)
(c) the set of negative integers (Z–)
(d) the set of integers (Z)
(e) the set of rational numbers (Q)
(f) the set of irrational numbers

(g) the set of real numbers (R)

6. Answer the following questions.
(a) Is 0 a rational number? Why?

(b) Why 1.3333... is a rational number?
(c) Why 2.4527314... is an irrational number?
(d) Is π a rational number? Why?
(e) Is 3 an irrational number? Why?



Oasis School Mathematics-8 115599

Answer

1. (a) rational (b) rational (c) irrational (d) rational (e) rational (f) rational
(g) irrational (h) rational (i) rational
2. (a) 0.5, terminating (b) 0.8, terminating (c) 1.6666..., non-terminating
(d) 0.6666..., non-terminating (e) 3.3333..., non-terminating
3. (a) 3/5 (b) 5/6 (c) 1/3 (d) 3/5 4. (a) true (b) false (c) true (d) false (e) false
5. (a) N = {1,2,3,4,10} (b) W = {0, 1, 2, 3, 4, 10} (c) Z = {-1, -2, -3, -4, -10}
(d) Z = {-10, -4, -3, -2, -1, 0, 1, 2, 3, 4, 10}
(e) Q = {-10, -4, -3, -2, -1, 0, 1, 2, 3, 4, 10, 4/5, -1/3 2/3, 4/3}
(f) = { 3, 2, 13, 125 }
(g) R = {-10, -4, -3, -2, -1, 0, 1, 2, 3, 4, 10, 4/5, -1/3, 2/3, 4/3, 3, 2, 13, 125 }
6. Consult your teacher.

11.4 Operations on Irrational Numbers

3 is an irrational number

3 + 3 = 2 3 is an irrational number

but 3 + (– 3 ) = 0 which is a rational number.

∴ sum of two irrational numbers may not be an irrational

3 × 3 = 3 which is a rational number.

3 × 2 = 6 which is an irrational number.

∴ Product of two irrational numbers may not be irrational.

Let's discuss the operations on irrational number.

I. Rationalisation

Let's take a fraction 3 .
2

Here, 2 is an irrational number.

To convert 2 into rational number

Multiply both numerator and denominator by 2 .

Then 3 × 2 = 32
2 2 2

This is the rationalisation of denominator.

II. Conjugate of Binomial Irrational Numbers

Let's take a binomial irrational number 4 + 3 and its conjugate is 4 – 3 .
Conjugate of 5 – 3 is 5 + 3 .
Hence, the conjugate of a + b is a – b and vice versa.


160 Oasis School Mathematics-8

III. Rationalising factor

Take an irrational number 3
Now 3 × 3 = 3 which is rational.
∴ 3 is the rationalising factor of 3 .
Again, take an irrational number 3 – 1.
Now ( 3 – 1) × ( 3 + 1) = ( 3 )2 – 12 = 3 – 1 = 2, which is a rational number.
∴ 3 + 1 is the rationalising factor of 3 –1.

IV. Addition and Subtraction of Numbers Involving Radicals

It is same as the simplification of the algebraic expressions. Under the distributive
law, similar irrational numbers are added and subtracted. Rational numbers which
are not similar cannot be added or subtracted.

Example, 2 5 + 5 = 3 5
5 3 – 2 3 = 3 3 , etc.

Worked Out Examples



Example: 1

What are the rationalising factor for each of the following.

(i) 2 (ii) 2 +1

Solution:

The rationalising factor of 2 is 2

The rationalising factor of 2 + 1 is 2 – 1.

Example: 2 1
3
Rationalise the denominator of .

Solution:

1 = 1 × 3
3 3 3

= 3
3

Example: 3

Simplify : (3 2 +1) (3 2 –1)
Solution:

(3 2 + 1) (3 2 – 1)

= (3 2 )2 – (1)2
= 18 – 1 = 17

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Example: 4

Simplify : 48 – 6 +5 3
3
Solution:

48 - 6 + 5 3 = 48 − 6 +5 3
3 3

= 42 × 3 − 6 × 3 + 5 3
33

= 4 3 − 63 +5 3
3

= 4 3 −2 3 +5 3

= (4 − 2 + 5) 3

= 7 3.

Example: 5

Simplify: 3+1
3–1

Solution:

3 + 1 = 3 + 1 × 3+1 [Multiplying the numerator and the denominator
3 – 1 3 – 1 3+1 with the rationalising factor of 3 + 1.]

( ) 3 +1 2 =( 3 )2 + 2 3 + 1
=( 3 )2 − (1)2 3−1

= 3+2 3 +1= 4+2 3 = 2(2 + 3)
2 2 2

= 2+ 3.



Exercise 11.3

1. Write the rationalising factor for each of the following expressions.
(a) 2 3 (b) 2 + 3 (c) 3 2 – 2 3 (d) 5 + 2

2. Rationalise the denominators of the following. 5
(c) 3 2
(a) 1 (b) 4
3 1 2 1
1 (f) 7 − 3
(d) 2 + 1 (e) 3 +1

3. Simplify the following expressions:

(a) 2 3 + 5 3 (b) 6 2 – 5 2 (c) –3 7 – 5 7

(d) 4 8 + 5 8 – 3 8 (e) 3 + 2 5 – 3 3 + 3 5

(f) 4 10 – 3 10 + 5 10 (g) 5 7 + 3 8 – 4 7 + 2 8

162 Oasis School Mathematics-8

4. Simplify: (b) 2 2 × 4 3 (c) 2 3 × 2 × 5 3
(a) 5 × 3 (e) 5 3 × 2 2 × 3 6 (f) 5 10 ÷ 5 2
(d) 2 5 × 3 × 15 (h) 6 15 ÷ 3 5
(g) 12 6 ÷ 2 6

5. Simplify:

(a) ( 3 + 2 ) ( 3 – 2) (b) ( 5 – 3 ) ( 5 + 3 ) (c) (2 2 + 3 ) (2 2 – 3 )

(d) (5 3 + 2 2 ) (5 3 – 2 2 ) (e) (2 3 – 2 )2 (f) ( 5 + 2 )2

(g) (2 3 + 2 ) (2 3 – 2 ) (h) (2 5 – 3 ) (2 5 + 3 )

6. Simplify the following expressions:

(a) 48 (b) 3 7 + 343 (c) 3 6 + 216

(d) 2 3 + 27 (e) 8 + 32 – 2 (f) 45 – 3 20 + 4 5

(g) 4 12 – 50 – 7 48 (h) 12 – 147 + 192 (i) 4 3 – 3 12 + 2 75

7. Rationalise, (if necessary) and simplify:

(a) 3 + 4 (b) 4 + 5 (c) 271 – 57
3 5 2 2
5
(d) 5 3 +2 27 + 1 (e) 4 3 + 3 48 – 2 3
3

