Prime Mathematics 4

4. Perform the following addition:

(a) years months days (b) years months days

5 8 20 12 9 25

+3 9 35 +8 10 24

(c) 10 years 5 months 20 days + 8 years 10 months 18 days.
(d) 7 years 11 months 28 days + 6 years 9 months 12 days.

5. Perform the following subtractions:

(a) years months days (b) years months days

8 74 12 3 10

−3 5 10 −7 6 12

(c) 12 years 7 months 16 days – 8 years 10 months 24 days
(d) 15 years 3 months 15 days− 9 years 6 months 11 days

6. Solve the following problems:

(a) Abhisek stayed 3 years 3 months and 17 days in India and 5 years
9 months and 24 days in Pakistan and returned to Nepal. How long
did he stay abroad?

(b) Sonit took Japanese language training for 1 year 6 months and 20
days and then German language training for 2 years 8 months and
25 days. How long did he spend on language training?

(c) Tara is 14 years 3 months and 12 days old and Asha is 12 years 7
months and 24 days old. By how much is Tara older than Asha ?

(d) Tsering’s age is 14 years 9 months and 19 days. He will get his
citizenship exactly when he will be 16 years old. After how long
can he apply for his citizenship?

96 Prime Mathematics Book − 4

Unit Revision Test

Time

1. Write the given times as a.m. or p.m.:

(a) 6:15 in the morning (b) 4:00 in the evening

(c) 11:45 at night (d) 2 o’clock at night

2. Complete the following:
(a) It is 11:00 p.m. It will be …………….. after 3 hours.
(b) It is 1:00 p.m. It will be ……………. after 4 hours.
(c) There are …………….. weeks in a year.
(d) 12 months is same as ………………. days.

3. Add : 4 hours 45 minutes and 17 hours 25 minutes.

4. Perform the given subtraction:

hours minutes

6 10

− 2 45

5. Convert 4 weeks 6 days into days.

6. Convert 75 months into years and months.

7. Perform the given sum:
years months days
10 8 20
+8 7 22

8. It is 2:40 p.m. What will be the time after 2 hours 65 minutes?

9. A father’s age is 46 years 7 months and 16 days and his son’s age is 17
years 8 months and 25 days. By how much is the father older than
his son?

Prime Mathematics Book − 4 97

Read the follwoing: Money
Coins Notes

Study and learn the following: paisa = p
Rupee = Re.
• 100 paisa = 1 rupee Rupees = Rs.
Also we write
100 p = Re. 1

• To convert rupees into paisa, we multiply by 100.
• To convert paisa into rupees we divide by 100.

÷ 100

paisa Rupees

× 100

Workedout Examples
Example 1: Convert Rs. 12 into paisa.
Solution: Rs. 12

= 12 × 100p
= 1200p
Therefore, Rs. 12 = 1200p

Example 2: Convert Rs. 7 and 50 paisa into paisa.
Solution: Rs. 7 and 50 p.

= Rs. 7 + 50 p
= 7 × 100p + 50p
= 700p + 50p
= 750p
Therefore, Rs. 7 and 50 paisa = 750 paisa.

Example 3: Convert 600 paisa into rupees.
Solution: 600p

= Rs. (600 ÷100)
= Rs. 6
Therefore, 600 paisa = Rs. 6

98 Prime Mathematics Book − 4

Example 4: Convert 875 paisa into rupees and paisa. 100) 875 (8
Solution: 875 paisa -800
75
= 875 ÷100 rupees.
= 8 rupees and 75 paisa
= Rs. 8 and 75 paisa.

Example 5: Add : Rs. 15 and 75 paisa and Rs. 9 and 50 paisa.
Solution: Rs. P.

15 75
+9 50
24 125
= Rs. 24 + 100p + 25p
= Rs. 24 + Re.1 + 25p
= Rs. 25 and 25 paisa.
Example 6: Perform the following subtraction.
Rs. P
23 40
-11 65

Solution: Rs. P.

22 =1100P 140 [ 100 + 40 = 140 and 140 – 65 = 75]

23 40

−11 65

11 75
= Rs. 11 and 75 paisa.

Exercise: 4.3

1. Convert into paisa:
(a) Rs. 16 (b) Rs. 6.25 (c) Rs. 9.50 (d) Rs. 50 and 50 paisa

2. Convert into rupees: (b) 320 paisa
(a) 32000 paisa (d) 12345 paisa
(c) 8660 paisa

3. Convert into rupees and paisa:
(a) 9876 paisa (b) 8640 paisa
(c) 19290 paisa (d) 280 paisa

4. Add the following:
(a) Rs. 15.85 and Rs. 165.50
(b) Rs. 24.50 and Rs. 2.75
(c) Rs. 70 and 60 paisa and Rs. 80 and 90 paisa.
(d) Rs. 120 and 30 paisa and Rs. 239 and 70 paisa.

Prime Mathematics Book − 4 99

5. Perform the following additions:

(a) Rs. p (b) Rs. p

10 65 75 50

+18 45 +9 90

(c) Rs. 7.05 (d) Rs. 24.86
+ Rs. 5.95 + Rs. 5.64

6. Subtract:
(a) Rs. 5 and 75 paisa from Rs. 28 and 45 paisa.
(b) Rs. 240 and 70 paisa from Rs. 302 and 25 paisa.
(c) Rs. 600.80 from Rs. 850.15
(d) Rs. 29.95 from Rs. 92.59

7. Perform the following subtractions:
(a) Rupees paisa (b) Rupees paisa
58
75 65 96 85

−24 85 − 65

(c) Rs. 125.05 (d) Rs. 24.75
− Rs. 5.50 − Rs. 13.90

Multiplication and division involved with money

Study the following examples:

Example 1: Multiply Rs. 24 and 45 paisa by 6.

Solution: Rs. p
24 45

×6
144 270
= Rs. 144 and 270 paisa
= Rs. 144 + 200 paisa + 70 paisa
= Rs. 144 + Rs. 2 + 70 paisa
= Rs. 146 and 70 paisa
= Rs. 146.70

100 Prime Mathematics Book − 4

Again, let’s see the same example in this way.

Rs. 24.45 • Multiply as usually in decimal system.
×6 • Locate the position of decimal after two

Rs. 146.70 digits from right to left.

Example 2: Divide Rs. 84 by 40.
Solution: Here, Rs. 84 = 84 × 100p

= 8400p
Now, Rs. 84 ÷ 40

= 8400p ÷ 40

= 8214000p
40
= 210p
= 200p + 10p
= Rs. 2 and 10 paisa.

Example 3: Divide Rs. 84 and 60 paisa by 20.

Solution: Here, Rs. 84 and 60 paisa Alternative method I
= 84 × 100 paisa + 60 paisa Rs. 84 and 60 paisa ÷ 20
= 8400 paisa + 60 paisa
= 8460 paisa Rs. P.
4 23
Now, Rs. 84 and 60 paisa ÷ 20
20) 84 60
423
8460 paisa −80
= 20 Rs. 4 60p
= 400p + 60p
= 423 paisa = 460p
= 400p + 23p
= Rs. 4 and 23p. 40
60
60
∴ (Rs. 84 and 60p) ÷ 20
= Rs. 4 and 23 paisa.

Alternative method II 20)84.60(4.23
(Rs. 84 and 60 paisa) ÷ 20 −80
46
= Rs. 84.60 ÷ 20 40
= Rs. 4.23 60
∴ (Rs. 84 and 60 paisa) ÷ 20 6×0

= Rs. 4.23 Prime Mathematics Book − 4 101

Exercise: 4.4

1. Perform the following multiplication:

(a) Rs. p (b) Rs. p

8 40 18 46

×6 ×5

(c) Rs. p (d) Rs. p
234 90 104 60
× 14 × 24

2. Multiply: (b) Rs. 54 and 25 paisa by 25
(a) Rs. 60 and 45 paisa by 20 (d) Rs. 145.75 by 12
(c) Rs. 36.50 by 8

3. Perform the following divisions: (b) 5)Rs.68 40P
(d) 15)Rs.48 30P
(a) 4)Rs.16 76P

(c) 10)Rs.81 20P

4. Divide: (b) Rs. 75 by 50
(a) Rs. 84 by 100 (d) Rs. 20 and 40 paisa by 4
(c) Rs. 72 by 24 (f) Rs. 84 and 60 paisa by 6
(e) Rs. 15 and 75 paisa by 5 (h) Rs. 95.40 by 10
(g) Rs. 47.10 by 3

5. Solve the following problems:
(a) The cost of a pencil is Rs. 8.40. Find the cost of 8 such pencils.
(b) The cost of 1 mathematics book is Rs. 142.50. Find the cost of 12
such maths books.
(c) The cost of 1 litre of petrol is Rs. 85.40. Find the cost of 10 litres
of petrol.
(d) If the cost of 16 pens is Rs. 60.15, find the cost of 1 pen.
(e) If the cost of 10 litres of milk is Rs. 184, find the cost of 1 litre of
the milk.

