Prime Mathematics 4

Example 4: Simplify 2 – 130. Example 5: Simplify 4 – 3 + 1 .
7 7 7
3 2 × 10 3 4 3 1
Solution: 2 – 10 = 1 × 10 – 10 Solution: 7 – 7 + 7

= 20 – 3 = 4 – 3 + 1
10 10 7
20 – 3 5 – 3
= 10 = 7

= 17 10)17(1 = 2
10 −10 7
= 1170 7

Exercise: 6.4

1. Subtract:

(a) 1 from 3 (b) 1 from 3 (c) 2 from 7 (d) 5 from 5
4 4 8 8 5 10 12 6

(e) 1 from 2 (f) 5 from 7 (g) 7 from 11 (h) 9 from 13
4 4 8 8 9 9 16 16

2. Perform the following substractions :

(a) 3 − 1 (b) 3 – 2 (c) 7 – 5 (d) 9 – 7
4 4 5 5 8 8 12 12

(e) 1 – 1 (f) 9 – 2 (g) 11 – 3 (h) 5 – 2
2 4 15 5 16 8 6 5

(i) 9 – 7 – 2 (j) 7 – 1 – 2 (k) 8 – 1 (l) 3 – 4
2 3 3 9 6 18 9 18 7 21

3. Simplify:

(a) 2 – 1 (b) 5 – 4 (c) 6 – 3 (d) 9 – 23
7 11 2 4

(e) 4 – 2 1 (f) 3 2 – 1 1 (g) 4 – 1 1 (h) 3 3 – 1130
4 3 6 5 5

4. Simplify:

(a) 3 + 5 – 2 (b) 3 + 9 – 2 (c) 1 + 2 – 1
4 4 4 12 12 12 2 3 6

(d) 7 + 1 – 5 (e) 2 2 – 1 1 + 3 1 (f) 3 3 – 2 1 – 1 3
9 9 9 3 3 3 4 4 4

(g) 4 3 – 2 1 + 3 1 (h) 3 − 2 + 11 (i) 5 − 5 + 3
8 8 8 4 3 12 8 16 4

146 Prime Mathematics Book − 4

Word problems on addition and subtraction of fractions:
2s14ug akrg doifd ssuhgearb uyyesatletrodgaeyth aenrdi n3t21w kogd oafy ssu? gar
Example 1: Upasana brought
today. How much
214 + 321 9 7
Solution: = 4 + 2

= 9 + 7 × 2 = 9 + 14 = 9 + 14 = 23
4 2 × 2 4 4 4 4
= 534 kg ∴ She bought 543 kg of sugar in two days.
3
Example 2: A man sold 4 of his land, how much land is left ?
Solution: 3
Total land = 1, Sold land = 4

∴ Remaining land = 1 – 3 = 1 – 3 = 1×4 – 3
4 1 4 1×4 4

= 4 – 3 = 4–3 = 1
4 4 4 4

∴ Now 1 of the land is left.
4

Exercise: 6.5

1. A farmer has cultivated bcarroep.sHinow52 of his land, vegetables in 1 of the
land and remaining left much land is used? 3

2. Aainnfdaam2d14ailyyl?inteeersdsin2t12heliterveesnoifngm.iHlkoiwn the morning, 3 litres in the afternoon
much milk is consumed by the family

3. ST2thi38hdeeektsdmrioisa.ftnWaagnhltcerae.itabnisegttlewheeaerdenis4tta28hneccpmela,bce6et38wAceamenndatnBhdeis1p413la41ccmekm.AF,aitnnhddeCtph?leacpeerBimaentdeCr is
4. of

5. t34heofcltahses students of a class are girls. What fraction of the students of
are boys?
AHoswhompkuecehpcelrohthaeds1i6s43lemftewtrieths 641
6. of clothes. He sold metres of clothes,
him?
A7iIfs34shuhmesoesapdonk,leddfei5n3p51d56ertmkhmgeloioxlnefegndtghp5teihe23mcoekifsxgtesohdfoewfrwiircriieecre,es hwlweoefiwttrhe.mc4uo12cnhnkgercioctefedriis.cIlefef8ot21fwamintohotfhhteimhret?ywpiree.
7.
8.

Prime Mathematics Book − 4 147

Multiplication of a fraction by a whole number
1
Let’s consider the product 3 × 5

We know 3 × 1 = 1 + 1 + 1 = 1+1+1 = 3
5 5 5 5 5 5

1 + 1 + 1
5 5 5

= 3 [from the diagram]
5

∴ Product of a whole number and a fraction

= Whole number × numerator
denominator

Again let’s consider 1 of 6
3

1 of 6 flowers = 2
3
1
∴ 3 of 6

= 1 × 6 = 6 =2
3 3

Workedout Examples
P4e×rf29orm the multiplication 4 × 29 .
Example 1:
Solution :

= 4 × 2
9
8
= 9

Example 2: Calculate 1 of 12.
Solution: 3
1 of 12
3

= 1 × 12
3

= 12 =4
3

148 Prime Mathematics Book − 4

Multiplication of a fraction by another fraction:
1 1
Let’s consider the multiplication 2 × 5 1
5
1 =
5
Representing the fraction let’s draw a

rectangle. Divide into 5 equal parts and 1
1 1 1 1 10
shade one part. Now, 2 × 5 is 2 of 5 =

i.e. half of 1/5. So divide shaded part 1/5 into two equal parts and again
shade one out of two. Divide the whole parts in same equal small parts.

Thus, 1 of 1 is 110.
2 5

∴ 1 × 1 = 1 × 1 = 1
2 5 2 × 5 10

Thus, to multiply a fraction by another, multiply numerators and denominator
separately.

Note:

The resulting fraction should be in lowest term.

Example 3: Perform the multiplication 4 × 83 .
5
4 3
Solution: 5 × 8

= 4 × 3
5 × 8

= 12 [dividing numerator and denominator by 4]
40

= 3
10

Prime Mathematics Book − 4 149

Exercise: 6.6

1. Perform the following multiplication:
2 1 3 1
(a) 2 × 5 (b) 3 × 6 (c) 4 × 8 (d) 5 × 3

(e) 6 ×3 (f) 1 ×3 (g) 2 ×2 (h) 3 ×2
8 4 5 4

2. Calculate : 3 8 1
1 4 2 5 3 9 3 6
(a) 2 of 5 (b) 3 of (c) 4 of (d) of 7

(e) 1 of 4 (f) 2 of 6 (g) 3 of 12 (h) 4 of 20
2 3 4 8

3. Simplify:

(a) 2 × 3 (b) 3 × 6 (c) 2 × 7 (d) 8 × 2
7 5 2 7 3 5 3 3

(e) 9 × 22 (f) 13 × 16 (g) 1 × 2 (h) 1 × 2
11 18 4 26 2 3 6 5

4. Draw figure to represent
1 1 3 1 1 2
(a) 2 of 3 (b) 2 of 4 (c) 8 of 4 (d) 4 × 3

5. Simplify : 10
4 15 1 1 6 2 12 8 24
(a) 5 × 16 × 3 (b) 4 × 9 × 3 (c) 15 × 18 ×

(d) 1 × 2 × 5 (e) 31 × 14 × 42 (f) 23 × 13 × 22
2 3 8 3 5 9 4 11 7

6. Find the value of
1 2 1
(a) 3 of 9 (b) 3 of Rs. 15 (c) 5 of 20kg

(d) 3 of Rs. 100 (e) 3 of Rs. 32 (f) 2 of 27 apples
10 4 9

7. Solve the following problems:
H52 iomfa5n0i
(a) month. students are boys how many boys are there? Rs. 8000 in a
(b) sHpoewndms u43cohfdhoeersmshoentshpleynidn?come. If she earns

(c) Out of 3 dozens of mangoes, 1 are rotten. How many are rotten
and how many are good? 12

(d) Orouptaonfie1s2irsolpeaftniuenscoufltaivlaatnedd,?43 of the land is cultivated. How many

(e) How much is half of half of an object?
3 52?
(f) How much is 4 of

150 Prime Mathematics Book − 4

Division of Fractions

Division of a fraction by a whole number:
1
Let's consider the division 2 ÷ 2. 1
2
This implies that one half of an object is again =

divided into 2 equal parts. The shaded part is 1 of
the whole part. The half 1 2
2 is again divided into

two equal parts and one part is again shaded. = 1
1 1 4
∴ 2 ÷ 2 = 4
1 1 1 1
It shows that 2 ÷ 2 = 2 × 2 = 4

Thus 1 ÷ 2 = 1 × 1
2 2 2
1
= 2 × reciprocal of 2

Note : Two numbers a and b are such that a × b = 1 then a and b are said
to be reciprocal to each other.

b is reciprocal of a. a is reciprocal of b.
H∴era1e,isbre=ca1iprocal of a ∴He1brei,sare=c1biprocal of b.

