Prime Math 6 - Final rabi_final

2. Addition and subtraction of unlike fractions.

Addition and subtraction of unlike fractions is same as in the case of addition
and subtraction of like fractions. First, we have to convert unlike fractions to
like fractions and then add or subtract.

To add 3 and 1
4 6
3 +
4 + 1 31 2 4, 6
6 46 2, 3
3×3 1 × 2
= 4 ×3 + 6 × 2 L.C.M. = 2 × 2 × 3 = 12

= 9 + 2
12 12

= 9+2
12

= 11 = 9 + 2 = 11
12 12 12 12

To subtract 1 from 2
2 3
2 1
3 - 2

= 2 × 3 - 1 × 3 2 1
3 × 2 2 × 3 3 2

= 4 - 3 [ L.C.M. of 3 and 2 is 6] 4 3
6 6 = 6 6
1
= 4 - 3 = 6
6

= 1
6

3. Addition and subtraction involved with mixed fractions.
- Convert mixed fractions to improper fractions
- If the fractions are unlike, reduce to like fractions
- Add or subtract the numerators with common denominator.

Prime Mathematics Book - 6 145

e.g. To add: 1 1 and 2 3
2 4

1 1 + 2 3
2 4

= 2 × 1 + 1 + 4 × 2 + 3
2 4

= 3 + 11
2 4

= 3 × 2 + 11
2 × 2 4

= 6 + 11 [ L.C.M. of 2 and 4 is 4]
4 4 [ L.C.M. of 4 and 2 is 4]

= 6 + 11
4

= 17
4

= 4 1
4

To subtract 1 1 from 2 3
2 4

2 3 - 1 1
4 2

= 4 × 2 + 3 - 2 × 1 + 1
4 2
11 3
= 4 - 2

= 11 - 3 × 2
4 2 × 2

= 11 - 6
4 4

= 11 - 6 = 5 = 1 1
4 4 4

Example 1: Illustrate the given addition or subtraction by suitable diagram.
Solution:
(a) 3 + 1 (b) 3 - 1
8 2 4 6

(a) 3 + 1
8 2
+= +

= = 7
8

146 Prime Mathematics Book - 6

(b) 3 - 1
4 6

-= -

= = = 7
12

Example 2: Perform the following addition or subtraction.

(a) 1 + 2 (b) 1 3 + 2 1 (c) 7 - 2 (d) 2 1 - 1 3
2 3 4 2 9 9 2 4

Solution: (a) 1 + 2 (c) 7 - 2
2 3 9 9

= 1 × 3 + 2 × 2 [ L.C.M. of 2 and 3 is 6] = 7 - 2
2 × 3 3 × 2 9

= 3 + 4 = 5
6 6 9

= 3 + 4
6

= 7 = 1 1
6 6

(b) 1 3 + 2 1 (d) 2 1 - 1 3
4 2 2 4

=4 × 1 + 3 + 2 × 2 + 1 =2 × 2 + 1- 4 × 1 + 3
4 2 2 4

= 7 + 5 = 5 - 7
4 2 2 4

= 7 + 5 × 2 = 5 × 2 - 7
4 2 2 2 × 2 4

= 7 + 10 = 10 - 7 = 10 - 7 = 3
4 4 4 4 4 4

= 7 + 10
4

= 17 = 4 1
4 4

147 Prime Mathematics Book - 6

Exercise 7.2

1. Illustrate the following additions or subtractions in diagram.

(a) 2 + 1 (b) 1 + 1 (c) 6 - 2 (d) 8 - 4
5 5 2 8 9 9 9 9

2. Add:

(a) 2 and 1 (b) 4 and 8
3 2 5 15

(c) 3 2 and 1 1 (d) 2 5 and 2 3
5 4 7 14

3. Carryout the following subtractions.

(a) 7 - 3 (b) 1 - 9 (c) 9 - 1
10 10 2 8 12 2

(d) 9 - 10 (e) 2 2 - 1 1 (f) 7 5 - 4 7
14 21 3 2 6 12

4. Simplify:

(a) 1 + 2 - 3 (b) 3 + 2 + 1 (c) 7 - 5 + 1
2 3 4 15 15 10 12 12 12

(d) 7 - 5 + 1214 (e) 6 1 + 2 2 - 1 3 (f) 6 1 + 4112 - 3 1
8 12 2 5 4 6 24

5. Solve the following word problems:

(a) Kanchan can finish 2 of a work in a day and Kabita can finish 3 of the
5 10
same work in a day. What part of the work can be finished in a day, if

they work together?

(b) In an election candidate A got 3 and B got 1 of the total votes and rest
5 5

by C. What part of total votes did C receive?
part of the pole is coloured white?

Prime Mathematics Book - 6 148

Multiplication and division of Fraction

Study the following:

1. Multiplication of a fraction by a whole number:

Consider a multiplication 3 × 1
4

Since 3× 1 = 1 + 1 + 1 ++=
4 4 4 4

= 1 + 1 + 1
4
3 1 + 1 + 1 = 3
= 4 4 4 4 4

Also, 1 ×3= 3
4 4
1 =
= 4 of 3 or

Out of 3 Out of 1
(Supposing 3 as whole) (Supposing 1 as whole)

Thus, we observed that

3 × 1 = 1 × 3 = 3 × 1 = 3 or 1 × 3 = 3
4 4 4 4 4 4

\ Whole number × fraction = Whole number × numeritor
denominator

2. Multiplication of fraction by another fraction:

Consider a multiplication 2 × 2
3 5

Since, 2 × 2 = two third of 2
3 5 5

= two third of

= = 4
15

Thus, we observed that 2 × 2 =145
3 5

i.e. 2 × 2 = 2 × 2 = 4 Thus, fraction × fraction = Product of numerators
3 5 3 × 5 15 Product of denominators

Prime Mathematics Book - 6 149

3. Division of a whole number by a fraction:

Let’s consider division 3 ÷ 1
2

which shows the number of halves in 3

111

2 halves + 2 halves + 2 halves = 6 halves

To show it in number line

Count 1 2 3 4 5 6

1 1 111 1
2 2 222 2

0123

we observed that there are 6 halves in 3.
1
i.e. 3 ÷ 2 =6
It shows
that 3 ÷ 1 = 3×2 =6
2 1

4. Division of a fraction by a whole number:

Let’s consider a division 1 ÷ 3
2
1
Half 2 is divided into 3 equal parts and one part is the result.

1 is divided into 3 equal parts and one part is 1
2 6

i.e. 1 ÷ 3 = 1
2 6
1 1 1 1
we observed that 2 ÷ 3 = 2 × 3 = 6

5. Division of a fracton by another fraction.

Let’s consider the division21 ÷ 1
4

This gives number of 1 in 1
4 2

[How many 1 's are there in 1 ?]
4 2

150 Prime Mathematics Book - 6

11 in 1 , there are two 1 's.
44 2 4
11
44 \ 1 ÷ 1 = 2
2 4

This, shows that 1 ÷ 1 = 1 × 4 = 4 = 2
2 4 2 1 2

Thus, a number (whole number or fraction) ÷ fraction

= the number (whole number or fraction) × reciprocal of the divisor

Note: The new fraction formed by interchanging numerator and denominator of a

fraction is called its reciprocal.

\ Reciprocal of a is b 1 a
b a a 1
Similarly, reciprocal of a is 1
a and reciprocal of is i.e. a

Product of two reciprocal terms is 1.

