Prime Mathematics 7

3. Division Method: In this method the given numbers are arranged in a row and
divided successively by common prime factors which should be common to at least
two numbers. The product of divisors and remainder is the L.C.M.
e.g. to find the L.C.M. of 20, 24 and 30.

2 20, 24, 30 2 is common to all the term.
2 10, 12, 15 2 is common to 10 and 12.
3 5, 6, 15 3 is common to 6 and 15.
5 5, 2, 5
5 is common to 5 and 5
1, 2, 1

\ L.C.M. = 2 × 2 × 3 × 5 × 2 = 120

Relation between H.C.F and L.C.M. of two numbers.

If two numbers (integer) be a and b

H.C.F. of a and b be denoted by H.C.F. (a,b)

L.C.M. of a and b be denoted by L.C.M. (a,b)

We have an important theorem from number theory.

H.C.F. (a,b) × L.C.M.(a,b) = a b

i.e. product of two numbers = product of their H.C.F. and L.C.M.
Lets consider two numbers 12 and 16
Here,

12 = 2 × 2 × 3
16 = 2 × 2 × 2 × 2

\ H.C.F. = product of common factors.
=2×2 =4
And L.C.M. = product of common factors and remaining factor
= 2 × 2 × 2 × 2 × 3 = 48
Now, product of two numbers = 12 × 16 = 192
And, product of H.C.F. and L.C.M. = 4 × 48 = 192
Thus we observed,
Product of two numbers = product of their H.C.F. and L.C.M.
Thus
H.C.F. (a,b) × L.C.M.(a,b) = a b
ab
H.C.F. (a,b) = L.C.M (a,b)

L.C.M.(a,b) = ab b)
H.C.F (a,

a = H.C.F(a, b) × L.C.M.(a, b)
b

And b = H.C.F.(a, b) × L.C.M.(a, b)
a

144 Whole Number

Example 1. Find the L.C.M of 6, 8 and 12 using the method of multiples.
Solution :
Multiples of 6 are 6, 12, 18, 24, 30 , 36, 42, 48, 54, 60, .....
Multiples of 8 are 8, 10, 24, 40, 48, 56, 64,.....
Multiples of 12 are 12, 24, 36, 48, 60, 72,.......
Where the lowest common multiple (L.C.M.) = 24.

\ L.C.M. of 6, 8 and 12 is 24.

Example 2. Find the L.C.M. of 12, 18 and 24 using prime factorization method.
Solution:
Here, given numbers are 12, 18 and 24
Where 12 = 2 × 2 × 3

18 = 2 × 3 × 3
24 = 2 × 2 × 2 × 3

\ L.C.M = Common factors × remaining factors

= 2×3×2×2×3 = 72

Note : Common in all three 2 × 3 12 = 2 × 3 × 2
2 18 = 2 × 3 × 3
Common in all two 2 × 3 24 = 2 × 3 × 2 × 2
Remaining factors

Example 3. Find the L.C.M. of 28, 42, 56 using division method.
Solution:

Here, the given numbers are: 28, 42 and 56
2 28, 42, 56

2 14, 21, 28

7 7, 21, 14

1, 3, 2

\L.C.M. = 2 × 2 × 7 × 3 × 2 = 168

Example 4. Find the smallest number which is exactly divisible by 15, 20 and 30.
Solution:
Here, the given numbers are: 15, 20 and 30
The smallest number exactly divisible by the number is their L.C.M.

2 15, 20, 30
5 15, 10, 15
3 3, 2, 3

1, 2, 1

\L.C.M. = 2 × 5 × 3 × 2 = 60
\ The smallest number exactly divisible by 15, 20 and 30 is 60.

Prime Mathematics Book - 7 145

Example 5 : Find the least number which when divided by 20, 25 and 30 leaves
remainder 1 in each case.

Solution:
Here, the given numbers are 20, 25 and 30.
The required number is the sum of L.C.M. of 20, 25 and 30 and 1.
2 20, 25, 30
5 10, 25, 15

2, 5, 3

\L.C.M. = 2 × 5 × 2 × 5 × 3 = 300
\ The required number = L.C.M. + 1 = 300 + 1 = 301

Example 6: Bells of a monastery, a school and a temple in the vicinity ring at intervals
of 40, 45 and 60 minutes respectively. They all ring together at 6 am, at
what time will they ring together again ?

Solution:
Here, Monastery bell rings it an intervals of 40 minutes

School bell rings at an interval of 45 minutes.
Temple bell rings at an interval of 60 minutes.
The least interval of time after which they ring together is the L.C.M. of 40, 45 and 60 .

2 40, 45, 60

2 20, 45, 30
3 10, 45, 15
5 10, 15, 5

2, 3, 1

\L.C.M. = 2×2×3×5×2×3 = 360

The three bells ring together after 360 minutes.
i.e. 360/60 hour = 6 hours.

\ The time they will ring together again = 6 am = 6 hr = 12 noon.

Example 7 : H.C.F and L.C.M. of two number are 6 and 36 respectively. If one of the
number is 12, find the other.

Solution
Here. H.C.F. (a,b) = 6, L.C.M. (a,b) = 36, one of the number (a) = 12
other number (b) = ?
We have
a × b = H.C.F. (a,b) × L.C.M. (a,b)
or 12 × b = 6 × 36

or b = 6 × 36
12

\ b = 18
\ The other number is 18.

146 Whole Number

Exercise 6.4

1. Find the L.C.M. of the following numbers by listing some multiples.
(a) 3, 4 (b) 4, 6 (c) 5, 12, 16 (d) 10, 15, 20

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

(a) 6, 8 (b) 14, 21 (c) 16, 24 (d) 18, 32

(e) 12, 15, 18 (f) 24, 48, 64 (g) 48, 64, 72 (h) 32, 36, 40

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

(a) 6, 14 (b) 8,18 (c) 8,10 (d) 24, 36

(e) 18,24, 32 (f) 25, 30 75 (g) 32, 40, 48 (h)20, 30 , 40

4. (a) Find the least number which is exactly divisible by 20, 25, 30
(b) Find the least capacity of a drum which can be filled exactly by the buckets
of capacity 10 ltr, 15 ltr and 20 ltr.
(c) Find the smallest number which when divided by 24, 30 and 40 leaves
remainder 3 in each case.
(d) Find the least number of chocolates that can be distributed equally among
14 women, 21 girls and 35 boys.
(e) Three bells ring at intervals of 8, 12 and 15 minutes respectively. If they all
ring together at 9 a.m., at what time will they all ring together again?

5 (a) The H.C.F. and L.C.M. of two numbers are 9 and 54 respectively. If one of
numbers is 18, find the other number.

(b) If H.C.F. (a,b) = 4, L.C.M. (a,b) = 416, a = 52, find b.
(c) The product of two numbers is 432. If their H.C.F. is 6, find their L.C.M.
(d) The product of two numbers is 1944. If their L.C.M. is 54, find their H.C.F.

Prime Mathematics Book - 7 147

6.5 Number Systems

Written numbers are called numerals. For expressing numbers in written form, symbols
are adopted and arranged in order. In the history of development of mathematics, we
find the following systems.

1. Grouping System. In this system symbols representing certain number were put in
groups in their respective position.

A. Simple Grouping System: Under this system comes Egyptian Hieroglyphic numeral
system which is based on scale of 10 (base 10)

They adopted the symbols for 1 and products of 10 as

1 = (a stroke) 104 = (a pointing finger)

10 = (a heel bone) 105 = (a tadpole)

102 = (a coil of rope) 106 = (a man in astonishment)

103 = (a lotus flower)

Any number were expressed using these symbols additivity each symbol being repeated
the required times.
e.g They wrote 23415 as

2(104) + 3(103) + 4(102) + 1(10) + 5 =

(Written from right to left)

Roman numeral System was also simple grouping system (also base 10). They adopted
the symbols

I, X, C, M for 1, 10 , 102, 103, and V, L, D for 5, 50, 500 respectively.