(f) 216 –5 6 + 294 – 3 (g) 3 147 – 3 7 + 7
6 3 3

8. Rationalise the denominators and simplify:

(a) 1+ 3 (b) 5− 2 (c) 2+ 3 (d) 1
1− 3 5+ 2 2− 3 5− 3


Answer

1. a. 3 b. 2– 3 c. 3 2 + 2 3 d. 5 5 – 2 3

2. a. 3 b. 2 2 c. 5 2 d. 2–1 e. 3 -1 f. 3 + 7
3 6 2 4

3. a. 7 3 b. 2 c. –8 7 d. 6 2 e. 5 2 – 2 3 f. 6 10 g. 7 + 5 8

4. a. 15 b. 8 6 c. 30 2 d. 30 e. 180 f. 5 g. 6 h. 2 3

5. a. 1 b. 2 c. 5 d. 67 e. 14–4 6 f. 7+2 10 g. 10 h. 17

6. a. 4 3 b. 10 7 c. 9 6 d. 5 3 e. 5 2 f. 5

g. – 5 2 – 20 3 h. 3 3 i. 8 3

7. a. 7 3 b. 8 5 + 25 2 c. 7 d. 34 3 e. 91 3 f. 15 6 g. 203 3
3 10 2 3 6 2 9

8. (a) –2– 3 b. 7 –2 1 0 c. 7 + 4 3 d. 3+ 5
3 2

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11.5 Scientific Notation of Numbers

Large numbers which are expressed as a × 10n where, (1 ≤ a < 10) and n=0, 1, 2, … are
known as the numbers written in scientific notation or in standard form.

Worked Out Examples



Example: 1

Express in scientific notation.

(i) 35,00,00,000 (ii) 53,000,000,000

Solution:
(i) Here, 350,000,000 = 35 × 10000000 = 3.5 × 100000000 = 3.5 × 108
(ii) Here, 53,000,000,000 = 53 × 1000,000,000 = 5.3 × 10000000000 = 5.3 × 1010

Example: 2

Express the standard form into usual forms 9.3 × 107

Solution:

Here, 9.3 × 10000000 = 93 × 1000000 = 93000000.

Example: 3

Hair of a man grows by 0.000004 cm per second. Write it in scientific notation.

Solution:

Here, 0.000004 cm = 4 cm = 4 cm = 4 × 10–6 cm
1000000 106

Example: 4

Simplify: (i) (6.8 × 105) + (3.7 × 108) (ii) (4 × 10–5) (9 × 104)
6 × 107
Solution:

(i) (6.8 × 105) + (3.7 × 108) (ii) (4 × 10–5) (9 × 104)
= (6.8 × 100000) + (3.7 × 100000000) 6 × 107
= (68 × 10000) + (37 × 10000000)
= 680000 + 370000000 [ [=
= 370680000 4×9 10-5×104
6 107

= 6[10-5 × 104 × 10-7]

= 6 × 10-8

= 3.7068 × 108

Example: 5

Convert the following scientific notation in usual form.

(i) 2.5 × 10–4 (ii) 3.75 × 10–5 (iii) 3.567 × 10–7 (iv) 7.003 × 10–10

164 Oasis School Mathematics-8

Solution: = 2.5 = 25 25 = 0.00025
(i) 2.5 × 10–4 104 105 = 100000
(ii) 3.75 × 10–5 10307050000 = 0.0000375
(iii) 3.567 × 10–7 = 3.75 = 375 =
(iv) 7.003 × 10–10 105 107

= 3.567 = 3567 = 0.0000003567
107 1010

= 7.003 = 7003 = 0.0000000007003
1010 1013

Exercise 11.4

1. Express the following in scientific notation.

(a) 3,700 (b) 39,000 (c) 3,000000 (d) 99000000

2. Express the following in scientific notation.

(a) 3.5 (b) 0.0057 (c) 0.000098 (d) 0.00307

3. Express the following standard form in usual forms.

(a) 2.5 × 105 (b) 3.6 × 107 (c) 0.5 × 103 (d) 2 × 107

4. Express the numbers in the statements below in scientific notation.

(a) There are 2250000 children in Nepal.

(b) A cold drink industry made a profit of Rs. 76200000 in a year.

(c) Approximately 257800000 litre of water is contained in Ranipokhari.

(d) The average distance of the sun from the earth is 149760000 km. approximately.

5. Convert the number in scientific notation in the statements below into the general
form.

(a) The velocity of light is approximately 2.98 × 108 m/sec.

(b) The diameter of the sun has been approximated as 1.39 × 1012 metre.

(c) T he volume of the moon is approximated to be 3.7 × 1013 cubic metre.

6. Express the numbers used in the following statements in scientific notation.

(a) The diameter of the hole of needle is 0.0007 cm.

(b) The height of a man grows by 0.000000035 cm. per second.

(c) The mass of an electron used to be approximated 0.000000375 grams.

Oasis School Mathematics-8 116655

7. Simplify and express in scientific notation.

(a) (7 × 10–7) (2.5 × 1011) (b) (3.5×10–6) (6.3×10-7)
5 × 1010 4.9 × 10-15

(c) 0.275×0.005 (d) 85000 × 0.00002
0.00000625×0.00125 0.0000034

8. Simplify and express in general form.

(a) (3.75 × 10 – 7) + (4.25 × 10 – 3) (b) (8 × 10-7)+(1.6 × 10-10)
3.2 × 10-7
(c) (5.25 × 10-6) – (2.25 × 10-7)
3 × 10-9

Answer

1. a. 3.7 × 103 b. 3.9 × 104 c. 3 × 106 d. 9.9×107 2. a. 3.5×100 b. 5.7×10-3 c. 9.8×10-5
d. 3.07×10-3 3. a. 250000 b. 36000000 c. 500 d. 20000000 4. a 2.25×106
b. 7.6×107 c. 2.578 × 108 litres d. 1.4976 × 108 km. 5. a. 298000000 m/second
b. 1.390000000000 meter c. 37000000000000 m
6. a. 7 × 10-4 cm. b. 3.5 × 10-8cm/second c. 3.75 × 10-7gm 7. a. 3.5 × 10-6 b. 4.5 × 102
c. 1.76 × 10-5 d. 5 × 105 8. a. 0.004250375 b. 2.5005 c. 1825

166 Oasis School Mathematics-8

Unit Ratio and
12 Proportion

12.1 Ratio

Let us suppose that marks obtained by Lakpa in Mathematics is 80 and the marks
obtained by Dorje is 60.

Let's compare their marks,

Lakpa got (80-60) = 20 marks more than Dorje.

Let's compare their marks by division.

Marks obtained by Lakpa = 80 = 80 = 4 = 4:3
Marks obtained by Dorje 60 60 3

Hence the ratio is the comparison of two quantities having same unit by division.
We use the symbol : to express ratio.

Write Say
a:b 'a' is to 'b'

or

'a' to 'b'
Terms of ratio:
In the ratio 4 : 5

4 is antecedent and 5 is consequent

The first term of the ratio is antecedent and the second term is consequent.

Remember !

• In a ratio, both the quantities should be of the same kind and should have the
same unit.

• A ratio is a pure number and has no unit.

