Prime Mathematics 5

Lowest Common Multiple (L.C.M.)

Let's take two numbers 6 and 9.
Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, …..
Multiples of 9 = 9, 18, 27, 36, 45, 54, 63, ……
Some common multiples of 6 and 9 are 18, 54, …...
Among the common multiples 18, 54, ….. and so on, 18 is the lowest common
multiple. So, 18 is the L.C.M. of 6 and 9.

To find L.C.M. using method of listing the multiples
Example: Find the L.C.M. of 8 and 12.

Solution: The multiples of 8 = 8, 16, 24, 32, ….
The multiples of 12 = 12, 24, 36, ….
The L.C.M. of 8 and 12 is 24.

To find the L.C.M. by prime factorisation method.
Example: Find the L.C.M. of 12 and 18.

Solution: Let's find prime factors of 12 and 18.

Here, 2 12 2 18
26 39

3 3

∴ 12 = 2×2×3

∴ 18 = 2×3×3

L.C.M. = Product of the common factors and remaining factors.

= 2×3×2×3
= 36.

46 Prime Mathematics Book - 5

To find the L.C.M. by division method.

This is most convenient method to find the L.C.M. of given numbers. First the
given numbers are divided by prime numbers. While dividing at least two
numbers must be divisible by the prime number. The process of division is
repeated until there are two numbers exactly divisible by a prime number.

Example: Find the L.C.M. of 30, 40, and 50.

Solution: 2 30, 40, 50 [All these numbers are even numbers so, they are
5 15, 20, 25 divided by 2.]

[All of them are divisible by 5.]

3, 4, 5 [No two numbers are exactly divisible by a number]

L.C.M. = 2 × 5 × 3 × 4 × 5 [L.C.M of two prime numbers is the product of

= 600 given two prime numbers]

Exercise: 3.9

1. Find the L.C.M. of the following numbers by the method of listing
their multiples:

a. 5 and 6 b. 10 and 12

c. 9 and 12 d. 8 and 12

e. 15 and 20 f. 20 and 30

g. 16 and 20 h. 18 and 24

i. 12 and 16 j. 20 and 25.

2. Find the L.C.M. of the following numbers by the method of prime
factorisation:

a. 24 and 48 b. 12 and 16

c. 40and 50 d. 60 and 80

e. 15, 20 and 30 f. 12, 16 and 24

g. 36, 40 and 48 h. 30, 45 and 60

i. 60, 80 and 90 j. 100, 150 and 200

3. Find the L.C.M. of the following numbers by the method of division:

a. 50 and 70 b. 30 and 35

c. 20, 30 and 50 d. 25, 40, and 60

e. 100, 125 and 150 f. 30, 45 and 60

g. 48, 72 and 80 h. 160, 170 and 200

i. 72, 90 and 120 j. 60, 70 and 80

Prime Mathematics Book - 5 47

Unit Revision Test

1. Write the greatest and smallest numbers formed by the following
digits. Also write them in local place value system and international
place value system in words:
a. 1,8,5,7,9,0,2. b. 1,3,4,6,7,8,9,0

2. Write the given numbers in numerals. Also in local place value system
and international place value system in words:
a. Greatest number of nine digits. b. Smallest number of ten digits.

3. Write the first five prime numbers.
4. Find the factors of 450.
5. Find the first five multiples of 12.
6. Find the prime factors of 180.
7. Write two numbers which are exactly divisible by 11.
8. What is the square of 25?
9. What is the square root of 324?
10. What is the cube of 11?
11. What is the cube root of 9261?
12. Find the H.C.F. of 48,60 and 92.
13. Find the L.C.M of 72,80 and 100.
14. What is the H.C.F of any two prime number.
15. What is the L.C.M of any two prime number.

Answers:

Exercise: 3.1 to Exercise: 3.5 [Show to your teacher]
Exercise: 3.6
1. a) 81 b) 144 c) 196 d) 625 e) 1156 f) 2304
g) 225 h) 121 i) 5625 j) 10000 k) 46225 l) 164025
2. a) 8 b) 12 c) 14 d) 11 e) 10 f) 9
g) 18 h) 15 i) 6 j) 20 k) 30 l) 21
3. 64 4. 24 5. 256

Exercise: 3.7
1. a) 216 b) 729 c) 1000 d) 1728 e) 3375 f) 5832
g) 13824 h) 17576 i) 27000 j) 42875 k) 373248 l) 1000000
2. a) 12 b) 5 c) 7 d) 10 e) 11 f) 6
g) 9 h) 16 i) 18 j) 14 k) 21 l) 17

b) 3 Exercise: 3.8 e) 4 f) 10
1. a) 3 h) 36 c) 5 d) 3 k) 10 l) 20
g) 15 b) 6 i) 25 j) 5 e) 10 f) 14
h) 15 c) 21 d) 18 k) 25 l) 64
2. a) 15 i) 25 j) 12
g) 32 e) 60 f) 60
e) 60 f) 48
1. a) 30 Exercise: 3.9 e) 1500 f) 180
g) 80 b) 60 c) 36 d) 24
h) 72 i) 48 j) 100
2 a) 48 b) 48 c) 200 d) 240
g) 720 h) 180 i) 180 j) 600
b) 210 c) 300 d) 600
3. a) 350 h) 13600 i) 360 j) 1680
g) 720

48 Prime Mathematics Book - 5

Un4it Fundamental Operations

Estimated periods − 27

Objectives
At the end of this unit, the students will be able to:
• perform the operations of addition, subtraction, multiplication and division.
• make associations between addition and subtraction as well as with multiplication
and division.
• perform the rules of (+) and minus ( - ) signs in the four basic operations.
• perform the operations of numbers according to DMAS rule.
• simplify the problems on the basic of BODMAS rule.

Teaching Materials
Place value chart, number cards, charts of showing the order of operations etc.

Activities
It is better to:
• practice the problems of addition, subtraction, multiplication and division on the basis of
previous knowledge.
• establish the relation between addition and subtraction, multiplication and division.
• discuss the rules of plus (+ ) and minus ( - ) signs in the basic operations.
• discuss orders of operation i.e. DMAS rule and BODMAS rule of simplification.
• let the students write mathematical expression of given problems and simplify it through
the discussion.
• organizae a mathematical race
• make a place value chart

Prime Mathematics Book - 5 49

Fundamental operation

Historical fact

Egyptians Multiplied numbers by a process called duplation and mediation.
e.g. to multiply 42 × 33.

42 33 Process
21 66*
10 132
5 264* • Successively half 42 and double 33
2 528
1 1056* • In the doubling column, add those multiples of
= 1386 33 corresponding to odd numbers in the halving
column.

or,

33 42* Process
16 84
8 168 • Successively half 33 and double 42

4 336 • In the doubling column, add those multiples
2 672 of 42 corresponding to odd num bers in the
1 1344* halving column.
= 1386

The four fundamental operations between the given numbers are addition,
subtraction, multiplication and division for which, we use the signs “+”, “-”,
“×” and “÷” respectively.
Addition
Addition is one operation among four fundamental operations which is used
to get a new number by adding the given numbers. We use “+” sign between
two numbers to be added. In the addition operation, we should add the digits
of respective places with carryover if necessary.
We can add the given number in two ways i.e. vertical and horizontal
arrangement method. But, in both the arrangement, the digits at the same
place value are added only.

Example 1. Add : 245694 and 59743

Solution: Vertical arrangement method:

111 1

245 6 94
+ 59 7 43

305 4 37

Horizontal arrangement method

1111

245694+059743=305437

50 Prime Mathematics Book - 5

Example 2: Ram had Rs. 25043. His friend Rohit gave him Rs. 17943.
Now how much money does he have ?

Solution: Ram had 1

His friend gave : 25043
: + 1 7 9 4 3
Total money with him : 4 2 9 8 6

∴ Now he has Rs. 42986.

1. Add the following: Exercise 4.1 c) 8 3 7 6 8 3 2
a) 1 5 4 0 6 643256
+4387 b) 2 6 5 7 6 3
+478446 +4529634
d) 4 9 7 6 2
4657 e) 7 3 2 4 8
6837
+36372
+53682

2. Add the following: b) 326546 + 3297 + 46985
a) 34365 + 6945 d) 40 + 45278 + 876549
c) 436456 + 8763 + 7469457

3. Solve the following:
a. Find the sum of the greatest and smallest numbers formed by using
the digits 1,4,6,3,5 and 8 at once .
b. Find the sum of the greatest and smallest numbers of 7 digits.
c. The number of vehicles in three towns are 32605 , 45637 and
34728. Find the total number of vehicles of the towns.
d. In the given pictures, the price of three buildings are given. What
is the total cost of three buildings ?