Example 1: Perform the multiplication 3 × 2
5
3
Solution: 5 × 2

= 3 × 2
5
6 = 115
= 5

Example 2: Multiply 1 2 by 3
3 5

Solution: 1 2 × 3
3 5

= 3× 1+ 2 × 3 Convert mixed fraction to improper
3 5 fraction

= 3 + 2 × 3 Multiply numerators to obtain the
3 5 numerator and multiply denominators
to obtain the denominator of the result.
= 5 × 3
3 5 Divide or convert the result into the
lowest term.
= 15 =1
15

Prime Mathematics Book - 6 151

Example 3: Perform the division: (i) 10 ÷ 23, (ii) 1 1 ÷ 4.
Solution: 3

(i) 10 ÷ 2 (ii) 1 1 ÷ 4
3 3
3 =3 × 1 + 1
= 10 × 2 3 ÷ 4

= 10 × 3 = 3 + 1 ÷ 4
2 3

= 30 = 4 × 1
2 3 4
4 × 1 4 1 1
= 15 = 3 × 4 = 12 3 = 3

Example 4: Perform the division: 4 ÷ 3 .
Solution: 5 4

4 ÷ 3 1
5 4 15
15) 16 (1
= 4 × 4
5 3 -15
4 × 4
= 5 × 3 1

= 16 = 1 1
15 15

Example 5: An artist can accomplish 1 of his painting in a day. What part of
12

the painting does he accomplish in 4 days?

Solution: In 1 day he can accomplish 1 of the painting
12
1
So, in 4 days he can accomplish 12 × 4 of the painting

= 1×4 of the painting
12

= 4 of the painting
12

= 1 of the painting
3

So, he can accomplish 1 of the painting in 4 days.
3

Example 6: Samfelma got 3 of a cake. She gave 2 of the cake that she had to
4 3

Tsering. What fraction of the cake did Tsering get?

Solution: Samfelma got 3 of a cake
4

Tsering got 2 of 3 = 2 × 3 = 1
3 4 3 4 2

So, Tsering got 1 of the cake.
2

152 Prime Mathematics Book - 6

Example 7: How many jars each of capacity 2 1 litres are needed to empty 45
2
litres of milk contained in a vessel?

Solution: Here, capacity of the vessel = 45 litres.

Capacity of the jar = 2 1 litres
2
1
\ No. of jars = 45 ÷ 2 2

= 45 ÷ 2 ×2 +1
2

= 45 ÷ 5
2
= 945 2
× 5 = 18

So, 18 jars are needed.

Example 8: A kurta requires 2 1 metres of cloth. How many such kurtas can be
2
3
made from 24 4 metres of cloth?

Solution: Here, total cloth = 24 3 m.
4
1
Cloth required for a kurta = 2 4 m.

\ No. of Kurtas = 24 3 ÷ 2 1
4 4

= 4 × 24 + 3 ÷ 4×2+1
4 4
99 9
= 4 ÷ 4

= 99 11 4
4 9
×

= 11
So, 11 Kurtas can be made from the given cloth.

Exercise 7.3

1. Multiply:

(a) 1 by 3 (b) 4 by 3 (c) 1 by 1 (d) 1 by 1 1
4 4 5 2 2 3

2. Perform the following multiplications:
1 1 1 1 1 12 1
(a) 5 × 3 (b) 2 × 3 (c) 6 × 3 (d) × 2 5

(e) 116 × 2 (f) 1 of 1 (g) 7 of 96 km (h) 1 of 800kg
2 6 12 10

Prime Mathematics Book - 6 153

3. Divide:

(a) 3 by 1 (b) 4 by 2 (c) 12 by 1 1 (d) 1 by 3
2 3 4 3

(e) 1 1 by 5 (f) 3 by 2
2 8 4 3

4. Perform the following:

(a) 1 ÷ 3 (b) 2 ÷ 2 (c) 2 2 ÷ 3 2 (d) 4 ÷ 1
5 3 3 3 3

(e) 3 ÷ 1 (f) 1 ÷ 1
6 2 4

5. Write the reciprocals of: 1 (d) 2152
1 4
(a) 2 (b) 5 (c) 3

6. Solve the following problems:

(a) How much is half of one fourth of an object?

(b) Out of 5 dozens of apples 1 are rotten. How many rotten apples are
there ? 12

(c) The price of a metre of pipe is Rs. 5221 . What is the price of 5 7 metres
of the pipe ? 15

7. Solve the following problems:
(a) How many bottles each of 112 litres can be filled from 12 litres of oil?

(b) 5 1 kg of flour is divided into 32 equal parts. What is the weight of each
3
part?

(c) 1 of a cake is divided equally to 5 children. What fraction of a cake
2
will each child get?

(d) Area of a rectangle is 30m2. If breadth is 421 m, find the length of the
rectangle.

154 Prime Mathematics Book - 6

Simplification of fractions

Study the following:

Simplification is the process of making a mathematical term or expression look
simple. Simplification of fractions may include single operations converting to lowest
term (cancellation), addition, subtraction, multiplication, division or mixed operations.

Complex form Simplification Simple form

(1) 36 = 36 3 = 3
48 48 4 4

(2) 1 = 1 × 4 2 = 2
2 2 3 3
3
4

(3) 1 + 3 = 1 × 2 + 3 = 2 + 3 = 2 + 3 = 5 = 114
2 4 2 2 4 4 4 4 4

(4) 132 - 1 = 3 × 1 + 2 - 1 = 5 - 1
6 3 6 3 6

= 5 × 2 - 1 = 10 - 1 = 121
3 × 2 6 6 6

= 10 - 1 = 9 3 = 3
6 6 2 2

(5) 4 × 334 = 4 × 343 = 4 × 4×3+3 =3
5 5 5 4

= 4 × 15 3 =3
5 4

(6) 1 ÷ 1 = 1 ÷ 1 = 1
4 2 4 2 2

= 1 × 2 = 1
4 1 2
2

(7) 3 of Rs. 300 = 3 × Rs. 300 = Rs. 225
4 4
900 225 155
= 4 = 225

Prime Mathematics Book - 6

The mixed operations of fractions are also simplification in the order of BODMAS

rule.

BODMAS (6) Subtraction

(5) Addition

(4) Multiplication

(3) Division

(2) of (multiplication)

(1) Brackets [{( )}]

Note: Addition and subtraction, any one can be peformed first

3×2-4+3 or 3 × 2 - 4 + 3
=6-4+3 =6-4+3
=9-4 =2+3
=5 =5

Example 1: Simplify:143 × 3 + 3 ÷ 3
14 8 10

Solution: 1 3 × 3 + 3 ÷ 3
4 14 8 10
10 5
= 4 × 1 + 3 × 3 + 3 × 3
4 14 84
1
7 × 3 + 5
= 4 14 4
2

= 3 + 5
8 4

= 3 + 10 = 13 = 1 5
8 8 8

Example 2: Simplify: 3 ÷ 123 of 335
5

Solution: 3 ÷ 132 of 335 [mixed fractions converted to improper fractions]
5

= 3 ÷ 132 × 353 [of = × and operation of 'of' is performed before ÷]
5
6
= 3 ÷ 5 × 18
5 3 5