Thus 2934 was written as:

MMDCCCCXXXIIII (Ancient additive way)

MMCMXXXIV (Modern subtractive way)

B. Multiplicative Grouping System: Under this system comes traditional Chinese
numeral system. They adopted the symbols

1= ! 7 =Q % They wrote 5678 as
2= L 8 =( ? 5(1000) + 6(100) + 7(10) + 8 as
3=# 9 =) &
4=$ 10 = * = They wrote vertically with group of
5=% 102 = = Q two multiplicatively
6=& 103 = ? *
(

148 Whole Number

2. Ciphered Numeral System : Under this system comes Greek alphabetic System . In
this system they adopted alphabets for 1 to 9 and all the ciphered number (number
with zero)

1 a (alpha) 10 i (iota) 100 r (rho)

2 b (beta) 20 k (kappa) 200 s (sigma)

3 g (gamma) 30 l (lambda) 300 t (tau)

4 d (delta) 40 m (mu) 400 u (upsilon)

5 e (epsilon) 50 n (nu) 500 f (phi)

6 s (digamma) 60 x (xi) 600 c(chi)

7 z (zeta) 70 o (omicron) 700 y (psi)

8 h (eta) 80 p (pi) 800 w (omega)

9 q theta 90 j (koppa) 900 l sampi

Thus 248 was writen as 200 + 40 + 8 = s mh

3. Positional Numeral System: Under this system comes our number system (decimal,
binary, hexadecimal system e.t.c). In this system after base (b) is selected, basic
symbols 0,1,2, ........... , b-1 are adopted. Thus there are b basic symbols called digits
and any number (N) can be writen uniquely as N = anbn+an-1bn-1+ ........+a2b2+a1b1+a0

Hindu Arebic numeral System (Decimal System) has adopted the digits
0,1,2,3,4,5,6,7,8,9.
Let’s consider a decimal number 3456

Positional weight. Tens Ones
Thousands Hundreds 56

34

6×1 = 6

5×10 = 50

4×100 = 400

3×1000 = +3000

3456

The distinguishing characteristics of a number system (in position number system) is
its base. The base of a number system is defined by the number of different digits
(symbols) used in the system.
Depending upon the base we have the following type of number system.

(1) Decimal number system
(2) Binary number system
(3) Quinary number system
(4) Octal number system
(5) Hexadecimal number system

Prime Mathematics Book - 7 149

Hindu numerals (Hindu Arabic Numerals) with positional system along with zero was
developed some time between 250 A.D. to 800 A.D. from Hindu Brahmi numerals which
was ciphered system. It has dominated all the numerals previously developed. After
development of computer it became necessary to use other systems like binary, octal
and hexadecimal systems.
Binary System: A computer consists numerous loops of circuits. A circuit can be either
on or off. Corresponding to these two actions, there are two digits 1 and 0. On or flow
of current means 1 and off or off circuit or no flow of current means 0.
Binary number system is a base 2 number system. It uses only two digits 0 and 1. The
digits 0 and 1 are called Binary Digits (Bits). Each one of the digits in a binary number
has a place value or weight as in the decimal number system. 6 or 8 bits form a letter
or number called byte we express the positional weights in decimal system as power
of 10 (base 10) similarly in binary system, positional weights are expressed as power
of 2(base 2).

Let’s consider a binary number 10011
Positional weights.
24 23 22 21 20
10011

1×20 = 1
1×21 = 2
0×22 = 0
0×23 = 0
1×24 = +16

19

Decimal to Binary Conversion:

To convert a decimal number to binary, we express the decimal number as the sum
of the possible binary positional weights (place values) in expanded form. Since
positional values are obvious in positional numeral system, removing the positional
weights (place values), the remaining face values in order gives binary equivalent.
e.g. to convert 2510 to binary.

Binary positional weights (place value) 25 24 23 22 21 20

Decimal equivalent 32 16 8 4 21

2510 16 8 0 0 1
In terms of binary place values 1×24 1×23 0×22 0×21 1×20

Binary equivalent 1 1 0 01

\ 2510= 110012

150 Whole Number

Alternative method :
The decimal number is divided by 2 successively.

The quotient and remainder are noted down in each step.
The quotient of one step is divided by 2 in the next step.
The process is repeated until the quotient becomes zero.
The first remainder is the least significant bit (L.S.B.) and the last remainder is
the most significant Bits (M.S.B.).
Thus the digits (bits) in the remainder arranged from MSB to L.S.B. gives the binary
equivalent.

e.g to convert 2510 to binary.

Quotient Remainder or simply
1 L.S.B. 2 25 remainder
25 ÷ 2 = 12 0 2 12 1
12 ÷ 2 = 6 0 26 0
6÷2 = 3 1 23 0
3÷2 = 1 1 M.S.B. 21 1
1÷2 = 0 01

\2510 = 110012
Binary to decimal conversion.

To convert a binary number to decimal, various bits (digits) are multiplied by their
respective positional weights (place Values) and their sum given the decimal equivalent.

e.g. to convert 1100121to decimal. 0 0 1
Binary number 1
Bit position 5th 4th 3rd 2nd 1st
(from right to left)
Positional weights 24 23 22 21 20
(Place value)
Decimal equivalent = 1 × 24 +1 × 23 + 0 × 22 + 0 × 21 +1 × 20
= 16 + 8 + 0 + 0 + 1 = 25

\ 110012 = 2510

Quinary number system.
Quinary number system is base-5 number system consisting 5 digits 0,1,2,3,4.
Lets consider a quinary number 12345

Positional weights 5 50 (L.S.D.) Decimal equivalent
(M.S.D.)53 52 3 4
4 × 50 = 4
12

3 × 51 = 15
2 × 52 = 50
1 × 53 =+125
194

Prime Mathematics Book - 7 151

Decimal to Quinary Conversion:

To convert a decimal number to quinary, we express the decimal number as the sum
of the possible quinary positional weights (place values) in expanded form. Since
positional values are obvious in positional number system, removing the positional
weights, the remaining face values in order gives quinary equivalent.
e.g to covert 201410 to quinary.

Quinary Positional weight 55 54 53 52 51 50
Decimal equivalent 3125 625 125 25 5 1
In term of quinary place value201410 1875 125 0 10 4
3×54 1×53 0×52 2×51 4×50
Quinary equivalent 310 2 4

\ 201410 = 310245

Alternative method.
The decimal number is divided by 5 successively.

- The quotient and remainders are noted down in each step.
- The quotient of one step is divided by 5 in next step.
- The process is repeated until the quotient becomes zero.
- The first remainder is the least significant digit (L.S.D.) and the last remainder

is the most significant digit (M.S.D.).

Thus the digits remainders the romanidor arranged in order from M.S.D. to L.S.D. gives
the quinary equivalent.

e.g. To convert 201410 to quinary.

Quotient Remainder or simply
4 L.S.D. remainder
2014 ÷ 5 = 402 2
402 ÷ 5 = 80 0 5 2014
80 ÷ 5 = 16 1 5 402 4
16 ÷ 5 = 3 3 M.S.D. 5 80 2

3÷5 = 0 5 16 0
53 1

03

\ 201410 = 310245

Quinary to Decimal Conversion.
To convert a quinary number to decimal the various digits are multiplied by their
respective positional weights and their sum gives the decimal equivalent.
e.g. To convert 203452to decimal.
Quinary number 0 3 4
Position of digits 4th 3rd 2nd 1st
(from right to left)
Positional weights 53 52 51 50

152 Whole Number

\Decimal equivalent = 2 × 53 + 0 × 52 + 3 × 51 + 4 × 50

= 250 + 0 + 15 + 4
= 269

\ 20345 = 26910

Note: Quinary number system is introduced only for practicing any arbitrary
base numerals. It is not an existing system.

Example 1. Convert 5110 to its binary equivalent.
Solution:

Here, given denary number = 521510 24 23 22 21 20
Binary positional weight :
Decimal equivalent : 32 16 8 4 2 1
In terms of bin5a1r1y0 = 32 16 0 0 2 2
Place values : 1×25 1×24 0×23 0×22 1×21 1×20

Binary equivalent : 1 1 0 0 1 1

\ 5110 = 1100112

Alternative method:

2 51 R
2 25 1
2 12 1

26 0
23 0
21 1

01

\ 5110 = 1100112 0
1st
Example 2. Convert 110102 to its decimal equivalent. 20
Solution 0×20

Given binary number : 1 1 0 1
(spread form)
Bit position : 5th 4th 3rd 2nd

(from right to left)
Positional weights : 24 23 22 21
Decimal equivalents : 1×24 1×23 0×22 1×21

\ 110102 = 1×24 +1×23 + 0×22 + 1×21 + 0×20
\ 110102 = 2610 = 16 + 8 + 0 + 2+ 0 = 26

Prime Mathematics Book - 7 153

Example 3. Convert 6410 to its quinary equivalent.
Solution

Quinary positional weights 53 52 51 50
Decimal equivalent
64 125 25 5 1
In terms of quinary positional values
64 50 10 4

2×52 2×51 4×50

Quinary equivalent 224

6410 = 2245

Alternative method.