Rs. 9540000 Rs. 7290600 Rs. 6470900

Prime Mathematics Book - 5 51

Historical fact

John Heinrich Rahn (1622 - 1676, Switzerland) used ÷ for division which was for a long
used in Europe to indicate subtraction.

Subtraction
Subtraction is one operation among four fundamental operations which is
used to get a new number by subtracting a number from another number.
We use “-” sign for subtraction. In the operation of subtraction, we should
subtract the digits at the same places.

In the subtraction operation, while
subtracting a greater digit form a smaller one,
we borrow from the next higher place and add

it to a smaller digit. Then we subtract.

We can subtract a number from another number in two ways i.e. vertical
and horizontal arrangement method. In both methods, the digits of the same
place value are subtracted.

Example 1: Subtract 53746 from 78329.

Solution: Vertical arrangement method:

12 7 87 3122 212 9 - 5 3 7 4 6 = 2 4 5 8 3
7 2 12

78329
-53746

24583 Horizontal arrangement method:

Example 2: The sum of two numbers is 85726. One of them is 32943.
Find the other number.

Solution: We find the other number by subtracting given number from the
sum of two numbers.

16
4 6 12

85726
-32943

52783
∴ The another number is 52783.

52 Prime Mathematics Book - 5

Exercise 4.2

1. Subtract the following:

a) 5 3 2 0 1 5 b) 7250432 c) 8 3 0 6 0 2 5 0 4
-364526436
-261521 -5627346

2. Subtract the following:

a) 630257 - 476539 b) 204002 - 17865

c) 6203415 - 3524789 d) 70020325 - 45342968

3. Solve the following:
a. Find the difference of the greatest and the smallest number of 7
digits.
b. Find the difference of the greatest and smallest numbers formed
by using the digits 2,5,0,6,8 and 4 at once.
c. A shopkeeper bought a radio for Rs. 6075 and a watch for Rs. 3590. By
how much is the price of the radio more expensive than the watch ?
d. The sum of two numbers is 73584. One of them is 47658. Find the
other number.
e. What should be added to 473659 to get 621057 ?
f. What should be subtracted from 725432 go get 562376.

4. Solve the following :
a. Subtract 590856 from the sum of 256489 and 473257.
b. A man bought a table for Rs. 5725 and a chair for Rs.2365. He sold
both the items for Rs. 9200. How much money did he gain ? Find it.
c. A man bought three buses for Rs. 9694820. One bus costs Rs. 3527500.
Another bus cost Rs. 3029870. Find the cost of the third bus.
d. The population of a town is 526729. If the male population and
female population of the town are 210568 and 265289 respectively,
excluding children, find the number of children in the town.

Prime Mathematics Book - 5 53

Historical fact

William Oughtred (1574 - 1600, England) used the cross (×) or dot (.) for multiplication.
Leibniz used cap (∩) for multiplication.

Multiplication

Multiplication is one operation among four fundamental operations which is
used to get a new number by multiplying the given numbers. We use “×” sign
for multiplication operation. The new number so obtained by multiplication
is called the product of the given numbers.

For examples: The given two numbers are 236 and 8.

The product of 236 and 8 = 236×8= 1888.

Let’s recall the following:

i) 12567 When a number is
×1 multiplied by 1, the product

is the number itself.

12567 When a number is
multiplied by 0, the
ii) 2 5 6 7
×0 product is 0.

0000 When a number is
multiplied by 10, the
iii) 2 7 5 6 5 product is the number itself
×10
with one 0 more.
275650
iv) 3 6 0 0 The product of 3600
and 400 is the product of 36
×400 and 4 with 0000 at the end of
1440000
the product.

Example 1: Find the product of 3465 and 24.

Solution: 3465 Simple way
3465
×24 20 + 4 ×24
13860 3465 × 4
13860
+69300 3465 × 20 +69300
83160 3465 × 24
83160

54 Prime Mathematics Book - 5

Example 2: Hari Shankar earns Rs. 17520 in a month. How much money
does he earn in a year ?

Solution: We know that, 1 year = 12 months

17520
×12

35040
+175200

210240

∴ He earns Rs. 210240 in a year.

Exercise 4.3

1. Multiply the following: c) 2 4 5 6 0 2 d) 7 6 9 3 8 7
a) 3 4 6 9 b) 5 8 9 4 2 ×486

×32 ×246 ×520

2. Multiply : b) 63500 × 300 c) 14000 × 5000
a) 17300 × 20 e) 231975 × 271 f) 256835 × 420

d) 6500 × 8000

3. Solve the following :
a. Find the product of the greatest and smallest numbers of 4 digits.
b. Find the product of the greatest and smallest numbers formed by
using the digits 2, 4, 3 and 6 at once.
c. The cost of one science book is Rs. 375. What is the cost of 56 such
books ?
d. Find the product of the greatest number of 5 digits and the smallest
number of 4 digits.

4. Solve the following :
a. Mohamad Khan earns Rs. 32540 in a month. How much money does
he earn in 3 years ?
b. 25 Tankers contain equal quantity of water. If one tanker contains
6500 litres of water, how much water is contained in all tankers ?
c. The cost of a television set is Rs. 45575. Find the cost of 30 such
television sets.
d. The cost of a radio and watch are Rs. 6500 and Rs. 4250 respectively.
Find the total cost of 25 radios and 15 watches together.

Prime Mathematics Book - 5 55

Division

Division is one operation among four fundamental operations which is used
to get a new number by dividing the given number by another number. We
use “÷” sign for division operation . The new number obtained by division is
called the quotient of the given number.

For example: 1232 ÷17

Solution: 17) 1232 (72
-119
42 ∴ 1232 ÷ 17 = 72 + 8
-34 17
8

72 is quotient and 8 is remember.

Let's recall the following: When a number is
i) 256 ÷ 1 = 256 divided by 1, the quotient
ii) 0 ÷ 25 = 0
iii) 6400 ÷ 80 = 80 is the number itself.
When 0 is divided by any
Example 1: Divide: 8368 ÷ 16 number, the quotient is 0.

Solution: 16) 8368 (523 Equal number of 0 of divisor
-80 and dividend are cancelled. Then
36 divide in the remaining part.
-32
48
-48
Example 2: 18 tan×kers ∴ Quotient = 523 of water. If the total
contain equal quantity
quantity of water contained by them is 44280 liters, how
much water is contained by each tanker?

Solution: Here, number of tankers = 18

Total water = 44280 litres

56 Prime Mathematics Book - 5

18) 44280 (2460 Once drop a digit
-36 division must be done
82
-72

108
-108

000

0 ∴ Each tanker contains 2460 litre of water.
x

Example 3: find the smallest number that should be subtracted form
7631 so that the result is exactly divisible by 25

Solution: 25) 7631(35

-75

131

-125
6

∴ Required no is 6 which is subtracted form 7631 so that the result is
exactly divisible by 25.

1. Divide: Exercise: 4.4
a) 800 ÷ 2
c) 81000 ÷ 900 b) 1400 ÷ 70
d) 64000 ÷ 1600

2. Divide: b) 481 ÷ 13 c) 9287 ÷ 37
a) 172 ÷ 4 e) 85070 ÷ 362 f) 87165 ÷ 447
d) 6526 ÷ 251
g) 1251264 ÷ 2352

3. Solve the following:
a. The product of two numbers is 2340. If one of the numbers is 52,
find the other number.
b. The cost of 18 television sets is Rs. 101700. Find the cost of 1
television set.
c. A man’s income is Rs. 366000 in two years. Find his income of a month.
d. Find the quotient when the greatest number of 4 digits is divided
by the greatest number of 2 digits.

4. a. Find the smallest number that should be subtracted from 6847
so that the result is exactly divisible by 27.

b. Find the smallest number that should be added to 22141 so that
the result is exactly divisible by 35.

Prime Mathematics Book - 5 57

Rules of plus (+) and minus (-) signs
Let's recall the following rules using plus (+) and minus (-) signs and learn to
operate (+) and (-) signs in the given problems.

Addition rule:

The numbers with the (+) + (+) ⇒ addition (+)
same signs are always (-) + (-) ⇒ addition (-)

added.

For example: (+3) + (+7) = 3 + 7 = 10
(-5) + (-4) = − 9


The numbers with the (+) + (-) ⇒ subtracting
different signs are always (- ) + (+) ⇒ subtracting
and the result has the sign
subtracted. of the greater number.

For example : (+4) + (-2) = 4 – 2 = 2
(-9) + (+5) = -9 + 5 = -4

Multiplication rule:

The product of the (+) × (+) ⇒ product (+)
numbers with same (-) × (-) ⇒ product (+)

signs are always
positive (+).

For example: (+5) × (+3) = 15
(-4) × (- 3) = 12

The product of the (+) × (-) ⇒ product (-)
numbers with the (-) × (+) ⇒ product (-)
different signs are always

negative (-).