I. Properties of ratio

Take a ratio a : b i.e. a .
b
Let's multiply both terms by 3, a × 3 : b × 3

= 3a : 3b

=a:b

Oasis School Mathematics-8 116677

Hence the ratio remains same if both of its terms are multiplied by a non zero con-
stant number.

Again, let's divide both terms of a ratio a : b by 2

a÷2 : b÷2 = a/2 = a ×b2 = a : b.
b/2 2

Hence the ratio remains same if each term of a ratio is divided by a non zero con-

stant number.

II. Compound Ratio

A new ratio formed by taking the product of two or more ratios is called compound

ratio. Thus, if there are any two ratios a and dc , then
b
a c ac
the compound ratio = b × d = bd = ac : bd.

For example, if 2:3 and 4:5 are any two ratios, their compound ratio

= 2 × 4 = 8 = 8 : 15
3 5 15

Worked Out Examples



Example: 1

Find the ratio of 75 cm and 1 m.
Solution:

Here 1 m = 100 cm.

∴ Ratio of 75 cm and 1 m

= 75cm = 3
100cm 4
Example: 2

Convert the ratio 1 : 2 in the form of a : b. Multiply both terms by the L.C.M
2 3 of denominators

Solution:

1 : 2 = 1 × 6 : 2 × 6
2 3 2 3

= 3 : 4

Example: 3

Find the compound ratio of 3:4, 4:5 and 5:6
Solution:

The compound ratio of 3:4, 4:5 and 5:6

= 3 × 4 × 5 = 3 = 1 = 1:2
4 5 6 6 2

168 Oasis School Mathematics-8

Example: 4

If a:b = 4:5, b:c = 3:4, find a : c.
Solution:

Given, a : b = 4 : 5

b : c = 3 : 4

Taking compound ratio

a × bc = 4 × 3
or, b 5 4

a = 3
c 5

∴ a : c = 3:5.

Example: 5

The ratio of the number of boys and girls in a school is 3 : 4. If there are 120 boys, find
the number of girls.

Solution:

Let the number of girls be x, Alternative method

Number of boys = 3 Let no. of boys be 3x and no. of
Number of girls 4 girls be 4x.
Then, we have
1x20 = 43 or, 3x = 120
3x = 120 × 4 or, x = 120/3 = 40
∴Number of girls = 4x =
x = 120 × 4 4 × 40 = 160
∴ 3
x = 160

Number of girls = 160.

Example: 6

Divide Rs. 7,200 in the ratio 3:4 : 5:6.
Solution:

Let four parts of the sum be 3x, 4x, 5x and 6x

Here, 3x + 4x + 5x + 6x = 7200

or, 18x = 7200

or, x = 400
∴ 3x = 3 × 400 = Rs. 1200

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∴ 4x = 4 × 400 = Rs. 1600

or, 5x = 5×400 = Rs. 2000

∴ 6x = 6×400 = Rs. 2400

Hence four parts of money are Rs. 1200, Rs. 1600, Rs. 2000 and Rs. 2400.

Example: 7

Two numbers are in the ratio 3:4. If 4 is added to each of them, their ratio becomes 4:5,
find the numbers.

Solution:

Let, the required two numbers be 3x and 4x. From the given condition,

3x + 4 = 4
4x + 4 5

or, 5(3x + 4) = 4(4x + 4)

or, 15x + 20 = 16x + 16

or, 16x – 15x = 20 – 16

∴ x = 4

∴ Required numbers are 3x = 3 × 4 = 12 and 4x = 4 × 4 = 16

Example: 8

The ratio of the present ages of a father and his son is 5:2. Five years ago, the ratio of
their ages was 3:1. Find their present ages.

Solution:

Let the present ages of the father and his son be 5x years and 2x years respectively. Then,

5 years ago, age of the father = (5x – 5) years.

5 years ago, age of the son = (2x – 5) years.

By the question, 5x – 5 = 3
2x – 5 1

or, 1(5x – 5) = 3(2x – 5)

or 5x – 5 = 6x – 15

or, 6x – 5x = 15 – 5

∴ x = 10

Hence, the present age of the father = 5x = 5 × 10 years = 50 years
and the present age of his son = 2x = 2 × 10 years = 20 years

170 Oasis School Mathematics-8

Exercise 12.1

1. Convert the given ratios into the lowest terms.

(a) 1255 (b) 3428 (c) 24
64

2. Find the ratios of the following and reduce them to their lowest terms.

(a) 75 cm and 125 cm (b) Rs. 4 and 50 paisa

(c) 4 months and 2 years (d) 200 gm and 2 kg.

3. Convert the given ratios in terms of a:b.

(a) 12 : 1 (b) 2 : 1 (c) 2 : 3 (d) 1 : 2
3 5 4 3 5 2 3

4. Find the compound ratio of the following ratios.

(a) 2:3 and 3:5 (b) 4:5 and 9:16

(c) 4:7 and 3:8 (d) 3:4 and 7:12

5. Find the value of x in each of the following.

(a) x = 2 (b) 4 = 2
4 3 x 3

(c) 4 = x (d) 2 = 8
7 21 3 x

6. (a) If a : b = 5 : 8 and b : c = 4 : 3, find: a : c

(b) If a : b = 2: 3 and b : c = 4 : 5, find a : c.

(c) If a : b = 2:3, b:c = 4:5 and c : d = 1:2, find a : d.

7. (a) A ratio is equal to 4:5. If antecedent is 36, find the consequent.

(b) If a ratio is 2:3 and its consequent is 45, find its antecedent.

8. (a) Divide Rs. 300 between A and B in the ratio 2 : 3.

(b) Divide Rs. 1,500 in the ratio 1:2.

(c) Divide Rs. 9,000 between A, B and C in the ratio 2 : 3 : 4.

(d) An alloy contains copper, zinc and tin in the ratio of 2 : 3 : 5. Find the mass of
each metal in 120 gm of alloy.

9. (a) If the angles of a triangle are in the ratio 2 : 3 : 5, find the size of each angle.

(b) If the ratios of the angles of a quadrilateral are 1:3:5:6, find the size of each
angle.

10. (a) The ratio of the number of boys and girls in a class is 5 : 3. If there are 35 boys,
find the number of girls.

(b) The ratio of the ages of two sisters is 3:5. If the elder sister is 25 years old, find
the age of the younger sister.

Oasis School Mathematics-8 117711

(c) The ratio of monthly income of Imran and Salman is 4:5. If the monthly income
of Salman is Rs. 60,000, find the monthly income of Imran.

11. (a) Two numbers are in the ratio 4:5. When 5 is added to each of them, their ratio
becomes 5:6. Find the numbers.

(b) Two numbers are in the ratio 3:5. When 5 is subtracted from each of term, their
ratio becomes 1:2, find the numbers.

(c) Two numbers are in the ratio 3:4. If 10 is added to both numerator and
denominator, the new ratio becames 4:5. Find the numbers.

12. (a) The ratio of the present ages of a father and his son is 5:2. Ten years hence, the
ratio of the ages will be 2:1. Find their present ages.

(b) The ratio of the present ages of a mother and her daughter is 3:1. Six years ago,
the ratio of their ages was 6:1. Find their present ages.