102 Prime Mathematics Book − 4

Unit Revision Test
Money

1. Convert into paisa: (b) Rs. 8 and 40 paisa
(a) Rs. 100

2. Convert into rupees and paisa:
(a) 6400 paisa (b) 4560 paisa

3. Perform the following task: (b) Rs. P
123 25
(a) Rs. P −12 95
20 40

+35 80

4. Add Rs. 23.60 and Rs. 142.75.

6. Subtract Rs. 9.45 from Rs. 108.60.

7. Perform the following task: (b) 5)Rs. 42 30p

(a) Rs. P
42 25
×6

8. If the cost of an eraser is Rs. 2.25, find the cost of 5 such erasers.
9. If the cost of 12 kg of sugar is Rs. 667.20, find the cost of 1 kg of

sugar.

Prime Mathematics Book − 4 103

Length

Study and learn the following:

Metric table of length
millimetre centimetre decimetre metre decametre hectometre kilometre

mm cm dm m dam hm km

× 10 × 10 × 10 × 10 × 10 × 10

• Every successive hiegher unit is 10 times its
proceeding unit. Thus, metric system is decimal
system. [In Latin, decem = 10]

a ruler • For conversion from bigger to smaller unit multiply
by 10 or multiple of 10 corresponding to the number
of steps from a unit to next lower unit.

measuring • For conversion from smaller to bigger unit divide
tape by 10 or multiples of 10 corresponding to the
number of steps from a unit to next higher unit.

Millimetre, Centimetre, Metre and Kilometre are very useful in daily life.

10mm = 1cm mm cm dm m dam hm km
100cm = 1m 10 10 10 10 10 10
1000m = 1km
10,00,000mm = 1km 1 step = 10
mm cm dm m dam hm km
104 Prime Mathematics Book − 4
10 10 10 10 10 10
2 steps = 100

mm cm dm m dam hm km
10 10 10 10 10 10

3 steps = 1000
mm cm dm m dam hm km

10 10 10 10 10 10

6 steps = 1000000
}

Example 1: Convert 42mm into centimetres.

Solution : 42mm 10mm = 1cm
For conversion from smaller to
bigger unit, we divide.
= 42 cm ∴
10 ∴

= 4.2cm ∴
∴ 42mm = 4.2cm ∴

Example 2: Convert 85cm into metres. ∴

Solution : 85cm 100cm = 1m
For conversion from smaller
= 85 m unit to bigger unit, we divide.
100

= 0.85m
∴ 85cm = 0.85m

Example 3: Convert 2591m into kilometre.

Solution : 2591m 1000m = 1km
For conversion from smaller
= 2591 km to bigger unit, we divide.
1000

= 2.591km

∴ 2591m = 2.591km

Example 4: Convert 34mm into centimetres and millimeters.

Solution : 34mm

= 34 cm [ 10mm = 1cm]
10

= 3.4cm

= 3cm 4mm

∴ 34mm = 3cm 4mm

Example 5: Convert 3215mm into metres.

Solution : 3215mm mm cm dm m dam hm km
10 10 10 10 10 10
= 3215 m
1000

= 3.215m 3 steps = 1000
∴ 3215mm = 3.215m Conversion from smaller unit to bigger
unit, we divide.

Prime Mathematics Book − 4 105

Example 6: Convert 4km 342m into metre.

Solution : 4km 342m

= 4 × 1000m + 342m [ ∴ 1000m = 1km]
∴
= 4000m + 342m

= 4342m

∴ 4km 342m = 4342m

Example 7: Convert 5498m into kilometres and metres.

Solution : 5498m

= 5000m + 498m [ 1000m = 1km]

= 5000 km + 498m
1000

= 5km 498m

∴ 5498m = 5km 498m

Exercise: 4.5

1. Convert the following: (b) 3m into millimeters.
(a) 6m into centimetres (d) 5km into centimetres.
(c) 9km into metres. (f) 0.5km into millimeters.
(e) 12cm into millimeters.
(b) 214mm into metres
2. Convert the following: (d) 324cm into metres
(a) 16mm into centimetres (f) 53cm into metres.
(c) 123400mm into kilometres
(e) 87650cm into kilometres

3. Convert the following: (b) 19cm 6mm into millimetres.
(a) 4cm 5mm into millimetres. (d) 11m 68cm into centimetres.
(c) 6m 45cm into centimetres. (f) 10km 672m into metres.
(e) 2km 254m into metres.

4. Convert the following:

(a) 91mm into centimentres and millimetres.

(b) 74mm into centimetres and millimetres.

(c) 150cm into metres and centimetres.

(d) 1475cm into metres and centimetres.

(e) 2500m into kilometres and metres.

(f) 7827m into kilometres and metres.

106 Prime Mathematics Book − 4

5. Perform the following additions:

(a) cm mm (b) cm mm (c) m cm
5 4 +136 6 25 47
+ 4 8 7 + 59 63

(d) m cm (e) km m (f) km m
7 60 3 256 50 12
+ 94 50 + 9 974 28 66
74
+ 122

(g) km m cm (h) km m cm
120 870 85 5 200 64
10 500 74
+ 205 630 75 850 52
+ 75

6. Perform the following subtractions:

(a) cm mm (b) cm mm (c) m cm
12 4 73 3 25 28
− 4 7 − 50 4 − 5 68

(d) m cm (e) km m (f) km m
120 84 45 73 100 79
18
− 59 48 − 13 − 68 24

(g) m cm mm (h) m cm mm
54 82 8 243 57 5
64 9 64 7
− 39 − 226

(i) km m cm (j) km m cm
70 120 30 10 84 49

− 34 488 94 − 6 223 57

7. Perform the following task:
(a) 21m 26cm + 18m 96cm
(b) 24km 576m + 15km 584m
(c) 38km 765m 76cm + 8km 435m 14cm.
(d) 18m 37cm – 17m 69cm
(e) 43km 174m – 27km 540m

Prime Mathematics Book − 4 107

8. Solve the following word problems:
(a) A road under construction is 18km 760m long. If 13km 540m is
further constructed, what is the total length of the road now?
(b) The thread of a kite is 214m 80cm long. If another 124m 65cm long
thread is connected, what is the total length of the thread of the
kite now?
(c) Rajesh is 1m 44cm tall and Nitu is 1m 36cm tall. By how much is
Rajesh taller than Nitu?
(d) The distance between two places is 32km 860m. If 15km 970m is
already travelled, how much distance is left to travel?

Mass (Weight)

Study and learn the following:

milligram centigram decigram gram decagram hectogram kilogram
dag hg kg
mg cg dg gm

× 10 × 10 × 10 × 10 × 10 × 10

• Weight of the rice in the given sack is 30 kg. This means the quantity of
rice contained in the sack is 30 kilograms.

• The quantity of the substance contained in an object is
called mass.

• Weight and mass are quite different things but laymen
understand the mass as weight.

• Mass/weight is measured in milligram, gram, kilograms
etc. In metric system there are seven different units.
Milligram (mg), centigram (cm), decigram (dg), gram (gm), decagram
(dag), hectogram (hg), kilogram (kg)

• Milligram is the smallest unit and kilogram is the greatest unit. Each
greater unit is 10 times the previous unit. Here we will use only milligram
(mg), gram (gm) and kilogram (kg).

1000mg = 1gm ∴ 1mg = 1 gm
1000gm = 1kg ∴ 1gm = 10100 kg
1000

108 Prime Mathematics Book − 4

Example 1: Convert 12 grams into milligrams.
Solution : Since,

1 gram = 1000 milligrams
∴ 12 grams = 12 × 1000 mg

= 12000 mg.
Therefore, 12 grams = 12000 milligrams.

Example 2: Convert 6 kg into grams.
Solution : Since,

1 kg = 1000 grams
∴ 6 kg = 6 × 1000 grams

= 6000 grams.
Therefore, 6 kg = 6000 grams.

Example 2: Convert 12300 milligram into grams.

Solution : Since

1 mg = 1 grams
1000

∴ 12300 mg = 1 × 12300 gm
1000

= 12300 = 123 = 12.3 grams.
1000 10

Therefore 12300 mg = 12.3 grams.

Example 3: Convert 2340 gm into kg.