= 3 ÷ 6
5 1
3 1
= 5 × 6 2

= 1
10

156 Prime Mathematics Book - 6

Example 3: Simplify:161 ÷ 223 - 134 + 1 × 10
2 21

Solution: 116 ÷ 232 - 134 + 1 × 10
2 21

= 6 × 1 + 1 ÷ 3 × 2 + 2 - 4 × 1 + 3 + 1 × 10
6 3 4 2 21

If you are confident this step can be steped up

= 7 ÷ 8 - 7 + 1 × 10
6 3 4 2 21

= 7 ÷ 8 - 7 + 1 × 2 × 10
6 3 4 2 × 2 21

= 7 ÷ 8 - 7 + 2 × 10
6 3 4 4 21

= 7 ÷ 8 - 7+2 × 10
6 3 4 21

= 7 ÷ 8 - 9 × 10
6 3 4 21

= 7 ÷ 8 × 4 - 9 × 3 × 10
6 3 × 4 4 × 3 21

= 7 ÷ 32 - 27 × 10
6 12 12 21

= 7 ÷ 32 - 27 × 10
6 12 21

= 7 ÷ 5 × 10
6 12 21

= 7 × 12 2 102 = 4 = 113
6 5 213 3
×

Example 4: What is the result when 4 of the difference of 3 and 3 is divided
by 7? 9 4 4

Solution: 4 of the difference of 3 and 1 divided by 7 can be writen
9 4 6

mathematically as

4 of 3 - 1 ÷7
9 4 6

= 4 of 3×3-2×1 ÷7
9 12

= 4 of 9-2 ÷7
9 12

Prime Mathematics Book - 6 157

= 4 of 7 ÷7
9 12

= 4 ×3172 ÷7
9

= 7 ÷ 7
27

= 7 × 1
27 7

= 1
27

Example 5: Distance between Hetauda and Phaparbari is 45 km. Mr. Tamang

travelled 2 of the distance by bus, 1 of the distance by motorbike
3 5
and the remaining distance he walked. How many kms did he walk?

Solution: We can write the given problem mathematically as

45 - 2 + 1 of 45
3 5

= 45 - 2×5+3×1 of 45
15

= 45 - 10 + 3 of 45
15

= 45 - 13 × 45
15

= 45 - 13 × 45 3
15

= 45 - 39 = 6

\ Mr. Tamang walked 6 km.

Exercise 7.4

1. Simplify:

(a) 1 + 3 × 2 (b) 4 × 5 - 1
4 4 3 5 8 3

(c) 2 1 + 213 × 1 - 3 (d) 1 1 + 4 × 3 - 1
2 14 4 3 9 8 6

158 Prime Mathematics Book - 6

(e) 2 1 × 3 - 1 - 1 (f) 1 2 ÷ 6 2 × 1 1 + 5
3 4 2 4 3 3 9 9

(g) 2 + 7 ÷ 7 × 114 (h) 3 - 2 ÷ 2 4
3 9 10 4 5 5

2. Simplify:

(a) 3 of 2 2 × 4 (b) 2 21 - 1 ÷ 1 of 1
4 3 3 2 3

(c) 5 of 3 ÷ 7 × 121 (d) 2 1 ÷ 1 1 of 3 - 2 × 141
6 4 8 3 6 4 5

3. Write the following statements mathematically and simplify.

(a) Subtract the sum of 321 and 223 from 665 .

(b) Reema gave two third of a bread to her son and half of the remaining
bread to her daughter. What fraction of the bread is left with her?

(c) Find the product of the sum and difference of 2 and 1 ?
3 6

Prime Mathematics Book - 6 159

Decimals

Recall the following
- Decimal means system based on tens or tenths.

- Decimal fraction are the fractions with denominators powers of 10

e.g. 1 , 2 , 3 , etc
10 100 1000

- Decimal fractions are expressed in other ways as

2 (two tenth) = 0.2 (zero point two)
10

12 (twelve hundredths) = 0.12 (zero point one two)
100

25 (twenty five thousandths) = 0.025 (zero point zero two five) etc which
1000

are simply called decimals.

Place values in decimals follow as

Thousands 123 4.567 thousandths
hundredths
Hundreds Decimal part tenths
decimal point
Tens

Ones

Whole number part

To convert fractions into decimal fractions and decimals: If the denominators of
the fractions are sub multiples (factors) of numbers which are powers of 10, they
can be converted into decimal fraction by converting the denominator into powers
of 10.

e.g. 1 = 1×5 = 5 =0.5
2 2×5 10

1 = 1×25 = 25 = 0.25.
4 4×25 100

160 Prime Mathematics Book - 6

Direct division method to convert a fraction into decimals:

e.g. 1 = 1.0 Steps
2 2
Divide 1 by 2 i.e. 1.0 by 2
2) 1.0 (0.5
(i) Whole number 1 being smaller than the divisor2,
0 quotient is 2 × 0 = 0 put 0 as quotient and product 0
below 1 and subtract.
10
(ii) Remainder 1 and drop 0.(after decimal, we can take
-10 as many zeros as we need), since number after decimal
× is dropped, put decimal after 0 in the quotient and
divide 10 by 2 as usual.
\ 1 = 0.5
2

Similarly consider a fraction 1
4
1 1.00
4 = 4 Divide 1 by 4 i.e. 1.00 by 4

= 4) 1.00 (0.25 1. First, divide the number in whole number part. Whole number
0
10 1 being smaller than the divisor 4, quotient is 4 × 0 = 0 (put o
at quotient and product 0 below 1 and subtract.)

- 8 2. Remainder is 1 and drop 0. As number after decimal point is
20 dropped, put decimal in the quotient. Divide 10 by 4 it is
- 20 4 × 2 = 8 put 2 next to decimal in the quotient and 8 below 10
× and subtract.

\ 1 = 0.25 3. Remainder is 2 and drop next zero (there are so many zeros after
4 decimal as we need) it becomes 20,now divide 20 by 4, it is
4 × 5 = 20 so put 5 next to 2 in the quotient and 20 below 20 and
subtract. No remainder left, so the process ends.

To convert decimals into decimal fraction:

e.g. 0.7 (7 tenths) = 7
10

0.25 (twenty five hundredths) = 25
100

0.124 (124 thousandths) = 124
1000

Write the decimal number as whole number and put in numerator and put 1 with

as many zeros after it as there are digits after decimal, in denominator (if possible

decimal fraction can be converted to lowest term to convert into simple fraction).

Prime Mathematics Book - 6 161

Addition of decimals

Consider the addition of 12.07 and 18.934.

12.07 + 18.934 Arrange vertically such that position of decimals
and digits in corresponding places be in a straight
11 1 line (vertical).
start adding from right to left with carry over if
12.07 any.
+18.934
=31.004

Subtraction of decimals:

Consider subtraction of 3.762 from 7.06

7.06 - 3.762 - Arrange the numbers vertically making decimal
point and corresponding digits of same places
15 along same vertical lines.
9 10
6 10 5 10 - For 7.06 thousandths place is 0

7.060 - Subtract from right to lift as usual process in
whole numbers.
-3.762
=3 . 2 9 8

Non decimal fractions or Vulgar fractions:

We know that the fraction having denominators powers of 10 are called decimal
fractions. e.g. 1/10, 3/100, 37/1000 etc. The decimals having denominators powers
of 2 or 5 or their products can be converted into decimal fractions.

e.g 1 = 1 × 5 = 5
2 2 × 5 10

2 = 7×4 = 28
25 25 × 4 100

Fraction having denominators other than powers of 2 or 5 or 10 or their products
cannot be converted into decimal fraction.

e.g. 2 , 7 , 43 , etc
3 12 56

Such fractions are called non-decimal fractions or Vulgar fractions. Such fractions

give non-terminating but recurring decimals which we will discuss in higher classes.