For example: (+7) × (-2) = -14
(-5) × (+2) = -10

58 Prime Mathematics Book - 5

Division rule:

The quotient after (+) ÷ (+) ⇒ quotient (+)
dividing a number by (-) ÷ (-) ⇒ quotient (+)
another number with the
same signs are always (+) ÷ (-) ⇒ quotient (-)
(-) ÷ (+) ⇒ quotient (-)
positive (+).

For example: (+10) ÷ (+2) = 5

(-15) ÷ (-5) = 3

The quotient after
dividing a number by another
number with different signs are

always negative (-).

For example: (+20) ÷ (-4) = -5
(-18) ÷ (+6) = -3

Exercise: 4.5

1. Simplify: (b) (+8) + (+13) (c) (+9) + (+15)
(a) (+3) + (+7) (e) (-7) + (-5) (f) (-12) + (-5)
(d) (-5) + (-3) (h) (+12) + (-8) (i) (-19) - (+7)
(g) (+15 ) + (-7) (k) (-19) + (+25) (l) (+14) - (-5)
(j) (-9) + (+5)

2. Simplify: (b) (+12) × (+7) (c) (+15) × (+6)
(a) (+8) × (+3) (e) (-11) × (-6) (f) (-14) × (-7)
(d) (-5) × (-9) (h) (+10) × (-7) (i) (+9) × (-4)
(g) (+6) × (-4) (k) (-13) × (+6) (l) (-9) × (12)
(j) (-7) × (+5)

3. Simplify: (b) (+30 ) ÷ (+6) (c) (+96) ÷ (+12)
(a) (+16) ÷ (+8) (e) (-49) ÷ (-7) (f) (-56) ÷ (-8)
(d) (-24) ÷ (-6) (h) (+84) ÷ (21) (i) (+91) ÷ (-7)
(g) (+64) ÷ (-4) (k) (-144) ÷ (+6) (l) (-36) ÷ (+9)
(j) (-72) ÷ (+8)

4. Simply: (b) (+12) + (-8) + (+4)
(a) (+5) + (+7) + (+8)
(c) (-9) + (-3) + (+5) (d) (+4) + (+9) + (-8) + (-7)
(e) (+17) + (-8) + (+6) + (-5) (f) (-6) + (-7) + (+5) + (-4)
(g) (-13) + (-8) + (+6) + (-3) + (+19)

Prime Mathematics Book - 5 59

Simplification:

The problem with mixed fundamental operations (addition subtraction,
multiplication and division) is called the simplification.

We should simplify such problems by using the proper order of fundamental
operations. The proper order of fundamental operations in the simplification
are performed in the following order.

First work of bracked (B), of (O), division Therefore, the rule
(D), then work of multiplication (M), of simplification is
then work of addition (A) and last work
of subtraction (S). BODMAS rule.

Example 1: Simplify: 7+63 ÷ 9 × 4 -24

Solution: Here,

7 + 63 ÷ 9 × 4 - 24

= 7 + 7 × 4 - 24 First work the division, 63 ÷ 9 = 7

= 7 + 28 - 24 Then work the multiplication, 7×4 = 28

=35 - 24 Then work the addition, 7+28 = 35

=11 At last work the subtraction, 35 -24 = 11

Example 2: Simplify: 6 × 7 - 47 + 12 × 45 ÷ 9
Solution: Here,

6 × 7 - 47 + 12 × 45 ÷ 9
= 6 × 7 - 47 + 12 × 5
= 42 - 47 + 60
= 42 + 60 - 47
= 102 - 47
= 55

Example 3: Subtract 17 from the quotient after dividing 128 by 16 and add
19 to the remainder. Rewrite this sentence in mathematical
form and simplify it.

Solution: The given statement can be written in mathematical form as:
128 ÷ 16 - 17 + 19
Now, 128 ÷ 16 - 17 + 19
= 8 - 17 + 19
= 27 - 17
= 10

60 Prime Mathematics Book - 5

Example 4: Add the quotient after dividing 98 by 7 to the product of 5
and 9 then subtract 39 from the sum. Rewrite this sentence
in mathematical expression and simplify it.

Solution: The given statement in mathematical form is:
98 ÷ 7 + 5 × 9 - 39
Now, 98 ÷ 7 + 5 × 9 - 39
= 14 + 5 × 9 - 39
= 14 + 45 - 39
= 59 - 39
= 20

Exercise: 4.6

1. Simplify:

(a) 25 - 19 + 6 (b) 37 - 12 - 7 (c) 7 + 9 - 3 × 5

(d) 5 × 7 - 2 × 5 + 9 (e) 15 ÷ 3 + 7 (f) 18 ÷ 6 - 2

(g) 12 ÷ 4 × 6 (h) 28 ÷ 7 × 5 + 19 - 11 (i) 34 + 15 ÷ 3 - 29

(j) 30 × 5 + 96 ÷ 8 (k) 56 ÷ 8 × 5 + 4 - 9

2. Simplify: (b) 5 × 10 ÷ 5 + 12 - 20
(a) 31 - 28 ÷ 7 × 3 + 6 (d) 144 ÷ 24 × 13 - 48
(c) 40 ÷ 8 + 3 × 12 - 28 (f) 7 × 6 - 17 + 13 × 16 ÷ 8
(e) 11 × 9 + 72 ÷ 12 - 65 (h) 112 ÷ 8 × 2 - 108 ÷ 12
(g) 18 ÷ 9 × 4 + 21 ÷ 7
(i) 96 ÷ 8 - 12 × 15 ÷ 5 + 32

3. Write the following statement in mathematical form and then
simplify:
a. Subtract 9 from the sum of 17 and 25.
b. Add 6 to the product of 9 and 12.
c. Multiply by 6 the quotient which is obtained after dividing 91 by 7.
d. Divide 135 by 15, then subtract 5 form the quotient.
e. Subtract the product of 8 and 13 from the sum of 121 and 83.

4. Write the following statements in mathematical form and then
simplify:
a. Add the quotient which is obtained after dividing 96 by 4 to the
product of 7 and 12.

Prime Mathematics Book - 5 61

b. Subtract 15 from the quotient which is obtained after dividing 296
by 8.

c. Subtract 12 from the quotient which is obtained after dividing 144
by 16 and add 21 to the remainder.

d. Add the quotient after dividing 98 by 14 to the product of 7 and 13
then subtract 45 from the sum.

e. The quotient of 145 divided by 5 is subtracted from the product of
8 and 14, then the remainder is added to 29.

5. Make the mathematical expression by using the following machines
and find the output by simplification.

(a) Input (b) -27÷3+17 Output (c) Input
12 Input
+7 37 24
×↓4 +21
-1↓8 ↓
×112
↓
÷7
↓
-97

Output Output

Simplification with brackets:
There are three types of brackets which are used in simplification. The four
brackets are ( ), { } and [ ] and (-). ( ) → small bracket, { } → curly bracket
and [ ] → big bracket. If the problems contain the brackets then we have to
first perform operation within 2nd with small brackets ( ), 3rd perform in the
order of with the curly bracket { } and last perform the big bracket [ ]. We
perform the DMAS rule inside all three brackets. Study and learn the use of
brackets in the following examples.

Example 1: Simplify : 7 × (19 - 12) + 4
Solution: Here, 7 × (19 - 12) + 4

=7×7+4
= 49 + 4
= 53

62 Prime Mathematics Book - 5

Example 2: Simplify: 38 - 6{45 ÷ (13 - 4)} + 8

Solution: Here, 38 - 6{45 ÷ (13 - 4)} + 8

= 38 - 6 {45 ÷ 9} + 8 ← Work the small brackets

= 38 - 6 × 5 + 8 ← Work the curly brackets

= 38 - 30 + 8 ← Work of multiplication

= 46 - 30 ← Work of addition

=16 ← Work of subtraction

Example 3: Simplify: 6 ÷ [5 - 14 ÷ {5 + 8 ÷ (6 - 2)}]

Solution: Here, 6 ÷ [5 - 14 ÷ {5 + 8 ÷ (6 - 2)}]

= 6 ÷ [5 - 14 ÷ {5 + 8 ÷ 4}] ← work the small brackets

= 6 ÷ [5 - 14 ÷ {5 + 2}] ← work of division inside curly brackets

= 6 ÷ [5 - 14 ÷ 7] ← work the curly brackets

= 6 ÷ [5 - 2] ← work of division inside big brackets

=6÷3 ← work the big brackets

= 2 ← work of division

Example 4: 5 times the difference between 12 and 8 is subtracted
from 7 times the sum of 9 and 5. Rewrite this statement in
mathematical form and then simplify.
Solution: Here
5 times the difference between
12 and 8 = 5 × (12 - 8)
According to question 7 times the sum of 9 and
7 × (9 + 5) - 5(12 - 8) 5 = 7 × (9 + 5)