Solution : Since

1 gm = 1 kg
1000
1
∴ 2340 gm = 1000 × 2340 kg

= 2340 kg
1000

= 2.340 kg

= 2.34 kg

Therefore, 2340 gm = 2.34 kg.

Prime Mathematics Book − 4 109

Example 4: Change 4320 gm into kilograms and grams.

Solution : Since, 1gm = 1 kg o4r320gm = 4000gm + 320gm
1000 = 4 × 1000gm + 320gm
4320 = 4kg + 320gm
4320 gm = 1000 kg = 4.320 kg = 4kg 320gm

= 4 kg 320 gm

Example 5: Change 5 kg 543 gm into gram.
Solution : Since

5 kg 543 gm = 5 kg + 543 gm = 5 × 1000 gm + 543 gm
= 5000 gm + 543 gm
= 5543 gm

Example 6 : Perform the following addition: kg gm mg
Solution : 625 840
5 550 735

+4

kg gm mg 1000 gm = 1 kg
840 1000 mg = 1 gm
5 625 735
1575 Thousands digits are
+4 550 -1000 carried over.
=9 1175
= 575
=9 +1
= +1 1176 575 mg
= 10 kg -1000

176 gm

110 Prime Mathematics Book − 4

Example 7 : Subtract 4 kg 650 gm from 9 kg 225 gm.
Solution :
kg gm
98 1 gm1222255
∴ 1 kg = 1000 gm
−4 =1000

650

4 575

= 4 kg 575 gm

Exercise: 4.6

1. Convert the following: (b) 14 kg into grams.
(a) 8 kg into grams. (d) 25 grams into milligrams.
(c) 6 grams into milligrams. (f) 7500 gm into kg.
(e) 34600 gm into kg.

2. Convert the following:
(a) 21 kg 432 gm into grams.
(b) 64 kg 614 gm into grams.
(c) 2520 gm into kilograms and grams.
(d) 5980 gm into kg and gm.
(e) 3465 mg into gm and mg.
(f) 10205 mg into gram and milligrams.
(g) 36 gm 269 mg into milligrams.
(h) 8 gm 450 mg into milligrams.

3. Perform the following additions:

(a) gm mg (b) gm mg (c) kg gm
21 250 31 200 9 413
+ 17 350 + 41 900 + 17 777

(d) kg gm mg (e) kg gm mg
5 285 50 35 618 820
715 450 381 180
+6 + 24

(f) kg gm mg (g) kg gm mg
3 250 540 2 219 230
5 350 340 8 130 356
400 250 420 527
+7 +3

Prime Mathematics Book − 4 111

4. Perform the following subtractions:

(a) gm mg (b) gm mg (c) kg gm
45 300 100 100 210 516
− 18 480 − 50 456 − 145 745

(d) kg gm (e) kg gm mg (f) kg gm mg
124 224 80 438 200 67 430 440
− 23 648 537 349 − 50 570 350
− 16

5. Add:
(a) 16 kg 230 gm and 74 kg 850 gm
(b) 234 gm 467 mg and 200 gm 783 mg
(c) 24 kg 413 gm 666 mg and 34 kg 587 gm 444 mg
(d) 437 kg 622 gm 950 mg and 333 kg 788 gm 150 mg

6. Subtract:
(a) 20 gm 340 mg from 48 gm 660 mg
(b) 217 kg 508 gm from 429 kg 348 gm
(c) 400 kg 289 gm from 560 kg
(d) 286 kg 812 gm 715 mg from 400 kg 300 gm 600 mg

7. Solve the following problems:
(a) Sheetal’s weight is 35 kg 460 gm. She is carrying a bag weighing 4
kg 800 gm. What is her weight including the bag?
(b) A bag contains 13 kg of rice, 2 kg 470 gm of pulses and 1 kg 775 gm
of sugar. Find the total weight of the bag.
(c) Siddhika’s weight is 29 kg 455 gm and Siddanta’s weight is 36 kg
820 gm. What is their combined weight?
(d) A shopkeeper had 200 kg of rice. He sold 76 kg 815 gm. How much
rice is left to sell ?
(e) How much is 5 kg more than 2 kg 780 gm ?
(f) What is the difference between 52 kg 200 gm and 36 kg 40 gm ?

112 Prime Mathematics Book − 4

Capacity 300
250
Learn the following: 200
This beaker can hold 300 millilitres of liquid. The amount of liquid 150
that a vessel can hold is called its capacity where as the space 100
occupied by the liquid in a vessel is the volume of the liquid. 50
0 ml

25ml

250 ml 250 ml

Beaker Measuring Conical Measuring Burette Pipette
cylinder flask flask

These are the different laboratory apparatus for measuring volume of
liquids.

Consider a cubical vessel of side 1 cm 1cm1cm
Its capacity is 1 cm × 1 cm × 1 cm = 1 cm3 1cm

or, 1 c.c or 1 millilitre (ml)
1000 ml make a litre

Thus, we have According to 1963 weight and measurement act (of
1000 ml = 1 litre SI Unit). 1 litre is defined as volume of 1 kg of pure
water at 4ºC and 760 mm atmospheric pressure
which is equivalent to 1000.028cc.

Example 1: Convert 18 litres into millilitres.
Solution : We have

1 litre = 1000 ml
∴18 l = 18 × 1000 ml
∴ 18 l = 18000 ml

Example 2: Convert 5320 ml into litres.

Solution : 5320 ml

= 5320 l [ 1 litre = 1000 ml ]∴
1000

= 5.320 l

Prime Mathematics Book − 4 113

Example 3: Convert 15 l 215 ml into milliliters.
Solution : 15 l 215 ml

= 15 l + 215 ml
= 15 × 1000 ml + 215 ml
= 15000 ml + 215 ml
= 15215 ml

Example 4: Convert 3214 ml into litres and milliliters.
Solution : 3214 ml

= 3214 l
1000

= 3.214 l
= 3 l and 214 ml.

Example 5: Add 65 l 675 ml and 9 l 650 ml.
Solution :
l ml Alternative method:
65 675 l ml
650
+9
74 1325 65 675
= 75 l 325 ml 650
+9 1325

74

= 74 l 1325 ml

= 74 l + 1000 ml + 325 ml

= 74 l + 1 l + 325 ml

= 75 l 325 ml

Example 6: Subtract 75 l 840 ml from 124 l 230 ml.
Solution :

l ml

123 1 ml 1230 ∴ 1 l = 1000 ml
∴ borrow 1 l from the column of
124 =1000 230
l = 1000 ml
− 75 840 1000 + 230 = 1230 ml.
48 390
= 48 l 390 ml

114 Prime Mathematics Book − 4

Exercise: 4.7

1. Convert into millilitres:

(a) 7l (b) 24l (c) 125l
(f) 2l 54ml
(d) 5l 345ml (e) 40l 930ml

2. Convert into litres:

(a) 634ml (b) 2340ml (c) 7500ml
(f) 100ml
(d) 43ml (e) 8015ml

3. Convert into litres and millilitres:

(a) 8642ml (b) 1298ml (c) 2300ml
(f) 1200ml
(d) 18004ml (e) 4020ml

4. Perform the following task:

(a) l ml (b) l ml (c) l ml
14 70 24 728 213 426
+ 9 + 45 422 + 87 154
980

(d) l ml (e) l ml (f) l ml
8 478 27 127 56 12
− 5 283 − 16 354 − 9
436

(g) 18l 240ml + 36l 912ml (h) 200l 744ml + 314l 842ml
(i) 16l 46ml – 12l 196ml (j) 375l 375ml – 274l 416ml

5. Solve the following problems:
(a) A tank contains 148l 876ml of water. If a bucket full of water
containing 12l 848ml is added to it, how much water will be there
in the tank ?
(b) A milkman sells 18l 560ml of milk in the morning and 14l 684ml of
milk in the evening. How much milk does he sell altogether in a
day ?
(c) A buffalo gives 7 litres of milk and a cow gives 12l 400ml of milk.
How much milk is collected from the two animals?
(d) The capacity of a tank is 200l 425ml. If there is78l 688ml of water,
how much water is to be added to fill the tank ?
(e) Nitesh had 8l 568ml of juice. If he drinks 2l 800ml of the juice,
how much juice is left ?

Prime Mathematics Book − 4 115

Unit Revision Test
Length, Mass (weight), Capacity

1. Convert:
( a) 12 Km into meters
( b) 18 kg 425 gm into grams

2. Convert:
( a) 346gm into gm and mg
( b) 234500 mm into km

3. Convert:
( a) 4 l 65 ml into millilitres
( b) 3045 ml into litres and ml

4. Preform the following additions:

( a) l ml ( b) kg gm mg
34 140 8 670 480
360 555 735
+ 10 +6

5. ( a) Subtract 27 m 79 cm from 43m 24cm.
( b) Subtract 12l 536ml from 64l 24ml.