Division of a whole number by a fraction :
tnhuemdbievrisoiofn212's÷in122.
Let's consider Next process:
This gives the

1111 2 ÷ 1
2222 2
2 2
= 1 = 2 × 1

0 1 23 2

NW∴uh2mic÷bhe21srho=ofw221s×tc2hoan=tt2a2i×n÷er12deci=inp4r2o=ca4l = 2 × 2
1 1

of 1 =14 = 4
2

Prime Mathematics Book − 4 151

Example 1 : Write the reciprocal of 5.
1
Solution : Reciprocal of 5 is 5

Alternative method
Let x be the reciprocal of 5, then x × 5 = 1

Or, x = 1
5
1
∴ Reciprocal of 5 is 5

Example 2 : How many 1 are there in 2 ?
3

Solution : Number of 1 in 2
3
1 3
= 2 ÷ 3 = 2 × 1 = 6

∴ There are 6, 1 's is in 2.
3
1
Example 3 : Simplify: 3 ÷ 4 .

Solution : 3 ÷ 1 = 3 × 4 = 12
4 1
1
Example 4 : Simplify: 3 ÷ 2.

Solution : 1 ÷ 2 = 1 × 1 = 1
3 3 2 6

Division of a fraction by another fraction:

Let's consider the division 3 ÷ 1
4 8
1 3
This gives the number of 8 in 4 3 = 3
A circle is 4 4
divided into 4 equal parts and is shaded.
1
Now, the circle is divided into eight equal parts one is 8 .

Thus, there are 6, 1 's in 3 . 1
3 1 8 4 81
4 ÷ 8 = 6
∴ 8
11
This shows that, 8 118
3 1 3 24
4 ÷ 8 = 4 × 8 = 4 = 6 88

152 Prime Mathematics Book − 4

Workedout Examples

Example 1: Simplify: 2 ÷ 1 . Example 2: How many 1 are there in 1 ?
3 3 4 2
2 1 1 1
Solution : 3 ÷ 3 Solution : Number of 4 contained in 2

= 2 × 3 = 1 ÷ 1
3 1 2 4
6 1 4 4
= 3 =2 = 2 × 1 = 2 = 2

∴ There are 2, 1 's in 1 .
4 2

Exercise: 6.7
1. Write the reciprocals of the following:

(a) 5 (b) 2 (c) 10 (d) 25

(e) 1 (f) 7 (g) 5 (h) 3
2 11 13 8

2. Show the following division in figure:

(a) 1 ÷ 2 (b) 1 ÷ 3 (c) 2 ÷ 1
4 2 4
1 1 1 2 1
(d) 3 ÷ 5 (e) 2 ÷ 4 (f) 3 ÷ 6

3. Simplify:

(a) 1 ÷ 3 (b) 2 ÷ 2 (c) 3 ÷ 1 (d) 4 ÷ 1
5 3 2 3
4 1 1 3 2 1 5 2
(e) 5 ÷ 5 (f) 4 ÷ 4 (g) 5 ÷ 3 (h) 7 ÷ 3

(i) 21 ÷ 42 (g) 54 ÷ 27
2 4 4 8

4. Solve the following:

(a) A metal wire is14 2 m long. How many pieces of wire each 3 m
5 5

long can be cut from it ?

(b) 5 1 kg of floor is divided into 8 equal parts. What is the weight
3
of each part ?

(c) The product of two numbers is 1 1 . If one of the numbers is
(d) H1o51wfminadntyhecaontsheear cnhum1 b12elri.tres 5 from 6 litres of oil

can be filled ?

Prime Mathematics Book − 4 153

Decimal Numbers

Whole number Decimal number

Thousands 1234.567 Thousandths
Hundreds Hundredths
Decimal point Tenths
Tens
Ones

Let’s divide a rectangle into 10 equal parts. We

write 1 shaded part in fraction as 1 (read as
10
one tenth) and is also written as 0.1 (read as zero

point one). Similarly, 3 out of 10 is 3 (read as
10
3 tenths) and is also written as 0.3 (read as zero

point 3).

Again, let’s divide a rectangle into 100 equal parts.

If we consider 1 part out of 100, it is written in
1
fraction as 100 (read as 1 hundredth) and is also

written as 0.01 (read as zero point zero one).

Similarly 2 out 0o.f021.0012isou1t020of(120h0uinsd11r0e20dt(htsw) ealnvde
also written as

hundredths) and also written as 0.12 (read as zero

point one two [not zero point twelve].

In the same way:

1 is 1 thousandths and is also written as 0.001 (read as zero point zero zero one)
1000
5
1000 is 5 thousandths and also written as 0.005. (read as zero point zero zero five)

24 is 24 thousandths and also written as 0.024 (read as zero point zero
1000

two four) etc.

Fractions whose denominators are powers of 10 are called decimal
fractions.

154 Prime Mathematics Book − 4

Mixed decimal fractions: Let’s consider a mixed fraction.

2130 = 23 = 20 + 3 = 2 tens +3 ones = 2 tens + 3 one
10 10 10 10 10
20
10 + 0.3 = 2 + 0.3 = 2.3 (read as 2 point 3 or 2 and 3 tenths)

In the given figure, the length of the pencil

= 4cm + 4 cm = 4140 cm
10
= (4 + 0.4)cm
= 4.4cm

[1cm is divided into 10 equal parts]

Workedout Examples

Example 1: Write in decimal numbers:

(a) 1 (b) 1
2 4
1 × 5 5
Solution: (a) 1 = 2 × 5 = 10 = 0.5 (zero point five)
2
1 1 × 25 25
(b) 4 = 4 × 25 = 100 = 0.25

(zero point two five)

Example 2: Express the given decimal numbers as fraction:
(a) 0.7 (b) 0.24 (c) 3.4

Solution: (a) 0.7 = 7 (0.7 = 7 tenths)
10
24
(b) 0.24 = 100 (0.24 is 24 hundredths)

(c) 3.4 = 3 and 4 tenths = 3140

Example 3: Make a place value table and write 12.5 in expanded form:
Solution : 12.5

Place value table
Tens Ones Tenths

125

∴12.5 = 1 × 10 + 2 × 1 + 5 × 1
10

Prime Mathematics Book − 4 155

Example 4: Write 2 + 3 + 5 in decimal form.
10 100

Solution: 2 + 3 + 5
10 100

= 2 × 100 + 3 × 10 + 5
1 × 100 10 × 10 100

= 200 + 30 + 5 = 200 + 30 + 5 = 235
100 100 100 100 100

= 2 35 = 2.35
100

Comparision of decimal numbers:
Start comparing the digits of higher place value. The number with greater
digit in higher place value is bigger one.
Example 5 : Compare 23.456 and 23.46
Solution:
Here, digits in Tens, ones and tenths place are equal. The digit in hundredths
place 6 > 5. So 23.456 < 23.46.

Exercise: 6.8

1. Write in decimal form: 2
3 4 7 10
(a) 10 (b) 10 (c) 10 (d)

(e) 2 (f) 5 (g) 12 (h) 54
100 100 100 100

(i) 2130 (j) 45 (k) 311030 (l) 235
10 100

2. Write in decimals: 9
1 1 50 7
(a) 5 (b) 2 (c) (d) 20

(e) 12 (f) 3 (g) 3 (h) 13
25 50 4 20

(i) 9 (j) 19
5 20

3. Write as fraction:
(a) 0.6 (b) 0.9 (c) 3.2 (d) 5.6
(e) 0.5 (f) 0.24 (g) 0.72 (h) 1.25
(i) 2.05 (j) 4.82

4. Write the following numbers in place value table.
(a) 3.57 (b) 23.7 (c) 5.86 (d) 12.34
(e) 16.287 (f) 64.38 (g) 15.112 (h) 38.237
(i) 123.45 (j) 654.236 (k) 2468.135 (l) 100.45

156 Prime Mathematics Book − 4

5. Write in expanded form: (d) 9.45
(a) 0.86 (b) 0.64 (c) 5.68 (h) 25.103
(e) 2.04 (f) 12.005 (g) 0.642
(i) 0.847 (j) 6.789

6. Write in decimal form:

(a) 4 + 2 + 3 (b) 0 + 3 + 5 (c) 7 + 0 + 8
10 1000 10 100 10 100

(d) 24 + 7 + 4 (e) 2 + 1 + 4 + 5
10 100 10 100 1000

(f) 12 + 2 + 0 + 7
10 100 1000

7. Compare the following decimal numbers:
(a) 1.05 and 1.04 (b) 20.3 and 20.5 (c) 23.45 and 23.35

(d) 354.82 and 354.78 (e) 642.123 and 642.32

Fundamental operations on decimal numbers

Addition and subtraction of decimal numbers:
Process: Arrange the numbers vertically so that the digits are at the same
place and decimal points be in the same column.
Follow the process of addition and subtraction as in addition and subtraction
of whole numbers.

Example 1: Add : 0.468, 0.57 and 0.6
Solution:
0.468
0.570
+0.600
=1.638

Learn the following:

NGumivbeenrn5umis bgeivre5n. No. because place of 5
Put zero to its left as 05. Does is not changed.

it make any differences? Yes, because place of
Put zero to its 5 is changed.

right as 50. Does it make any
difference?