= 7 × 14 - 5 × 4

= 98 - 20

= 78

Example 5 26 + { 9 x 15 ÷ 5 - ( 4 - 3 + 3 )}
Solution: = 26 + { 9 x 15 ÷ 5 - (1 + 3 ) }
= 26 + { 9 x 15 ÷ 5 - 4 }
= 26 + { 9 x 3 - 4 }
= 26 + { 27 - 4 }
= 26 + 23
= 49

Prime Mathematics Book - 5 63

Exercise: 4.7

1. Simplify:

a. 18 + (25 - 21) b. 6 × (18 ÷ 6) - 10

c. 48 ÷ (3 × 2) - 5 d. 17 + (25 - 5) ÷ 4

e. 21 ÷ 7 + (18 ÷ 6 × 2)

2. Simplify:

a. 23 - {7 - (2 + 3)} b. 5 + {12 ÷ (6 ÷ 2)}

c. 35 - 2[48 ÷ (4 × 2) - 3] d. 30 - 7[42 ÷ (56 ÷ 8)] + 14

e. 7 + 12 × {(8 - 2) + 3} ÷ 18 f. 14 ÷ {10 - 2 × (7 - 6 - 3)}

g. 20 + [18 - {22 - 10 - 2 × 2}] ÷ 3

3. Simplify:
a. 26 + {9 × 15 ÷ 5 - (12 - 3 + 7)}
b. 40 - [24 ÷ {20 ÷ (6 + 4) × 2}]
c. 90 + [11 × {17 - (18 ÷ 6)}]
d. 19 - [90 - 5{9 - (15 - 13)}] ÷ 11
e. 28 + 3[{12 + 2(14 – 4) ÷ 5} - 6] ÷ 3
f. 27 + 3[{14 - 3(15 - 5) ÷ 5} - 5] - 19
g. 20 + [18 -{22-10-2x2}] ÷ 3

4. Write the following statements in the mathematical form and then
simplify:
a. Multiply the sum of 19 and 5 by 3.
b. Divide the difference of 46 and 11 by 7.
c. 6 times the difference of 19 and 7 is divided by 18
d. Subtract 19 from the product of 12 and 7, then the result is divided
by 13.
e. Add 4 times the difference of 13 and 5 to the product of 6 and 3.
f. The sum of 21 and 12 is divided by the difference of 21 and 10.
g. 6 times the difference between 8 and 3 is subtracted from 3 times
the sum of 45 and 15.

64 Prime Mathematics Book - 5

Unit Revision Test

1. Complete the following:

(a) 3 7 8 5 6 4 8 (b) 982145 (c) 84694
×26
2374356 -690236

+478249 (b) 73526 × 372

2. Do the following task :
(a) 869421 – 684514
(c) 81926 ÷ 23

3. Solve the following:
a. The height of the mountain is 8848m. If the height of a base camp
above the sea level is 4856m, what is the height of the peak from
the base camp?
b. The cost of an egg is Rs. 7.50. if one crate contains 30 eggs, find
the cost of 19 crates eggs.
c. The cost of 78 watches is Rs. 15990. What is the cost of one
watch?

4. Solve the following:
a. The sum of two numbers is 30,42,753. One of them is 18,58,961.
Find the other number.
b. The product of two numbers is 6625. If one of them is 53, find the
other.
c. Find the least number that should be subtracted from 371283 so
that the result is exactly divisible by 253.

5. Simplify:
(a) 144 ÷ 24 - 15 × 4 + 70
(b) 19 × 11 - 12 × 10 + 405 ÷ 27 - 75 ÷ 15
(c) 43 - 6{78 ÷ (13 × 2) + 2}
(d) 19 + [6 + {16 - (4 × 12 ÷ 4)}]
(e) 10 - [90 - 5 {5 - (15 - 9 + 4)}] ÷ 11

6. Solve the followings:
(a) Add the quotient of 35 and 7 to the product of 3 and 8.
(b) Find the product of 30 and the difference of 70 and 58.

Prime Mathematics Book - 5 65

1. (a) 19793 (b) 744209 Answers: (d) 90791 (e) 133767
2. (b) 41310 (b) 376828 (d) 921867
3. (a) 999999 (b) 10999999 Exercise: 4.1 (d) Rs. 23301500
(c) 13549722
(c) 7914676
(c) 112970

1. (a) 270494 (b) 1623086 Exercise: 4.2 (d) 24677357 (e) 147398
2. (a) 153718 (b) 186137 (c) 466076068 (d) 25926
3. (a) 8999999 (b) 660852 (c) 2678626
(c) Rs. 2485 (d) 50872
(f) 163056 (b) Rs. 1110
4. (a) 138890 (c) Rs. 3137450

1. (a) 111008 Exercise: 4.3 (d) 373922082
(b) 14499732 (c) 127713040 (d) 52000000

2. (a) 346000 (b) 19050000 (c) 70000000 (d) 999990000
(d) Rs. 226250
(e) 62865225 (f) 107870700

3. (a) 9999000 (b) 15089472 (c) Rs. 21000

4. (a) Rs. 1171440 (b) 162500 lit. (c) Rs. 1367250

1. (a) 400 (b) 20 (c) 90 Exercise: 4.4
(d) 40

2. (a) 43 (b) 37 (c) 251 (d) 26 (e) 235 (f) 195 (g) 532
(f) 14
3. (a) 45 (b) Rs. 5650 (c) Rs. 15250 (d) 101 (e) 16

1. (a) 10 (b) 21 (c) 24 Exercise: 4.5 (e) -12 (f) -17 (g) 8
(h) 4 (i) -26 (j) -4 (d) -8 (f) 98 (g) -24
(c) 90 (f) 7 (g) -16
2. (a) 24 (b) 84 (j) -35 (k) 6 (l) 19 (f) -12 (g) 1
(h) -70 (i) -36 (c) 8
(j) -9 (d) 45 (e) 66
3. (a) 2 (b) 5 (c) -7
(h) 4 (i) -13 (k) -78 (l) -108

4. (a) 20 (b) 8 (d) 4 (e) 7

(k) -24 (l) -4

(d) -2 (e) 10

1. (a) 12 (b) 18 Exercise: 4.6 (e) 12 (f) 1 (g) 18
(h) 28 (i) 10 (c) 1 (d) 34 (f) 51 (g) 11
(j) 162
2. (a) 25 (b) 2 (c) 13 (k) 30
(h) 19 (i) 8
(d) 30 (e) 40
3. (a) 33 (b) 114
4. (a) 108 (b) 22 (c) 78 (d) 4 (e) 100
5. (a) 22 (b) 45 (c) 18 (d) 53 (e) 112
(c) 263

1. (a) 22 (b) 8 (c) 3 Exercise: 4.7 (e) 9
2. (a) 21 (b) 9 (c) 29 (d) 22
3. (a) 37 (b) 34 (c) 244
4. (a) 72 (b) 5 (c) 4 (d) 2 (e) 13

66 (d) 14 (e) 38 (f) 17
(f) 3
(d) 5 (e) 50 (g) 150

Prime Mathematics Book - 5

Un5it Measurements

Estimated periods − 27

Objectives
At the end of this unit, the students will be able to:
• perform fundamental operations (addition / subtraction / multiplication and division)
involving length, mass, capacity, time and money.
• solve the simple problems related to lengths , mass , capacity, time and money.

Teaching Materials
Measuring devices like scale, tape, weights and balance, measuring vessels, watch,
calendar etc.

Activities
It is better to:
• make discussion of conversion of units of length, mass, capacity, time and money.
• discuss about the four fundamental operation on the units of measurements.
• discuss and solve simple word problems related to the measurements.
• organize a bame about identifying the instruments used in measuring.
• orgnize a bame using playcards with measurment values and let students match the cards in
groups.

Prime Mathematics Book - 5 67

Length

Let's recall the following: Relation Conversion to
mm
Units of length in metre system 1 10 mm 1
10 millimeter (mm) = 1 centimeter (cm) 10 × 10 10 cm 10
1 dm 1
10 cm = 1 decimeter (dm) 10 cm 10
10 dm = 1 metere (m) 1 dm × 10
10 m = 1 decameter (dam) 10 m × 10 10 m 1
10 dam = 1 hectometer (hm) 1 dam × 10 10
10 hm = 1 kilometer (km) 10 hm × 10
1 km × 10 10 dam 1
Most commonly used units of length 10 10
1
10 10 hm 1
10

10 km 1
10

10mm = 1 cm

100 cm = 1m

1000 m = 1 km

Four fundamental operations related to lengths
Addition:
• Arrange vertically in the columns of respective units.
• Add the numbers along each column.
• Convert the sum in the column of smallest unit to next higher unit and

the unit itself. Add the number converted to the higher unit to the next
column.
• Perform the same process in other columns.