162 Prime Mathematics Book - 6

Example 1: Convert into decimals:

(a) 3 (b) 9 (c) 27
10 100 1000

Solution: (a) 3
10
In denominator, 10 has one zero so
= 3.0 = 0.3 position of shifts one digit to left.
10

By direct division 10) 3 (0 ones 3 tenth = 0.3
0
= 10) 3.0 (0.3 30 3 = 30 tenths
-0 30
30 ×
-10
×

= 0.3

(b) 9 In denominator, 100 has two zeros
100 so the position of decimal point is
9.00 shifted two digits to left.
= 100

= 0.09

By direct division

9 = 100) 9.00 (0.09
100
-0
90 100) 9 ( 0 ones o tenths 9 hundredths
0
-0 90 9 = 90 tenths = 0.09
900 0
-900 900 90 tenths = 900 hundredths
× 900
×

= 0.09

Prime Mathematics Book - 6 163

(c) 27 In denominator, 1000 has three zeros
1000 so the position of decimal point
shifts three digits to the left.
= 0.027
1000

= 0.027

By direct division

27 = 1000) 27.000 (0.027
1000
-0

270

-0

2700

-2000

7000

-7000

×

= 0.027

Example 2: Convert into decimals.

(a) 3 (b) 2 1 (c) 1 (d) 1
4 2 5 25

Solution: (a) 3 = 3 × 25 = 75 = 0.75
4 4 × 25 100

By direct division 4 ) 3 ( 0 ones 7 tenths 5 hundredths = 0.75
0
3 = 4) 3.00 (0.75 3 0 3 = 30 tenths
4 -0 28
2 0 2 tenths = 20 hundredths
30 20
×
- 28
= 0.75
20

- 20
×

= 0.75

164 Prime Mathematics Book - 6

(b) 221 = 2×2+1 = 5 = 5×5 = 25 = 2.5
2 2 2×5 10

By direct division

(b) 212 = 2×2+1 = 5 = 2) 5.0 (2.5
2 2
-4

10

-1 0

×
=2.5

(c) 1 = 1 × 2 = 2 = 0.2
5 5 × 2 10

By direct division

1 = 5) 1.0 (0.2
5
-0

10

-1 0

×

= 0.2

(d) 3 = 3×4 = 12 = 0.12
25 25 × 4 100

By direct division

3 = 25) 3.00 (0.12
25
-0

30

-25

50

-50

×

= 0.12

Prime Mathematics Book - 6 165

Example3: Convert the following into fractions:

(a) 0.8 (b) 1.45 (c) 0.056

Solution: (a) 0.8 = 8 [8 is in place value of tenths ]
10

= 84
10 5

= 4
5

(b) 1.45 = 145 Last digit 5 is in the place value of
100 hundredths

= 145 29
100 20

=1290

(c) 0.056 = 56 Last digit 6 is in the place
1000 value of thousandths

= 56 28 14 7 125
1000 500 250

= 7
125

Example 4: Add 0.5, 1.025 and 0.09.

Solution: 0.500

1.025

+ 0.090

1.615

\ The sum is 1.615

Example 5: Subtract: 13.07 form 62.7.

Solution: 5 12 6 10

62.70

-13.07

49.63

166 Prime Mathematics Book - 6

Example 6: Simplify

180.4 + 52.60 – 4.19

Solution: 11

180.40

+ 52.60

233.00

Again 233.00

-4.19

228.81

\ 180.4 + 52.60 - 4.19

= 233.00 - 4.19

= 228.81

Exercise 7.5

1. Convert the following fractions into decimals.

(a) 7 (b) 9 (c) 37 (d) 145
10 100 1000 1000

(e) 143 (f) 1234 (g) 503 (h) 7891
10 1000 10000 100

2. Convert the following fractions into decimal fractions and then into decimals.

(a) 1 (b) 1 (c) 5 (d) 12
4 2 16 25

(e) 2 (f) 5
5 8

3. Convert the following decimals into fractions and reduce them to their
lowest terms:

(a) 0.2 (b) 0.04 (c) 0.005 (d) 1.4

(e) 0.15

Prime Mathematics Book - 6 167

4. Carry out the following addition or subtraction.

(a) 3.05 (b) 1.425
+ 1.47 + 0.068

(c) 5.67 (d) 21.282
-3.18 -13.075

(e) 3.056 + 2.560 (f) 64.7 - 13.08

5. Simplify: (b) 0.94 + 0.63 - 0.69
(a) 8.12 – 7.9 + 0.006 (d) 0.135 -1.2 + 3.246
(c) 1.204 - 0.8 - 0.07

6. Solve the following problems:

(a) The price of a book is Rs. 215.45 and price of a pen is Rs. 67.65. Find
the total price of these two things.

(b) What should be subtracted from 86.420 to get 12.037?

(c) What should be added to 0.357 to get 23.456?

(d) Dixshant buys a football for Rs. 275.75 and a rakit for Rs. 370.85. If he
gives Rs. 1000 note to the shopkeeper, how much will he get back as
return?

168 Prime Mathematics Book - 6

Multiplication and division of decimals

Multiplication of decimals by 10 or powers of 10:

Let's consider the multiplication 10 × 1.234

Here, 10 × 1.234 = 10 × 1234 Conclusion:
1000
10 × 1.234 = 12.34. Decimal
= 12340 point shifts after one digit
1000 to the right.

= 1234
100

=12.34

Again consider the multiplication 100 × 1.234

100 × 1.234 = 100 × 1234 = 1234 = 123.4
1000 10

Conclusion:
100 × 1.234 = 123.4
Decimal point shifts after two digits to the right

Again consider 1000 × 1.234

1000 × 1.234 = 1000 × 1234 =1234
1000

Conclusion:

1000 × 1.234 = 1234.0

Decimal point shifts after three digits to the right.

Thus, while multiplying a decimal number by 10 or powers of 10, the position of
decimal shifts to the right after a number of digits as there are number of zeroes
after 1.

Multiplication of decimals by whole numbers other then 10 or powers
of 10

Consider multiplication 2.34 × 6 Here, 234 × 6
= 234
2.34 × 6 = 234 × 6 ×6
100
= 1404
= 1404 = 14.04
100

Prime Mathematics Book - 6 169

In case of multiplication of a decimal by a whole number other than 10 or powers
of 10, multiply the decimal by the given whole number as in whole numbers and
take the position of decimal point as many decimal places as in the given decimal.

Multiplication of a decimal number by another decimal number

Consider the multiplication 12.34 × 1.6

Here, 12.34 × 1.6 The product 1234
×16
= 1234 × 16
100 10 7404
1234
= 19744 19744
1000

= 19.744

- Multiplication is as usual process
- No. of digits after decimal point in the multiplicand = 2
- No of digits after decimal point in the multipler = 1
- Therefore, no of digits after decimal point in the product is = 2 + 1 = 3

Thus in case of multiplication of decimal number by another decimal number,
multiply two numbers as usual. Count the total digits after decimal in two numbers
and locate decimal point according to this sum .