Prime Mathematics Book − 4 157

Let’s consider a decimal
number 0.5. Put zero to its right.

Does it make any difference?
0.50 No, place of 5 is
not altered.

0.5, 0.50, 0.500, 0.5000 …………. are same
value and measure but it is written as 0.5

Example 2: Subtract 24.305 from 30.46.
Solution:
30.460
− 24.305
6.155

Example 3: 4.37 + 1.86 – 3.45 4.37 6.23
Solution: 4.37 + 1.86 – 3.45 + 1.86 − 3.45
= 6.23 – 3. 45
= 2.78 6.23 2.78

Example 4: A jar contains 12.345 liters of milk and another jar contains
6.838 liters of milk. Find the total quantity of milk in two
jars.

Solution: Here, 12.345 l
+ 6.838 l
19.183 l

∴ Total quantity of milk in two jars is 19.183 litres.

Exercise: 6.9

1. Add:

(a) 28.37 and 72.46 (b) 32.56 and 54.92

(c) 14.67, 16.42 and 12.66 (d) 0.216, 0.32 and 0.8

(e) 4.63, 42.246 and 562.7 (f) 8.587, 0.843 and 92.83

158 Prime Mathematics Book − 4

2. Subtract : (b) 15.814 from 32.72
(a) 5.572 from 9.43 (d) 8.300 from 824.236
(c) 0.864 from 2.246 (f) 12.118 from 20.3
(e) 5.832 from 8.415

3. Perform the following tasks:

(a) 6.30 (b) 0.5 (c) 14.2 (d) 84.75
− 3.7 − 14.65
+ 4.75 + 0.6

(e) 8.65 (f) 432.256 (g) 5.342 (h) 192.85
4.23 − 195.442 0.176 −14.23

+ 1.08 + 1.852

4. Simplify: (b) 0.115 + 0.877 – 0.725
(a) 4.169 + 2.556 – 3.817 (d) 9.4 + 3.924 – 6.54
(c) 0.6 – 0.825 + 0.241 (f) 24.6 – 7.67 – 6.542
(e) 0.842 – 0.639 + 0.785

5. Solve the following problems:
(a) Find the sum of 16.17 and 14.13.
(b) By how much is 70.85 greater than 66.77?
(c) By how much is 50.25 less than 90.56?
(d) By how much is the sum of 31.42 and 8.38 more than 11.28?
(e) Kusum bought a copy for Rs. 30.75 and a pencil for Rs. 8.85. If she
gave Rs. 100 note to the shopkeeper, how much money will she get
back?
(f) The perimeter of a triangle is 12.4cm. If two sides are 3.4cm and
4.6cm, find the third side.

Prime Mathematics Book − 4 159

Multiplication and division of decimal numbers:
A. Multiplication of decimal numbers by 10, 100, 1000 etc.
In the given ruler,
10 × 0.5 = 5.0 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5

Similarly, we get mm 1 cm 2 34 5 67
100 × 0.5 = 50.0
Centimeters

And 1000 × 0.5 = 500.0
We observed that on multiplying a decimal number by 10, the decimal
point shifts to the right by one place, on multiplying by 100, the decimal
point shifts to the right by two places and on multiplying by 1000, the
decimal point shifts to the right by 3 places.

Thus, 10 × 0.1234 = 1.234

100 × 0.1234 = 12.34

1000 × 0.1234 = 123.4

B. Multiplication of decimals by whole numbers:
In the given figure,
1 × 1.4 = 1.4
mm 1 cm 2 3 4 5 6 7

2 × 1.4 = 2.8 Centimeters

3 × 1.4 = 4.2 Thus, on multiplying a decimal number by a whole
Similarly, number, place of number after decimal doesn’t
change.
10 × 1.4 = 14.0
11 × 1.4 = 15.4 Process of multiplication is as usual in multiplication
of whole numbers.

C. Multiplication of decimal number by another decimal number.
Let’s consider the multiplication of 0.2 and 0.3.
2 3 6
0.2 × 0.3 = 10 × 10 = 100 = 0.06 [Six hundredth]

And 0.3 × 0.4 = 3 × 4 = 12 = 0.12
10 10 100

Again let’s see the multiplication 0.4 × 0.34

0.4 × 0.34 = 4 × 34 = 136 = 0.136
10 100 1000

Thus, we observed that the multiplication of a decimal number by another
decimal number is same as in multiplication of whole numbers. The number
of digits after decimal point is the sum of the digits after decimal points in
multiplicand and multiplier.

160 Prime Mathematics Book − 4

Division of decimal number by 10, 100, 1000 etc.
Let’s consider the division 123.4 ÷ 10
Here, 123.4 ÷ 10
1
123.4 × 10

= 1234 × 1
10 10

= 1234
100

= 12.34

Again, consider 123.4 ÷ 100 1 1234 1
100 10 100
Here, 123.4 ÷ 100 = 123.4 × = ×

= 1234 = 1.234
1000

Also, consider 123.4 ÷ 1000

123.4 ÷ 1000 = 1234 × 1 = 1234 = 0.1234
10 1000 10000

Thus on dividing a decimal number by 10, the decimal point shifts to the left
by one place. Dividing by 100, the decimal point shifts to the left by two
places and dividing by 1000, the decimal point shifts to the left by 3 places.

123.4 ÷ 10 = 12.34
123.4 ÷ 100 = 1.234
123.4 ÷ 1000 = 0.1234

Workedout Examples
Example 1: Multiply 4.3215 by 10, 100 and 1000.
Solution: 4.3215 × 10 = 43.215

4.3215 × 100 = 432.15
4.3215 × 1000 = 4321.5

Example 2: Multiply 2.25 by 6.
Solution: 2.25 × 6

= 225 × 6 = 1350 = 13.50
100 100

Prime Mathematics Book − 4 161

Example 3: Perform the multiplication 2.34 × 1.2.
Solution: 2.34 × 1.2

= 234 × 12 = 2808 = 2.808
100 10 1000

Example 4: Divide 1574.6 by 10, 100 and 1000.

Solution : 1574.6 ÷ 10 = 157.46

1574.6 ÷ 100 = 15.746

1574.6 ÷ 1000 = 1.5746

Example 5: Divide 2.48 ÷ 4.

Solution : 2.48 ÷ 4 1 248 1 248 ÷ 4 62
4 100 4 400 ÷ 4 100
= 2.48 × = × = = = 0.62

Exercise: 6.10

1. Multiply the following by 10, 100 and 1000 separately:
(a) 0.0004 (b) 0.0123 (c) 3.4567 (d) 20.2343

2. Multiply: (b) 20 and 0.19 (c) 6 and 9 tenth
(a) 1.3 and 4 (e) 3.2 and 1.2 (f) 8.576 and 6
(d) 4.2 and 0.4

3. Divide :
(a) 25.7 by 100 (b) 987.65 by 100 (c) 3.8 by 10
(d) 126.5 by 1000 (e) 4567.8 by 1000 (f) 56.78 by 10
(g) 1.8 by 0.2 (h) 3.6 by 0.9 (i) 2.4 by 0.3
(j) 4.8 by 1.2

4. Find the quotient: (c) 0.576 ÷ 9 (d) 4.23 ÷ 3
(a) 0.548 ÷ 4 (b) 0.426 ÷ 2 (g) 2.436 ÷ 6 (h) 0.873 ÷ 3
(e) 0.354 ÷ 6 (f) 8.785 ÷ 5

5. Solve the following problems:
(a) If 10 kg of potatoes cost Rs. 345.5, find the cost of 1 kg of potatoes.
(b) If the cost of 1 pen is Rs. 40.75, find the cost of 10 pens.
(c) If 4 metres of clothes cost Rs. 523.6, find the cost of 1 m of clothes.
(d) A rope 173.61m long is cut into 3 equal pieces. Find the length of
each piece.
(e) The cost of 1gm of sweet is Rs. 7.4. What is the cost of 4.2gm of
the sweet?

162 Prime Mathematics Book − 4

Use of Decimals 1
100
We know, 100 P. = Re. 1 so, 1 p = Rs.

10mm = 1cm so, 1mm = 1 cm
10

100cm = 1m so, 1cm = 1 m
100

1000m = 1km so, 1m = 1 km
1000

1000gm = 1kg so, 1gm = 1 kg
1000

1000ml = 1litre so, 1ml = 1 litre
1000

Thus, conversion of such units are completely decimal system.

To change bigger to small one do multiply (x)
small to bigger do divide (÷)

Example 1: Convert 64 paisa (p) into rupees (Rs).