Example 1: Add: cm
km m 85
12 680 70
+4 565

Solution: km m cm
12
+4 680 85
16
16km 565 70
1
17km 1245 155

1km 245m 1m 55cm

1

246m 55cm

∴ The sum in 17km 246m 55cm.

68 Prime Mathematics Book - 5

Example 2: Add: 80m 78cm 8mm and 72m 67cm 7mm.

Solution: m cm mm

80 78 8

+ 72 67 7

152 145 15

152m 1m 45cm 1cm 5mm

11

153m 46cm 5mm

∴ The sum is 153m 46cm 5mm.

Subtraction:
• Arrange vertically in column of respective units.
• Perform subtraction in each column. If necessary borrow 1 from the

column of heigher unit.

Example 3: Subtract: 5km 720m 82cm from 16km 450m 68cm.

Solution: km m cm

=10010kmm 1449

15 449 168

16 450 =11m00cm 68
− 5 720 82
10 729 86

∴ The difference in 10km 729m 86cm.

Multiplication
• Arrange the multiplicand in columns of respective units.
• Multiply numbers in each units.
• Convert the quantities in the columns of smaller units to next higher

units and the units itself.
• Add the number converted to higher unit to next column.

Prime Mathematics Book - 5 69

Example 4: Multiply 9km 624m 45cm by 5.

Solution: km m cm

9 624 45

×5

45 3120 225

45km 3km 120m 2m 25cn

3km 2m

48km 122m 25cm

∴ The product is 48km 122m 25cm.

Division
• Arrange the parts of dividend in the columns of the respective units.
• Put the divisor to the left. Divide the number in the column of biggest unit.

Convert the remainder to the next smaller unit and add to the number in
the next column.
• Repeat the same process in next column.

Example 5: Divide: 8km 852m 40cm by 6

Solution: km m cm

1 475 40
6 8 852 40
+200
6 +2000 240
2 2852 24

−24 0
45 0
×
−42
32

−30
2

∴ The quotient is 1 km 475m 40cm.

70 Prime Mathematics Book - 5

Exercise 5.1

1. Add the following:
a) 20 km 560m 80cm and 32km 720m 45cm
b) 90km 240m 80cm and 65km 420m 75cm
c) 150m 50cm 7mm, 30m 40cm 4mm and 6m 60cm 8mm
d) 45m 45cm 6mm, 23m 80cm 8mm and 2km 951m 20cm 9mm
e) 70km 81dam 46dm and 49km 48dam 28dm
f) 34km 25dam 16dm 61mm and 84km 81dam 98dm 87mm

2. Subtract:
a) 20km 700m 40cm from 42km 425m 70cm
b) 34km 450m 95cm from 57km 32m 36cm
c) 4m 25cm 6mm from 10m 20cm 4mm
d) 16m 75cm 8mm from 45m 35cm 6mm
e) 31dam 3m 3dm from 72dam 8m 4dm
f) 12hm 330dm 72mm from 19hm 450dm

3. Multiply: b) 45km 320m 50cm by 8
a) 12km 12m 32cm by 6 d) 20m 38cm 7mm by 5
c) 45m 15cm 8mm by 9
e) 8km 384m by 17

4. Divide: b) 15km 358m by 7
a) 16m 44cm by 3 d) 8km 350m 32cm by 6
c) 6m 25cm 6mm by 4
e) 75dam 25dm 44mm by 12

5. a) A roll of film is 7m 56cm 8mm long. Find the length of film in 15
such rolls.

b) Sharad drives 67km 820m every day. What distance does he drive
in a month?

c) A silk worm produces 245m 65cm of silk in a day. Find the total
length of silk fibre produced by it in a week.

d) A motor bike travels 766km 920m in 14 hours. How far does it
travel in an hour?

Prime Mathematics Book - 5 71

e) A snail moved 1m 52cm 5mm in 5 minutes. How far does it move
in a minute?

f) 18m 66cm 4mm long ribbon is divided among 8 girls. How much
will each get?

6. a) Two places A and B are 7km 360m apart and B and C are 5km 905m
apart. Find the distance between A and C.

b) Find the perimeter of the triangle 8cm 3mm A 10cm 4mm
given alongside. 12cm 8mm
B C
c) On a day tour Megh drove 37km
578m and Rajendra drove next
57km 968m. How far did they drive
altogether?

d) A rope is 17m 89cm 8mm long and other is 23m 17cm 5mm long,
by how much is the second rope longer than the first?

e) In an experiment, a 2m 63cm 7mm long magnesium ribbon is taken
for burning. If 97cm 8mm of the ribbon is left, what length of the
ribbon was burnt?

Capacity 1 liter is the
Liquids do not have their own shape. But volume of 1 kilogram of
liquid takes the shape of the vessel where
it is kept. The volume of the liquid that a water at 4oC which is
vessel can hold is called its capacity. equal to 1000cc.

Capacity of this Capacity of this Capacity of this
bottle of coke in 1 polythene tank milk bottle is
250 ml.
litre. is 1000 l.

72 Prime Mathematics Book - 5

Units of capacity are litre and millilitres which are related as
1000 millilitres (ml) = 1 litre (l)

∴ 1ml = 10100l

Example 1: Express 750ml as fraction or decimal of a litre.
Solution:

We have, 1000ml = 1l
∴ 1ml = 10100l

∴ 750ml =410100 × 3 l

750

= 43l

Example 2: Convert 2.4 litre to millilitre.
Solution:

We have,
1l = 1000ml

∴ 2.4l = 2.4 × 1000 ml = 2400 ml
∴ 2.4 litres = 2400ml.

Addition of capacities:

• Arrange the quantities vertically in the columns of litres and millilitres.

• Add the numbers in each column.

• Convert the quantity in the column of ml into l and ml.

• Add the quantity converted to l in the column of litre.

Example 3: Find the sum of 92l 720ml and 205l 386ml.

Solution: Litres Millilitres

92 720 1106ml = 11010006× l
+ 205 386
1106
297

297l 1l 106ml =1.106l = 1l 106ml
1

298l 106ml

∴ The sum is 298l 106ml.

Prime Mathematics Book - 5 73

Alternate method:
Convert into decimal of litre and add.
92l 720ml = 92.720l
205l 386ml = 205.386l

Sum = 298.106l
∴ The sum is 298.106l = 298l 106ml

Subtraction of capacities:
• Arrange vertically in the columns of litres and millilitres.
• Subtract along each column.

If necessary, we can borrow 1 unit (1 litre) from the column of higher unit.

Example 4: Subtract: 892l 750ml from 1027l 224ml.

Solution: l ml

1026 =110l00ml 1224
1027 224

- 892 750

134 474

∴ The difference is 134l 474ml.

Alternative method:
• Convert into decimal of litres
• Subtract as usual

1027l 224ml = 1027.224l
892l 750ml = 892.750l
Difference = 134.474l
∴ The difference is 134.474l = 134l 474ml

Multiplication of capacities:
• Arrange in the columns of l and ml
• Multiply the numbers in each column by the given number.
• Convert the quantity in the column of ml into l and ml if possible and

add the quantity converted to l to the number in the column of litre.

74 Prime Mathematics Book - 5

Example 5: Multiply 12l 554ml by 7.

Solution: l ml

12 554

×7 [ 1000ml = 1l]

84 3878

84l 3l 878ml

3l

87l 878ml

∴ The product is 87l 878ml.

Alternative method:

Convert into decimal of litre and multiply usually as in decimal numbers.

Solution: 12l 554ml = 12.554l

New 12.554l

×7

87.878l

∴ The product is 87.878l = 87l 878ml.

Division of capacities:
• Arrange the dividend in the columns of l and ml.
• Devide by the given number (divisor). First divide the number in the col-

umn of l, convert the remainder if any, to ml. Add to the number in the
column of ml and divide the sum.

Example 6: Divide 15l 240ml by 6.

Solution: l ml

2 540

6 15 240

-12 +3000

3 +1000 3240

-30

24

-24

0 ∴ The quotient is 2l 540ml.

−0 Prime Mathematics Book - 5 75

×

Alternative method:
• Convert the dividend into decimal of litre.
• Divide the decimal usually as in decimal system.

Solution: 15l 240ml = 15.240l

Now 2.540
6 15.240l
-12
32
-30
24
-24
0
-0

∴ The quotient is 2.540l or 2l 540ml.