Given denary number = 64
5 64 R
5 12 4
522

02

\ 6410 = 2245

Example 4. Convert 20145 to its decimal equivalent.
Solution.

Here, given quinary number = 20145
Arranging in quinary place value table

Quinary positional value : 53 52 51 50

Given quinary number : 2 0 1 4

Decimal equivalent : 2×53 0×52 1×51 4×50

\20145 = 2×53 + 0×52 + 1×51 + 4×50

= 250 + 0 + 5 + 4

= 259

\20145 = 25910

154 Whole Number

Exercise 6.5

1. Show the following numbers in their respective place value charts.

(a) 1012 (b) 11102 (c) 1010102 (d) 1110112
(e) 245 (f) 2345 (g) 12405 (h) 430215

2. Convert the following decimal into their binary equivalent.

(a) 3 (b) 5 (c) 9 (d) 10

(e) 17 (f) 34 (g) 40 (h) 100

(i) 125 (j) 144 (k) 302 (l) 425

3. Convert the following binary numbers into their denary equivalent.

(a)102 (b)112 (c)1012
(d)1102 (e)1112 (f)11002
(g)10012 (h)11112 (i)10112
(j)11012 (k)110012 (l)101002
(m)101012 (n)111012 (o)111002
(p)10101112 (q)1100112 (r)10110112

4. Convert the following denery numbers to their quinary equivalent.

(a) 28 (b) 44 (c) 84 (d) 120

(e) 144 (f) 240 (g) 464 (h) 568

(i) 900 (j) 1260 (k) 1324 (l) 4290

5. Convert the following quinary numerals to their denary equivalents.

(a) 145 (b) 245 (c) 325 (d) 405
(e) 1425 (f) 125 (g) 11025 (h) 12305
(i) 24345 (j) 12415 (k) 41205 (l) 201305

6. Convert the following Binary numbers into Quinary numbers.
a. 10112 b) 11012 c) 11112

7. Convert the following Quinary numbers into Binary.
a) 1235 b) 2345 c) 1345

Prime Mathematics Book - 7 155

Unit Revision Test

1. Find the square root of
(a) 576 by using factorization method.
(b) 12769 by using division algorithm.

2. Simplify:
(a) 2205 - 605
(b) 162 + 480

3. Find the square root of 24 correct to two decimal places.

4. Find the least number by which 7623 should be divded to make it a perfect
square.

5. Simplify:
5 3 320 - 3 1715 + 3 1024

6. A cubical tank is 1.2 m long. Find the capacity of the tank in litres.

7. Find the H.C.F. of
(a) 18, 24, 30 by factorization
(b) 112 and 128 using Euclidian Algorithm.

8. Find the greatest number that divides 72, 85 and 101 leaving respectively 2, 1 and
3 remainders.

9. Find the least number which is exactly divisible by 20, 25 and 40.

10. Convert :
(a) 125 into its binary equivalent.
(b) 41205 into its denary equivalent.

156 Whole Number

Answers

Unit: 6 Whole Numbers
Exercise: 6.1
1. a) 529 b) 6561 c) 11664 d) 23409
2. a) 4 b) 12 c) 14 d) 24
f) 252 g) 108 h) 105
e) 84 b) 15 c) 37 d) 53
3. a) 13 f) 113 g) 153 h) 234
d)8 3
e) 83 b) 6 11 c) 9 5 h) 120 3
4. a)3 5
f) 84 g) 36
e)10 5
i) 189 j) 10 3

5. a) 1 b) 3 3 c) 1 d) 7 e) 21 f) 42
4 5 c) 1.5 3 2
6. a) 0.6 c) 8.49
7. a) 1.73 b) 1.2 c) 64m d) 0.48
8. a) 12.49cm b) 4.9 c) 3 d) 11.18
9. a) 2 b) 36cm c) 2 d) 112m
10. a) 3 b) 10 c) 271 d) 7
11. a) 171 b) 7 d) 5
b) 9

1. a) 64 b) 1728 Exercise: 6.2 d) 125000
2. a) 3 b) 5 c) 17576 d) 7
f) 9 c) 6 h) 11
e) 8 j) 16 g) 13
i) 12 7
3 4 d) 10
2 b) 8 c) 9 d) 3 10
3. a) 3 f) 0.5
d) 60m
e) 0.2 b) 12 3 2 + 3 3 c) 13 3 5 +8 3 2 d) not, by 7
4. a) 8 3 3
f) 6 g) 11
e) 15 b) 1728 liters 12
5. a) 13824cm3 b) Yes
6. a) not, by 2 c) 9cm
c) not, by 2

Prime Mathematics Book - 7 157

b) 12 Exercise: 6.3
1. a) 6 b) 9 c) 8 d) 14
2. a) 8 b) 14 c) 6 d) 8 e) 15 f) 12
3. a) 8 f) 20 c) 12 d) 9
b) 10 liters g) 25 h) 12
e) 15 e) 14 c) 36; oranges 2, bananas 3, lichies 5 each
4. a) 18

d) 12

1. a) 12 b) 12 Exercise: 6.4
2. a) 24 c) 240 d) 60
b) 42 c) 48 d) 288
e) 180 f) 192 g) 576 h) 1440
3. a) 42 b) 72 c) 40 d) 72
f) 150 g) 480 h) 120
e) 288 b) 60 litres c) 123 d) 210
4. a) 300 d) 36
b) 32 c) 72
e) 11 am
5. a) 27

1. Show to your teacher. Exercise: 6.5

2. eiaea)))))12171111011002010110212 b) 1113100201000110102010002 02 c) 5119101010000101012000112 102 dhldh)))))111611150100011100001210000212
3. f) g) l) 2010
j) k) p) 8710
b) c)
f) g) d) 4405
h) 42335
i) 1110 j) 1310 k) 2510 l) 1141305
d) 20
m) 2110 n) 2910 o) 2810 h) 190
l) 1290
q) 5110 r) 9110

4. a) 1035 b) 1345 c) 3145

e) 10345 f) 14305 g) 33245

i) 121005 j) 200205 k) 202445

5. a) 9 b) 14 c) 17

e) 47 f) 7 g) 152

i) 369 j) 196 k) 535

6. a) 211005 1102 b) 210305 01012 c) 310015 1002
7. a) b) c)

158 Whole Number

UNIT INTEGERS, RATIONAL AND Objectives:

7 IRRATIONAL NUMBERS

Estimated periods 11

At the end of this unit the student will be able to
● understand operations of integers.
● understand sign rules.
● introduce rational numbers.
● understand the properties of rational numbers.
● introduce irrational numbers.
● understand four fundamental operations of real numbers.

p = 3.142.... +7

-3

(-3) + (+7) = (+4)

Teaching Materials:
Graphs, square papers, chart showing tree of numbers.

Activities:
It is better to
● discuss about different types of numbers in decimal system and their

properties.
● clarify the rule of sign in multiplication of integer by using pattern.
● discuss about irrational numbers.

7.1 Integers

German mathematician Leopold Kronecker had said "God made the natural numbers,
all the rest is the work of man".
In fact all the other numbers are developed from natural numbers according to the
needs. You know that natural numbers are not closed under subtraction. As 5-5=0 which
is not a natural number, thus including 0 in the natural numbers, came whole numbers
(W). Also, whole numbers are not closed under subtraction. As 5-7=-2 is a negative
whole number which is not a whole number, thus including negative whole numbers
came integers (Z).
The set of all the numbers both positive and negative whole number including zero is
called integers.
The set of integers is denoted by I or Z.
Z came from German word “Zahlen” means to count.