Division of decimal numbers:

A) Division of a decimal number by 10 or powers of 10.

Consider the division 543.2 ÷ 10.

Here, 543.2 ÷ 10 \ 543.2 ÷ 10 = 54.32

= 5432 × 1 On dividing a decimal number by
10 10 10, position of decimal point
shifts after a digit to the left.
= 5432
100

= 54.32

170 Prime Mathematics Book - 6

Again consider the division 543.2 ÷ 100

Here, 543.2 ÷ 100 \ 543.2 ÷100 = 5.432

= 5432 × 1 On dividing a decimal number by
10 100 100, position of decimal point
shifts after two digits to the left.
= 5432
1000

= 5.432

Thus, on dividing a decimal number by 10 or powers of 10, the position of
decimal point is shifted to the left after as many digits as there are zeros after
1 in the divisor.

B) Division of a decimal number by a whole number other than 10 or powers
of 10

Consider the division 2.48 ÷ 4

Here, 2.48 ÷ 4 400) 248.00 (0.62
-0
= 248 × 1 2480
100 4
248 -2400
= 400 800
-800
= 0.62 ×

\ 2.48÷4 = 0.62

4) 2.48 (0.62
= -0

24
-24

8
-8
×

Prime Mathematics Book - 6 171

C) Division of a decimal number by another decimal number.

Consider the division 8.765 ÷ 2.5 5) 17.530 (3.506
-15
Here, 8.765÷2.5 2.5
-2.5
= 8.765 × 10 03
2.5 × 10 -0
30
= 87.65 -30
25 ×
(cancel by 5)
= 17.53
5

= 3.506

- On dividing a decimal number by another decimal number, convert the divisor
to whole number by multiplying numerator and denominator by 10 or powers
of 10 and divide as in division of decimal by whole number

Example 1: Multiply: 0.5432 by 10, 100 and 1000.

Solution: 0.5432 × 10 = 5.432. On multiplying a decimal by 10
or powers of 10, the digits in the
0.5432 × 100 = 54.32 multiplicand remain same and
position of decimal shifts after
And 0.5432 × 1000 = 543.2 as much digits as there are zeros
after 1 in the multiplier.
Otherwise
(5432 thousandth )
0.5432 × 10 = 5432 ×10
10000

= 5432
1000

= 5.432

0.5432 × 100 = 5432 ×100
10000

= 5432 (5432 hundredth )
100

= 54.32

0.5432 × 1000 = 5432 ×1000
10000

= 5432 (5432 tenth)
10

= 543.2

172 Prime Mathematics Book - 6

Example 2: Multiply: 5.432 by 6

Solution: 5.432×6 - Usual multiplication

= 5.432 - Position of decimal point in the
×6 product is same as in the
multiplicand
32.592

Otherwise
5.432 × 6

= 5432 × 6 (decimal converted to decimal fraction )
1000 1

= 32592 = 32.592
1000

Example3. Multiply 104.32 by 2.4

Solution: 104.32 × 2.4 10432
×24
= 250.368 total decimal place 2 + 1 = 3
(250368 thousandth ) 41728
Alternate method 208640
104.32 × 2.4 250368

= 10432 × 24
100 10
250368
= 1000

= 250.368

Example4. Divide 8765.4 By 10,100 and 1000.
Solution: 8765.4 ÷10

= 876.54
8765.4 ÷ 100
= 87.654
And 8765.4 ÷ 1000
= 8.7654

Prime Mathematics Book - 6 173

Alternate method:

8765.4÷10 = 87654 × 1 = 87654 = 876.54
10 10 100

8765.4 ÷ 100 = 87654 × 1 = 87654 = 87.654
10 100 1000

And 8765.4 ÷ 1000 = 87654 × 1 = 87654 =8.7654
10 1000 10000

Example 5: Divide 32.144 by 8
Solution: 32.144÷8

= 8) 32.144 (4.018
-32
1
-0
14
-8
64
-64
× =4.018

Example 6: Divide 2.676 by 1.2

Solution: 2.676 ÷ 1.2

= 2.676 × 10
1.2 × 10

= 26.76 (to convert denominator to whole number)
12

= 2.23 12) 26.76 (2.23

-24

27

-2 4

36 Division can be applied
-36 after cancellation.

×

174 Prime Mathematics Book - 6

Exercise 7.6

1. Multiply the following decimals by 10, 100 and 1000.

(a) 34.567 (b) 8.27 (c) 0.004

2. Perform the following multiplications.

(a) 3.154 × 6 (b) 43.21 × 16 (c) 0.1234 × 60

3. Multiply
(a) 3.24 by 8.2 (b) 5.75 by 6.21 (c) 1.8 by 0.03

4. (a) How many metres are there in

(i) 0.12 km ? (ii) 4.5 km?

(b) How many days are there in 2.5 months?

(c) If the cost of a pen is Rs. 74.75, find the cost of 8 such pens.

(d) If a shirt requires 1.25 meters of cloths, how much cloths is required for
12 such shirts?

(e) What is the area of a rectangle of length 4.2cm and breadth 2.4cm?

5. Divide the following by 10, 100 and 1000:

(a) 12.3 (b) 0.15 (c) 24.6

6. Carry out the following divisions.

(a) 0.1326 ÷ 3 (b) 7.2 ÷ 8 (c) 97.29 ÷ 9

7. Perform the following divisions.

(a) 0.5784 ÷ 0.8 (b) 600 ÷ 2.5 (c) 8 ÷ 0.125

8. Simplify: (b) (4.5 - 2.0) × 1.2 (c) 4.5 × (3.9 - 1.7)
(a) 0.2 × (2.3 + 1.2)

9. (a) If the cost of a dozen of exercise books is Rs289.8,what is the cost of an
exercise book ?

(b) The perimeter of a square is 13.48 cm find lengthof its side.

Prime Mathematics Book - 6 175

Rounding off of numbers

If we divide Rs. 225 among 8 persons, each will get Rs. 28.125. In expanded form
1 2 5
Rs. 28.125 = Rs. 2 × 10 + Rs. 8 × 1 + Rs. 10 + Rs. 100 + Rs. 1000

Here, Rs. 1 is 10 paisa, Rs 2 is 2 paisa
10 100

but Rs. 5 which is 0.5 p
1000

and not so significant. Thus for our
convenience we express Rs. 28.125
approximately i.e. nearly about Rs.28.13.

If a fruit seller shows weight of apple to

be 4kg and 997 grams, customers will accept
it as 5 kgs .

If somebody asks you time and your watch is 03:10:04 PM
showing 3 : 10 : 04 P.M i.e. 3 O'clock, 10 minutes
04 seconds, you will of course say 3 O'clock.

Such process of expressing quantities or numbers in the nearest convenient units is
called rounding off (approximation).

1. Whole numbers are rounded off to the nearest tens, hundreds, thousands,

etc. far

near

62 60 60 61 62 63 64 65 66 67 68 69 70
(nearest to tens) far near

190 200
(nearest to hundreds) 100 110 120 130 140 150 160 170 180 190 200
or » Stands for approximately equal to.

Process:
i. Determine to which place to approximate, tens or hundreds or thousands etc.
ii. Drop (omit) the number or numbers in the lower places.
iii. If the first digits of drop is less than 5, just drop (omit) (rounding off down).

176 Prime Mathematics Book - 6

(iv) If the first digit of the drop is 5 or greater than 5, add 1 in the rounded digit
(rounding off up).