Solution: We know, 1 p = Rs. 1
100
1
∴ 64p = Rs. 64 × 100

= Rs. 64 = Rs. 0.64
100

Example 2: Convert 0.354kg into gm.
Solution: We know, 1kg = 1000gm

∴ 0.354kg = 1000 × 0.354gm
= 354gm

Example 3: Convert 475 m into Km.
1
Solution: 1m = 1000 km

475m = 475 km
1000

∴ 0.475km

Prime Mathematics Book − 4 163

Exercise: 6.11

1. Convert into paisa: (c) Rs. 30 (d) 64.50
(a) Rs. 5.40 (b) Rs. 12.25

2. Write in rupees: (c) 75 paisa
(a) 10 paisa (b) 5 paisa (e) Rs. 23 and 25 paisa.
(d) Rs. 7 and 75 paisa

3. Convert into millimetres:

(a) 9cm (b) 15cm (c) 6.70cm (d) 2.25cm

4. Write in centimetres:

(a) 4m (b) 8.50m (c) 14.75m (d) 25mm

(e) 123.4mm

5. Convert the following: (b) 2468cm into metres.
(a) 2475m into kilometres. (d) 5km into metres.
(c) 1020m into kilometre (f) 20103 millilitres into litres.
(e) 1020m into kilometres. (h) 1.2 litres into milliliters.
(g) 30 litres into millilitres

Percentage
Let’s divide a rectangle into 100 equal parts, then

1 part out of 100 parts is written in fraction as

1 and is also called 1 percent or 1%; 5 out of
100 is as called 5 percent or 5 %.
100 5 and is also
100

Similarly, 25 out of 100 is written in fraction as

25 and is also called 25 percent or 25%.
100

Thus, percentage is the fraction whose

denominator is 100. (Any part taken from 100 parts)

Note : % = 1
100

164 Prime Mathematics Book − 4

2 implies that out of 5 is 2
5
2 × 100
∴ out of 1 is 5

∴ out of 100 is 2 × 100 Fraction Percent
5
2 ÷ 100
i.e. 5 × 100%

∴ Fraction × 100 gives percent.

3 Workedout Examples
10
Example 1: Convert into percentage.
Solution: 3
10 Or, 3
10
3 ×10
= 10 × 10 = 3 × 100% [∴ Fraction × 100 percentage]
10
30
= 100 = 30%

= 30%

Example 2: Convert 0.25 into percentage.
Solution: 0.25
Or, 0.25
= 0.25 × 100 = 0.25 × 100%
1× 100 = 25%

= 25 Fraction × 100 = %
100
Decimal × 100 = %
= 25%

Example 3: Write 16 % into fraction. 2 16 2 100
Solution : 16% 2 8 2 50
2 4 5 25
= 16 = 2 × 2 × 2 × 2 25
100 2 × 2 × 5 × 5
4 ∴16 = 2×2×2×2 ∴100 = 2×2×5×5
= 25

Example 4: Out of 40 students 12 students failed in mathematics. Find
the percentage of the failed students.
Solution: Total number of students = 40
Number of failed students = 12
12 12 ÷ 4 3
∴ Failed students as fraction = 40 = 40 ÷ 4 = 10

Now, 3 = 3 × 100% = 30%
10 10

Prime Mathematics Book − 4 165

Therefore, 30% students have failed.
or
Shortly,
Total students = 40
Number of failed students = 12
∴ Failed students as percentage
12
= 40 × 100%

= 30%

Example 5: How much is 25% of Rs. 2000?
Solution: 25% of Rs. 2000

= 25 × Rs. 2000
100

= Rs. 50000
100

= Rs. 500
∴ 25% of Rs. 2000 is Rs. 500

Exercise: 6.12

1. Convert the following fractions into percentage:

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

(f) 1 (g) 3 (h) 9 (i) 3 (j) 14
10 10 10 20 20

(k) 3 (l) 8 (m) 18 (n) 3 (o) 45
25 25 25 50 50

2. Convert the following percentages into fractions:

(a) 8% (b) 3% (c) 16% (d) 24% (e) 40%

(f) 50% (g) 72% (h) 80% (i) 90% (j) 95%

3. Find the following : 10% of 200 students (c) 50% of 50kg
(a) 45% of Rs. 60 (b) 60% of 75 apples.
(d) 36% of Rs. 400 (e)

166 Prime Mathematics Book − 4

4. Convert the following decimals into percentage:
(a) 0.06 (b) 0.08 (c) 0.15 (d) 0.28 (e) 0.48
(f) 0.66 (g) 0.78 (h) 1.25 (i) 1.06 (j) 2.20

5. Solve the following problems:
(a) There are 72 oranges in a basket. If 25% of them are bad, find the
number of bad oranges.
(b) 34 out of 50 students of a class have passed, what percent of the
students have passed?
(c) Mohan got 14 marks out of 25 full marks in maths in a monthly
test. Convert his mark into percentage.
(d) The price of a shirt is listed to be Rs. 300. If 20% discount is allowed,
find how much discount was given?
Unitary Method

• Unit literally means single or individual thing.
• A pen costs Rs. 52

∴ 5 such pens cost Rs. 52 + Rs. 52 + Rs. 52 + Rs. 52 + Rs. 52
= 5 × Rs. 52
= Rs. 260
Value of more units is more so we multiply.

• 8 Chocolates cost Rs. 48.
1 Unit costs less so we divide.
∴ 1 Chocolate costs Rs. 48 ÷ 8 = Rs. 6

Prime Mathematics Book − 4 167

Thus, if the value of 1 object is known, we can find the value of any number
of objects and if the value of certain number of objects is known we can find
the value of one (unit) object. This process is called unitary method.

Workedout Examples

Example 1: If the cost of 1 book is Rs. 125, find the cost of 5 such
books.
Solution: 1 book cost Rs. 125
More books cost more
∴ 5 books cost Rs. 125 × 5 so we multiply.
= Rs.625

Example 2: The cost of half dozen of bananas is Rs. 18. Find the cost of
a banana.
Solution: 1 dozen = 12
12 Less number of
∴ half dozen = 2 = 6 bananas cost less.
Now, So, we divide
The cost of 6 bananas = Rs. 18
6)18(3
∴ The cost of 1 banana = Rs. 18 ÷ 6 −18
= Rs. 3 ×

Example 3: If the cost of 12 mangoes is Rs. 72, find the cost of 8
mangoes.
Solution : Cost of 12 mangoes is
given so, we first find the cost
of 1 mango and then find the
cost of 8 mangoes. 12)72(6
−72
×
The cost of 12 mangoes = Rs. 72
∴ The cost of 1 mango = Rs. 72 ÷ 12 1 mango costs less so
we should divide
= Rs. 6

∴ The cost of 8 mangoes = Rs. 6 × 8 cost of 8 is more than the cost
Therefore, 8 mangoes cost Rs. 48 of one, so we can multiply

Exercise: 6.13

Item Rate Quantity Cost
Notebook
Pencil Rs. 22 2 Rs. 22 × 2 = Rs. 44
Ruler
Eraser Rs. 8 4
Total
Rs. 12 12

Rs. 6 21

168 Prime Mathematics Book − 4

2. (a) If the cost of a pen is Rs. 42, what is the cost of 6 such pens?
(b) The weight of a bag of rice is 72kg, find the weight of 5 such
bags.
(c) A car moves 16km with 1 litre of petrol. How far does the car move
with 12 litres of petrol?
(d) A bottle of sun flower oil contains 1.5 litres. Find the capacity of
6 bottles.

3. Complete the following table:

Item Total Cost Quantity Cost
Pen Rs. 42 6 Rs. 42 ÷ 6 = Rs. 7
Bag Rs. 72 5
Notebool Rs. 175 7
Pencil Rs. 84 12
Total

4. (a) The cost of 8 eggs is Rs. 54. Find the cost of an egg.

(b) 9 packets contain 108 chocolates. How many chocolates will be
there in a packet?

(c) Cost of 1 dozen pencils is Rs. 96. What is the cost of a pencil?

(d) Anu earns Rs. 2380 in a week. How much does she earn in a day?

5. (a) If the cost of a dozen of Notebook is Rs. 288, find the cost of 3 such
Notebooks.

(b) If 7 jars contain 168 litres of oil, how much oil is contained in 20
such jars?

(c) 14 buses can carry 910 passengers. How many passengers can be
carried by 5 such buses?

(d) The cost of 8 litres of petrol is Rs. 680, what is the cost of 15 litres
of petrol ?