(c) Ages of two brothers are in the ratio 3:4. 5 years ago, the ratio of their ages was
2:3. Find the their present ages.


Answer
1. (a) 3/5 (b) 2/3 (c) 3/8 2. (a) 3:5 (b) 8:1 (c) 1:6 (d) 1:10 3. (a) 3:2
(b) 8:5 (c) 10:9 (d) 3:4 4. (a) 2:5 (b) 9:20 (c) 3:14 (d) 7:16
5. (a) 8/3 (b) 6 (c) 12 (d) 12 6. (a) 5:6 (b) 8:15 (c) 4:15 7. (a) 45 (b) 30
8. (a) Rs. 120, Rs. 180 (b) Rs. 500, Rs. 1000 (c) Rs. 2000, Rs. 3000, Rs. 4000 (d) 24gm, 36gm, 60 gm
9. (a) 360, 540, 900 (b) 240, 720, 1200, 1440 10. (a) 21 (b) 15yrs (c) Rs. 48,000
11. (a) 20, 25 (b) 15, 25 (c) 30, 40 12. (a) 50 yrs, 20 yrs, (b) 30 yrs, 10 yrs, (c) 15 yrs, 20 yrs.

12.2 Proportion

Let's take two ratios 8:12 and 6:9.

Here, 8:12 = 8 = 2 and 6:9 = 6 = 2 .
12 3 9 3

i.e. 8:12 = 6:9. Such equality of two ratios is called a proportion.

Hence, four quantities a, b, c and d are in proportion, if a = c i.e. a : b = c : d.
b d

A proportion is also written as a : b : : c : d

In a proportion, first and fourth terms are called extremes whereas second and third
terms are called means.

In a = dc , a × d = b × c.
b

The product of extremes = the product of means.

172 Oasis School Mathematics-8

Continued Proportion
Let's take any two ratios 3:6 and 6:12.

Here, 3 = 1 = 1 : 2
6 2

and 6 = 1 = 1 : 2
12 2

∴ 3 : 6 = 6 : 12

Hence, 3, 6, 6 and 12 are in proportion.

i.e. 3, 6 and 12 are in proportion.
This type of proportion is continued proportion.

Three quantities are said to be in continued proportion if the ratio of the first to the
second is equal to the ratio of second to the third.

i.e. three quantities a, b and c are said to be in continued proportion if a : b = b : c,
where b is called as mean proportional.

Remember !

• a, b, c, d are said to be proportional if a:b :: c:d.

• 'd' is said to be fourth proportional to a, b, c if a = c .
b d
• If a, b, c, d are in proportion, first and last are extremes and second and third
are means.
• In proportion, product of extremes = product of means.

Worked Out Examples

Example: 1

Test whether the numbers 2, 3, 8 and 12 are in proportion or not.
Solution:

Here, ratio of first two = 2 : 3

ratio of last two = 8:12 = 8 = 2 = 2:3
12 3

Since the ratio of the first two and the last two numbers is equal given from terms

are in proportion.

Example: 2

The first, second and third terms of proportion are 4, 22 and 6 respectively. Find the
fourth proportional.
Solution:

Let fourth proportional be x. Then, 4, 22, 6 and x are in proportion.

Oasis School Mathematics-8 117733

Thus, we have, 4 : 22 = 6 : x
or,
4 = 6
22 x

or, 4x = 22 × 6

or, x = 224×6 = 33

Hence, the required fourth proportional is 33.

Example: 3

Find the mean proportional between 2 and 18.
Solution:

Let the mean proportional between 2 and 18 be x.

Then, 2, x, 18 are in continued proportion.

∴ 2 = x
x 18

or, x² = 36

or, x = 6

∴ Mean proportional is 6.

Example: 4

What number must be added to each of the terms 4, 9, 17 and 35 to be in proportion?
Solution:

Let, the required number be x.

then, (4 + x), (9 + x) (17 + x) and (35 + x) are in proportion.

i.e. 4+x = 17 + x
9+x 35 + x

(4 + x) (35 + x) = (17 + x) (9 + x)

or, 140 + 4x + 35 x + x² = 153 + 17x + 9 x + x²

or, 140 + 39x = 153 + 26x

or, 39x – 26 x = 153 – 140

or, 13x = 13

or, x = 1

∴ The required number is 1.

174 Oasis School Mathematics-8

Exercise 12.2

1. Identify whether the given numbers are in proportion or not?

(a) 1, 2, 6, 8 (b) 6, 8, 10, 20

(c) 25, 35, 50, 70 (d) 4 cm, 6cm, 12, cm, 18 cm

(e) 5kg, 10kg, 20kg, 40 kg

2. Find the values of x in the following proportions.

(a) 2 : 5 = 6 : x (b) 7 : 5 = x : 35 (c) 4 : x = 16 : 20 (d)x : 7 = 16 : 28

3. Find the fourth proportional in the following:

(a) 1, 3, 5 (b) 2, 4, 8 (c) 15, 18, 30 (d) 10, 8, 5

4. Find the mean proportional between the following:

(a) 4 and 9 (b) 4 and 16 (c) 9 and 16 (d) 5 and 20

5. (a) If 3, x, 12 are in continued proportion, find the value of x.

(b) If 4, 12, x, 15 are in proportion, find the value of x.

(c) Two extremes of a proportion are 10 and 60. If the first mean is 15, find the sec-
ond mean.

(d) Two means of a proportion are 12, 8. If the first extreme is 6, find the last extreme.



Answer
1. (a) Not in proportion (b) Not in proportion (c) Proportion (d) Proportion (e) Proportion
2. (a) 15 (b) 49 (c) 5 (d) 4 3. (a) 15 (b) 16 (c) 36 (d) 4
4. (a) 6 (b) 8 (c) 12 (d) 10 5. (a) 6 (b) 5 (c) 40 (d) 16

12.3 Types of proportion

I. Direct proportion
Let the cost of 1 copy = Rs. 10, then the cost of 2 copies = Rs. 2 × 10 = 20
Now,

Number of copy Cost (Rs.)

1 10
2 20

Oasis School Mathematics-8 117755

Here, more the number of copies, more the cost.
Again, ratio of copies = 1 : 2
ratio of cost = 10:20 = 1:2
such type of proportion is direct proportion

II. Indirect Proportion (Inverse proportion)
Suppose 1 man can do a work in 20 days.
then 2 men can do the same work in 10 days.
Now,

Men Days

1 20
2 10

Here, more the number of men, less the number of days.
Ratio of men = 1:2
Inverse ratio of days = 20:10 = 2:1
Such proportion is indirect proportion.

Remember !

• In direct proportion, if one quantity increases another also increases in the
same ratio.

• In indirect proportion, if one quantity increases another decreases in the same
ratio.

Worked Out Examples



Example: 1

If two ratios 4:5 and x:20 are in direct proportion, find the value of x.
Solution:

Here, 4:5 and x : 20 are in direct proportion,

4 = 2x0
5

or, 5x = 80

or, x = 850 = 16
∴ x = 16.