Decimal numbers are rounded off to the nearest tenths, hundredths etc.

e.g. to round off 21.0748 to the nearest tenths (1 decimal place)

21.0748 21.1 For rounding off to 1 decimal place ( nearest tenth)
digits 74 and 8 are to be dropped

The digit immediately after tenth place is 7 which
is greater than 5,so 1 is added to the digit at tenth
place i.e. 0. Thus 0 + 1 = 1.

To round off 1.2345 to the nearest hundredths (2 decimal place)

1.2345 1.23 For rounding off to 2 decimal place (nearest hundredth),
the last two digits 4 and 5 are to be dropped.

The digit immediately after hundredth places is 4
which is less than 5. So they are just dropped.

To round off 14.6 to the nearest ones.

14.6 15 For rounding off to the nearest ones, the digit immediately
after it is 6 which is to be dropped.

The digit to be dropped 6 is greater than 5 so 1 is added
to the ones places.

Example 1. Round off 21.5063 to (a) the nearest ones (b) 1 decimal place
(c) 2 decimal places (d) 3 decimal places.

Solution: (a) 21.5063 to the nearest ones (0 decimal place) 22.
(b) 21.5063 to 1 decimal place 21.5
(c) 21.5063 to 2 decimal place 21.51
(d) 21.5063 to 3 decimal place 21506

Prime Mathematics Book - 6 177

Exercise 7.7

1. Round off the following decimals to the nearest ones.

(a) 2.61 (b) 0.59 (c) 24.429 (d) 60.128

2. Round off the following decimals to one decimal place.

(a) 5.63 (b) 12.68 (c) 8.05 (d) 0.97

3. Round off the following to the nearest hundredths.
(a) 14.597 (b) 214.568 (c) 0.3421 (d) 1.105

4. Round off the following to 3 decimal places
(a) 82.5963 (b) 19.9998 (c) 0.2461 (d) 213.0107

5. Find the product and write the answer to 2 decimal places.

(a) 0.78 × 1.12 (b) 1.054 × 2.5

6. Find the quotient to 4 decimal places and round off to 3 decimal places.

(a) 2 (b) 22
3 7

Unit Revision Test I

1. Add: (a) 2 and 1 (b) 3 52 and 1 1
3 2 4

2. Carry out the subtractions: (a) 9 - 1 (b) 765 - 4172
8 2

3. Simplify: 6 1 + 2 2 - 1 3
2 5 4

4. What should be added to 4 7 to make 8 3 ?
12 8

5. Multiply: 1 by 2
6 5

6. Divide: 114 by 1
2

178 Prime Mathematics Book - 6

7. Evaluate the following:

(a) 1012 ÷ 1041 (b) 7 of 108 km
12

8. How many bottles each of 1 1 litres can be filled from 12 litres of oil?
2

9. Simplify: 221 + 231 × 1 - 3
14 4

10. Simplify: 232 of 5 ÷ 1 ÷ 212 1 - 1
6 4 2 4

Unit Revision Test II

1. Convert the following fractions into decimals.

(a) 5 (b) 11
10 20

2. Convert into fraction and reduce to their lowest term.

(a) 0.04 (b) 1.235
3. Perform the following:

(a) 21.083 (b) 14.00
9.084 -8.24

+ 0.056

4. Write the reciprocals of:

(a) 1 (b) 25
4

5. What should be subtracted from 84.42 to get 12.037?

6. Perform the following:

(a) 0.03 × 1.8 (b) 97.29 ÷ 9 (c) 8.4 ÷ 0.021

7. Area of a rectangle is 247.632cm2 and breadth is 9.24cm, find the length
of the rectangle.

8. Round off:

(a) 14.568 to the nearest hundredth

(b) 0.2463 to 3 decimal places

9. Find the product 1.054 × 2.5 and write the answer to 2 decimal place.
22
10. Find the quotient 7 to 4 decimal places and round off to 3 decimal places.

Prime Mathematics Book - 6 179

Estimated periods – 6

Objectives

At the end of this unit, the students will be able to:
define ratio and proportion
solve problems related to ratio and proportion
define percentage
convert fractions or decimals into percentage and percentage into fraction and
decimal
solve the problems related to percentage

24 = 24 out of 100 = 24 percentage = 24%
100

Teaching Materials Activities

Graphs, grids, charts to show fractions It is better to:
discuss about percentage with the help
of graph of hundredth
clarify the concept of ratios as a way of
comparisons of two quantities of same
kind
clarify the concept of proportion by
giving example of direct varying
quantities

Ratios, Proportion and Percentage

Ratios

How we compare two quantities?

Lets consider weight of two goats 20 kgs and 10 kgs . Let the weight of bigger goat
is 20 kgs be denoted by A and that of smaller goat 10 kgs be denoted by B.

Expressing these quantities as fraction

A = 20kg
B 10 kg

or, A = 2
B 1

or, A = 2B (cross multiplying)

A is twice B.

Also if we take

A = 2
B 1

or, 2B = A

or, B = A
2

B is half A.

Expressing two quantities of same units as a fraction is called a ratio .

A = 2 is the ratio A to B which is represented as A:B = 2:1(read as A is to B )
B 1

= 2 is to 1.

And B = 1 is the ratio B to A which is represented as B:A = 1:2
A 2

(read as B is to A = 1 is to 2).

Process to write the ratio of two quantities. 181
1. Convert two quantities to the same unit.
2. Take the quanties as fraction.
3. Reduce to the lowest term if possible.

Prime Mathematics Book - 6

The ratio of Rs. 5 to 75 p is

= Rs.5 = 500p = 500100 20
75p 75p 75 15 3

= 20 or 20:3
3

Note: - Colon (:) is used to denote ratio which is read as 'is to'.
- Ratio is the comparison of two quantities.
- Two terms of a ratio are called antecedent and consequence. In the ratio 1:2, 1
is antecedent and 2 is the consequence.

Example 1: Find the ratio: (a) 6 to 18, (b) 500 ml to 4l

Solution: (a) Ratio 6 to 18 (b) The ratio 500 ml to 4l

= 61 = 500ml
18 3 4l
1
= 3 = 1:3 = 500ml
4000ml
51
= 40 8

= 1 = 1:8
8

Example 2: Write the ratio 151 : 3 to its simple form.
Solution: 115 : 3

6
5
= 3

= 6 2× 1
5 3

= 2 = 2:5
5

Example 3: In a class there are 24 boys. If the ratio of number of girls to the
Solution: number of boys is 3:4, find the number of girls in the class.
Here, ratio of number of girls to the number of boys = 3:4
No. of boys = 24
Let the number of girls = x, then

182 Prime Mathematics Book - 6

No. of girls : No. of boys = 3:4

or, No. of girls = 3
No. of boys 4

or, x = 3
24 4

or, x = 3 × 24 6
4

\ x = 18

\ The number of girls is 18.

Example 4: There are 30 students in class VI. If the ratio of boys to the girls
is 2:3, find the number of boys and girls.

Solution: Here, total number of students = 30.

Ratio of number of boys to the number of girls = 2:3

2 parts out of (2 + 3) = 5 parts is boys.