-7
+5

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

Thus, Z = { .................... -3, -2, -1, 0, 1, 2, 3, ..................}
Z+ = {1, 2, 3, 4, ..........................} are positive integers.
Z- = {..............., -5, -4, -3, -2, -1} negative integers.

→ Zero (0) is a neutral number
→ In a number line positive integers are put to the right and negative

integers to the left of origin or zero (0)
→ In a number line every number is greater than any integer on its left side 2 > 1 > 0

> -1 > -2 and so on.
→ +1, -1, +2,-2, +3,-3 etc are called opposite integers or additive inverse numbers to

each other.
→ If a and b are two integers, then either a > b or a < b or a = b.
→ If a > b then a is to the right of b, if b> a, b lies to the right of a and if a=b, a and

b represent the same point on the number line. These laws are called the laws of
trichotomy.

7.1.1 Absolute value of an integer

Sometimes, positive and negative numbers create confusion. Let a place P is 3 km from a
place 0 and the place Q is 3km to the opposite direction i.e. -3 km. It seems that distance
between these two places P and Q is (+3 km)+ (-3 km) = 0 km which is impossible.
Here, we need to consider the distances irrespective of direction which we call absolute
value.

160

PQ

km -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 km

Taking + 3 km = 3 km and - 3 km = 3 km
Then distance PQ = 3 km+3 km = 6 km
Absolute value of the integer (+3), denoted by

|+3| = 3
and absolute value of the integer (-3), denoted by

|-3| = 3
The absolute value of an integer is only the numerical value of the integer.

Exercise 7.1.1

1. List the members of the sets.
a) Z1 = {integers between +5 and -5}

b) Z2 = {x : -1 ≤ x ≤ 1, x ∈ Z}
c) Z3 = {x : -5 < x ≤ 2, x ∈ Z}

2. Answer the following.
a) Write the smallest positive integer and the greatest negative integer.

b) What is the successor of the integer -1 ?

c) Write the natural number which is not the successor of any natural number.

d) What types of integers are there to the right and left of origin (O) in a
number line?

3. Join the two integers with appropriate signs “<” or “=” or “ >”

a) 0 -1 b) -2 -7 c) -5 +5

d) 1 -10 e) 0 -15 f) 5 (- 2+7)

4. Write the additive inverse of each the following integers.
a) -3 b) -7 c)+5 d) +12

5. Simplify: b) |-4| + |+4|- |-2|
a) |+5| + |-2|+|-7| d) |+12| +|-12| + |-1|
c) |-10|- |+3| - |+7|

6. a) On a winters day, the minimum temperature of a place is -10°c and maximum
b) temperature is 9°c. Find the difference between the maximum and minimum
temperature of the day.
A volcanic island is 200 m high above the sea level where as its height below
sea level is 3000 m. Find its total height from the sea bed.

Prime Mathematics Book - 7 161

7.1.2 Operations on integers:

Now we will observe rules and properties of fundamental operations on integers.

Addition of integers

a) If the integers to be added are of same signs, their numerical values are added and
their sign remain same.
e.g. (+5) + (+2) = + (5+2)
= +7
and (-3) + (-2) = -(3+2)
= -5

b) If the integers to be added are of opposite signs, we find the difference of their

numerical values and take the sign of the integer of greater numerical value.
e.g = (+5)+(-3) = +(5-3) = +2
and 3 +(-7) = -(7-3) = -4

Properties of addition of integers Closure Property
(Existence of kind): If a and
a) Addition of integers is closed b are two integers then a+b is
e.g. (+2)+(+3) = +5 which is also an integer. also an integer.
(-4)+(+1) = -3 which is also an integer.
Commutative Property
b) Addition of integers is commutative (Change in Position): If a and b are
e.g. (+7)+(+5) = +12 two integers, then a+b = b+a.
(+5) + (+7) = + 12
Associative Property
\ (+7) + (+5) = (+5)+(+7) (Order of grouping): If a, b
and c are three integers then
c) Addition of integers (more than three, is (a+b)+c=a+(b+c)
associative.)
e.g. [(+5) + [(+6)+(+1)] = (+5)+(+7) = + 12
[(+5) + (+6)] + (+1) = (+11) + (+1) = +12

d) Existence of additive inverse
For every integer, there exists another integer such that their sum is zero. In such case
the integers are said to be additive inverse of each other.
e.g. (+5) + (-5) = 0
(-25) + (+25) = 0 Additive inverse makes zero when added.

e) Existence of additive identify:
For every integer, there exists another integer Additive identify
such that their sum is the integer itself. Zero Under operation, identify not lost.
(0) is the additive identify or identify element If a is an integer and b is an other
for any integer. integer such that a+b=a
e.g. (+1)+0 = +1
(-2) + 0 = -2 then b=a-a=0

162 Integer, Rational and Irrational

Subtraction of integers:
Subtraction of an integer is same as addition of the additive inverse of the integer to
be subtracted.
e.g. (+5) - (+3) = (+5) + (-3) = +(5-3) = +2

(+7) - (-3) = (+7) + (+3) = + (7+3) = + 10

Example 1 : Add the following. c) (-3) + (-4) d) (+7) - (+6)
a) (+6) + (+3) b) (+6) + (-4)

Solution:
a) (+6) + (+3) = +(6+3) = +9
b) (+6) + (-4) = +(6-4) = +2
c) (-3) + (-4) = -(3+4) = -7
b) (+7) - (+6) = (+7) + (-6) = + (7-6) = +1

Example 2 : Show the following addition in a number line.
a) (+4) + (+3) b) (+5) + (-3) c) (-6)+ (+2) d) (-4) + (-5) e) (-2) - (-3)

Solution:
a) (+4) + (+3)

= +(4 + 3)
= +7
b) (+5) + (-3)
= (+5) - (+3)
= + (5-3)
= +2
c) (-6) + (+2)
= - (+6) + (2)
= -4
d) (-4) + (-5)
= -(+4) - (+5)
= -(4+5)
= -9
e) (-2) - (-3)
= -(+2)+(+3)
= +(-2+3)
= +1

Prime Mathematics Book - 7 163

Exercise 7.1.2

1. Show the operations in number lines. And simplify.
a) (+2)+ (+4) b) (+5) + (-3) c) (-7) + (+4) d) (-6)+(-3)
e) (+6) - (+3) f) (-7)-(+5) g) (+6)-(-2) h) (-3)-(-4)

2. Copy the following. Write the sum and simplify.

a)

b)

c)
d)

e)

f)

3. Simplify. b) (-4) - (+6) + (+7) c) (+7) + (-8) + (+5)
a) (+3) + (+5) + (+4) e) (-1) + (-4) + (-7) f) (-5) - (-2) - (-6)
d) (-2) - (-4) + (-11) h) (-12) - (-5) + (+7) - (+3)
g) (+14) + (-9) - (+3) + (-2) j) (+3) + (-16) - (-12) - (+10)
i) (-20) - (-10) + (-5) - (+8)

4. Verify the following: b) (+9) + (-3) = (-3) + (-9)
a) (+7) + (+3) = (+3) + (+7) d) [(-8) + (-3)] + (+10) = (-8) + [(-3) + (+10)]
c) [(+4) + (+3)] + (+5) = (+4) + [(+3) + (+5)]

5. Write the additive inverse of : a) (+3) b) (-8)

6. Write the additive identity element of: a) (-5) b) (+10)

7.a) A rhino walks 7 km towards east and then walks 4 km towards west. Show this
information in a number line and find how far and in which direction is it from the
starting point ?

b) Temperature of a place was -3°c in the morning. It increased by 10oc in the day and
decreased by 4°c in the evening. Calculate the temperature of the place recorded
in the evening.

164 Integer, Rational and Irrational

7.1.3 Multiplication and division of integers.

Historical fact:

Hindu Mathematician Brahmagupta (628A.D.) used the rule:
Positive divided by positive or negative divided by negative is positive. Positive
divided by negative is negative and negative divided by positive is negative.