176 Oasis School Mathematics-8

Example: 2

If two ratios 2 : 3 and a : 50 are in indirect proportion, find the value of 'a'.
Solution:

Here, 2:3 and a : 50 are in indirect proportion,

Hence, 2 = 5a0
3

or, 2a = 50 × 3

or, a = 502× 3

∴ a = 75.

Example: 3

Find the value of x in the given case.

Weight (in kg.) Cost (in Rs.)

5 400
8x

Solution:

Since weight and cost are in direct proportion,

5 = 4x00 Weight (in kg.) Cost (in Rs.)
or, 8
or, 5 400
5x = 8 × 400 8 x

x = 8×400
5

x = 8×80

∴ x = Rs. 640.

Example: 4

If 40 metres of a cloth costs Rs. 1200, how many metres can be bought for Rs. 870 ?

Solution:

Let, cloth bought for Rs. 870 be x metres'

Length of cloth(m) Cost in (Rs.)

40 1200
x 870

Here, length of cloths and cost are in direct proportion,

so, 40 : x = 1200 : 870

or 40 = 1200
x 870

Oasis School Mathematics-8 117777

or, 1200 × x = 40 × 870

or x = 40 × 870 = 87 = 29 meters
1200 3

Hence, cloth bought for Rs. 870 is 29 metres.

Example: 5

A garrison of 600 men had provisions for 36 days. However, a reinforcement of 300 men
arrived. For how many days will the food last now?

Solution:

Let's suppose the food will now last for x days,

Number of men Number of days
600 36
x
600 + 300 = 900

Clearly, more men require less days for provisions to last. So, it is a case of inverse
proportion.

∴ 600 : 900 = x : 36
or,
600 = x
900 36

or, 9x = 6 × 36

or, x = 6 × 36
9

= 24 days.

Hence, the food will now last for 24 days.

Exercise 12.3

1. (a) If the following ratios are in direct proportion, find the value of x.

(i) 3 : 5 and x : 25 (ii) 12 : x and 24 : 36 (iii) 15 : 20 and x : 40

(b) If the following ratios are in indirect proportion, find the value of x.

(i) 4 : 5 and x : 16 (ii) x : 10 and 20 : 50 (iii) 5 : 10 and x : 25

2. Identify whether the given variables are in direct or indirect proportion.

(a) The quantity of rice and cost (b) The number of book and cost

(c) The number of workers and the time taken to finish a certain work

(d) The number of days worked and the quantity of work

(e) Interest on money and the amount invested

178 Oasis School Mathematics-8

3. Find the value of x in each of the following cases.

(a) Weight (in Cost (b) Number Days (c) Number Work
of work-
kg) (in Rs.) of work- ers 1
3
ers 55
x
5 120 30 x 10

12 x 50

4. (a) If 8 pens cost Rs. 64, what is the cost of 12 pens ?

(b) The cost of 5 books is Rs. 625. How many books can be purchased for Rs. 1000?

(c) The cost of 20 books or 30 pens is Rs. 360. What is the total cost of 16 books and
24 pens?

5. (a) A bus covers a distance of 120 km in 2 hours. How long will it take to run 600
km with the same speed?

(b) If the cost of painting 4 walls having area 250 m2 is Rs. 3500, what will be the
cost of painting the walls having an area of 1000 m2?

6. (a) If 18 men can do a piece of work in 36 days, how many men can do it in 54
days ?

(b) If 36 men can do a piece of work in 25 days, in how many days will 15 men
do it?

7. (a) A garrison has provisions for 240 persons for 30 days. If 60 more persons join
the garrison, how long will the provisions last ?

(b) A hostel has enough food for 150 boys for 30 days. After 10 days 50 boys leave
the hostel. Find how long will the remaining food last.

(c) A garrison had provision for 300 men for 90 days. After 20 days 50 more men
joined the garrison. How long will the remaining food last at the same rate?



Answers

1. (a) (i) 15 (ii) 18 (iii) 30 (b) (i) 20 (ii) 25 (iii) 50
2. (a) Direct variation (b) Direct variation (c) Indirect variation (d) Direct variation
(e) Direct variation 3. (a) Rs. 288 (b) 3 days (c) 2/3 4. (a) Rs. 96 (b) 8 books
(c) Rs. 576 5. (a) 10 hrs. (b) Rs. 14000 6. (a) 12 men (b) 60 days
7. (a) 24 days (b) 30 days (c) 60 days

Oasis School Mathematics-8 117799

Unit

13 Percentage

13.1 Introduction:

Per cent means out of every hundred or per hundred. For example, a boy got 80 percent
marks in Maths means he got 80 marks out of 100. The symbol for percent is % or p.c.

A fraction with 100 as its denominator is called percentage.

e.g. 35 = 35%, 65 = 65% and so on.
100 100

Some operations related to the percentage

(i) Conversion of percentage into fraction

If the given percentage is divided by 100,

then, it converts into fraction.

Percentage Fraction

20% 20 = 1
100 5

(ii) Conversion of fraction into percentage
If a given fraction is multiplied by 100, then it converts into percentage.

Fraction Percentage

1 1 × 100% = 20%
5 5

(iii) Conversion of decimal into percentage

If a given decimal is multiplied by 100, then it converts into percentage.

Decimal Percentage

0.25 0.25×100% = 25%

(iv) Calculation of certain percentage of a number

Let a number be x. If we have to find the value of 10% of x. 10% of x means 10 out of

100 of x.

∴ 10 % of x = 11000 of x

= 11000x = x .
10

180 Oasis School Mathematics-8

(v) Conversion of one quantity as the percentage of another quantity

If the unit of two quantities is the same, then one quantity can be expressed as the
percentage of another quantity.

If the rate of the first quantity is to be expressed as the percentage of another, the
ratio of the first to the second quantity is multiplied by 100 to convert one quantity
as the percentage of another.

For example:

What per cent of 60 is 15?

Here,

Required percent = 15 × 100 % = 1 × 100% = 25%
60 4

Worked Out Examples



Example: 1

If 1 of the total students of a school are girls, find the percentages of boys and girls.
5

Solution:

Let, Total students of a school = 100%.
Then,
The percentage of girls = 1 × 100% = 20%
5

The percentage of boys = 100% – 20% = 80%.

Example: 2

What per cent of Rs. 5 is 25 paisa?
Solution:

Here, Rs. 5 = 500 Paisa.
∴
Required percentage = 25 × 100 %
500
Example: 3
= 5 %

20% of a number is 60. Find the number.
Solution:

Let the required number be x.

Then 20% of x = 60

or 20 × x = 60
or 100
60×100
x = 20 = 300

∴ Hence, required number is 300.

Oasis School Mathematics-8 118811

Example: 4

The monthly income of Kalpana is Rs.15,000. If she spends 30% of her income on food,
15% on house rent, 10% on education and 20% on miscellaneous, find her

(a) expenditure on each item. (b) saving in a month.

Solution:

(a) Here, the monthly income of Kalpana = Rs.15,000.