6. ( a) A bag contains 7kg of rice, 4 kg 560 gm of pulses and 2 kg 875 gm
of sugar. Find the total weight of the bag.

( b) A jar contains 3l 800ml of juice. If 1l 025ml of juice is drunk, how
much juice is left in the jar?

116 Prime Mathematics Book − 4

Answers:

Exercise: 4.1

1. Show to your teacher. 2. Show to your teacher 3. Show to your teacher.

4. a) 1 pm b) 6:30 am c) 10 am d) 8 am e) 5 pm f) 3 pm

5. a) 1800 seconds b) 2 minutes 10 seconds c) 7200 seconds d) 740 seconds

e) 220 seconds

6. a) 420 minutes b) 135 minutes 1 hour and 30 minutes c) 1 minutes and 35 seconds

d) 9015 seconds.

7. a) 9 hour 55 minutes b) 14 hour c) 45 minutes c) 13 hours 41 minutes 46 second

d) 16 hours 16 minutes 20 seconds

8. a) 4 hours 20 minutes b) 4 hours 20 minutes c) 10 hours 20 min 10 second

d) 6 hours 54 min 40 second

9. a) 5:30 pm b) 3 hour 15 minutes c) 1 hour 45 minutes d) 9 hours 15 minutes

Exercise: 4.2

1. a) 35 days b) 210 days c) 1095 days d) 30 days e) 255 days f) 2380 days

2. a) 60 months b) 29 months c) 144 months d) 3 months

3. a) 1 year 6 months b) 10 year 10 months c) 3 year 2 months 20 days d) 2 years 4 months

4. a) 9 years 6 months 15 days b) 21 year 8 months 19 days c)19 years 4 months 8 days

d) 14 years 9 months 10 days

5. a) 5 years 1 month 24 days b) 4 years 8 months 28 days c) 3 years 8 months 22 days

d) 5 years 9 months 4 days

6. a) 9 years 1 month 11 days b) 4 year 3 month 15 days c) 1 year 7 months 18 days

d) 1 year 2 months 11 days

Exercise: 4.3

1. a) 1600 Paisa b) 625 Paisa c) 950 Paisa d) 5050 Paisa
d) Rs. 123.45
2. a) Rs.320 b) Rs. 3.20 c) Rs. 86.60 d) 2 Rupees 80 Paisa
d) Rs.360
3. a) 98 Rupees 76 Paisa b) 86 Rupees 46 Paisa c) 192 Rupees 90 Paisa d) Rs.30.50
d) Rs.62.64
4. a) Rs. 181.35 b) Rs.27.25 c) Rs. 151.50 d) Rs.10.85

5. a) Rs.29 10 Paisa b) Rs.85 40 Paisa c) Rs.13

6. a) Rs.22.70 b) Rs.61.55 c) Rs.249 35 Paisa

7. a) 50 Rupees 80 Paisa b) 30 Rupees 73 Paisa c) Rs. 119.55 Paisa

Exercise: 4.4

1. a) Rs. 50.4 b) Rs.92.3 c) Rs.3288.6 d) Rs. 2510.4
2. a) Rs. 1209 b) Rs. 1356.25
3. a) Rs. 4.19 b) Rs. 13.68 c) Rs. 292 d) Rs. 1749
4. a) Rs. 0.84 b) Rs. 1.5
h) Rs. 9.54 c) Rs. 8.12 d) Rs. 3.22
g) Rs. 15.7 b) Rs. 1710
5. a) Rs. 67.2 c) Rs. 3 d) Rs. 5.1 e) Rs. 3.15 f) Rs. 14.1
e) Rs. 18.4
c) Rs. 854 d) Rs. 3.75

Prime Mathematics Book − 4 117

Exercise: 4.5

1. a) 600 cm b) 3000 mm c) 9000 m d) 500000 cm e) 120 mm f) 500000mm

2. a) 1.6 cm b) 0.214m c) 0.1234 km d) 3.24 m e) 0.8765 km f) 0.53 m

3. a) 45 mm b) 196 mm c) 645 cm d) 1168 cm e) 2254 m f) 10672 m

4. a) 9 cm 1 mm b) 7 cm 4 mm c) 1 m 50 cm d) 14 m 75 cm e) 2 km 500 m

f) 7 km 827 meters

5. a) 10 cm 2 mm b) 20 cm 3 mm c) 85 m 10 cm d) 102 m 10 cm

e) 13 km 230 m f) 200 km 152 m g) 326 km 501 m 60 m h) 91 km 551 m 90 cm

6. a) 7 cm 7 mm b) 22 cm 9 mm c) 19 m 60 cm d) 61 m 36 cm

e) 32 km 55 m f) 32 km 55 m g) 15 m 17 cm 9 mm h) 16 m 92 cm 8 mm

i) 35 km 631 m 36 cm j) 3 km 860 m 92 cm

7. a) 40 m 22 cmb) 40 km 16 m c) 47 km 200 m 90 cm d) 68 cm

e) 15 km 634 m

8. a) 32 km 300 m b) 339 m 45 cm c) 0.08 cm d) 16 km 890 m.

Exercise: 4.6

1. a) 8000 grams b) 14000 grams c) 6000 mg d) 25000 mg e) 34.6 kg f) 7.5 kg

2. a) 21432 grams b) 64614 gm c) 2 kg 520 gm d) 5 kg 980 gm e) 3 gm 465 mg

f) 10 grams 205 mg g) 36269 mg h) 8450 mg

3. a) 38 gm 600mg b) 73 gm 100 mg c) 27kg 190 gm d) 12 kg 500mg e)60 kg

f) 16 kg 1 gm 130mg g) 13 kg 770gm 113 mg

4. a) 26 gm 820mg b) 49 gm 644 mg c) 64 kg 771gm d) 100 kg 576gm e) 63kg 900 gm 891 mg

f) 16 kg 860 gm 90 mg

5. a) 91kg 080gm b) 435gm 250 gm c) 59 kg 1gm 110gm d)771kg 411gm 100mg

6. a) 28gm 320mg b) 211 kg 840gm c) 159 kg 711 gm d) 113kg 487 gm 885mg

7. a) 40kg 260 gm b) 17 kg 245 gm c) 66kg 275 gm d) 123 kg 185 gm e) 2 kg 220gm

f) 16kg160 gm

Exercise: 4.7

1. a) 7000 ml b) 24000 ml c) 125000 ml d) 5345 ml e) 40930 ml f) 2054 ml
f) 0.1 l
2. a) 0.634 l b) 2.34 l c) 7.5 l d) 0.043 l e) 8.015 l f) 1l 200ml
f) 46l 576ml
3. a) 8l 642ml b) 1l 298ml c) 2l 300ml d) 18l 4ml e) 4l 20ml

4. a) 24l 50ml b) 70l 150ml c) 300l 580ml d) 3l 195ml e) 10l 773ml

g) 55l 152ml h) 515l 586ml i) 4l 264ml j) 100l 959ml

5. a) 161l 724ml b) 33l 244ml c) 19l 400ml d) 121l 737ml e) 5l 768ml

118 Prime Mathematics Book − 4

U5nit Mensuration
(Perimeter, Area, Volume)

Objectives Estimated periods − 10

At the end of this unit, the students will be able to:
• find the perimeter of plane figures.
• calculate the perimeters of regular plane figures like rectangles, squares, equilateral triangles

using formula.
• find the area of figure by counting the squares.
• find the area of regular figures like squares and rectangles using formula.
• find the volume of cubes and cuboids using formula.

Teaching Materials
Square papers, graph papers, cubical blocks, models of cuboids and cubes.

Activities

It is better to:
• let participate the students measure the length of school / house compounds.
• demonstrate the activities of measuring the boundaries of plane figures and find their sum

to give the concept of perimeter.
• demonstrate the counting whole units and half units of graph to clear the concept of area

and perform the activities related to area of square and rectangle.
• demonstrate counting cubical blocks and use formula v = l × b × h to clear the concept of

volume of cuboidal and cubical blocks.
• use play cards with mensuration formulas organize a pair matching game for students.

Perimeter and Area

Study and learn the following:

What will be the length of the fence to

surround the land ABCDEF ?