Observe the following:

i) Multiply (+3) and (+2)
(+3) × (+2) = +6
Count a unit (+2) three time to the
right (to the direction of (+2) )

ii) Multiply (-3) × (+2) +2
Count a unit (+2) three times
to the left of 0 (opposite
direction of (+2))
= (-3) × (+3) = -6

iii) Multiply (+3) × (-2) Count a unit
(-2) to its same direction i.e.
three times to the left.
= (+3) × (-2) = -6

iv) Multiply (-3) × (-2)
From 0, count a unit of (-2) thrice
to its opposite direction i.e. to the
right.
= (-3) × (-2) = +6

Note:

(+3) × (+2) means (+3) times (+2) means count a unit of (+2) three times to the same direction of it.
(+3) × (-2) means (+3) times (-2) means count a unit of (-2) to its direction three times.
(-3) × (+2) means (-3) times (+2) means count a unit of (+2) to its opposite direction three times.
(-3) × (-2) means (-3) times (-2) means count a unit of (-2) to its opposite direction three times.

Rule of Signs (in multiplication)

Since the positive integers are also whole numbers and product of whole numbers is also
a whole number, the product of positive integers is a positive integer.

\ (+3) × (+2) = +6
\ (+) × (+) = +

Prime Mathematics Book - 7 165

Follow the Pattern Again Again
(+3) × (+2) = + 6 (+2) × (+3) = +6 (-2) × (+3) = -6
(+2) × (+2) = + 4 (+2) × (+2) = +4 (-2) × (+2) = -4
(+1) × (+2) = + 2 (+2) × (+1) = +2 (-2) × (+1) = -2
(0) × (+2) = 0 (+2) × (0) = 0 (-2) × (0) = 0
(-1) × (+2) = -2 (+2) × (-1) = -2 (-2) × (-1) = +2
(-2) × (+2) = -4 (+2) × (-2) = -4 (-2) × (-2) = +4
(-3) × (+2) = -6 (+2) × (-3) = -6 (-2) × (-3) = +6

\ (-) × (+) = - \ (+) × (-) = - \ (-) × (-) = +

\+×+=+ \ Division being opposite operation of multiplication, the rule

-×+=- holds in division also.
+×-=- i.e.
-×-=+
+÷+=+; +÷-=-; -÷+=- ;-÷-=+

Properties of multiplication of integers:

1. Closure property: Multiplication of two integers is closed.

For example :
(+3) × (+2) = +6 which is also an integer
(+3) × (-4) = -12 which is also an integer
(-4) × (+5) = -20 which is also an integer
(-5) × (-6) = +30 which is also an integer

2. Commutative property: The products of two integers remain unchanged if their

positions are interchanged. If a and b are any two integers
For example: then a × b is also an integer.
(+2) × (-3) = -6

(-3) × (+2) = -6

\ (+2) × (-3) = (-3) × (+2)

3. Associative property: Product of three or more than three integers remain unchanged

if the order in which they are grouped is changed.

For example: = -24 If a and b are any two integers
[(+2) × (+3)] × (-4) = (+6) × (-4) = -24 than a × b = b × a
(+2) × [(+3) × (-4)] = (+2) × (-12)

\ [(+2) × (+3)] × (-4) = (+2) × [(+3) × (-4)]

4. Distributive property: If the sum or difference of two or more integer is multiplied
by an integer, the result is equal to the sum or difference of products of the integer
and the terms of the sum or difference.

166 Integer, Rational and Irrational

For example : If a, b and c are any three inte-

(+2) × [(+3) + (+4)] = (+2) × (+7) = +14 ger then (a × b) × c = a × (b × c)

(+2) × (+3) + (+2) × (+4) = +6 + (+8) = + 14

\ (+2) × [(+3) × (+4)] = (+2) × (+3) + (+2) × (+4)

5. Existence of multiplicative identity: To any integer there exists an integer such

that the product is the integer itself.

For example: If a, b and c are any three
(+5) × (+1) = (+5) integer than a(b + c) = ab + ac

\ (+1) is an identity element

Note: Since division of integer by another integer is not closed, division is not an

operational part of integers.

Exercise 7.1.3

1. Show the following operations in number line.
a) (+3) × (+2) b) (-3) × (+3) c) (+5) × (-2)
d) (-3) × (-2) e) (-3) × (-4) f) (+4) × (-3)

2. Find the products. c) (-7) × (+4) d) (-4) × (-8)
a) (+2) × (+6) b) (+6) × (-5) g) (-9) × (+75) h) (-100) × (-6)
e) (+3) × (+5) f) (+25) × (-8)

3. Simplify. b) (-3) × (+2) × (-7) c) (+3) × [(+4) + (+5)]
a) (+2) × (+3) × (+4) e) [(-5) + (+8)] × (+4) f) [(-12) - (-17)] × (-9)
d) (-4) × [(+2) - (+5)]

4. Verify the following relations.
a) (-5) × (+4) = (+4) × (-5)
b) [(+6) × (-2)] × (-3) = (+6) × [(-2) × (-3)]
c) (+2) × [(+6) + (-8)] = (+2) × (+6) + (+2) (-8)
d) (-3) × [(+4) - (+3)] = (-3) × (+4) - (-3) × (+3)

Prime Mathematics Book - 7 167

7.2 Rational numbers

We have seen that N ⊂W ⊂Z are not closed under division. e.g. let’s consider two

integers 2 and 4 there.

4 ÷ 2 = 2 (2 is an integer)
( )2
÷ 4 = 2 = 1 1 is not an integer
4 2 2

This indicates the need of another set of numbers which can also include such members.
The set of such numbers is called rational numbers.
p
The numbers which can be expressed in the form of q where p and q are integers and

q ≠ 0 are called rational numbers. The set of rational numbers is denoted by the Q.
The word rational comes from the word ratio and the letter Q stands for quotient.
5 9 3 5 1 1 5
\ Q = { ......... , -3, - 2 , - 4 , -2, - 2 , - 4 , -1, - 2 , 0, 2 , 1, 4 , 2, ..........}

Rational numbers include, integers, fractions, terminating decimals and non-terminating
but recurring decimals.

Integers are ratios with consequence (denominator) 1.

e.g. 5 = 5 , -3 = - 3
1 1
p
To convert terminating decimals into q form.

0.5 = 0.5 = 5 = 1 A fraction p in its lowest term is a terminating
1.0 10 2 q when its denominator (q) can be
1
0.25 = 0.25 = 25 = 4 decimal only
1.00 100

0.375 = 0.375 = 375 = 15 = 3 expressed as q = 2p.5q where (p, q =0,1,2.......)
1.000 1000 40 8 form:
p
To convert non-terminating recurring decimals into q

e.g. 0.333 ...... = 0.3ú (read as zero point 3 recurring)
(multiply by 10 both sides)
Let 0.333 ..... =x

i.e. x = 0.3ú ............. (i)

\ 10 x = 3.3ú ................. (ii)

Subtract (i) from (ii) we get

9x = 3
3 1
x = 9 = 3

Note: A fraction p in its lowest term is a recurring decimal only when its denominator
q

(q) has a prime factor other than 2 or 5.

168 Integer, Rational and Irrational

Example : To convert 0.8333.... into p form.
Solution :
q

Given decimal 0.8333 ... Note:

Let x = 0.8333...... • Every terminating decimal can
i.e. x = 0.83ú be converted into a
decimal fraction.
\ 10x = 8.3ú .... (i) (multiply by 10)
Again 100x = 83.3ú .. (ii) (multiply by 10) • Every recurring decimal cab be
converted into a vulgar fraction.
Subtracting (i) from (ii) we get

90x = 75 5 5
75 6 6
x = 90 = \ 0.8333 ...... =

Properties of rational numbers.

Let Q be the set of rational numbers and a and b be any members of Q, then the
following results hold.

1. a, b ∈ Q ⇒ a + b ∈ Q

For example, 5 + 3 = 35+3 = 38 which is a rational number.
7 7 7

2. a, b ∈ Q ⇒ a - b ∈ Q

For example, 1 - 2 = 5-6 = - 1 which is rational number.
3 5 15 15

3. a, b ∈ Q ⇒ ab ∈ Q

For example, 1 × 2 = 2 which is a rational number.
a, b 3 7 a 21
∈ Q, b ≠ 0 b
4. ⇒ ∈ Q

For example, 1 + 2 = 1 × 7 = 7 which is a rational number.
3 7 3 2 6

Thus, the set of rational numbers is closed under all the four fundamental operations
of arithmetic.