Her expenditure on food = 30% of Rs. 15,000

= 30 × 15000 = Rs. 4,500
100

Her expenditure on house rent = 15% of Rs.15,000

= 15 × 15,000 = Rs. 2,250
Her expenditure on education 100
Her expenditure on miscellaneous
= 10% of Rs.15000 = 10 × 15,000 = Rs.1,500
100

= 20% of Rs. 15000 = 20 × 15,000 = Rs. 3,000
100

∴ Total expenditure in a month = Rs. 4,500 + Rs. 2,250 + Rs. 1,500 + Rs. 3,000

= Rs. 11,250

(b) Now, her monthly saving = Rs. 15,000 – Rs. 11,250

= Rs. 3,750

Example: 5

In an election, a candidate secured 40% of the votes polled and lost the election by 6000
votes. Find the total number of votes cast.

Solution:

Let, the total number of votes cast be x.

Then, v otes secured by losing candidate = 40% of x = 40 × x = 2x
100 5

( )∴ Votes secured by winning candidate = 2x = 3x
x– 5 5

Difference of votes = 3x – 2x = x
5 5 5

But, by question, x = 6000
5

or x = 6000 × 5 = 30000.

Hence, the total number of votes cast is 30000.

182 Oasis School Mathematics-8

Example: 6

The population of Tamghas Bazaar in 2070 was 1,20,000. In 2071 it was increased by
10% and one year later it was decreased by 5%. What is the population of the town in
2072?
Solution:

Here, P opulation of Tamghas in 2070 = 1,20,000.

Increment in the population = 10 % of 1,20,000

= 10 × 1,20,000
100

= 12,000

∴ Population of the town in 2071 B.S. = 1,20,000 + 12,000

= 1,32,000

Again,

Reduction in the population = 5 % of 1,32,000
= 1050 × 1,32,000
= 6600

∴ Population of the town in 2072 B.S. = 1,32,000 – 6,600

= 1,25,400

Example: 7

A milkman mixes 20 litres of milk containing 25% water with 50 litres of milk containing
10 % of water. What is the percentage of water in the mixture?
Solution:

Here,

Amount of water in the first lot of milk = 25 % of 20

= 25 × 20
100

= 5 litres

Amount of water in the second lot of milk = 10 % of 50
10
= 100 × 50

= 5 litres.

Total amount of water in the mixture = (5 + 5) litres

= 10 litres.

Total amount of mixture = (20 + 50) litres

= 70 litres.

∴ Percentage of water in the mixture = 10 × 100 %
70
2
= 14 7 %

Oasis School Mathematics-8 118833

Exercise 13.1

1. Express the following fractions or decimals in percentage.

(a) 15 (b) 3 (c) 0.05 (d) 2.5
10

2. Express the following percentages in fraction and reduce them to their lowest

terms.

(a) 35% (b) 30% (c) 50% (d) 85%

3. Convert the following percentages into decimals.

(a) 35% (b) 40% (c) 45% (d) 0.5%

4. Convert the following decimals into percentage.

(a) 0.75 (b) 0.01 (c) 0.2 (d) 0.5

5. Find the value of each of the following.

(a) 25% of 300 (b) 20% of 1000 (c) 2 1 % of Rs. 800 (d) 5% of 100
2

6. (a) What percent of 150 is 75?

(b) What percent of Rs. 5 is 50 paisa?

(c) What percent of 8 kg is 500 gm?

(d) What percent of 2m is 5cm?

7. (a) If 25% of a number is 800, find the number.

(b) If 20% of a sum is Rs. 600, find the sum.

(c) If 15% of a number is 240, find the number.

8. (a) A number becomes 190 when it is reduced by 5%. What is the number?

(b) A number becomes 55 when it is increased by 10%. What is the number?

(c) If a number is increased by its 20%,it becomes 360. Find the number.

9. (a) 7 of the total number of students in a school are girls. Find the percentages of
10

girls and boys

(b) If 1 of examinee failed in an examination, calculate the percentage of students
4
who passed?

(c) 53 of the population of a village is female. Find the percentage of male.
10. (a) 60% of the students in a school are boys and the number of girls is 300. Find the

number of boys.

(b) In an examination, 60% of the total examinees passed. If 120 students failed,
find the total number of examinees.

(c) A man spends 60% of his income and saves Rs. 4,000, find his income.

184 Oasis School Mathematics-8

(d) A man spends 80% of his salary. If he saves Rs. 1500 per month, find his salary
per month.

11. (a) Gunpowder contains 70% nitre, 15% sulphur and rest charcoal. Find the
amount of charcoal in 12 kg of Gunpowder.

(b) A man earns Rs. 8000 in a month and spends 30% on food 15% on cloth, 20%,
on education, 10% on house rent and 15% on others,

(i) find the expenditure on each item.

(ii) the saving in a month.

12. (a) Among 2,500, only 1000 voters cast their votes in an election.

(i) What per cent of the total voters cast their votes?

(ii) What per cent of the voters did not cast their votes?

(b) Among 1,200 students in a school, 200 were absent, find the

(i) percentage of absent students.

(ii) percentage of present students.

13. (a) The price of petrol is increased from Rs.800 to Rs. 1,000. Find the percentage
increase in the price.

(b) The price of an article is decreased from Rs. 500 to Rs. 400. Find the percentage
decrease in the price.

14. (a) The population of Beni Bazaar increases by 10% annually. If the present
population is 2,000, what will be its population after 2 years?

(b) In 2012 A.D., the number of tourists visiting Nepal was 150,000. In 2013 A.D., it
increased by 10% and in 2015 A.D., it is decreased by 10%. How many tourists
visited Nepal in 2015 A.D?

15. A milkman mixes 300 litres of milk containing 10 % water with 200 litres of milk
containing 5 % water, what is the percentage of water in that mixture?



Answer

1. (a) 20% (b) 30% (c) 5% (d) 250% 2. (a) 7/20 (b) 3/10 (c) 1/2 (d) 17/20

3. (a) 0.35 (b) 0.4 (c) 0.45 (d) 0.005 4. (a) 75% (b) 1% (c) 20% (d) 50%
1
5. (a) 75 (b) 200 (c) Rs. 20 (d) 5 6. (a) 50% (b) 10% (c) 6 4 % (d) 2.5%

7. (a) 3200 (b) Rs. 3000 (c) 1600 8. (a) 200 (b) 50 (c) 300 9. (a) 70%, 30% (b) 75% (c) 40%

10. (a) 450 (b) 300 (c) Rs. 10,000 (d) Rs. 7500 11. (a) 1800gm (b) (i) Food-Rs.2400,

Cloth-Rs. 1200, Education-Rs.1600, Other item-Rs. 1200, House rent-Rs. 800 (ii) Rs. 800
2 1
12. (a) (i) 40% (ii) 60% (b) (i) 16 3 % (ii) 83 3 % 13.(a) 25% (b) 20% 14. (a) 2420 (b) 1,48,500 15. 8%

Oasis School Mathematics-8 118855

Unit

14 Unitary Method

14.1 Introduction

The method of finding the value of articles using unit value is called unitary method.
In unitary method, we come across two types of variations namely direct variation
and indirect variation. We have already discussed the direct and indirect variation
in the chapter of Ratio and Proportion.