Total length all around the land is A 5m B
AB + BC + CD + DE + EF + FA 7m

= 5m + 7m + 10m + 5m + 15m + 12m

= 54m 12m C D
10m 5m
E
Here, what we calculated is the length F 15m
D
all around which is called perimeter. a
C
Thus, total distance around a closed figure is called perimeter.
B
∴ Perimeter = Sum of the lengths of the sides all around. b
C
For the pentagon ABCDE, AB
Perimeter (p) = AB + BC + CD + DE + EA

For the triangle ABC. A E D C
Perimeter (p) = AB + BC + AC C A

B

Perimeter of equilateral triangle ABC of side ‘a’ a C
= AB + BC + AC B
= a + a + a = 3a

∴ Perimeter of an equilateral triangle = 3 × length of side. a

ABCD be a square of side ‘a’ then, A
Perimeter = AB + BC + CD + DA a

= a + a + a + a = 4a

∴ Perimeter of square = 4 × length of a side. A Ba
Let’s consider a rectangle ABCD with length b l
AB = CD = l and breadth BC = DA = b, D
Perimeter = AB + BC + CD + DA l

=l +b+l +b
= 2l + 2b
= 2(l + b)
= 2 × sum of length and breadth.

120 Prime Mathematics Book − 4

Workedout Examples

Example 1: Calculate the perimeter of the triangle ABC given in the
figure.
Solution: Here, sides AB = 3cm, BC = 5cm and A

AC = 4cm 4cm
∴ Perimetre of ∆ABC = AB + BC + AC 3cm
= 3cm+ 5cm + 4cm 2cm
= 12cm
B 5cm C

Example 2 : Calculate the perimeter of the pentagon PQRST given in the
figure.
Solution: Here, lengths of the sides 5cm T 4cm
PQ = 3cm, QR = 6cm, RS = 2cm, P
ST = 4cm and TP = 5cm S

∴ Perimeter = PQ + QR + RS + ST + TP 3cm

= 3cm + 6cm + 2cm + 4cm + 5cm

= 20cm. Q 6cm R

Example 3: Find the perimeter of an equilateral triangle of side 4cm.
Solution : Here,

Length of side of equilateral triangle = 4cm
∴ Perimeter of equilateral triangle = 3 × length of side

= 3 × 4cm = 12cm.

Example 4: Find the perimeter of a square of side 6cm.
Solution : Here,

Length of side of square (a) = 6cm
We have,
Perimeter of a square = 4 × length of side = 4a

= 4 × 6cm
= 24cm
∴ The perimeter of the square is 24cm.

Example 5: Find the perimeter of the rectangle given in the figure.
Solution: Here, D C
B
Length of the rectangle (l) = 100cm 50cm
Breadth of the rectangle (b) = 50cm.
We have, A 100cm
Perimeter of the rectangle = 2(l + b)
= 2(100cm + 50cm)
= 2 × 150cm = 300cm
∴ The perimeter of the rectangle is 300cm.

Prime Mathematics Book − 4 121

Exercise: 5.1

1. Find the perimeter of the following figures:

(a) A (b) P 4cm Q (c) P 9cm W
3cm
3cm 5cm 5cm 3cm

B 7cm 6cm Q 3cm R U 3cm V
C S 7cm R 4cm 4cm

S 3cm T

2. Find the perimeter of the following regular figures:

(a) (b) (c)

5.4cm 4cm

5.4cm 6cm 4.5cm
24cm
3. Find the perimeter of the following:

(a) (b)

Perimeter = Sum of sides 60cm

90cm 18cm
(c)
(d)

80cm 32cm

80cm 32cm

4. Find the perimeter of the following:

(a) A rectangle of length 6cm and breadth 4cm.

(b) A rectangle of length 18cm and breadth 12cm.

(c) A rectangle of length 24.4cm and breadth 20.6cm.
(d) A square of side 6 cm.
(e) A square of side 15 cm.
(f) A square of side 8cm.

122 Prime Mathematics Book − 4

Area

Learn the following:

• Why do we say that the footprint
of an elephant is bigger than
that of a goat? It is because,
foot of an elephant occupies
more surface where as foot of a
goat occupies less surface.

• The surface occupied by an foot of an elephant foot of a goat
object or figure is called area.

• Let’s consider a square of side 1cm. We say, that the square occupies a
surface of length 1cm and breadth 1cm.

Mathematically, we write area = 1cm × 1cm 1cm
= 1cm2

= 1 square cm (sq.cm). 1cm

Three such squares occupy an area 3 × 1cm2 = 3cm2 1sq.cm 1sq.cm 1sq.cm
= 3 sq.cm.

If area of a square is 1 sq.cm, we can find the area of
given figure in the grid by counting the number of unit
squares.

Example 1: If the area of a square is 1sq.cm, find the area of the shaded
region.

Solution : Here, area of 1 square is 1sq.cm. There are 5
small shaded squares.
∴ Area of the shaded region = 5 × 1 sq.cm
= 5 sq.cm.
∴ Area of shaded region = 5sq.cm

Example 2: If the area of a small square is 1cm2, find the area of the
shaded region.
Solution : Here, area of 1 square is 1cm2
∴ area of 10 complete squares is 10cm2
1
And area of 4 half squares = 4 × 2 cm2 = 2cm2.

∴ Area of the shaded region = Area of 10 complete
squares + area of 4 half squares
= 10cm2 + 2cm2
= 12cm2
∴ The area of shaded region is 12 sq.cm.

Prime Mathematics Book − 4 123

Area of rectangular and square regions:
Let’s consider a rectangle 6cm in length
and 3cm in breadth. We can divide the
rectangle into 18 squares of side 1cm.
So area of the rectangle is 18cm2. Let’s
multiply length and breadth.

6cm × 3cm = 18cm2 In a square
Thus, we can take Length = breadth = l
Area of rectangle = length × breadth ∴ Area = length × breadth
Area of rectangle = l × b
= l × l = l2
∴ Area of square = (side)2

Note :
• length and breadth both should be of the same unit.
• If length and breadth both are in cm, area is in cm2 or sq.cm.
• If length and breadth both are in m, area is in m2 or sq.m.

Example 1: Find the area of rectangle given in the figure.

Solution : Here

Length (l) = 10cm 8cm

Breath (b) = 8cm

We have

Area of rectangle = l × b 10cm

= 10cm × 8cm

= 80cm2

∴ The area of the rectangle is 80 sq.cm.

Example 2: Find the area of the square of side 4.2m. 4.2
Solution : Length of side of square (l) = 4.2m. × 4.2

We have, 84
Area of square = l2 +1 6 8

= (4.2m)2 = 4.2m × 4.2m 17.64
= 17.64m2
∴ Area of the square is 17.64 sq.m.

124 Prime Mathematics Book − 4

Exercise: 5.2

1. Find the area of the following figures (Area of each small square = 1cm2):

( a) ( b) ( c)

( d) ( e) ( f)

2. Find the area of the shaded parts of the following figures:

( a) ( b) ( c)

( d) (d) (e)

3. Find the area of the following: 32.5cm
( a) ( b)
4cm

( c) 6cm ( d) 58cm
21cm 50cm

21cm 50cm

Prime Mathematics Book − 4 125

( e) (f)

20cm 80cm

35cm

80cm

4. Find the area of the rectangles with given length and breadth:

(a) length = 5cm, breadth = 3.5cm

(b) length = 8cm, breadth = 6cm

(c) length = 20cm, breadth = 10cm

(d) length = 125cm, breadth = 60cm

5. Find the area of the squares having following lengths:

(a) 5cm (b) 13cm (c) 34.5cm (d) 50cm

6. (a) A rectangular piece of land is 20 m long and 18 m broad. Find
the area of the land.

(b) A carpet is 800 cm long and 600 cm broad. What is the area of
the carpet ?

(c) A square tower is of side 125 cm. Find its area.

Volume

Learn the following:
Fill a trough full with water. Put a brick into it.
You will observe that some water over flows.
It is because objects occupy space. The space
occupied by water is taken by the brick.

Matters (solid/liquid/gas) occupy space.

The space occupied by an object is called volume.

126 Prime Mathematics Book − 4

Let’s consider a cube of side 1cm. It occupies 1cm
a space of length 1cm, breadth 1cm and 1cm 1cm
height 1cm. We also say, it occupies a space of
dimension 1cm × 1cm × 1cm [read as 1cm by 3cm
1cm by 1cm] 2cm
4cm
Mathematically we take
Volume (V) = 1cm × 1cm × 1cm

= 1cm3
= 1cubic cm
= 1 c.c

Let's arrange 24 cubes of volume 1cm3 as in the
figure. The volume of the cuboid so formed is
24 × 1cm3 = 24cm3.