Unit Revision Test
3
1. Write three fractions equivalent to 8 .
2. Convert:
68
(a) 2 3 into improper fraction (b) 11 into mixed fraction.
5

3. Compare the following fractions:
3 7
(a) 5 and 9 (b) 54.43 and 54.19

Prime Mathematics Book − 4 169

4. Simplify : 1
9 12
(a) 12 + (b) 1 − 1 (c) 2 × 3 (d) 4 ÷ 13
2 6 5

5. Write the following in decimals:

(a) 12 (b) 25 (c) 1234
10 100 100

6. Simplify : 6.25 + 8.4 – 3.25

7. Convert the following : (b) 32 into percentage.
1000
(a) 25 % into fraction.
8. Find 32% of Rs. 500.

9. The cost of 1 gram of sweets is Rs. 12.40. What is the cost of 6.3
grams?

10. Out of 32 students of a class, 4 students are absent. Find the
percentage of present students.

11. If the cost of 15 toys is Rs. 930, find the cost of one toy.

12. The cost of 10 litres of petrol is Rs. 850. Find the cost of 12 litres of
pertrol.

Answers:

21..aS)h13ow tbo)th25e teacch)e1r30. d) 5 e) 10 Exercise: 6.1
8 15

43..aa))I21mprbo)p23er c) 51Properd) 5 e) P83 roperf) 4 g) 72Improphe) r270 f) 5 fraction
b) 6 c) 10 d) e)
M12ixed
f) Mixed fraction g) Proper h) Improper i) Mixed fraction j) Proper
5. a) 152 b) 112 c) 123 d) 216 e) 217 f) 135 g) 214 h) 124 i) 225 j) 197

6. a) 14 b) 7 c) 23 d) 9 e) 11 f) 17 g) 13 h) 19 i) 13 j) 15
4 3 5 5 2 7 8 5 2 7

1. Show to the teacher. Exercise: 6.2

2. Show to the teacher. 3 1 1 E52quiuail)en145t
3. a) E41 quibu)a13lent c) 23Equiuadl)en51t e) f) g) h) j) 3
4. a) b) c) N4 ot d) 3Not e) N2 ot f) 7

5. a) like fraction b) like fraction c) unlike fraction d) unlike fraction e) like fraction
6. Show to the teacher.
7. Show to the teacher.

Exercise: 6.3
3 4 8 6 15 15
1.a) 4 b) 5 c) 12 d) 7 e) 21 f) 18

2. a) 5 b) 3 c) 7 d) 13 e) 23 f) 5 g) 5 h) 138
6 10 8 16 24 6 6

3.a) 116 b) 13 c) 1 13 d) 13 e) 14109 f) 37 g) 7 h) 5 i) 23 j) 1274
15 30 20 56 12 12 30

170 Prime Mathematics Book − 4

4. a) 232 b) 451 c) 332 d) 243 e) 454 f) 3 g) 543 h) 7 i) 414

3 5 Exercise: 6.4 1
1 1 10 12 1 1 4 4
1.a) 2 b) 4 c) d) e) 4 f) 4 g) 9 h)

2.a) 1 b) 1 c) 1 d) 1 e) 1 f) 1 g) 5 h) 13 i) 121 j) 1 k) 5 l) 5
2 5 4 6 4 5 16 30 2 6 21

3.a) 176 b) 4 7 c) 421 d) 314 e) 143 f) 2 1 g) 245 h) 2 3
11 2 10

4. a) 121 b) 5 c) 1 d) 1 e) 423 f) -14 g) 538 h) 1 i) 1 1
6 3 16

1. 11 2. 743 3. 585 4. 1187 5. Exercise: 6.5 8. 4 9
15 6. 1021 7. 631 20

d) 123 Exercise 6.6
4 1 c) 112 e) 214 3 4 h) 121
1.a) 5 b) 2 e) 2 f) 4 g) 5
e) 1
2.a) 2 b) 2 c) 2 d) 2 f) 4 g) 9 h) 10
5 5 3 7

3.a) 6 b) 172 c) 14 d) 197 f) 2 g) 1 h) 1
35 15 3 15

4. Show to the fraction.
1 1 4 5 1 5
5. a) 4 b) 9 c) 27 d) 24 e) 135 45 f) 21 4

6. a) 3 b) 10 c) 4 kg d) Rs.30 e) Rs. 24 f) 6 apples e) 1 f) 3
7. a) 20 b) Rs. 6000 c) 3 rotten, 33 good d) 9 ropanies cultivate, 3 ropanies uncultivated 4 10

Exercise: 6.7

Q1. a) 1 b) 1 c) 1 d) 1 e) 2 f) 11 g) 13 h) 8
5 2 10 25 7 5 3

Q 2. Show to the teacher.

Q3. a) 1 b) 1 c) 6 d) 12 e) 4 f) 1 g) 6 h) 15 i) 4
15 3 c) 1 3 5 14

Q4. a) 24 b) 2 d) 4
3
c) 0.7 Exercise: 6.8
Q.1 a) 0.3 b) 0.4 l) 2.35 d) 0.2 e) 0.02 f) 0.05 g) 0.12 h) 0.54 i) 0.23
j) 4.5 k) 3.13

Q. 2 a) 0.2 b) 0.5 c) 0.18 d) 0.35 e) 0.48 f) 0.06 g) 0.75 h) 0.65 i) 1.8 j)
0.95
b) 910 c) 3120 d) 5160 e) 510 f) 12040 g) 17020 h) 112050 i) 210050 j) 418020
Q.3 a) 610
c) 5 + 160 + 1080
Q. 4 Show to the teacher. b) 0 + 160 + 1040 f) 10+2+ 100 + 1000 + 10500
e) 2 + 100 + 1040
Q.5 a) 0 + 180 + 1060

d) 9 + 140 + 1050

Prime Mathematics Book − 4 171

g) 0 + 160 + 1040 + 1020 h) 20 + 5 + 110 + 1000 + 13000
i) 0 + 180 + 1040 + 17000 j) 6 + 170 + 1080 + 19000

Q. 6 a) 4.203 b) 0.35 c) 7.08 d) 24.74 e) 2.145 f) 12.207

Q.7 Show to the teacher.

Exercise 6.9

Q.1 a) 100.83 b) 87.48 c) 43.75 d) 1.336 e) 609.576 f) 102.26
Q.2 a) 3.858 b) 16.906 c) 1.382
Q. 3 a) 11.05 b) 1.1 c) 10.5 d) 815.936 e) 2.583 f) 8.182
Q.4 a) 2.908 b) 0.267 c) 0.016
Q.5 a) 30.30 b) 4.08 c) 40.31 d) 70.10 e) 13.96 f) 236.814 g) 7.370 h) 178.62

d) 6.784 e) 0.988 f) 10.388

d) 28.52 e) 60.40 f) 4.4cm

Exercise 6.10
Q.1 a) 0.004, 0.04, 0.4 b) 0.123, 1.23, 12.3 c) 34.567, 345.67, 3456.7

d) 202.343, 2023.43, 20234.3

Q.2 a) 5.2 b) 3.8 c) 5.4 d) 1.68 e) 3.84 f) 51.456
Q.3 a) 0.257 b) 9.8765
Q.4 a) 0.137 b) 0.213 c) 0.38 d) 0.1265 e) 4.5678 f) 5.678 g) 9h) 4 i) 8 j) 4
Q.5 a) Rs. 34.55
c) 0.064 d) 1.41 e) 0.059 f) 1.757 g) 0.406 h) 0.291

b) 407.5 c) 130.9 d) 57.87m e) 31.08

Exercise 6.11

Q.1 a) 540 p b) 1225 p c) 3000p d) 6450 p

Q.2 a) Rs. 0.1 b) 0.05 c) Rs. 0.75 d) Rs. 7.75 e) Rs. 23.25

Q.3 a) 90mm b) 150mm c) 67mm d) 22.5mm

Q.4 a) 400cm b) 850cm c) 1475cm d) 2.5cm e) 12.34cm

Q.5 a) 2.475 km b) 24.68m c) 1.02 kmd) 5000 m e) 1.02 km f) 20.103 l g) 30,000 ml h) 1200 ml

Q.1 a) 40% b) 50% c) 25% Exercise 6.12
k) 12% l) 32% m) 72% d) 75% e) 20% f) 10% g) 30% h) 90% i) 15% j) 70%
n) 6% o) 90%

Q.2 a) 225 b) 1030 c) 245 d) 265 e) 52 f) 21 g) 2158 h) 54 i) 190 j) 2109

Q.3 a) 27 b) 20 c) 25 d) 144 e) 45

Q.4 a) 6% b) 8% c) 15% d) 28% e) 48% f) 66% g) 78% h) 125% i) 106% j) 220%

Q.5 a) 18 b) 68% c) 56% d) Rs. 60

172 Prime Mathematics Book − 4

U7nit Bill and Budget

Objectives Estimated periods − 7

At the end of this unit, the students will be able to:

• know what bills are

• understand the informations from the given bill

• prepare the simple bill

Teaching Materials
bill, menu, price list, etc.

Activities
It is better to:
• collect different bills and discuss the information on it
• show price list, menu etc and let the students find costs
• provide sample bills and let them prepare simple bills

Bill

RAJ ENTERPRISES

Birgunj, Nepal Date : 2073/07/10

Bill No: 142

Name of customer : ..M....r.....D..e...v...S..h..r..e..s..t.h...a.....................................................
Address : ...S..h..r..e..e..p..u...r.,..B...i.r.g...u..n...j......................................................................

S.N. Particulars Quantity Rate (Rs.) Amount (Rs.)

1. Pens 12 25.00 300.00

2. Books 40 110.00 4400.00

3. Geometry boxes 15 60.00 900.00

4. Pencils 30 15.00 450.00

5. Chart papers 50 7.00 350.00

Total 6400.00

In words : .S.i.x...t.h..o..u..s.a..n..d..a..n..d...f.o..u..r..h..u.n..d..r.e..d...r.u..p..e..e..s..o..n.l.y......................