5. The set of rational numbers is ordered i.e. If a and b are any two rational numbers
then either a>b or a<b or a=b.
6. If a and b are any two different rational numbers, then there is a rational number
a+b
between them i.e. if a<b then a< 2 <b. There exist infinitely many rational

numbers between two different rational numbers.

Prime Mathematics Book - 7 169

12 1 + 2 11
3 5 30
For example, if 3 and 5 be two different rational numbers, then = is
2 2
a rational number which lies between 1 and 5 .
3
4 5
Example: Insert a rational number between 5 and 7 .
Solution:
A rational number between 4 and 5
5 7

= 4 + 5 = 28+25 = 53
5 7 70 70
2

Comparing 4 and 5
5 7

Since 4 × 7 > 5 × 5

\ 4 > 5
5 7
5 53 4
Now, the numbers in ascending order are 7 , 70 , 5

Irrational numbers

We observed that rational numbers are closed under four fundamental operations
(Addition, Subtraction, Multiplication and Division). But we have some other operations
too, like square roots, cube roots, summing series e.t.c.

Let’s find the square root of 3 by division algorithm.

13 1.732

+1 - 1

27 200

+7 -189

343 1100

+3 - 1029

3462 7100

+2 - 6924

3464 176

\ 3 = 1. 73205 ...

We find that 3 is non terminating and non recurring decimal which certainly cannot
p
be converted to q form. Such members are called irrational numbers.

If the rational numbers are not perfect squares, their square roots are irrational. If
the rational numbers are not perfect cubes, their cube roots are irrational. Thus if a

170 Integer, Rational and Irrational

rational number is not the nth power of a rational number where n∈N, n>1, then n a
is an irrational. Some other irrational numbers are p, e, etc.

p (the ratio of circumference to diameter of a circle)
= 3.1415926 ...
1 1 1 1
e (sum of infinite series 1 + 1! + 2! + 3! + 4! + .......)
e = 2.7182 ........

The set of numbers including rational and irrational numbers is called real numbers.

Set of real numbers are denoted by R. The number system where we use real numbers
is known as real number system.

Does there exist Yes, there exist unreal or
unreal number system ? imaginary numbers too.
We will study about imaginary
numbers in higher classes.

Exercise 7.2

1. Convert the following decimals into fractions.
a) 0.125 b) 0.625 c) 2.072 d) 0.0025

2. Without actual division, state whether the following are terminating or recurring.
1 17 2 5 196
a) 4 b) 40 c) 9 d) 6 e) 250

3. Express the following recurring decimals as vulgar fraction.
a) 0.6 b) 0.3 c) 0.15 d) 0.16

4. Insert a rational numbers between the given numbers and arrange the numbers in
ascending order.
2 3 2 3 2 4 3 5
a) 9 and 8 b) 3 and 4 c) 5 and 7 d) 8 and 9

5. Insert two rational numbers between the given numbers and arrange the numbers
in descending order.

a) 1 and 1 b) - 3 and 2 c) 1 and 1 d) 2 and 3
3 4 4 3 5 6 7 10

6. Write whether the following are rational or irrational. e) 3 8
1 j) 4 81
a) - 3 b) 2 c) 625 d) 0.14
f) 3 12 g) 0.666.. h) 0.14287142871.. i)5 5

Prime Mathematics Book - 7 171

7.3 Order of operations

We have discussed that the set of rational numbers is closed under all the four
fundamental operations, addition, subtraction and division. But when they are to be
performed in mixed form with rational numbers or expressions, they should be in
certain order.
We use a well known rule called BODMAS rule.
i.e. perform the operations in the following order.

B stands for ‘Braces’
O stands for ‘Of’ which is multiplication (×).
D stands for ‘Division’
M stands for ‘Multiplication’
A stands for ‘Addition’
S stands for ‘Subtraction’

Operations with brackets also must be in order

1. − vinculum / semi brackets.

2. ( ) small brackets / round bracket / parenthesis.

3. { } middle brackets / braces.

4 [ ] Large brackets / square brackets.

Addition or subtraction, any one can be preformed first. But if order is broken.
e.g 5 + 3 - 2 = 8 - 2 = 6
4+ 1 of 16
or 5 + 3 - 2 = 5 + 1 = 6 2

‘of’ is for into (×) but must be performed before division. =4+ 1 × 16
2
e.g 4 + 1 + 4 1
2 of 16 = 4 8 = 8 = 2 = 8 × 16 = 128.

Operation of number within bracket with number just out which is wrong.
of it is same as ‘of’

e.g 4 + 2 (5-2) = 4 + 2 of 2 = 4 + 4 = 1

The order of operations holds for irrational and real numbers.

Example 1: Simplify : 12 - 5 × 3 + 12 + 3
Solution: Here,

12 - 5 × 3 + 12 + 3

= 12 - 5 × 3 + 4
= 12 - 15 + 4
= 16 - 15 = 1

172 Integer, Rational and Irrational

Example 2 : Simplify : [14 + {27 - (12 + 11)}] + 6
Solution: Here,

[14 + {27 - (12 + 11)}] + 6

= [14 + {27 - 23}] + 6 = [14 + 4] + 6 = 18 + 6 = 3

Example 3 : Write mathematically and Simplify.
Double of difference of 9 and 3 is
divided by 24.

Solution:
Expression in word = Double of different of 9 and 3 is divided by 24.

Mathematical expression = 2 of (9-3) + 24

= 2 of 6 + 24 = 12 + 24

=2124 = 21

Exercise 7.3

1. Simplify: b) 10 ÷ 2 x (9 – 2 x 3)
a) 20 ÷ 2 (6 x 2 – 7)

c) 45 - 2 x (9 + 2 x 3) d) 36 ÷ 3 x (4 x 2 – 5)

e) 20 + 2 (18 + 12 + 6) f) 16 × 3 - 4 × 21 + 7

g) 30 - {23 - 3(3-5)} h) 24 - 3 (5 + 2 × 5 + 10)
1
i) 3 - 24 + [1-5{7 + (7-15)}] j) 70 - {10 + 3 (18 + 3 × 2)}
k) 5 × 12 + 2 [-9 + {15 + (-9)}]
l) 14 + [3 + 8 + {18 - 8 + (6 - 2)}]

m) 45 + 3 [90 + 2 {6 + 3 (16 - 13)}] n) 72 + 9 [44 + {16 - (4 - 10)}]

o) 30 ÷ [4 + 8 ÷ {3 + 5 ÷ (3 + 4 ÷ 4 - 2)}] p) 252 ÷ 4[24 + {28 ÷(9 + 3 - 5) - 7}]

2. Write following expressions mathematically and simplify.
(a) 5 times the sum of 6 and 4 is subtracted from 100.
(b) Divide 24 by thrice the difference of 9 and 5.
(c) Subtract 9 from the product of 4 and 3.
(d) Add the double of the difference of 12 and 7 to the product of 2 and 5.

3. Make mathematical expressions and simplify.
(a) Sneha had Rs 58. She bought 2 pencils costing Rs 11 each and a copy costing
Rs 24. How much money is left with her ?
(b) Sangdup had Rs 18. Mother gave him Rs 32. In a gambling he bet all his money
and trebbled it. He gave Rs 25 each to his 2 sisters. How much money is left
with him now ?
(c) Rahul had 32 marbles but 4 marbles were lost. The remaining marbles were
divided among 4 including him. How many marbles are there with him now ?

Prime Mathematics Book - 7 173

Unit Revision Test
Unit Revision Test

1. (a) Write the smallest positive integer and the greatest negative integer.
(b) Write the natural number which is not the successor of any natural number.

2. Copy the figure, write the sum and simplify.

-10 -9 -8

3. Verify: [(+4) + (+3)] + (+15) = (+4) + [(+3) + (+5)]
4.
5. Express 0.12 as vulgar fraction. 1 and 1 and arrange in ascending order.
Insert two rational numbers between
6. Show (+3) × (+2) in a number line. 4 3

7. Simplify : [(-5 + (+18)] × (+4).
8. Simplify : 5 × 12 + 2 [-9 + {15 + (-9)}]
9. Write the expression mathematically and simplify.