Exercise 5.2

1. Convert the following as indicated:

a) 8 litres (into millilitres or ml) b) 36 litres (into millilitres)
d) 414 l (into ml)
c) 0.24l (into millilitres) f) 12000ml (into l)
e) 4l 448ml (into ml)

g) 12345ml (into l) h) 15l 250ml (into litre or l )

2. Add the following:
a) 12l 810ml and 18l 614ml
b) 260l 925ml and 118l 815ml
c) 405l 705ml and 513l 415ml
d) 96l 375ml; 65l 556ml and 24l 764ml
e) 81l 240ml; 60l 720ml and 80l 328ml

3. Subtract: b) 82l 819ml from 487l 223ml
a) 215l 218ml from 428l 86ml d) 12l 432ml from 16l 124ml
c) 564l 219ml from 828l 116ml
e) 7l 812ml from 9l

76 Prime Mathematics Book - 5

4. Multiply: b) 210 litre 820 millilitres by 2
A) 79 litre 42 millilitres by 4 d) 15 l 325 ml by 6
c) 45 litre 340 millilitres by 12
e) 112l 145 ml by 14

5. Divide: b) 71l 586ml by 6
a) 35l 872ml by 8 d) 26l 496ml by 18
c) 40l 180ml by 20
e) 44l 226ml by 21

6. a) A fraud milk seller mixed 3l 450ml of water to 12l 795ml of milk.
What is the total quantity of milk now?

b) To empty kerosene contained in three tanks of capacity 200l 750ml,
150l 625ml and 110l 450ml, find the minimum capacity of single
tank.

c) A bottle contained 3l 500ml of alcohol. If only 1l 750ml of alcohol
is left, how much alcohol is used up?

d) In a bottle containing 4l 250ml of liquid chemical, if 2l 850ml is
left, how much is spilled?

7. a) If a person needs 10l 750ml of water per day, how much does he
need in a week?

b) A bucket holds 3l 400ml of water. If 6 such buckets are filled in a
pot, how much water is there in the pot?

c) 16 bottles contain total of 37l 520ml of soyabean oil. What is the
capacity of a bottle?

d) There is 2l 976ml of tea in a kettle. How much tea is served in a
glass, if 24 glasses of tea is prepared?

Prime Mathematics Book - 5 77

Mass (Weight)

Do you have Basmati Yes, how much do
rice, Sahuji? you need?

5 kilogram
please.

The quantity of substance contained in an object is called mass. In the above
dialogue the customer is buying 5 kilogram of Basmati rice. 5 kilogram of rice
means the quantity of rice is 5 kg which is the mass. Bigger quantities are
measured in kilograms while small quantities are measured in grams and further
smaller quantities are measured in milligram.

In metric system
10 milligrams (mg) = 1 centigram (cg)
10 centigrams (cg) = 1 decigram (dg)
10 decigrams (dg) = 1 gram (gm)
10 gram (gm) = 1 decagram (dag)
10 decagrams (dag) = 1 hectogram (hg)
10 hectograms (hg) = 1 kilogram (kg)

mg cg gd g dag hg kg

10 10 10 10 10 10 10
Note: Mass and weight are entirely different things. We however understand

mass as weight
• 1 kg of mass weighs about 10 N on the Earth surface. (N stands for

Newton).
• Mass is the quantity of substance where as weight is the force with

which a body is pulled by the Earth's gravity towards its centre.

78 Prime Mathematics Book - 5

Conversion of units of mass:
• Remember the relation

10 10 10 10 10 10
mg ÷ cg ÷ dg ÷ gm ÷ dag ÷ hg ÷ kg

××××××
10 10 10 10 10 10

Count the steps from one unit to another.

If conversion is from smaller unit to bigger units divide by as much tens as
there are steps. If conversion is from bigger to smaller unit, multiply by as
much tens as there are steps.

Example 1: Convert 225 gm into milligram.

Solution: Since 1 gm = 1000 mg

225 gm = 225 × 1000mg

= 225000 mg

∴ 225 gm = 225000 milligrams.
Example 2: Convert 1250 grams into kilograms.

Solution: Since 1000 gm = 1kg
1
1 gram = 1000 kg

1250 grams = 1 × 1250 kg = 12500 kg
1000 1000

∴ 1250 gm = 12.5 kg.

Addition and subtraction of masses:
• Arrange the terms vertically.
• Units in metric system follow decimal system. So, the carry over or borrow

is same as in decimal addition / subtraction.

Example 1: Add 10 kg 450g 340mg, 15kg 630g 600mg and 16kg 450g
500mg.

Solution: kg gm mg
Working raw 1 1 1 1 340

10 450

15 630 600

+16 450 500

42 531 440
∴ The sum is 42kg 531gm 440mg.

Prime Mathematics Book - 5 79

Example 2: Subtract 14kg 360gm 780mg from 20kg 240gm 650mg.

Solution: kg gm mg

Working row 19 =11000 1239 =11000 1650

20 240 650

-14 360 780

5 879 870

∴ The difference is 5kg 879gm 870mg.

Exercise 5.3

1. Convert the following: (into gm)
a) 9 kg 340gm (into mg)
b) 12gm 450mg (into hg)
c) 8kg 5hg (into dag)
d) 6hg 5dag (into mg)
e) 20gm 8cg 9mg (into cg)
f) 35hg 20gm 48cg (into dg)
g) 5dag 8g 4dg (into gm)
h) 2kg 4hg 25gm

2. Convert the following: (into gm)
a) 50 gm 250 mg (into cg)
b) 24 dg 7 cg (into gm)
c) 48gm 9 dg (into kg)
d) 16 kg 650gm (into kg)
e) 4kg 840gm 240mg (into kg)
f) 62kg 5hg 6dag (into kg)
g) 75kg 265gm 75cg (into mg)
h) 15dag 50dg 44mg

3. Add the following:
a) 12kg 600gm 210mg and 24kg 400g 850mg
b) 25kg 710gm and 18kg 515gm
c) 28kg 950gm and 12kg 550gm
d) 10kg 450gm 250mg; 12kg 450gm 600mg and 18kg 300gm 340mg
e) 19kg 200gm 340mg; 20kg 150gm 380mg and 17kg 870gm 650mg
f) 2kg 425gm 525mg; 13kg 840gm 780mg and 5kg 920gm 290mg

80 Prime Mathematics Book - 5

4. Sbutract:
a) 80kg 300gm 400mg from 125kg 525gm 600mg
b) 25kg 850gm 920mg from 40kg 280gm 450mg
c) 15kg 750gm 825mg from 482kg 850gm 205mg
d) 12kg 420gm 740mg from 20kg 220gm 560mg
e) 34kg 895gm 570mg from 400kg 300gm 200mg
f) 182kg 720gm 630mg form 210kg 500gm 360mg

Multiplication and Division of masses:
Multiplication
• Multiply each term by the given number by usual process ?
• Convert quantities exceeding the given unit to its higher unit and add to

the next higher unit.

Example 1: Multiply 32kg 460gm by 3.

Solution: kg gm

32 460

×3

96 100=01gkmg 1380
1 380

97 380

∴ The product is 97kg 380gm.

Example 2: Multiply 9gm 25cg 4mg by 6.

Solution: gm cg mg

9 25 4

×6

54 150 20=m2cgg 24
2 4

54 1=001cggm 152 4
1 52

55 52 4

∴ The product is 55gm 25cg 4mg.

Prime Mathematics Book - 5 81

Division:
• Divide the quantity in the greatest unit. Convert the remainder to the

next lower unit and add to its column.
• Proceed similarly in the column of next lower units.

Example 1: Divide 14kg 450gm by 5.

Solution: kg gm

2 890
5 14 450

10 +4000

4 4450

-40

45

-45

0

−0

×

∴ The quotient is 2kg 890gm.

Alternate method:
To divide 14kg 450gm by 5.

14kg 450gm = 14.45kg

Now 2.89kg

5 14.450kg

-10

44

-40

45

-45

0

−0

×

∴ The result is 2.89kg = 2kg 890gm.

82 Prime Mathematics Book - 5

Example 2: Divide 36gm 4dg 8cg by 8.

Solution: gm dg cg

4 5 6
8 36 4 8

-32 +40 +40

4 44 48

-40 -48

4×

∴ The result is 4gm 5dg 6cg.

Example 3: A packet of tea weighs 2kg 330gm. Find the weight of 8 such
packets.

Solution: The weight of 1 packet = 2kg 330gm

∴ The weight of 8 packets = (2kg 330gm) × 8

kg gm

2 330

×8

16 200=02gkmg 2640
+2 640

18 640

∴ The weight of 8 packets of tea is 18kg 640gm.

Example 4: A bag contains 50kg 750g of sugar. It is divided equally and
put into 7 packets. Find the quantity of sugar in each small
packet.