Multiply length (l), breadth (b) and height (h)
l×b×h
= 4cm × 2cm × 3cm
= 24cm3

∴ Volume of cuboid = l × b × h

Example 1: Each cube has volume of 1cu.cm. Find the volume of the
cuboid formed by the cubes by counting the number of
cubes.

Solution : Here number cubes = 24

Volume of a cube = 1cu.cm

∴ Volume of the cuboid = 24 × 1cu.cm

= 24cu.cm.

Example 2: If length (l) = 5cm, breadth (b) = 3cm and height (h) = 2cm,
Find the volume of the cuboid using formula.

Solution : Here, l = 5cm, b = 3cm, h = 2cm.

Volume = l × b × h

= 5cm × 3cm × 2cm 5 × 3 = 15 2cm
= 30cm3 15 × 2 = 30
∴ Volume is 30cu.cm. 5cm 3cm

Note : Length (l), breadth (b) and height (h) must be in same unit.
If l = b = h,
Volume = (l)3

Prime Mathematics Book − 4 127

Example 3: Find the volume of a cube of side 3cm.

Solution : Here,

Length of side (l) = 3cm 3×3×3 3cm
Volume = (l)3 =9×3 3cm 3cm
= 27
= (3cm)3
= 27cm3

∴ Volume of the cube is 27 cu.cm.

Exercise: 5.3
1. If the volume of each cube is 1cm3, find the volume of the following
figures by counting the number of cubes:

( a) ( b) ( c)

( d) (e )

2. Calculate the volume of the cuboid of the following sides using
formula:
(a) l = 2cm, b = 3cm, h = 1cm
(b) l = 4cm, b = 3cm, h = 2cm
(c) l = 5cm, b = 4cm, h = 2cm
(d) l = 6cm, b = 5cm, h = 1.5cm
(e) l = 10cm, b = 8.2cm, h = 5cm
(f) l = b = 6.5cm, h = 12cm
(g) l = 60cm, b = h = 45cm
(h) l = 12.4cm, b = 10.5cm, h = 6.5cm

3. Find the volume of the cubes with side:

(a) 2cm (b) 6cm (c) 10cm (d) 12cm
(h) 6.2cm
(e) 1.2cm (f) 3.5cm (g) 8.1cm

128 Prime Mathematics Book − 4

4. Find the volume of the following: 4cm
( a) ( b)

2cm

3cm4 cm 4cm 4cm
2cm
( c) ( d)

2cm 8cm 20cm
15cm 4cm 8cm

( f) 15cm

( e) DUS9TEcRm 6cm 5cm
5cm

5. Solve the following word problems: 9cm

(a) A cuboidal box is 50cm long, 40cm broad and 30cm high. Find its volume.

(b) The dimension of a geometry box is 16cm × 7.5cm × 1.5cm. Find the
volume.

(c) Applied Mathematics book IV is 34cm long, 18cm broad and 1cm thick.
How much space does it occupy?

(d) Find the volume of a cubical box of side 16cm.

Unit Revision Test

1. Find the perimeter of the following:

(a) A (b) A3cm F
5 cm 3cm
5cm D

E

B 8 cm C B 4cm
C
2. Find the perimeter of the following:

(a) (b)

2.5cm

3 cm 7.2cm

Prime Mathematics Book − 4 129

3. Find the area of the following:
(a) A rectangle of length 6 cm and breadth 3.5 cm
(b) A square of side 4.2 cm.

4. Define volume.
5. Find the volume of a cuboid in which l = 6cm, b = 4cm and h = 2.5cm.
6. Find the volume of the following:

(a) (b) 1.5cm
24cm
5.8cm 18cm

5.8cm 5.8cm

Answers:

Exercise: 5.1

1. a) 15 cm b) 22 cm c) 32 cm 2. a) 21.6 cm b) 20 cm c) 27 cm
3. a) 300 cm b) 84 cm
4. a) 20 cm b) 60 cm c) 320 cm d) 128 cm

c) 90 cm d) 24 cme) 60 cm f) 32 cm

Exercise: 5.2

1. a) 12 cm2 b) 10 cm2 c) 11 cm2 d) 13 cm2 e) 6 cm2 f) 9 cm2
2. a) 4 cm2
3. a) 24 cm2 b) 8 cm2 c) 8 cm2 d) 8 cm2 e) 18 cm2 f) 10 cm2
4. a) 17.5 cm2
5. a) 25 cm2 b) 1885 cm2 c) 441 cm2 d) 2500 cm2 e) 700 cm2 f) 6400 cm2
6. a) 360 cm2
b) 48 cm2 c) 200 cm2 d) 7500 cm2

b) 169 cm2 c) 1190.25 cm2 d) 2500 cm2

b) 480000 cm2 c) 15625 cm2

Exercise: 5.3

1. a) 4cm3 b) 5cm3 c) 4cm3 d) 12cm3 e) 14cm3

2. a) 6 cm3 b) 24 cm3 c) 40 cm3 d) 45 cm3 e) 410 cm3

f) 507 cm3 g) 121500 cm3 h) 846.3 cm3

3. a) 8 cm3 b) 216 cm3 c) 1000 cm3 d) 1728 cm3 e) 1.728 cm3

f) 42.875 cm3 g) 531.441 cm3 h)238.328 cm3

4. a) 12 cm3 b) 64 cm3 c) 240 cm3 d) 1200 cm3

e) 270 cm3 f) 360 cm3

5. a) 60000cm3 b) 180 cm3 c) 612cm3 d) 4096 cm3

130 Prime Mathematics Book − 4

U6nit Fraction, Decimal, Percentage
and Unitary Method

Objectives Estimated periods − 32

At the end of this unit, the students will be able to:
• know proper, improper, mixed fractions, like and unlike fractions.
• compare unlike fractions.
• add and subtract like and unlike fractions.
• express fractions into decimals and vice versa.
• add, subtract, multiply and divide the fractions.
• write the fraction and decimals into percentage and vice versa.
• find the value of 1 unit when value of big quantities are given and find the value of

big quantity when value of 1 unit is given.

Teaching Materials
Fraction models, fractional chart, graph paper, scale, etc.

Activities
It is better to:
• illustrate with diagram comparison, addition and subtraction of like and unlike
fractions.
• demonstrate the activities of converting fractions into decimals and decimals into
fractions, and their operations.
• explain the concept of percentage illustrating it with fractions and decimals.
• demonstrate the activities to classify the process of multiplication and division to find
the value of big quantity and small quantity in unitary method.
• prepare a quiz game about unitary method, decimal & percentage.
• divide the charts into three or four teams and play the game

Fraction
Study and learn the following:
• We often face problems of dividing any thing into equal parts in our daily

life.
• A rectangular piece of land is to be divided equally among 5 brothers.

How much will each get?
Each will get 1 part out of 5 parts.
We show it in figure as:

Mathematically we write, each gets 1 part of the whole land.
5

• A rectangle is divided into two equal parts.
The shaded part is ½ of the whole and unshaded part

• Aalscoirc12leofisthdeivwidheodleinptaor4t.equal parts.

Its one part is ¼ of the whole.
Its two parts is 2/4 of the whole.
Its three parts is ¾ of the whole.

A fraction is a quantity which expresses a part of the whole.
1
By fraction 5 we mean, a thing is divided into 5 equal parts and considered 1

part. We say one fifth of the whole.

Mathematically, a whole or single is written as 1.

∴ 1 of a whole = 1 of 1 = 1
2 2 2

Instead of writing 1 1
2 2
Half of whole + half of a whole = a whole or, of whole + of a whole = 1 whole,
1
We just write 1 + 1 =
2 2

Anything itself is a whole (single)

The shaded figure is 1, a whole

In comparison to the whole figure, the shaded part

is 1 of the whole.
4

132 Prime Mathematics Book − 4

In a fraction, the lower number, which shows into how many equal parts the
unit is divided, is called the denominator. The upper number, which shows
how many of these equal parts are taken to form a fraction is called the
numerator.

Types of fractions:
1. Proper fraction: A proper fraction is the fraction whose numerator is
less than its denominator.
3
e.g. 4 is a proper fraction since numerator 3 < denominator 4.

2. Improper fraction: An improper fraction is the fraction whose numera-
tor is equal or greater than its denominator.
e.g.

(a) 4 is an improper fraction, since numerator 4 > denominator 3.
3

(b) 4 is an improper fraction, since numerator 4 = denominator 4.
4

23, 34, 54, 57, ………. etc are examples of improper fraction.

3. Mixed fraction: A mixed fraction is a quantity consisting an integer and
141
a proper fraction. E.g. is a mixed fraction where 1 is an integer (z)

and 1 is a fraction.
4

141 is the combination of a whole 1 and a proper fraction 14.