\ No. of boys = 2 of 30
5
2
= 5 × 30 6

= 12

And 3 parts out 5 parts is girls

\ No. of girls = 3 × 30 6
5

= 18

\ No. of boys = 12 and No. of girls = 18

Alternate method: 183
Total No. of boys = 30
Ratio of No. of boys to the No. of girls = 2:3
Let the No. of boys = 2x and No. of girls = 3x
than
2x + 3x = 30
or, 5x = 30
or, x = 6
\ No.of boys = 2x = 2 × 6 = 12
\ No. of girls = 3x = 3 × 6 = 18

Prime Mathematics Book - 6

Exercise 8.1

1. Find the ratio of the following:

(a) 35 to 75 (b) 50 p to Rs. 3 (c) 5kg to 750 gram
1 (e) 3 dozen pens to 24 pens (f) 1200 m to 2 km
(d) 2 l to 750ml

2. Write the following ratios to simple form.

(a) 1 : 1 (b) 231 : 14 (c) 390 : 810
2 4
(e) 500 ml : 1 l (f) 2 : 213
(d) 750 gram: 2kg

3. (a) Price of a pant is Rs. 1500 and the price of a shirt is Rs. 750. Compare
the price of the pant and shirt.

(b) The ratio of ages of a father and son is 12:5. If the father age is 48 years,
find the age of the son.

(c) A milk man mixes milk and water in the ratio 5:1. How much water does
he mix in 14 litres of pure milk?

(d) In a class, there are 33 students. If the boys and girls are in the ratio
5:6, find the number of boys and the girls.

Proportion

Proportion means equality of two ratios.

Let’s consider, there are 15 boys and 20 girls in class VI and 18 boys and 24 girls in
15
class VII. In class VI, ratio of number of boys to the number of girls = 20

= 3 = 3:4. In class VII, ratio of number of boys to the number of girls = 18
4 24

= 3 = 3:4. We get, ratio of No. of boys to the No. of girls in two class are
4
same.

We write, 15:20 = 18:24 [Read as 15 is to 20 as 18 is to 24]
or, 15:20 :: 18:24

- When ratios in two or more cases are equal, the elements are said to be in
proportion.

If a:b :: c:d, the elements a, b, c, d are in proportion.

- In a proportion a:b :: c:d, a and d are called extremes and b and c are called
means.

184 Prime Mathematics Book - 6

a:b :: c:d

Means
Extremes

- If a:b :: c:d, the product of extremes = the product of means

i.e. a × d = b × c which is also called cross product.
a c
If b = d then a × d = b × c

Note: For the quantities to be in proportion, it is not necessary that all the quantities
should be of the same kind (unit), it is only necessary that quantities taken in
first ratio should be of same kind and the quantities taken in the second ratio
should be of same kind.

For example: Ram's height = 6 ft = 60 = 5
Shyam's height 4.8ft 48 4

Ram's weight = 65kg = 5
Shyam's weight 52kg 4

So, their heights and weights are in proportion

Example 1: Cheek whether the following are in proportion
Solution:
(a) 4, 12, 15, 45 (b) 6, 8, 10, 12

(a) Given elements 4, 12, 15, 45

If the number are in proportion
4:12 :: 15:45
Where product of extremes = 4 × 45 = 180
Product of means = 12 × 15 = 180
Since, product of extremes = product of means, the numbers
are in proportion

(b) If 2, 8, 10, 12 are in proportion
2:8 ::10:12
Where product of extremes = 2 × 12 = 24
and product of means = 8 × 10 = 80
Since, product of extremes is not equal to the product of means,
the numbers are not in proportion.

Prime Mathematics Book - 6 185

Example 2: If x:3 :: 6:9, find the value of x.

Solution: Here, x:3 ::6:9

or, x = 6 = 6923× 3
or, 3 × 9
x 3
3

or, x = 2

\x=2

Example 3: If 3, 5, 24 and x are in proportion, find the value of x.

Solution: Since 3, 5, 24 and x are in proportion

3 = 24
5 x

or, 3x = 5 × 24

or, x = 5 × 24
3

\ x = 40

Exercise 8.2

1. Check whether the following numbers are in proportion or not.

(a) 4, 14, 16, 56 (b) 3, 10, 17, 24
(c) 2, 6, 10, 16 (d) 4, 12, 16, 48

2. Find the value of x. (b) 6 : x :: 15 : 40
(d) 1 : 6 :: 18 : x
(a) x : 4 :: 12 : 48
(c) 2 : 3 :: x : 36

3. (a) If 5, 4, 13 and x are in proportion, find the value of x.
(b) If 3, 2, y and 20 are in proportion, find the value of y.
(c) If 18, P, 7 and 14 are in proportion, find the value P.
(d) Find the value of k, if k, 13, 14, 26 are in proportion.

4. (a) Number of boys and girls in class VI A and class VI B are proportional. If
the number of boys and girls in class VI A are 14 and 21 respectively and
the number of boys in class VI B is 18, find the number of girls in class
VI B.

(b) A bus can travel 180km is 3 hours. How many km will it travel in 5 hours?
(Hint: Ratio of time = ratio of distance)

186 Prime Mathematics Book - 6

(c) A,B,C and D have some apples. A has 12, B has 15, C has 21 apples. If
the ratio of apples with A and B is same as the ratio of apples with C
and D. Find the number of apples with D.

(d) Akbar's height is 5.6 ft and Birbal's height is 4.2 ft. If their height and
weight are proportional and Akbar's weight is 68 kg, find the Birbal's
weight.

Percentage 1 3
We have already learnt that the 2 4

fraction 1 means 1 out of 2 7
2 10

3 means 3 out of 4
4

7 means 7 out of 10
10

24 means 24 out of 100
100

In this way we express quantities with comparision. Generally we express quantities

comparing with 100 as percentage.

So, 24 = 24 out of 100 = 24 percentage = 24%
100

Expressing a quantity as percentage is standard comparison. Percentage
means per hundred or out of hundred.
In a fraction with denominator 100, numerator is the percentage.

Changing fraction into percentage:
1
Let's consider a fraction 4

Which means 1 out of 4

\ Out of 1 it is 1
4

\ Out of 100 it is 1 ×100
4

\ 1 = 25 out of 100 = 25 = 25%
4 100

\ A fraction × 100 is percentage

\ Fraction × 100 = %

Prime Mathematics Book - 6 187

Example 1: Express the following into percentage.
Solution:
(a) 9 (b) 11
20 24

(a) 9 Alternate method :
20
9
= 9×5 20
20 × 5
9 5
= 45 = 20 ×
100 100%

= 45% = 45%

(b) 11
24
25
= 11 ×
24 100%
6
275
= 6 %

= 45 5 %
6

Changing a decimal into percent:

Let's consider a decimal 0.45.

Since a decimal means out of 1.

0.45 out of 1 = 0.45 = 45 = 45%
1.00 100

or, 0.45 means \ A fraction × 100 is %
A decimal × 100 is %
\ out of 1 it is 0.45

\ out of 100 it is 0.45 × 100 = 45%

Example 2: Convert into percentage.

Solution: (a) 0.42 (b) 0.465

(a) 0.42 (b) 0.465
= 0.42 × 100% = 0.465 × 100%
= 42% = 46.5%

Changing a parentage into fraction:

Since, percentage means out of 100, we just divide the quantity in percentage by

100 to convert it into fraction.

i.e. % = decimal
100

188 Prime Mathematics Book - 6

Example 3: Convert 75% into fraction.

Solution: 75%

= 75 3 = 3
100 4 4

Example 4: Convert 25% into decimal.