Solution: Quantity of sugar in a packet is (50kg 750gm) ÷ 7

kg gm

7 250
7 50 750

-49 +1000

1 1750

-14

35

-35

0

−0

×

∴ The quantity of sugar in each packet is 7kg 250gm.

Prime Mathematics Book - 5 83

Exercise 5.4

1. Multiply:

a) kg gm mg b) kg gm
428 120
7 12 7 ×45

×6

c) 25kg 640gm by 4 d) 34kg 580gm by 8
e) 7gm 12cg 7mg by 6 f) 120gm 480mg by 5
g) 9hg 7dag 5g by 8 h) 12dag 16dg 15mg by 9

2. Divide: gm b) kg gm
a) kg 128 53 650
12 13

c) gm dg cg d) dag gm dg
8 35 44 9 28 98

e) 8kg 352gm by 6 f) 21kg 125gm by 13
g) 12gm 6804mg by 8 h) 14dg 9cg 6mg by 4

3. Solve the following word problems:
a) A family needs 4 kg 350gm of rice each day. What amount of rice
is needed for the family for 12 days?
b) A packet of potatoes is 10kg 420gm. What is the weight of 5 such
packets of potatoes?
c) The weight of an ink pot is 270gm. Find the weight a dozen of ink
pots.
d) The total quantity of tea contained in 7 boxes is 15kg 750gm. Find
the quantity of tea in a box.
e) Weight of a dozen of eggs is 2kg 350gm 440mg. Find the weight of
an egg.
f) Weight of 4 tablets of medicine is 4dg 9cg 6mg. Find the weight of
a tablet.

84 Prime Mathematics Book - 5

Time

Remember the following: The Earth takes 365 days to go
around the Sun.

The Sun

60 seconds = 1 minute
60 minutes = 1 hour
24 hours = 1 day

The Earth

7 days = 1 week
30 days = 1 month
12 months = 1 year
365 days = 1 year
52 weeks = 1 year
10 years = 1 decade
100 years = 1 century

Note: In a leap year there are 366 days. The leap year occurs in every 4 years.

You have already learned much about time in lower classes. In this class we
will learn about the fundamental operation on time.
Addition of time:
• Arrange vertically with columns of hours, minutes and seconds.
• Convert the number in small units to next bigger units leaving the number

of same unit in the same column, add the number changed to higher unit
to the next higher unit number.

Example 1: Add 6 hours 48 minutes 54 seconds and 10 hours 32 minutes
and 42 seconds.
Solution:
hours minutes seconds

6 48 54

+10 32 42

16 80 96

16 hrs. 1 hrs 20 mins 1 mins 36 sec.

11

17 21 36

∴ The sum is 17 hours 21 minutes and 36 seconds.

Prime Mathematics Book - 5 85

Example 2: Add 9 years 8 months 19 days and 6 years 7 months 20
days.
Solution:
years months days

98 19

+6 7 20

15 15 39

15 years 1 year 3 months 1 month 9 days

11

16 years 4 months 9 days

∴ The sum is 16 years 4 months and 9 days.

Subtraction of time:
Example 3: Subtract 8 hours 36 minutes 45 seconds from 24 hours 14

minutes 30 seconds.
• Arrange the numbers in respective columns.
• If necessary take borrow 1 unit to the next lower unit

column.
• Subtract in each column.

Solution: hours minutes seconds

24 14 3023 1 hrs=60 min 73 1 min=60 sec 90

-8 36 45

15 37 45

∴ The difference is 15 hours 37 minutes 45 seconds.

Example 4: Subtract 12 years 10 months 20 days from 20 years 11 months
14 days.

Solution: years months days

20 10 =1m30d 44

11 14

-12 10 20

8 00 24

∴ The difference is 8 years 24 days.

86 Prime Mathematics Book - 5

Multiplication of time:
• Arrange the numbers in the respective columns.
• Multiply the numbers in each column by the given number.
• Convert the number in smaller unit exceeding the given unit to next bigger

units and the same unit.
• Add the number converted to bigger unit to next column of bigger unit.

Example 5: Multiply 7 years 8 months 9 days by 6.

Solution: years months days 30) 54 (1
7 8 9 -30
24
×6
12) 48 (4
42 48 54 -48
×
42 yrs. 4 yrs. 0 m. 1 m. 24 d

+4 1

46 years 1 month 24 days

∴ The product is 46 years 1 month 24 days.

Example 6: Multiply 3 weeks 6 days 12 hours by 5.

Solution: weeks days hours 24) 60 (2
3 6 12 -48
×5 12

15 30 60 7) 30 (4
-28
15 weeks 4 weeks 2 days 2 days 12 hours 2

+4 2

19 weeks 4 days 12 hours

∴ The product is 19 weeks 4 days 12 hours.

Division of time
• Arrange the number in column of respective units.
• Divide the number in the biggest unit by the given number.
• Convert the remaining part to next smaller unit and add to the number in

respective column.
• Repeat the same process in the next columns.

Prime Mathematics Book - 5 87

Example 7: Divide 15 hours 21 minutes 24 seconds by 6.

Solution: hours minutes second

2 33 34
6 15 21 24
-12 +180 +180
3 3=hr1s80 min 201 204

-18 -18
= 1380misnec
21 24
-18 -24
3×
∴ The quotient is 2 hours 33 minutes 34 seconds.

Example 8: Divide 14 years 8 months 15 days by 5.

Solution: years months days

2 11 9

5 14 8 15

-10 4=4ye8amr on +48 1 mon +30
4 56 =30 days 45
-5 -45

6×
-5

1

∴ The quotient is 2 years 11 months 9 days.

Exercise 5.5

1. Add
a) 8 hours 34 minutes 25 seconds and 7 hours 48 minutes 55 seconds
b) 36 hours 42 minutes 50 seconds and 18 hours 48 minutes 35
seconds
c) 10 years 4 months 20 days and 18 years 9 months 15 days
d) 5 years 26 days, 14 years 10 months and 2 years 3 months 14 days
e) 3 years 40 weeks 5 days and 5 years 35 weeks 4 days
f) 12 years 50 weeks 6 days and 6 years 20 weeks 5 days.

88 Prime Mathematics Book - 5

2. Subtract:
a) 5 hours 30 mins 40 secs from 9 hrs 20 mins 10 secs
b) 8 hrs 43 mins 52 secs from 12 hrs 35 mins 40 secs
c) 2 years 7 months 20 days from 7 years 4 months 13 days
d) 10 years 8 months 27 days from 14 years 6 months 29 days
e) 7 weeks 5 day 20 hours from 15 weeks 4 days 14 hours.

3. Multiply:
a) 4 hours 45 minutes 30 seconds by 5
b) 8 hours 40 minutes 42 seconds by 12
c) 6 days 20 hours 15 minutes by 20
d) 6 months 15 days 14 hours by 15
e) 18 years 7 months 12 days by 10
f) 6 years 32 weeks 5 days by 4

4. Divide:
a) 5 hours 17 minutes 15 seconds by 3
b) 18 hours 15 minutes 12 seconds by 16
c) 15 days 22 hours 48 minutes by 12
d) 18 years 9 months 18 days by 8
e) 26 years 49 weeks 1 day by 8

5. a) This morning Khusi took 2 hours 48 minutes 38 seconds to do maths
homework and 1 hour 32 minutes 40 seconds to do science home
work. How much did she spend to complete her homework?

b) Jenisha taught in a school for 3 years 10 months and in college for
4 years 11 months. For how long has she been teaching?

c) Night bus takes 9 hours 45 minutes to complete a journey from
Kathmandu to Birganj where as day bus takes 11 hours 25 minutes.
By how much is the night bus faster than the day bus?

d) Lachhen is 12 years 10 months 22 days old. Today's date is 2068-9-
25 B.S. When is her birth date?

e) Dawa studies 4 hours 40 minutes each day. How long does he study
in a week?

f) A filling pipe can fill a water tank in 2 hours 45 minutes 32 seconds.
How long does it take to fill 6 such tanks?

g) A bus travels 30 hours 36 minutes in 9 days. How long does the bus
travel in a day?

Prime Mathematics Book - 5 89

Money

We are familiar with our currency and monetary system. In olden times there
used to be three units in Nepalese currency Rupaiya (Rupees), Paisa and
dam. But now our currency has two units Rupees (Rs.) and paisa (P) which
relate as

100 Paisa = 1 Rupee (Re)
Different coins and notes are in use. The highest denomination note in use
is Rs. 1000 note.

Addition and subtraction of money:

Example 1: Add Rs. 48 and 85 P and Rs. 125 and 65 P.
Method I
• Arrange vertically with columns of Rs. and P.
• Take number of hundred's place of paisa as carry over. Add as in decimal system.

Solution: Rs. P
48 85
+125 65
=173 1 150
174 50

∴ The sum is Rs. 174 and 50 P.