Add the double of the difference of 12 and 7 to the product of 2 and 5.

10. Sangdup had Rs 35. Mother gave him Rs 25. In a gambling he bet all his money and

doubled it. He gave Rs 10 each to his two sisters. How much money is with him

now ?

Unit Revision Test

1. Simplify:

a) 2 1 - 3 5 + 2 b) 4 1 ÷ 3 × 3
3 6 9 8 4

2

2. Simplify: 1- 1

1 + 1 1 1
+ 2

[ { ( )}]3. 1 3 1 1 1 1 1
Simplify: 1 3 - 4 ÷ 1 8 3 - 1 2 3 - 6

174 Integer, Rational and Irrational

4. In a school there are 450 students. If 2 of the students are boys, find the number
3

of girls.

5. A can do 1 work and B can do 1 of the work in a day. How many days will they
10 15

take to complete the work working together?
6. Convert the following into fraction.
a) 0.016 b) 0.6
7. Simplify: a) 6.78 × 0.37
b) 0.2184 ÷ 0.28
8. Simplify: (216.35+111.25)÷ 3.2
9. A rectangular field has length 42.75 m and breadth 32.24 m. Find its area and
perimeter.

10. What is the result when 2 is divided by the sum of 1 and 1 ?
3 2

Answers

Unit:7 Integers

Exercise: 7.1.1

1. a) z1={4,3,2,1,0 ,-1,-2,-3,-4} b) z2={-1,0,1}
c) 1 d) +ve and -ve respectiely
c) z3={-4,-3,-2,-1,0,1,2}

2. a) 1 and -1 b) 0

3. Show to your teacher. 4. Show to your teacher.

5. a) 14 b) 6 c) 0 d) 25

6. a) 10°c b) 3200m

Exercise: 7.1.2

1. Show to your teacher. 2. Show to yourteacher.

3. a) (+12) b) (-3) c) (+4) d) (-9)

e) (-12) f) (+3) g) 0 h) (-3)

i) (-23) j) (-11)

5. Show to your teacher. 6. Show to your teacher.

7. a)

3km east from the starting point.
b) 3° C

Prime Mathematics Book - 7 175

Show to your teacher. Exercise: 7.1.3
Exercise: 7.2

1 5 9 1
1. a) 8 b) 8 c)2 125 d) 400
d) reccuring
2. a) terminating b) terminating c) reccuring
d) rational
e) termination h) rational

2 1 c) 5 d) 1
3. a) 3 b) 3 33 6
2 17 3
2 43 3
4. a) 9 , 144 , 8 b) 3 , 24 , 4

1 7 13 1 3 -1 5 2
5. a) 3 , 24 , 48 , 4 b) - 4 , 24 , 16 , 3

6. a) rational b) irrational c) rational
e) rational f) irrational g) rational
i) irrational j) rational

1. a) 2 b) 15 Exercise: 7.3 d) 36 e) 60
c) 18 j) 52
f) 36
k) 54 g) 1 h) 6 i) -1
n) 666 l) 39 m) 405 d) 20
2. a) 50 o) 5 p) 3
3. a) Rs 12 b) 2 c) 3
b) Rs 100 c) 7

176 Integer, Rational and Irrational

8UNIT FRACTIONS AND
DECIMALS
Estimated periods 8
Objectives:

At the end of this unit students will be able to
● perform the operation on fractions.
● solve the word problems on fractions.
● identify terminating and non terminating but recurring decimal.
● simplify the expression involved with decimals.
● solve the word problem on decimals.

=+

Teaching Materials:
Fraction Charts, grids, graphs, etc.

Activities:
It is better to
● discuss about simplification of fractions.
● discuss about simplification of decimals.
● explain about the process of solving word problems.

8.1 Operations on fractions

Addition and subtraction of fractions:

Convert mixed fractions into improper fractions.

Convert each fractions to like fractions (having same denominators).

Write single common denominator and add or subtract the terms in the
numerator.

Example: To add 2 and 1
Solution: 9 12

2 + 1 L.C.M. of 9 and 12 3 9, 12
9 12 =3×3×4 3, 4
1 ×3
= 2 ×4 + 12 ×3 = 36
9 ×4
8 3
= 36 + 36

= 8 +3 = 11
36 36

The same process can also be performed as -

Convert mixed fractions into improper fractions if any.

Write the L.C.M of the denominators as common denominator.

In each term divide the L.C.M. by the denominator and multiply the numerator by
the quotient so obtained.
Add or subtract the terms in the numerators.
Solutions:

2 + 1
9 12

= 2 ×4 + 1 ×3
36
8+3 11
= 36 = 36 L.C.M. of 9 and 12 = 36
36 ÷ 9 = 4 and 36 ÷12 = 3

Example 1 : Simplify : 2 3 + 1 2 - 3 1
4 5 10
Solution:

2 3 + 1 2 - 3 1
4 5 10

= 11 + 7 - 31 = 11× 5 + 7 × 4 - 31 × 2
4 5 10 4× 5 5 × 4 10 × 2

= 55 + 28 - 62 = 55 + 28 - 62 = 83-62 = 21 = 1 1
20 20 20 20 20 20 20

178 Fraction and Decimal

Alternate method:

Solution: LCM of 4, 5 and 10 = 20
20 ÷ 4 = 5 and 5 × 11 = 55
2 3 + 1 2 - 3 1 20 ÷ 5 = 4 and 4 × 7 = 28
4 5 10 20 ÷ 10 = 2 and 2 × 31 = 62
11 7 31
= 4 + 5 - 10

= 5×11 + 4×7-2×31
20

= 55 + 28-62
20

= 83 - 62 = 21 = 1 1
20 20 20

Product of two fractions

To multiply two fractions:

Convert mixed fractions into improper fractions, if any.

Numerator of the result is the product of the numerators and denominator of the
result is the product of the denominators.

If the result is an improper fraction, convert it to mixed one.

Cancellation can be applied in any step.
3
Example 2 : Multiply 1 and 11
Solution: 4 14

1 3 × 11 = 7 × 11 = 11 = 1 3
4 14 4 14 8 8

Division of a fraction by another fraction:

Convert mixed fraction into improper fraction, if any.

Multiple the first fraction by the reciprocal of the divisor.

Example 3: Perform the division 1 1 ÷ 1 7
Solution : 8 20

1 1 ÷ 1270 = 9 ÷ 27 = 9 × 20 = 5
8 8 20 8 27 6

Complex fraction: If fraction occurs within a fraction, it is a complex fraction.
1
2
e.g. 3 , 1 etc
4
1 + 1
1
1 + 2

Prime Mathematics Book - 7 179

1 1
2 D
Here, 3 = 1 × 4 = 2 Complex fraction = N ×
2 3 3

4 1 1 1 1 1 1 3
1 1 3×1+2 5 5
And 1 = 2×1+1 = 1 + 3 = 2 = = =
+ 2 3
1 + 1 1 + 1 + 33
2
1 2

Example 4. Simplify :

4 1 ÷ 3 × 12
2 8 16

Solution:

9 ÷ 3 × 12 = 9 × 8 × 12 =9
2 8 16 2 3 16

Example 5. Simplify :

1- 1

1 + 1 1 1
+ 2

Solution: 1- 1 = 1- 1 = 1- 1
1+ 1 2
1 + 2 1 3 1 + 3
+1
22
3 5-3 2
= 1- 1 = 1- 1 = 1- 5 = 5 = 5
3+2 5
33

Exercise 8.1

1. Add the following:

(a) 2 + 1 (b) 3 + 4 (c) 2 + 5 (d) 3 + 5
8 8 7 7 3 6 8 4

(e) 2 + 4 (f) 3 + 4 (g) 2 + 5 (h) 3 + 1
3 5 5 7 3 12 4 20

2. Subtract the following:

(a) 7 - 3 (b) 6 - 1 (c) 1 - 1 (d) 5 - 1
8 8 7 7 3 9 7 14

180 Fraction and Decimal

(e) 4 - 2 (f) 3 - 5 (g) 2 - 4 (h) 3 - 1
15 9 10 6 3 7 4 6

3. Calculate the following:

(a) - 1 +1 4 (b) - 4 1 + 2 (c) - 2 5 +1 3 (d) 3 5 +2 2
8 5 5 3 6 8 6 3

(e) 1 -2 3 (f) 2 3 - 1 (g) - 5 + 3 4 (h) - 2 3 +4
6 4 4 12 7 5

4. Simplify:

(a) 2 + 3 + 7 (b) 3 + 1 - 5 (c) 2 1 - 3 5 + 2 (d) 3 1 -1 3 - 2112
5 10 20 4 2 6 3 6 9 4 8

(e) 3 × 1 × 5 1 (f) 2 ÷ 18 ÷ 9 (g) 2 4 × 2227 ÷ 1 7 (h) 4 × 1 ÷ 3 + 6
5 8 3 3 25 10 7 9 5 3 2 9

5. Simplify: 1 2 1 7
3 4 18
11 2 1

(a) 1 (b) 15 (c) 5 (d) 7
7 16 12
3 1 4

(e) 1- 1 (f) 1 (g) 3- 2
3
1 - 1 1 2 1 + 1 1 1 2- 4
- 3 + 3 3- 5

8.2 Word problems on fractions

We are familiar with the fundamental operations of fractions. Now we try to solve the
word problems related to fractions. For this,

Write the problem mathematically.
Simplify the expression so obtained.