• 114 can also be written as 1 + 1
4

• Mixed fraction can be expressed as improper fraction and im-
proper fraction can be expressed as mixed fraction.

Converting improper fraction into mixed fraction.

Example 1: To convert 5 into mixed fraction.
4
5 1
Solution: The fraction 4 means 5 times 4 .

Prime Mathematics Book − 4 133

Here, 5 times 1 = 1 whole and one 1
1 whole 4 114 4
5 = and one 1 =
∴ 4 4

Division method: 4)5(1
We know −4
1
5 = 5 ÷ 4
4

Where, quotient = 1 remainder
remainder = 1 divisor
divisor = 4

∴ 5 = 141 quotient
4

Converting mixed fraction into improper fraction.

Example 2: HToerceo,n1v14ermte1a14nsin1towihmolperoapnedrfrfaracctitoionn14..
Solution:

Fraction 1 means 1 part out of 4 parts.
4
4 such parts make a whole.
Thus, there will be 5 such parts.

= 5 times 1 = 5
4 4

(1 whole consists 4 times 14)

Short method: Process
Solution: 114 • Multiply whole number by
denominator and add the
= 1 + 1 numerator.
4
• Keep the result as numerator
= 4 × 1 + 1 and the denominator as it is.
4
4 + 1 1 5
= 4 1 4 = 4

= 5
4

134 Prime Mathematics Book − 4

Exercise: 6.1

1. Write the shaded parts of the given figures in fractions:
(a) (b) (c)

(d) (e) (f)

2. Shade the number of parts according to the given fractions:

(a) 1 (b) 3 (c) 7
4 8 10

(d) 3 (e) 2 (f) 3
4 3 8

3. Write fractions of the following:
(a) one half (b) two third (c) one fifth
(d) five sixth (e) three eighth (f) four tenth
(g) two seventh (h) seven twentieth. (i) five twelfth

4. Identify whether the following are proper fraction, improper
fraction or mixed fraction:
7 3 1 3 3
(a) 5 (b) 4 (c) 2 (d) 2 (e) 2 5

(f) 1 2 (g) 4 (h) 5 (i) 5 3 (j) 9
3 5 4 5 10

5. Convert the following improper fractions into mixed fraction:
7 3 5 13 15
(a) 5 (b) 2 (c) 3 (d) 6 (e) 7

(f) 8 (g) 9 (h) 6 (i) 12 (j) 16
5 4 4 5 9

6. Convert the following mixed fraction into improper fractions:
2 1 3 4 1
(a) 3 4 (b) 2 3 (c) 4 5 (d) 1 5 (e) 5 2

(f) 2 3 (g) 1 5 (h) 3 4 (i) 6 1 (j) 2 1
7 8 5 2 7

Prime Mathematics Book − 4 135

Equivalent fractions:

Let’s take 5 equal rectangles and divide:

(i) First rectangle into two equal parts and = 1
1 2
shade one of them which is 2 of whole.

(ii) Second rectangle into 4 equal parts and 2
2 4
shade two of them which is 4 of whole. =

(iii) Third rectangle into 6 equal parts and
3
shade three of them which is 6 of whole = 3
6
(iv) The fourth rectangle into 8 equal parts and
4
shade four of them which is 8 of whole.

(v) The fifth rectangle into 10 equal parts and = 4
5 8
shade five of them which is 10 of whole.

= 5
10

In the given figures, size of the shaded

parts are same whether they are

parted or not 1 2 3 4 Equivalent
2 4 6 8
∴ The fractions = = = = 150.

Such fractions are called equivalent

fractions.

In case of equivalent fractions, size of
whole and considered parts are same.
We have seen from above example that
1 2 3 4
2 = 4 = 6 = 8 = 5 etc.
10

We get, 2 = 1 × 2 , 3 = 1 × 33, 4 = 1 × 4
4 2 × 2 6 2 × 8 2 × 4

5 = 1×5
10 2×5

Also, we get 2 ÷ 2 = 21, 3 ÷ 3 = 12, 4 ÷ 4 = 12, 5÷5 = 1
4 ÷ 2 6 ÷ 3 8 ÷ 4 10 ÷ 5 2

We obtain equivalent fraction by multiplying or dividing the numerator and
denominator of the given fraction.

136 Prime Mathematics Book − 4

Workoedut Examples

Example 1: WWeritken ofiwv,e equivalent fractions of 32 .
Solution : 2 2 × 2 64, 2 2 × 3 96, 2 2 × 4 8
3 3 × 2 3 3 × 3 3 3 × 4 12
= = = = = =

2 = 2 × 5 = 1105, 2 = 2 × 6 = 12
3 3 × 5 3 3 × 6 18

Therefore, 46, 69, 182, 1105, 8 are equivalent fractions of 32.
12
12
Example 2: Write 3 equivalent fractions of 24 .
Solution : We know,
12 12 ÷ 2 12 12 ÷ 3
24 = 24 ÷ 2 = 6/12, 24 = 24 ÷ 3 = 4/8

12 = 12 ÷ 4 = 3/6
24 24 ÷ 4

Therefore, 162, 4 and 3 are the three equivalent fractions of 2142.
8 6

A fraction in its lowest term/order.

Consider a fraction 12
18
We have already learnt that by dividing the numerator and denominator of a
fraction by the same number, we get the equivalent fraction of lower order.

∴ 12 = 12 ÷ 2 = 6 , further 6 = 6 ÷ 3 = 2
18 18 ÷ 2 9 9 9 ÷ 3 3

Or, 12 = 12 ÷ 3 = 4 , further 4 = 4 ÷ 2 = 2
18 18 ÷ 3 6 6 6 ÷ 2 3

Or, 12 = 12 ÷ 6 = 2
18 18 ÷ 6 3

The fraction 2 is of the lowest order of 12 or 6 or 4
3 18 9 6
Converting a fraction to its lowest term:

Example 1: Reduce 18 to its lowest term.
30
Solution : Process 1:
• 18 and 30 both are divisible
18 = 18 ÷ 2 = 9 by 2. So, divide numerator
30 30 ÷ 2 15 and denominator by 2

Further 9 = 9÷3 = 3 • 9 and 15 both are divisible
15 15 ÷ 3 5 by 3. So, divide numerator
and denominator by 3.
∴ 18 in lowest term = 3
30 5 Prime Mathematics Book − 4 137

Directly, we obtained 18 = 18 ÷ 6 = 3
30 30 ÷ 6 5

This process is also called cancellation process.

18 9 3 = 3
3015 5 5

by 2

by 3

Process 2: Find the prime factors of 18 and 30
18 2 × 3 × 3 3 and cancel the common factors.
30 = 2 × 3 × 5 = 5

Example 2: Are 2 and 12 equivalent fractions?
5 30
Solution : Process I:
2 12
5 is in its lowest term. Converting 30 to the lowest term.

∴ 12 = 2×2×3 = 2 12 = 2 × 2 × 3
30 2×3×5 5 30 = 2 × 3 × 5

∴ 2 and 12 are equivalent fractions.
5 30

Process II:
2 = 1320,
If 5

2 × 30 = 5 × 12

Or, 60 = 60 which is true.

∴ 2 and 12 are equivalent fractions.
5 30

Like and unlike fractions
In each of the given figures, the
total number of parts is 4, only the
parts considered are different. Such
1 2 3
fractions are like fractions. 4 , 4 , 4 1 2 3
etc are like fractions.
A fraction having same denominator 4 4 4

are called like fraction.

138 Prime Mathematics Book − 4

In the figure given along side, unit or
whole are same but total number of
parts are not same, such fractions are
3 3 5
unlike fractions. 4 , 5 , 8 are unlike 3 3 5
fractions. 4 5 8

Note: (i) (ii)
1
For like fractions (i) unit (whole) should be same; (ii) Total number 4

parts should be same i.e. size of each part should be same. In the

given figure, total number of parts is same. But size of each part

is not same. In other words, units (wholes) are not same. Thus, 1
1 1 4
pictorially, 4 of a circle and 4 of a square are not like fractions.

Conversion of unlike fractions to like fractions:

11 1 = 5 1 = 4
45 4 20 5 20

1 = 1 × 5 = 5 Process:
4 4 × 5 20 Product of denominators 4 and 5 is 20.
Multiply numerator and denominator
1 = 1 × 4 = 4 by the number so as to get
5 5 × 4 20 denominators equal to 20.