……………………….. ……………………………
Customer’s sign. Raj Enterprises

Dev went to a shop to buy pens, books, pencils etc. and he bought those
things from a shop. The shopkeeper gave him a piece of paper shown above.
It is called a bill.
A bill is a written statement of sale and purchase. The bill tells us many things
like date, bill number, name of customer, address, items, quantity, rate etc.

Exercise: 7.1
1. Find the total cost by preparing bills (use the table shown below):

S.N. Particulars Quantity Rate (Rs.) Amount (Rs.)

Total

174 Prime Mathematics Book − 4

( i) 3 kg oranges at Rs. 50 per kg, 4kg of apples at Rs. 60 per kg, 7kg
of mangoes at Rs. 70 per kg.

( ii) 8 books at Rs. 120 per book, 10 pencils at Rs. 7 per pencil, 15 copies
at Rs. 12 per copy and 20 chart papers at Rs. 10 per piece.

( iii) 10 kg rice at Rs. 50 per kg, 7 kg wheat flour at Rs. 45 per kg, 5 kg
of sugar at Rs. 75 per kg, and 8 litres of oil at Rs. 95 per litre.

( iv) 3 pieces of T−shirt at Rs. 450 each, 4 shirts at Rs. 1100 each, 7
pants at Rs. 1500 each, 8 pairs of socks at Rs. 60 per pair.

2. Study the given bill:

JAY ENTERPRISES

New Road, Kathmandu Date : 2073/07/19

Bill No: 786

Name of customer : ..M....r.....A..m....u...M....a...h..a..r..j.a..n..................................................
Address : ...G..h...a..t..t.a..g...h..a...r.................................................................................

S.N. Particulars Quantity Rate (Rs.) Amount (Rs.)
1. Shirts 5 1200.00 6000.00
2. T−shirts 7 800.00 5600.00
3. Pants 3 1500.00 4500.00
4. Gloves 125.00 250.00
2 pairs 16350.00
Total

In words : .S.i.x..t.e..e..n...t.h..o..u.s..a.n..d...t.h..r..e.e...h..u..n.d..r..e.d...a..n..d...f.i.f.t.y...r.u..p..e..e.s...o.n..l.y......

……………………….. ……………………………
Customer’s sign. Jay Enterprises

Answer the following questions:
( i) What are the items purchased ?
( ii) When was the purchase made ?
( iii) Who is the purchaser ?
( iv) What is the cost of a shirt ?
( v) What is the bill number ?
( vi) What is the total amount paid to purchase those items ?
( vii) How many pants were purchased ?
( viii) How many shirts were purchased ?

Prime Mathematics Book − 4 175

3. Suresh Chaudhari of Kalanki bought one coat for Rs. 5250.00 two sweaters
at Rs. 950.00 each, 4 shirts at Rs. 1450.00 each, 7 pants at Rs. 1840.00
each and 3 T−shirts at Rs. 850.00 each from Jay Enterprises, New Road,
Kathmandu. Prepare a bill. (Sample bill is given below)

JAY ENTERPRISES

Bill No: 786 New Road, Kathmandu
Date : ................

Name of customer : .....................................................................................
Address : ........................................................................................................

S.N. Particulars Quantity Rate (Rs.) Amount (Rs.)

Total

In words : ...................................................................................

……………………….. ……………………………
Customer’s sign. Jay Enterprises

4. Mrs. Aashu bought 4 dozens of banana at Rs. 60.00 per dozen, 9 kg of
apples at Rs. 95.00 per kg, 12 kg of orange at Rs. 110.00 per kg, 10 kg
of grapes at Rs. 145.00 per kg from a fruit shop. Prepare a bill for Mrs.
Aashu and find the total payment made by her.

5. Sweta bought 3 ice−cream cones at Rs. 65.00 each, 7 cold drinks at
Rs. 35.00 each, 5 cans of juice at Rs. 75.00 each and 9 packets of
biscuits at Rs. 45.00 each. Make a bill for the goods bought by Sweta.

176 Prime Mathematics Book − 4

6. Raj went to restaurant with his Menu

friends to give them a treat on S.N Items Rate

the occasion of his birthday and 1. MoMo Rs. 95/plate

ordered the following: 2. Pizza (Full) Rs. 240

( i) MoMo : 3 plates 3. French fries Rs. 90/plate

( ii) French fries : 5 plates 4. Chowmein Rs. 120/plate

( iii) Chicken chilly : 4 plates 5. Thukpa Rs. 135
( iv) Mutton sekuwa : 2 plates
( v) Veg. pakoda : 7 plates 6. Chicken chilly Rs. 180/plate
How much money does Raj pay
for the bill prepared by the res- 7. Mutton sekuwa Rs. 200/plate

8. Veg. Pakoda Rs. 80/plate

taurant? 9. Green salad Rs. 50

Unit Revision Test

1. Find the total cost by preparing a bill. 6 books at Rs. 150 per book,
10 pencils at Rs. 8 per pencil, 12 copies at Rs. 20 per copy and 24
chart papers at Rs. 12 per piece.

2. Asmita Tondon from Dhapasi bought 4 dozens of banana at Rs. 65 per
dozen, 10 kg of apples at Rs. 105 per kg, 12 kg of orange at Rs. 90
per kg and 8.5 kg of grapes at Rs. 120 per kg form Yadav Fresh Fruits,
Chabahil, Kathmanu. Prepare a bill. Sample bill in shown below:

Yadav Fresh Fruits

Bill No: 715 Chabahil, Kathmandu
Date : ................

Name of customer : .....................................................................................
Address : ........................................................................................................

S.N. Particulars Quantity Rate (Rs.) Amount (Rs.)

Total

In words : ...................................................................................

……………………….. ……………………………
Customer’s sign. Prepare by

Prime Mathematics Book − 4 177

3. Study the given bill and answer the following questions:

Asha Stationary Date : 2073/08/02

Thimi, Bhaktapur

Bill No: 1435

Name of customer : ..M....r.....J.a..y....G..a...u..t..a..m.........................................................
Address : ...G..h...a..t..t.a..g...h..a...r.,..B...h..a..k..t..a..p...u..r............................................................

S.N. Particulars Quantity Rate (Rs.) Amount (Rs.)
1. Geometry box 4 125.00 500.00
2. Pencil case 5 140.00 700.00
3. Pens 18 95.00 1710.00
4. Notebook 24 20.00 480.00
5. Pencil 36 15.00 540.00
3930.00
Total

In words : .T..h.r..e..e..t..h.o..u..s.a..n..d...n.i.n..e...h..u..n.d..r..e.d...a..n..d..t..h..i.r.t.y...o..n..l.y..................

……………………….. ……………………………
Customer’s sign. Jay Enterprises

( a) Who is the purchaser and where is he from?
( b) Where did he make this purchase? And when?
( c) How much did he pay in total?
( d) How many pens did he buy?
( e) What is the cost of a notebook?

Answers:

Exercise 7.1

Q. Show to the teacher.

178 Prime Mathematics Book − 4

U8nit Statistics

Estimated periods − 9
Objectives
At the end of this unit, the students will be able to:
• collect the data and represent them in a table in ascending order.
• give and take informations from the table
• draw the bar graph of the given data in a graph paper.
• read and take information from the thermometer.

Teaching Materials
bar graph of different data, graph paper, thermometer.

Activities
It is better to:
• ask the students collect data of their heights, weights, marks etc.
• ask them to arrange the data in ascending order and represent the data in a table.
• ask them to represent the arranged data in a bar graph.
• teach them how to get information from the bar graph.
• show the thermometer to the students and teach them how to get the information.

Graph and ordered number pairs P
Where is the point P? Just pointing the point with
your fore finger is not enough to tell the position of
the point P in a plane paper. We need other bases to
locate the point.

1. Reference point with respect to which the point is to be located. The
reference point is called origin (O).

2. Horizontal and vertical distances of the point from the origin. The
horizontal distance in called horizontal component or X-component and
vertical distance in called vertical component or Y-component.

Graph paper Square paper

With the help of these two things, now we Y
can say the position of the point P. It is at a
distance of 4 units along horizontal and 3 units P
along vertical from O. We write the position
of P by an ordered pair number (4,3).

The point O is called origin. The horizontal O X
line drawn through O is called x-axis and the X
vertical line drawn though O is called y-axis. X’

To plot a point A(2,3), we start from O. Take Y’ Y
a point as origin (O). Draw horizontal axis OX
and vertical axis OY through O, count 2 units A
towards right along x-axis and then 3 units
upwards to reach the point.