Solution: 25%

= 25 [25 hundredth]
100

= 0.25

Example 5: How much is 25% of Rs. 500?

Solution: 25 % of Rs. 500

= 25 × Rs. 500 5
100

= Rs. 125

So, 25% of Rs. 500 is Rs. 125.

Exercise 8.3

1. Express the following fractions into percentage.

(a) 4 (b) 2 (c) 5 (d) 7
5 25 8 25

(e) 17 (f) 1 (g) 4 (h) 31
20 3 7 50

2. Express the following decimal numbers into percentage.

(a) 0.17 (b) 0.25 (c) 0.345 (d) 0.36

(e) 0.008 (f) 0.02 (g) 1.235 (h) 5.05

3. Express the following percentages into fraction and reduce to their lowest

terms.

(a) 24% (b) 50% (c) 40% (d) 8%

(e) 1% (f) 95% (g) 120% (h) 500%

4. Express the following percentages into decimals.

(a) 10% (b) 38% (c) 75% (d) 80%

(e) 125% (f) 1234% (g) 50% (h) 2%

Prime Mathematics Book - 6 189

5. Find the value of the following:

(a) 20% of 450 (b) 40% of Rs. 840

(c) 32% of 2kg (d) 75% of Rs. 600

(e) 50% of 700 students (f) 50% of 100

6. (a) What percentage of Rs. 500 is Rs. 100?
(b) What percentage of Rs. 8 is 24 paisa?
(c) Express Rs.20 as percentage of Rs. 500.
(d) What is the difference between 4/5 of Rs. 200 and 60% of Rs. 200?

7. (a) Sajina secured 16 marks out of 25 marks in a unit test in mathematics.
What percentage did she secure in mathematics?

(b) In a box containing 5 dozens of apple, 5% are rotten. How many apples
are good?

(c) In S.E.E examination of a year 47% students were passed. How many were
failed out of 4,00,000 students appeared in S.E.E.?

(d) 240 student of a school participated in a drill which is 40% of the total
students of the school. Find the number of students in the school.

Unit Revision Test I

1. Write the fraction represented by the shaded
portion in the figure and express it in decimal
fraction and percentage.

2. Express 74% into percentage.

3. Express 4 into fraction.
5

4. Calculate 8% of 300.

5. Covert the following fraction into the lowest terms

(a) 2 2 : 4 1 (b) 20cm : 5m.
4 8

6. If 5,8,y and 16 are in proportion, find the value of y.

190 Prime Mathematics Book - 6

7. The lengths of two squares are 6cm and 8cm .

(a) Find the ratio of their sides.

(b) Find the ratio of their area.

8. In a class of 50 students 17 students are failed. What fraction of the students
are passed?

9. Length and breadthof a room are in the ratio 3:2 If lengthis 7m, find its breadth.

10. The number of boys and girls in class V and VI are in proportion. The number
of boys and class V and VI are respectively 15 and 16. If the number girls in
class is 24 find the number of girls in class VI.

Unit Revision Test II

1. Find the ratio: (a) 35 to 75 (b) 60 paisa to Rs. 2

2. Write the given ratio in simple form: 1 : 2
3 9

3. Find the value of x when x :6 :: 24 : 36

4. If 5, 4, 13 and y are in proportion, find the value of y.

5. Express the following into percentage. (a) 5 (b) 0.25
8

6. Find 20% of 480.

7. What percentage of Rs. 1000 is Rs. 120 ?

8. The price of a pen is Rs. 120 and that of ball pen is Rs. 40, compare the price
of the pen and the ball pen.

9. Number of boys and girls in class VI A and class VI B are in proportion. Number
of boys and girls in class VI A are 18 and 24 respectively and the number of
boys in class VI B is 15, find the number of girls in class VI B.

10. Amrita secured 12% marks in an examination. Find her total mark out of 8
subjects each of full mark 100.

Prime Mathematics Book - 6 191

Estimated periods – 6

Objectives

At the end of this unit, the students will be able to:
get the introduction of profit and loss.
find the profit or loss when selling
price and cost price are given.
find the profit or loss in percentage.
find the selling price or cost price
when profit or loss is given.
solve the problems on profit and loss.

Teaching Materials Activities

Chart of market where the people are It is better to:
buying the things, price tag, chart of demonstrate the activities of buying and
formula etc. selling the things in the classroom to give
the concept of cost price and selling price.
display the chart of market where the people
are buying and selling the things to give the
concept of cost price, selling price, profit
and loss.
derive the formula to find the profit and loss,
selling price and cost price.
say the students to solve the problems by
using the formula.
discuss about the process of solving the
problems of profit and loss.

Introduction

The words profit and loss are related
to business. So, let us talk about the
adjoining figure.

In the figure, Bhaktaraj is a shopkeeper
who brought the things in his shop from
the dealer.

Mrs. Khatun is a customer who is buying
the things in his shop. It means
Bhaktaraj is selling the things in his shop. Mrs. Khatun is paying the money to
Bhaktaraj for the things.

Bhaktaraj brought the things in his shop after buying from the dealer and sold the
things in his shop to the customer (Mrs. Khatun). So, the money paid by Bhaktaraj
to the dealer for the things is called cost price. It is also written as C.P.

Similarly, for Bhakta Raj the money paid by Mrs. Khatun for the things is called the
selling price. It is also written as S.P.

Suppose Bhaktaraj bought the things for Rs. 550 from the dealer and he sold the
things to Mrs. Khatun for Rs.600. It means the cost price of the things is Rs.550 and
the selling price of the things is Rs.600. In this case selling price is more than the
cost price. So, Bhaktaraj has made the profit.

\ Profit = Rs.600 - Rs.550 = Rs.50.
\ Profit = Selling price – cost price

= S.P. – C.P.

Similarly, if Bhaktaraj bought the things for Rs.550 from the dealer and he sold the
things to Mrs. Khatun for Rs.500. In this case, the cost price of the things is Rs.550
and the selling price of the same things is Rs.500. Here, the cost price is more than
the selling price. So, Bhaktaraj made a loss.

\ Loss = Rs.550 – Rs.500 = Rs.50
\ Loss = Cost price – Selling price

= C.P. – S.P.

Prime Mathematics Book - 6 193

From the above examples, I understand that
the profit or loss is made on an investment (Cost

Price).

Ramu! Do you know about the percentage?

Yes, sir! Percentage means the result
calculated out of 100.

Ok! Let’s discuss about profit or
loss percentage.

In the above example, Bhaktaraj bought the things for Rs.550 and sold them for
Rs.600. In this case, he made a profit.

\ Profit = Rs.600 – Rs. 550. = Rs.50. Note:
his investment (C.P.) is Rs.550.
So, Rs.50 is profit on Rs.550. Profit (P) = S.P. - C.P.

Loss (L) = C.P. - S.P.
P
Profit % = C.P. x 100%

\ Rs. 50 is profit on Re.1. Loss % = L x 100%
550 C.P.

\ Rs. 555001×1 100 is profit on Rs.100

\ On Rs. 100, profit is Rs. 100 = 9111 .
11

So, 9111 is the result which is calculated out of 100.
Therefore in percentage, profit = 9 111%

Profit percent = 50 x 100 = Profit x 100%.
550 C.P.

Similarly, loss percent = Loss x 100%.
C.P.

194 Prime Mathematics Book - 6