∴ 5 and 4 are like fractions
20 20

Note: In the above example

• Units are same (Same circle) 1
20
• Dividing the circle into 20 equal sectors so that the small parts are equal in size, . In the first
circle the shaded part is 5 out of 20. In the second, the shaded part is 4 out of 20.

Example 1: Convert 2 and 3 into like fractions.
3 4
Solution : Product of 3 and 4 is 12.

Here, 2 = 2 × 4 = 8
3 3 × 4 12
3 2 × 3 9
4 = 4 × 3 = 12

Therefore, 8 and 9 are like fractions.
12 12

Prime Mathematics Book − 4 139

Comparison of fractions:
A. Comparison of like fractions: Two equal
rectangles are drawn, and divided each into 5 2
equal parts. In the first two parts are shaded and 5

in the second 3 parts are shaded. Obviously 3 3
parts out of five parts is greater than 2 parts out 5
of 5 parts. Thus, like fractions can be compared
by comparing numerators.

Example 2: Compare 2 and 3 .
5 5
2 3
Solution : The fractions 5 and 5 are like fractions (having same

denominators) and as numerator 2 < numerator 3.

∴ 2 < 3
5 5

B. Comparison of unlike fractions: In 2 of whole
case of unlike fractions, first they 3 thing
are converted into like fractions. 3 of the same
Then comparing numerators, we 5 thing
can compare the fractions.

Example 3: Compare the fractions 2 and 3 .
3 5
2 2 × 5 10
Solution : Here, 3 = 3 × 5 = 15

3 = 3 × 3 = 9
5 5 × 3 15

Since, 10 > 9

∴ 10 > 9 i.e. 2 > 3
15 15 3 5

Alternative Method:
As a short method, we compare the two given fractions by comparing the
product of numerator of first and denominator of second and numerator of
second and denominator of first fraction.
2 3
Example 1: Compare 3 and 4 .

Solution: Here, 2 3
3 4

2×4<3×3
8<9

∴ 2 < 3
3 4

140 Prime Mathematics Book − 4

Exercise: 6.2

1. Write any two equivalent fractions of the following:
3 2 1 4
(a) 4 (b) 5 (c) 7 (d) 7

2. Write any three equivalent fractions of the following:
5 6 7 8
(a) 9 (b) 11 (c) 10 (d) 15

3. Reduce the following fractions to their lowest terms:
2 3 4 4 6
(a) 8 (b) 9 (c) 6 (d) 20 e) 8

(f) 5 (g) 20 (h) 8 (i) 8 (j) 9
15 40 20 30 21

4. Determine whether the given pairs of fractions are equivalent or not:
1 4 2 6 3 1
(a) 2 and 8 (b) 5 and 15 (c) 4 and 3

(d) 2 and 3 (e) 3 and 12 (f) 2 and 14
7 8 5 30 5 35

5. Identify whether the following pairs of fractions are like or unlike:
5 4 9 10 4 3
(a) 9 and 9 (b) 12 and 12 (c) 8 and 7

(d) 15 and 13 (e) 1 and 2
14 12 3 3

6. Convert the following unlike fractions to like fractions:
5 3 2 1 3 2 2 3
(a) 9 and 5 (b) 5 and 3 (c) 4 and 3 (d) 5 and 10

(e) 5 and 4 (f) 5 and 3 (g) 3 and 4 (h) 7 and 8
6 9 12 8 4 5 8 9

7. Compare the following fractions:
2 1 3 5 2 4 3 1 2 3
(a) 3 and 3 (b) 7 and 7 (c) 9 and 9 (d) 4 and 3 (e) 5 and 4

(f) 5 and 4 (g) 5 and 4 (h) 10 and 2 (i) 6 and 12 (j) 5 and 7
8 5 15 12 12 3 13 26 18 36

Prime Mathematics Book − 4 141

Fundamental Operations on Fractions

Addition of like fractions:
Learn the following:

It follows that
2 1 2 + 1 3
5 + 5 = 5 = 5

2 + 1 ∴ 2 + 1 = 3
5 5 5 5 5
3
= 5

For adding like fractions, numerators are added keeping denominator
common.

Example 1: Add the fractions 1 and 2 .
4 4
1 2
Solution: 4 and 4 are like fractions.

∴ 1 + 2
4 4

= 1 + 2
4

= 3
4

Addition of unlike fractions: 1
1 3
Let’s see the addition of 2 and

= 3 =
1 6 12
2 36

+= + =
32 5
11 66 6
23

∴ 1 + 1 = 3 + 2 = 5
2 3 6 6 6

142 Prime Mathematics Book − 4

Thus, for the addition of unlike fractions, the fractions are converted to like
fractions and the add the numerators keeping the denominator same.

Example: Perform the addition 1 + 2 .
4 3
Solution : Product of 4 and 3 is 12. We can also proceed like this
3 1 2
∴ 1 = 1 × 3 = 12 4 + 3
4 4 × 3
1 × 3 2 × 4
And 2 = 2 × 4 = 8 = 4 × 3 + 3 × 4
3 3 × 4 12
3 8
Now, 1 + 2 = 12 + 12
4 3
3+8
= 3 + 8 = 12
12 12
11
= 3+8 = 12
13

= 11
12

Addition of fractions involved with mixed numbers:
If the addition of fractions is involved with mixed numbers, the mixed
numbers are to be converted to improper fractions and further if fractions
are unlike they are to be converted to like fractions and added with usual
addition process.

Example: Add 1 and 1 1 .
3 2

Solution: 1 + 112
3

= 1 + 3 [∴ 121 = 2 + 1 = 23]
3 2 2

= 1 × 2 + 3 × 3 [∴ Product of 3 and 2 is 6]
3 2 2 3

= 2 + 9
6 6

= 2 + 9
6

= 11
6

= 156 [11 ÷ 6, quotient is 1 and remainder 5]

Prime Mathematics Book − 4 143

Example: Perform the addition 3 + 34.
Solution:
3 + 4 All the whole numbers are 1
3 fractions with denominator

= 3 + 4
1 3

= 3 × 3 + 4
1 × 3 3

= 9 + 4
3 3

= 9 + 4
3
3)13(4
= 13 −12
3 1

= 4 1
3

Exercise: 6.3

1. Perform the following additions: 5 3
2 1 1 3 12 12
(a) 4 + 4 (b) 5 + 5 (c) +

(d) 1 + 2 + 3 (e) 4 + 5 + 6 (f) 5 + 7 + 3
7 7 7 21 21 21 18 18 18

2. Add the following: 1 1 9
1 2 1 10 3 8 1 16
(a) 6 and 3 (b) 5 and (c) 4 and (d) 4 +

(e) 7 and 9 (f) 1 and 2 (g) 3 and 1 (h) 21 , 3 and 1
12 24 2 6 4 12 4 8

3. Perform the following additions: 3
1 2 2 1 3 5 1 2 7 5
(a) 2 + 3 (b) 3 + 5 (c) 5 + 6 (d) 4 + 5 (e) 8 +

(f) 2 + 3 (g) 1 + 1 (h) 3 + 2 (i) 1 + 2 (j) 5 + 2
7 8 3 4 4 3 10 3 8 3

4. Perform the following additions: 2 4
2 1 3 3 5
(a) 2 + 3 (b) 5 + 4 (c) 3 + (d) 2 + 4 (e) 4 +

(f) 2 1 + 2 (g) 3 1 + 4 1 (h) 4 1 + 2 2 (i) 1 1 + 2 3
3 3 2 4 3 3 2 4

144 Prime Mathematics Book − 4

Subtraction of fractions: 3 3
6 7 7
From the figure, it is clear that 7 – =

It follows that 6
6 – 3 7
6 – 3 = 7 = 3 3
7 7 7 7
The process of subtraction of fractions is
same as in the case of addition.

Workedout Examples 6 3
2 4 7 7 3
Example 1: Subtract 5 from 5 . – = 7

Solution : 4 – 2
5 5
4 - 2
= 5 [ 4 and 2 being like fractions]
5 5
= 2
5
2 3
Example 2: Perform the subtraction 3 – 5 .
2 3
Solution : 3 – 5

= 2 × 5 – 3 × 3 [Product of 3 and 5 is 15]
3 × 5 5 3

= 10 – 9
15 15

= 10 – 9
15

= 1
15
Example 3: Subtract 452 from 732.
732 – 452 3 × 7 + 2 5 × 4 + 2
Solution: = 3 – 5

= 21 + 2 – 20 + 2 = 23 – 22
3 5 3 5

= 23 × 5 – 22 × 3 = 115 – 66
3×5 5×3 15 15

= 115 – 66 = 49
15 15

= 3145

Prime Mathematics Book − 4 145