Note: (i) (2,3) ≠ (3,2) 3
(ii) (2,3) can not be written as
{2,3} or [2,3] X’ O Y’ 2

180 Prime Mathematics Book − 4

Exercise: 8.1

1. In the given graph, fix origin and draw x-axis and y-axis and plot the
points with the given ordered pair numbers.
( a) A (3,1) ( b) B(2,3) ( c) C(4,5) ( d) D(5,4)

Y

2. Write the ordered pair numbers P S
representing the points P, Q, R
and S. Q R

Y X

X’ O

Y’

3. Plot the points P (1,6), Q (6,4), R
(1,2), S (4,7), T (5,0) in the given
graph. Join the points PQ, QR, RS, ST
and TP. Name the figure obtained.

X’ O Y’ X

Y

4. Study the given graph and write the A C
ordered pair number of the vertices B
of ∆ABC. X’ O
Y’ X
Y

A

CB L 5. From the graph given along side,
K write the ordered pair numbers
representing the vertices A, B,
D C, D, E, F, G, H, I, J, K and L of
J the given hexagram.

EF I X
H
Prime Mathematics Book − 4 181
X’ G
Y’

Simple Bar Diagrams

We have learnt about the bar diagram in class III too. Bar diagram is a pictorial
representation of numerical data. It gives immediate information at a glance.
We generally see bar diagrams in newspapers and magazines.
The marks obtained by Jay in a terminal exam are as follows:

Subjects English Mathematics Science Nepali Social Studies

Marks 75 95 85 60 70

Let’s represent it in a simple bar diagram.
Y

100

90

Marks obtained 80

70

60

50

40

30

20

10

O English Mathematics Science Nepali Social Studies X

Subjects

Steps -
(i) Draw horizontal axis OX and vertical axis OY.
(ii) On the horizontal axis take five points at equal intervals representing

English, Mathematics, Science, Nepali, Social studies.
(iii) Since maximum number of the student is 95, mark the vertical axis

equally from 0 to 100.
(iv) Raise the bars of suitable width on the horizontal axis representing

the different subjects of heights corresponding to the number of the
student.

Note : i) The width of the bars should be same.
ii) The distance between the bars should be same.

182 Prime Mathematics Book − 4

The bar graph given below shows the number of students enrolled in different
classes.

Number of students Y

50 I II III IV V X
45 Classes
40
35
30
25
20
15
10

5

O

Answer the following questions:
( i) Which class has the least number of students ? What is the number of

students ?
( ii) What is the number of students in class II ?
( iii) Which class has the maximum number of students ? Write the number.
( iv) What is the total number of students from class I to class V ?
( v) What is the number of students in class I ?

Exercise: 8.2

1. Represent the following data on bar graphs.
( i) The marks obtained by Jay in terminal exam are given below:

Subjects Nepali English Science Maths G.K.

Marks 70 90 85 100 95

( ii) The number of students enrolled in classes from 1 – 5 are as follows:

Classes I II III IV V

No. of students 30 40 20 35 50

( iii) The incomes of different people are given below:

Name Jay Avay Dipu Raj Mamata

Income (Rs.) 1500 2000 1000 2500 3000

( iv) The maximum temperature of Birgunj of 5 different months are as
follows :

Months Baisakh Jestha Asadh Shrawan Bhadra

Temp (ºC) 32ºC 40ºC 35ºC 33ºC 28ºC

Prime Mathematics Book − 4 183

( v) The number of students who failed in different subjects in a
terminal examination is as follows :

Subjects Nepali English Maths Science G.K.
No. of students 5 8 2 10 7

2. Observe the bar graph shown below and answer the following questions:
Students belonging to different houses.

Y

No. of students 250
225
200 Garnet Emerald Moonstone Sapphire X
175 Houses
150
125
100
75
50
25

X′ 0 O

Y′

( i) What is the number of students belonging to the Garnet house ?

( ii) Which house has the maximum number of students ? Write the

number.
( iii) Can you say the number of students belonging to the Sapphire house ?

( iv) Which house has the least number of students ? Write the number.

3. Study the given bar graph and answer the following questions:

No. of students graduated in SLC of a school in different years.

Y

No. of students 100
90
80 2060 2061 2062 2063 2064 X
70
60 Years in B.S
50
40
30
20
10

X′ O

Y′

184 Prime Mathematics Book − 4

( i) In which year, the least number of students graduated in SLC ? Write
the number.

( ii) What is the number of students who graduated in SLC in the year 2063 B.S ?
( iii) In which year the maximum number of students who graduated in SLC

? write the number.
( iv) What is the total number of students who graduated in SLC from the

year 2060 B.S to 2064 B.S ?

Measurement of Temperature Normal body
37º-98.6o
To measure the hotness or coldness of an object means
to measure its temperature. To measure the temperature temperature
of an object, an instrument is used and it is called
thermometer.

The standard units used in thermometer are degree
Fahrenheit (ºF) and degree Celsius (ºC). In Fahrenheit
scale, 32ºF is the freezing point of water and 212ºF is
the boiling point of water. In Celsius scale, 0ºC is the
freezing point and 100ºC is boiling point of water.
The thermometer used to measure the temperature of
human body is called clinical thermometer. The normal
temperature of our body is 98.6ºF or 37ºC.

Exercise: 8.3

1. Write the temperature shown by the following thermometers.

( i) 0 10 20 30 40 50 60 ºC ...................................

( ii) 0 10......50 60 70 80 90 100ºC ...................................

( iii) 0 10......50 60 70 80 90 100ºC ...................................

( iv) 30 40 50 60 70 80 90 100 .................... 212ºF ...........................

( v) 30 40 50 60 70 80 90 100 .................... 212ºF ...........................

( vi) 30 40 50 60 70 80 90 100 110 ............. 212ºF ...........................

Prime Mathematics Book − 4 185

2. The temperature of different places of Nepal in the month of Baisakh
is given below:

Places Maximum Places Maximum
Temperature Temperature

Kathmandu 20ºC Narayangrah 29ºC
32ºC
Pokhara 23ºC Janakpur 10ºC
34ºC
Biratnagar 28ºC Jomsom

Birgunj 30ºC Bhairahawa

Answer the following questions:

( i) Which is the coldest place in the month of Baisakh? What is it’s
temperature ?

( ii) Which place is the warmest place in the month of Baisakh ? What
is it’s temperature ?

( iii) Among Biratnagar and Birgunj, which place is warmer and by how
much in the month of Baisakh ?

( iv) What is the difference in the temperature of Pokhara and Bhai
rahawa in the month of Baisakh ?

( v) What is the difference in the temperature of Jomsom and Janakpur
in the month of Baisakh ?

Chart

Class I Class II Class III Class IV Class V

Girls Boys Total Girls Boys Total Girls Boys Total Girls Boys Total Girls Boys Total

10 30 40 15 20 35 20 25 45 18 23 41 28 15 43

It is very easy to get information by looking at the given chart. For example the
number of girls and boys in class III are 20 and 25 respectively. Similarly, the
total number of students of class II is 35. So, at a glance we get information
from a chart.

Exercise: 8.4
1. The chart given below shows the marks obtained by Jay in different

subjects.

Subjects English Maths Nepali Science Social Studies
98
Marks 90 50 75 88

186 Prime Mathematics Book − 4

Answers the following questions:
( i) In which subject Jay scored the highest mark ?
( ii) In which subject he scored the least mark ?
( iii) What is the mark obtained by him in English ?
( iv) How much did he score in Social Studies ?

2. The chart given below shows the price list of vegetables in a market.

S.N. Vegetable items Rate Per kg
1. Cauliflower Rs. 65.00
2. Tomato Rs. 75.00
3. Potato Rs. 50.00
4. Carrot Rs. 80.00
5. Peas Rs. 42.00

Answer the following questions;
( i) What is the price of tomato per kg ?
( ii) Which is the cheapest vegetable ?
( iii) Which vegetable is the most expensive one ?
( iv) What is the cost of peas per kg ?

3. Item Cold drinks Ice− cream Tea Lassi Milk

No. of students 105 120 25 90 100

The chart given above shows the number of students who prefer
different items. Answer these questions.
( i) Which item is liked by most of the students ?
( ii) Which item is liked least by the students ?

( iii) What is the number of students who like cold drinks ?
( iv) Mention the number of students who like milk ?

Unit Revision Test

1. Shreya bought some articles from a stationary shop and they are
given below.

Items Pencils Notebook Pens Erasers Books Geometry
No. of 16 22 8 32 12 18
items

Represent the given data in a bar graph. Prime Mathematics Book − 4 187

2. The bar graph shows the favorite subjects of the students of class 5.

Y Maths English Nepali Science Social X
Subjects Studies
55
50
45
40
35
30
25
20
15
10

5

X' 0
Y'

Study the bar graph given above and answer the following questions:
( i) Which is the most favorite subject of the students? What is the

number of students who like this subject?
( ii) What subject is liked by the least number of students and what is

the number of students who do not like this subject?
( iii) Which is the next favorite subject of the students? Can you mention

the number of students who like this subject?
( iv) What number of students like Science? What number of students

like Social Studies? Can you say the total number of students who
like Science and Social studies?