Worked Out Examples



Example: 1

The cost of 8 articles is Rs. 640. What is the cost of 20 articles?

Solution: Alternative method
Cost of 8 articles is Rs. 640

Cost of 1 article is Rs.6480 Articles Cost

Cost of 20 articles is Rs.6480× 20 8 640
20 x

∴ Required cost = Rs. 640× 20 Since the number of articles and
8 cost are direct variations

= Rs. 1600 20 = x
8 640
∴ Cost of 20 articles is Rs. 1600.

Cost of 1 article is less than or, 8x = 20×640
the cost of 8 articles or,
x = 20×640
Cost of 20 articles is more 8
than the cost of 1 article = Rs. 1600


Example: 2

If the cost of 2 of a piece of land is Rs. 5,00,000, what is the cost of 3 piece of the same
5 4
land?

Solution:

The cost of 2/5 of a piece of land = Rs. 5,00,000

The cost of 1 piece of land = Rs. 5020/0500

186 Oasis School Mathematics-8

= Rs. 500000×5 Alternative method

2 Land Cost

= Rs. 1250000 2 500000
5 x
The cost of 3 of a piece of land = Rs. 3 × 3
4 4 4

12,50,000 = Rs.9,37,500 Since the land and cost are direct
variations, 3
Hence, the required cost of 3 of a piece of
4

the same land is Rs. 9,37,500. x = 4
500000 2
5

Example: 3 x = 3 × 5 × 500000
4 2

The cost of 3 chairs and 2 tables is Rs. 1050. If x = Rs. 937500

the cost of one chair is Rs. 150, find the cost of

one table.

Solution:

Here,

Cost of 1 chair is Rs. 150

Cost of 3 chairs is Rs. 3×150 = Rs. 450

Now, Cost of 3 chairs + Cost of 2 tables = Rs. 1050

Rs. 450 + Cost of 2 tables = Rs. 1050

Cost of 2 tables = Rs. 1050 – 450 = Rs. 600

Cost of 1 table = Rs. 600 = Rs. 300.
2

Example: 4

48 men are employed to construct a building in 75 days. They wanted the building to be
ready in 60 days. How many more men should be employed to construct the building in
60 days?

Solution:

I n 75 days, 48 men can construct a building More men are required to
construct the building in 1 day.
I n 1 day, 48 × 75 men can construct the building
Less men are required to
In 60 days, 48 × 75 men can construct the building construct the building in 60 days.
60
48 × 75
∴ Men required to construct the building = 60 = 60.

Number of additional men = (60 – 48)men = 12 men

Oasis School Mathematics-8 118877

Alternative method

Days Men

75 48
60 48+x

Since, the men and days are indirect proportion,

48 + x = 75
48 60

or, 48 + x = 5
48 4

or, 4(48 + x) = 240

or, 192 + 4x = 240

or, 4x = 240 – 192

or, 4x = 48

∴ x = 48 = 12
4

Example: 5

A hostel has provisions for 45 students for 30 days. If 15 more students joined the
hostel after 6 days, how long will the provision last?

Solution:

Here, Total number of students = 45 + 15 = 60

Remaining number of days = 30 – 6 = 24
For 45 students remaining provision lasts for 24 days.

For 1 student remaining provision lasts for 24 × 45 days.

60 students have provision for 24 × 45 days = 6 × 3 days = 18 days.
60

H ence, the remaining provision will last for 18 days.

Alternative method

Here, the total number of students = 45 + 15 = 60

Remaining number of days = 3 0 – 6 = 24

Let us suppose t he remaining provision will last for x days.

Now, Student Days

45 24

60 x

As the number of students and days are indirect variations,
x 45
24 = 60

or, x = 45 × 24 = 18 days.
60

188 Oasis School Mathematics-8

Example: 6

20 men can do a piece of work in 12 days working 9 hours a day. In how many days 30
men can do the same work working 6 hours a day.
Solution:

Working 9 hours a day 20 men can do a work in 12 days.

Working 1 hour a day 20 men can do a work in 12 × 9 days.

Working 1 hour a day 1 man can do a work in 12 × 9 × 20 days.

Working 6 hours a day 1 man can do a work in 12 × 9 × 20 days.
6

Working 6 hours a day 30 men can do a work in 12 × 9 × 20 days.
6 × 30

∴ Required number of days = 126××93×020 days = 12 days.

Alternative method


Let's suppose 30 men can do a work in x days working 9 hours a day.

Men Hours Days

20 9 12

30 6 x

Men and days are inverse variations. Similarly, hours and days

are also inverse variations.

x = 9 × 20
12 6 30

x = 9×20×12 = 12
6×30

∴ Required number of days = 12 days.

Exercise 14.1

1. (a) If the cost of 20 pens is Rs 640, what is the cost of 12 pens?
(b) If the cost of 5m cloth is Rs. 1,200, what is the cost of 15 m cloth?
(c) If a dozen oranges cost Rs. 120, find the cost of 4 oranges.
(d) If the cost of 50 kg of sugar is Rs. 5,000, what is the cost of 2 quintals of sugar?
2. (a) Lochan earns Rs. 2,16,000 in 3 months! How much does he earn in 1 year?
(b) A worker earns Rs. 54,000 in 6 months. How much does he earn in 9 months?
(c) Cost of 18 kg rice is Rs. 900. How much rice can be bought with Rs. 500?
(d) If the cost of 25 pens is Rs. 246, how many pens can be bought for Rs. 984?
(e) A car consumes 8 litres of petrol in covering a distance of 120 km. How many

kilometers will it go with 15 litres of petrol?

Oasis School Mathematics-8 118899

3. (a) If 1 part of a property is Rs. 48600, find the 2 part of it.
(b) If 4 41mpoanretyoifstRhes.l4a0n,d00i0s,Rws.h8a0t ,i0s0041, 5
(c) If 1
the cost of find the cost of 2 part of the land.

1 part of a part of it?
3

4. (a) Cost of 5 pens and 7 books is Rs. 350. If the cost of one pen is Rs. 28, find the cost of

1 book.

(b) The cost of 3 tables and 8 chairs is Rs. 7000. If the cost of one table is Rs. 1000, find
the cost of 10 chairs.

5. (a) If 4 men can do a piece of work in 30 days, in how many days could 12 men do the same
work?

(b) 12 workers can do a piece of work in 20 days. How many workers should be added
to complete the work in 16 days?

(c) 15 men can do a piece of work in 80 days. How long will it take to finish the work if
5 men are added?

(d) 6 taps take 15 minutes to fill up a water tank. How many minutes will 10 taps take
to fill up the same tank?

6. (a) A garrison of 1500 men has provisions for 60 days. How long would the food last if
the garrison is reduced to 900 men ?

(b) 150 students of a hostel have food enough for 45 days. 25 students leave the hostel
after 10 days. How long will the remaining food last ?

(c) A garrison of 1200 men has food for 60 days. How long does it last if 300 men were
added after 10 days?