Method II
• Write as decimal of rupees.
• Arrange vertically and add.

Solution: Rs. 48 and 85 P = Rs. 48.85
Rs. 125 and 65 P = Rs. 125.65

Sum = Rs. 174.50
∴ The sum is Rs. 174.50

or Rs. 174 and 50 P

Example 2: Subtract Rs. 66 and 85 P from Rs. 213 and 28 P.
Method I
• Arrange vertically in column of Rs. and P.
• Subtract as in decimal system
• Borrow 1 from the column of Rs. to the column of paisa = 100 P

90 Prime Mathematics Book - 5

Solution: Rs. P

212 R1s.001P 128

213 28

-66 85

146 43

∴ The difference is Rs. 146 and 43 P.

Method II
• Write as decimal of rupees.
• Arrange vertically and subtract as usualy.

Solution: Rs. 213 and 28 P = Rs. 213.28
Rs. 66 and 85 P = Rs. 66.85
Difference = Rs. 146.43
∴ The difference is Rs. 146 and 43 P.

Multiplication and Division of money:

Example 3: Multiply Rs. 62 and 80 P by 7.

Method I

• Arrange in column of Rs. and P.

• Multiply each term by the given number. Take the number in hundred
place in the column of paisa as carry over.

Solution: Rs. P

62 80

×7 Re 1 = 100 paisa
434 560 Rs 5 = 5 x 100 = 500 paisa

+5 -500

439 60

Method II ∴ The product is Rs. 439 and 60 P.
• Write the amount of money as decimal of rupees.
• Multiply as usual.

Solution: Rs. 62 and 80P = Rs. 62.80
×7

Rs. 439.60
∴ The product is Rs. 439 and 60 P.

Prime Mathematics Book - 5 91

Example 4: Divide Rs. 185 and 40 P by 12.

Solution: Rs. P.
15 45
12 185 40 • Arrange in the column of Rs. and P.
-12 +500 • First divide Rs. and convert the
540
65 -48 remainder into paisa and add to the
-60 column of paisa
• Now divide paisa.

5 60

-60

×

∴ The quotient is Rs. 15 and 45P.

Method II

Rs. 185 and 40P = Rs. 185.40

Now, Rs. 15.45
12 185.40
• Convert the amount as decimal
-12 of Rs.
65
-60 • Divide as usual.
54
-48
60
-60
×

∴ The quotient is Rs. 15.45
or Rs. 15 and 45P.

Example 5: Farjana's father gave her Rs. 63 and 65P and her mother gave
her Rs. 40 and 85 P. If she spent Rs. 50 and 40 P from this
money, how much money does she have now?

Solution: She has Rs. 63.65 + Rs. 40.85 - Rs. 50.40
= Rs. 104.50 - Rs. 50.40
= Rs. 54.10

∴ She has Rs. 54 and 10P.

92 Prime Mathematics Book - 5

Example 6: The cost of 1 liter of petrol is Rs. 95 and 15 P. Find the cost
of 15 liters of petrol.

Solution: The cost of 1 liter of petrol = Rs. 95 and 15 P

∴ The cost of 15 liters of petrol = (Rs. 95 and 15 P) × 15

= Rs. P

95 15

×15

1425 225

+2

1427 25

∴ The cost of 15 liters of petrol is Rs. 1427 and 25 P.

Example 7: A family spends Rs. 3048 and 50 P in a week. Find the
expenditure of the family in a day.

Solution: Expenditure of a day is (Rs. 3048 and 50 P) ÷ 7

We perform the division as P
Rs.

435 50
7 3048 50

-28 +300

24 350

-21 -350

38 0

-35

3

∴ The daily expenditure of the family is Rs. 435 and 50P.

Exercise 5.6
1. Add the following:

a) Rs. 21 and 65 P and Rs. 72 and 80 P
b) Rs. 45 and 75 P and Rs. 88 and 50 P
c) Rs. 139 and 65 P and Rs. 294 and 75 P
d) Rs. 240 and 24 P; Rs. 24 and 80 P and Rs. 106 and 76 P
e) Rs. 40 and 42 p; Rs. 9 and 72 P and Rs. 98 and 36 P
f) Rs. 245.20; Rs. 127.85 and Rs. 72.77

Prime Mathematics Book - 5 93

2. Subtract:
a) Rs. 209 and 80 P from Rs. 483 and 35 P
b) Rs. 519 and 20 P from Rs. 851 and 60 P
c) Rs. 111 and 46 P from Rs. 344 and 16 P
d) Rs. 13.45 from Rs. 481.20
e) Rs. 6.27 from Rs. 232.07
f) Rs. 25.25 from Rs. 1000

3. Multiply: b) Rs. 196 and 83 P by 8
a) Rs. 62 and 30 P by 3 d) Rs. 76.34 by 5
c) Rs. 215and 72 P by 6 f) Rs. 4310.82 by 4
e) Rs. 345.96 by 7

4. Divide: b) Rs. 136 and 80 P by 12
a) Rs. 278 and 10 P by 5 d) Rs. 4426 and 20 P by 12
c) Rs. 79 and 60 P by 8

5. Solve the following problems:
a) Shahrukh earned Rs. 456 and 85 P on Monday, Rs. 710 and 25P on
Tuesday and Rs. 512 and 75P on Wednesday. Calculate the total
income made by him in these three days.
b) A mathematics book costs Rs. 180 and a science book costs Rs. 165
and 25P. Which book is more expensive and by how much?
c) Nitesh bought a pair of shoes for Rs. 570.50 and a pant for Rs. 305
and 75P, if he gives a Rs. 1000 note to the shopkeeper, how much
money will the shopkeeper return him?

d) If a packet of biscuit costs Rs. 18 and 50P, find the cost of 5 packets
of biscuits.

e) If 12 pens cost Rs. 665 and 40 paisa, what is the cost of a pen?

Unit Revision Test

1. Add the following :
(a) 12 kg 600 gm 210 mg and 24 kg 400 gm 850 mg
(b) 5 years 26 days, 14 years 10 months and 2 years 3 months 14
days.

2. Subtract :
(a) Rs. 66 and 85 paisa from Rs. 123 and 28 paisa
(b) 7l 812ml from 9l

94 Prime Mathematics Book - 5

3. Multiply:
(a) 9 km 624 m 45 cm by 5.
(b) 10 gm 25 cg 4 mg by 6.

4. Divide:
(a) 14 years 8 months 15 days by 5.
(b) Rs. 185 and 40 p by 12.

5. Solve the following problems:
(a) Weight of a dozen of eggs is 2 kg 350 gm 440 mg. Find the weight
of an egg.
(b) This morning Khusi took 2 hours 48 minutes 38 seconds to do the
maths homework and 1 hour 32 minutes 40 seconds to do science
homework. How much did she spend to complete her homework?

Answers:

1. (a) 53 km 281m 25cm Exercise: 5.1 (c) 187m 51cm 9mm
(b) 155km 661m 55cm

(d) 3km 20m 47cm 3mm (e) 120 km 29 dam 74 dm (f) 119km 7dam 15dm 48mm

2. (a) 21km 725m 30 cm (b) 22km 581m 41cm (c) 5m 94cm 8mm

(d) 28m 59cm 8mm (e) 41 dam 5m 1 dm (f) 7km 119dm 28mm

3. (a) 72km 73m 12cm (b) 362km 564m (c) 406m 42cm 2mm

(d) 101m 93cm 5mm (e) 142km 528m

4. (a) 5m 48cm (b) 2km 194m (c) 1m 56cm 4mm

(d) 1km 391 m 72cm (e) 6dam 27 dm 12mm

5. (a) 113 m 52cm (b) 2034km 600m (c) 1719m 55cm

(d) 54km 780m (e) 30cm 5m (f) 2m 33cm 3mm

6. (a) 13 km 265m (b) 31cm 5mm (c) 95km 546m

(d) 5m 27 cm 7 mm (e) 1m 65cm 9mm

1. (a) 8000ml (b) 36000ml Exercise: 5.2 (d) 4250ml (e) 4448l
(f) 12l (g) 12.345l (c) 240ml
(h) 15.250l

2. (a) 31l 642ml (b) 379l 740ml (c) 919l 120ml (d) 186l 695ml (e) 222l 288l

3. (a) 213l 850ml (b) 404l 404ml (c) 263l 897ml (d) 3l 692ml (e) 1l 188ml

4. (a) 317l 68 ml (b) 461l 640ml (c) 544l 80ml (d) 91l 950ml (e) 1570l 30ml

5. (a) 4l 484ml (b) 11l 931ml (c) 2l 9ml (d) 1l 472ml (e) 2l 106ml

Prime Mathematics Book - 5 95