3. Plot the given points in graph:
A (1,3); B (3,1); C (0,4); D (5,0)

Answers:

Exercise 8.1

1) Show to your teacher. 2) P (1,3), Q (2,1), R (5,1) & S (4,3) 3) Show to your teacher.
4) A (0,2), B (3,1) & C (4,4)
L(5, 6) 5) A(4, 8), B(3, 6), C(1, 6), D(2, 4), E(1, 2), F(3, 2), G(4, 0), H(5, 2), I (7, 2), J(6, 4), K(7, 6),

1. Show to your teacher. ii) Emerals = 250 Exercise 8.2 iv) Moonstone = 175
2. i) 225 ii) 80 students iv) 40 + 60 + 50 + 80 + 90 = 320 students
3. i) 2060, 40 students iii) 200
iii) 2064, 90

Exercise 8.3

1) i) 40oc ii) 70oc iii) 90oc iv) 90oF v) 60oF vi) 110oF

2) i) Jomsom ii) Bhairahwa iii) Birgunj is warmer. by = 30o - 28o = 2oc

iv) 34oc-23oc = 11oc v) 22oc

Exercise 8.4

1) i) Maths ii) Nepali iii) 90 iv) 88
2) i) Rs. 75 ii) Peas
3. i) Ice-cream ii) Tea iii) Carrot iv) Rs. 42

iii) 105 Students iv) 100 students

188 Prime Mathematics Book − 4

U9nit Set

Estimated periods − 9

Objectives

At the end of this unit, the students will be able to:

• define set.

• differentiate between well defined collection of objects and not well defined collection of
objects.

• say whether the given object belongs to or doesn't belong to the set.

• represent the set in diagrammatic method, description method and listing method.

Teaching Materials
chart showing collection of objects.

Activities
It is better to:
• ask the students collect objects of same kind.
• ask them to list the elements belonging to the set.

Set

It is a collection of birds. It is a collection of fruits.
It is a set of birds. It is a set of fruits.

It is a collection of animals.
It is a set of animals.

Thus, a set is a collection of objects.
Well defined collection of objects:

Asha Ashisa Alina

Alina is bigger than Asha but thinner than Ashisa.

Does Alina belong to No, Alina does not belong to the
the collection of big collection of big girls because
she is thinner than Alina.
girls?

Does Alina belong to the
collection of thin girls?

No, Alina doesn’t belong to the
collection of thin girls because

she is bigger than Asha.

Thus, a collection of thin girls or a collection of big girls is not well defined.
Similarly, a collection of girls dark in colour or a collection of girls fair in
colour is not well defined. If the collection is not well defined, then the
collection cannot be a set.

190 Prime Mathematics Book − 4

a, e, i, It is a collection of vowels. It contains only a, e, i,
o, u o, u. So, it is collection of well defined objects and
it is a set.

Thus, a set is a collection of well defined objects.

2, 4, 6, It is a collection of first five even numbers. It is a set
8, 10 of first five even numbers. The elements or members
of this set are 2, 4, 6, 8 and 10.

Thus, each object of the set is called a member or an element of the set.

Objects
A = {1,3,5,7,9}
B = {a, e, i, o, u}
C = {cat, dog, horse, cow}
Sets are usually denoted by English capital letter A, B, C ….. Z. Each member
of a set is separated by comma and enclosed by the curly brackets { }.
Here, ‘a’ is a member of set B. So ‘a’ belongs to set B.
Symbolically, it is can be written as a ∈ B.
Similarly, 5 is an element of the set A. So, 5 is an element of set A or 5 belongs
to set A. Symbolically, it can be written as 5 ∈ B.

The symbol ∈ is used in place of “is an element of” or “belongs to”.
‘e’ is not an element of set A. It means ‘e’ doesn’t belong to the set A.
Symbolically, it can be written as e ∉ A.
Similarly, 3 is not an element of set B. It means 3 doesn’t belong to the set
B. Symbolically, it can be written as 3 ∉ B.

Thus, the symbol ∉ is read as “is not an element of” or “doesn’t
belong to”.

Workedout Examples
1. Write whether the following collection of objects are well defined or

not:
( a) A collection of intelligent students.
Answer: Not well defined.
( b) A collection of 1st five even numbers.
Answer: Well defined.

Prime Mathematics Book − 4 191

( c) A collection of tall boys.
Answer: Not well−defined.
( d) A collection of students of class IV.
Answer: Well defined.
2. Write the following set notations in word.
( a) 3 ∈ {3, 6, 9, 12, 15}
( b) 4 ∉ {1, 3, 5, 7}
( c) a ∈ {a, e, i, o, u}
( d) 5 ∉ {0, 10, 20, 40}
Solution:
( a) 3 belong to the set {3, 6, 9, 12, 15}

or
3 is an element of the set {3, 6, 9, 12, 15}

( b) 4 doesn’t belong to the set {1, 3, 5, 7}
or

4 is not an element of the set {1, 3, 5, 7}

( c) a belongs to the set {a, e, i, o, u}
or

a is a member of the set {a, e, i, o, u}

( d) 5 doesn’t belong to the set {0, 10, 20, 40}
or

5 is not a member of the set {0, 10, 20, 40}.

Exercise: 9.1
1. Write whether the following collections are well defined or not.

( a) A collection of intelligent students of a class.
( b) A collection of vegetables.
( c) A collection of kind people.
( d) A collection of teachers of a school.
( e) A collection of beautiful flowers.

2. Write the following set notations in words.
( a) a ∈ B
( b) 1 ∉ {2, 4, 6, 8}
( c) e ∈ {a, e, i, o, u}
( d) Nepal ∉ {England, Ireland, Scotland, South Wales}
( e) 4 ∈ {2, 4, 6, 8}

192 Prime Mathematics Book − 4

3. Use the symbols ∈ or ∉ in the following statements.
( a) Apple belongs to {Apple, Orange, Grapes, Mango}
( b) Parrot doesn’t belong to {Lion, Tiger, Leopard, Bear}
( c) 2 belongs to set A.
( d) b doesn’t belong to the set D.
( e) 5 doesn’t belong to the set {0, 10, 20, 30}

Ways of Writing Sets
i) Diagrammatic method:

It is a set of birds. It is a set of animals.

Thus, the members of set can be shown inside an oval as in the above figure.
It is diagrammatic method.

ii) Descriptive method:
{The months of a year}; {the days of a week}; {the prime numbers less
than 10} and so on.
In this type of representation the elements of the set are described in
words.

iii) Listing method:
In this method the elements of a set are listed inside curly brackets.
The elements are separated by comma.
A = {a, e, i, o, u}
B = {2, 4, 6, 8}
C = {2, 3, 5, 7}

Exercise: 9.2

1) List the members of the following sets.
( i) A = {first five prime numbers}
( ii) B = {A set of five fruits}
( iii) C = {five consonants}
( iv) D = {first four even numbers}
( v) E = {five animals}

Prime Mathematics Book − 4 193

2. Write the following sets in descriptive and listing method.

( a) ( b)

1, 3, 5,
7, 9

( c ) ( d)

Unit Revision Test
1. Write whether the following collection of objects are well defied or

not.
( a) A collection of tall students of your class.
( b) A collection of buildings of your school premises.

2. Write the following set notations in words
( a) 5 ∈ {1, 2, 3, 4, 5, 6, 7}
( b) b ∉ {a, e, i, o, u}

3. Write the following statements in set notation.
( a) 2 belongs to {even number less than 10}
( b) Lion does’t belong to {parrot, pigeon, sparrow, peacock}

4. Write the following sets in listing method:
( a) A = { prime numbers between 10 and 20}
( b) B = { vowels of English alphabets}
( c) C = { four birds}

Answers: c) e belong to the set

Exercise 9.1
1) show to your teacher.
2) a) a belong in the set B b) 1 doesn’t belong to the set {2,4,6,8}

d) Nepal does not belong to the set e) 4 belongs to the set.
3) Show to your teacher.

Exercise 9.2

1) i) A = {1,3,5,7,11} ii) B = { Apple, Mango, Banana, Peach, Orange}

194 iii) C = { B,C,D,F,G} iv) D = { 2,4,6,8}

v) E = { Dog, Cat, Rat, Rabbit, Cow}

Prime Mathematics Book − 4

1Un0it Algebra

Estimated periods − 19
Objectives
At the end of this unit, the students will be able to:
• identify the variables and constants.
• find the value of algebraic expression by putting the given values of variables.
• recognize the base and coefficient of a term.
• recognize like and unlike terms .
• add and subtract the like terms.
• add and subtract the algebraic expressions.
• find the value of variables in the equations.

Teaching Materials
Files, blocks, etc.

Activities
It is better to:
• ask the students about the concept of variables and constants with suitable and simple
examples.
• ask the students about the concept of base and coefficient of a term with suitable examples.
• ask the students to find the value of algebraic expression by giving the values of variables.
• ask the students about the like and unlike terms with suitable examples.
• add and subtract like terms and algebraic expressions.
• use balance to give the concept of algebraic equation.
• ask the students to demonstrate the activities related to this lesson suggested in the
curriculum prescribed by CDC.