7. (a) 15 men reap 45 hectares of field in 6 days. How many hectares will 9 men reap in 4
days?

(b) 40 men can do a piece of work in 72 days if they work 8 hours a day. How many men
are required to finish the same work in 40 days working 9 hours a day?

(c) 28 men can dig a field in 18 days working 8 hours a day. In how many days can 16
men dig the same field working 9 hours a day?



Answer
1. (a) Rs. 384 (b) Rs. 3600 (c) Rs. 40 (d) Rs. 20,000 2. (a) Rs. 8,64,000 (b) Rs. 81,000
(c) 10 kg (d) 100 pens (e) 225 km 3. (a) Rs. 77760 (b) Rs. 1,60,000 (c) Rs. 30,000
4. (a) Rs. 30 (b) Rs.5,000 5. (a) 10 days (b) 3 (c) 60 days (d) 9 mins.
6. (a) 100 days (b) 42 days (c) 40 days 7. (a) 18 hectares (b) 64 men (c) 28 days.

190 Oasis School Mathematics-8

14.2 Time and Work

While solving problems on time and work, the Focus on:
following facts should be kept in mind.
• Number of workers to do
Let's discuss the relation of time and work.
the work

20 men can do a work in 15 days. • Duration of time to do the
1 work
20 men can do 15 work in 1 day.

i.e. less the number of days, less the work and • Amount of work done

more the number of days, more the work.



Worked Out Examples



Example: 1

A can do a piece of work in 10 days and B alone can do it in 15 days. How much time
will both take to finish the work ?

Solution:

A can do 1 work in 10 days.

A can do 1 work in 1 day.
10

B can do 1 work in 15 days.

B can do 1 work in 1 day.
15
( ) 1 1
(A + B) can do 10 + 15 work in 1 day.

(A + B) can do 3+2 work in 1 day.
30

(A + B) can do 5 work in 1 day.
30

(A + B) can do 1 work in 30 days.
5

∴ Required number of days = 30 days = 6 days.
5

Example: 2

A and B together can do a piece of work in 4 days

and A alone can do it in 6 days. In how many days A can do 1 work in 1 day.
can B alone do it? 6

Solution: B can do ( 1 – 1 ) work in 1 day.
4 6

(A + B) can do 1 work in 4 days. B can do (31–22) work in 1 day.

(A + B) can do 1 work in 1 day. B can do 1 work in 1 day.
4 12

A can do 1 work in 6 days.

B can do 1 work in 12 days.

Oasis School Mathematics-8 119911

Example: 3

A can do 1 of a certain work in 6 days and B can do 1 of the same work in 12 days. In
3 2
how many days will they together complete the work?

Solution:

A can do 1 work in 6 days. ( )(A + B) can do 1 + 1 work in 1 day.
3 18 24

A can do 1 work in 1 day. ( )(A + B) can do 4+3 work in 1 day.
3×6 72

A can do 1 work in 1 day. (A + B) can do 7 work in 1 day.
18 72

B can do 1 work in 12 days. (A + B) can do 1 work in 72 days.
2 7

B can do 1 work in 1 day. ∴ Required number of days = 72 days
24 days. 7
= 10 2
7

Exercise 14.2

1. (a) A can do a piece of work in 10 days and B can do it in 15 days. In how many
days will they do it together?

(b) Aadhya can do a piece of work in 20 days and Anasuya can do it in 30 days.
How long will both take to do it together?

2. (a) A and B can finish a work in 15 days. A alone would take 20 days to finish it.
How many days will be taken by B to finish it alone?

(b) A and B can finish a work in 12 days. B alone takes 30 days to finish it. In how
many days can A finish it alone?

(c) A and B can complete a piece of work in 8 days and A alone completes it in 12
days. In how many days would B alone complete the work?

3. (a) A can do 1 of a work in 10 days and B can do 1 of the same work in 20 days.
4 3
In how many days will they complete the work together?

(b) A can do 1 of a work in 2 days and B can do 2 of it in 2 days. In how many
4 5
days will they do the whole work together?

Answer
1. (a) 6 days (b) 12 days 2. (a) 60 days (b) 20 days (c) 24 days
3. (a) 24 days (b) 3113 days

192 Oasis School Mathematics-8

Unit

15 Profit and Loss

15.1 Profit and Loss (Introduction)

The term profit and loss is generally used by the businessmen in the transaction of goods
while buying or selling them. A person buys goods at a certain cost which is called its
cost price, in short C.P., and the cost at which s/he sells it is called its selling price, in
short S.P.

If the cost price is less than the selling price, there is profit and if the cost price is
more than the selling price, there is loss in the transaction. Hence, we can relate the
terms profit, loss, C.P. and S.P. as follows:

(i) If S.P. > C.P. then profit = S.P. – C.P.

(ii) If S.P. < C.P. then loss = C.P. – S.P.

Note:

(i) Overhead expenses like transportation, maintenance, repair and other
expenditures are always added with the cost price to get net cost price.

(ii) Profit or loss is calculated with the help of S.P. and C.P. of the same
number of articles.



Profit and Loss Percentage

A businessman always desires to expand the business by studying the market value
of goods in terms of his/her profit percentage. Profit or loss percentage always gives
an exact idea about the nature of demand and supply. Just by knowing the profit or
loss, one can't predict for the future plan of the business.

For example, if a calculator is bought for Rs. 500 and sold for Rs. 700, the profit is

Rs. 200. Also if a geometrical box is bought at Rs. 100 and sold for Rs. 300, the profit

is again Rs. 200. Rs. 200×100
Rs. 500
But, profit percent in first case = = 40%

Rs. 200×100
While, profit percent in second case = Rs. 100 = 200%

Hence, a businessman would better go for the business of the second type.

Note:
Profit percent or loss percent is always calculated out of the given cost price

unless and until it is stated otherwise.

Oasis School Mathematics-8 119933

Hence, we can write,

Profit % = Profit×100 % ........ (i)
C.P. ........ (ii)
........ (iii)
Loss % = Loss×100 % ........ (iv)
C.P.

Again, if profit % is given,

S.P. = C.P. + Profit % of C.P.

If loss % is given,

S.P. = C.P. – Loss % of C.P.

From (iii) and (iv) we can derive:

(i) S. P = (100 + profit per cent) × C.P. (ii) S. P = (100 – loss per cent) × C.P.
100 100

S.P × 100 S.P × 100
(iii) C. P = (iv) C. P =
(100+profit per cent)
(100–loss per cent)

Worked Out Examples



Example: 1

A man buys a motorbike for Rs. 1,50,000 and after some days he sells it for Rs. 1,30,000.
What is his gain or loss percentage?
Solution:

Here, C.P. = Rs. 1,50,000

S.P. = Rs. 1, 30,000

Since, C.P. > S.P.

Loss = C.P. – S.P.

= 1,50,000 – 1,30,000

= 20,000

We have, loss per cent = Loss × 100%
C.P.

20,000
= 1,50,000 × 100%

= 125 × 100%

= 13 1 %
3

194 Oasis School Mathematics-8