Example 1: Rahul ate 1 of a cake and his sister ate 2 of the cake. What
4 7

fraction of the cake is left ?

Solution:

Rahul ate 1 of the cake, His sister ate 2 of the cake
4 7

Prime Mathematics Book - 7 181

Remaining cake =1- 1 - 2
4 7

28 × 1 - 7 × 1 - 4 × 2
= 28

= 28 - 7 - 8 = 13 \ 13 of the cake is left.
28 28 28 the students
Example 2: There 2
many are girls? are 450 students in a school. If 3 of are boys, how

Solution:

Here, Total No. of students = 450

No. of boys = 2 of 450
No. of girls 3
2 2
= 450 - 3 of 450 = 450 - 3 × 450

= 450 – 2 × 150 = 450 – 300 = 150

\The number of girls in the school is = 150.

Exercise 8.2

1. a) 3 of a sum is Rs 1800. Find the sum.
b) 7 shopkeeper 3
sold 4 of the cloths and 350 m of cloth was left. How much
c) A
d)
2. a) cloths did he have in the beginning? 2
7
b) There are 35 oranges in basket. Out of which are rotten. How many good
oranges are there?
7
Out of 48 students in a class 12 are girls. Find the number of girls and boys.
In an examination, Nitesh of full marks
and Ritesh secured 3 and 4
4 5
respectively. If full mark is 200, who secured more marks and by how much?
2 1
Mr. Pradhan gave 25 of his money to his son and 10 to his daughter.

Who got more money?

3. a) A can do 1 of a work and B can do 1 of a work in a day. How much work
b) 10 15
will they finish working together in a day ?
1 1
A and B together can do 6 of a work in a day. If A can do 10 of the work

in a day, how much work can B do in 1 day?

182 Fraction and Decimal

c) If the machines x, y and z can do 1 , 1 and 1 of a work respectively
10 12 15
in 1 day. What part of work is done by them together in 1 day?
1
4. a) What is the result when 1 is divided by the sum of 1 and 2 ? 1 3
b) What is the quotient when 10 is divided by 1 2 4
c) Subtract the quotient when 2 is divided by the 1 3 of the sum of and ?
1
sum of 1 and 2 from 2 .
3 5

8.3 Decimals

Terminating and Non terminating but recurring decimals

In rational numbers we have two types of decimals

i) Terminating decimals ii) Non-terminating decimals

Terminating decimals
Let's consider the following fractions with their decimal forms

i) 5 = 2) 5 ( 2.5 ii) 2 = 5) 20 (0.4 iii) 3 =4 ) 3 ( 0.75
2 -4 5 20 4 0

10 0 30
28
-10 20
0
20
0

iv) 3 = 25 ) 3 ( 0.12 v) 1 =8 ) 1 ( 0.125
25 0 8 0 20
-16
30 10 40
25 -40
50 -8 0

50
0

vi) 3 = 125 ) 3 ( 0.024 vii) 7 = 64 ) 7 ( 0.109375
125 0 64
0
300 70 240
-0 64 192
300
250 60 480
500 -0 - 448
-500 600
0 -576 320
- 320

0

Prime Mathematics Book - 7 183

In each of the above cases, the process of division doesn't go infinitely but ends after
certain steps leaving no remainder. Such decimals are called terminating decimals.

The fractions having denominator powers of 2 or 5 or both 2 and 5 or their products
give terminating decimals.

Converting terminating decimals into fractions:

Given terminating decimal

Divide the given decimal number by 1 (i.e put 1 as denominator) with as many zeroes
after decimal as there are digits after decimal in the numerator.
Remove decimal in numerator and denominator and convert the fractions into
lowest term.

Example: 15012=

i) To convert 0.5 into fraction 0.5 = 0.5 = 1
1.0 2

ii) To convert 0.24 into fraction 0.24 = 01..0240= 1204062=5 6
25

iii) To covert 0.0305 into fractions 0.0305 = 0.0305 = 305 = 61
1.0000 10000 2000

Non- terminating but recurring decimals
Let’s consider the following fractions and their decimal forms.
2 1
i) 3 = 3) 2 ( 0.666... ii ) 7 = 7) 1 ( 0.142857142857...
-0
20 0 40 20
-18 10 - 14
20 -35
-18 7 60
20 30 50 -56
-18 28
2 -49 40
20 -35
14 10
-7 50
60 30 -49
56 -28
1

iii) 1 = 9) 1 ( 0.111... iv) 1 = 11 ) 1 ( 0.0909...
9 0 11 -0

10 10
-9
10 -0
100
-9 - 99
10
-9 10
-0
1 100

-99
1

184 Fraction and Decimal

In each of these cases, the process of division never ends which can go infinitely. But

we observe that same digit or a group of digits repeat again and again.
1
3 = 0.333 ……… the digits 3 goes on repeating

1 = 0.111………. the digits 1 goes on repeating
9

1 = 0.090909……… the group of digits 09 goes on repeating
11

1 = 0.142857142857……. the group of digits 142857 goes on repeating
7

Such decimals in which a digit or a group of digits repeat (recur) are called
non-terminating but recurring decimals. Fractions having denominators whose prime
contain digits other than 2 or 5 give non- terminating but recurring decimals. Recurrence
of digits or group of digits is denoted by putting dot over the digit or above the first
and last digit of group of digits which recur.

Thus,

0.333….. = 0.3

0.111………… = 0.1

0.090909…….. = 0.09

0.142857142857………. = 0.142857

Converting recurring decimals into fractions:
Given a recurring decimal say 0.3

Let, x = 0.3 ---------- (i)
10x = 3.3 ---------- (ii)

Subtracting (i) from (ii), we get

9x = 3

or, x = 3 Multiplying both sides by 10 a power of
9 10 so on to more the position of decimal
(point) after the digit/group of digits
or, x = 1 where recurrence occur.
3

\ 0.3 = 1
3

Prime Mathematics Book - 7 185

Historical fact:

1 is a non terminating but recurring decimal.
19

To divide 1 by 19

19 ) 1 ( 0.052631578947368421
0
100 150 130
-95 - 133 -114

50 170 160
- 38 - 152 -152

120 180 80
- 114 - 171 -76

60 90 40
- -57 - 76 -38

30 140 20
-19 -133 -19
110
70 1
-95 -57 Repetition starts.

To divide 1 by 19, thus we need to divide 18 times using usual division algorithm
forming a long-tail. Recurrence starts after 18 decimal places. Of course a 12 digited
calculator cannot show the point of termination after which same digits repeat.

Shankaracharya Bharati Krishna Tirthaji (1884-1960) in Vedic Ganita has explained
different rules and sub rules (Sutras and Upasutras) of mathematics. Among them a
rule "Ekadhikena Purvenah" is about recurring decimals.

"Ekadhikena Purvenah" means by one more than previous. This method is used to
1
change the fraction having ones digit of denominator 9 into decimals. To change 19

into decimals ones digit of denominator is 9; the digits previous to 9 is 1; one more
than previous is thus 1+1=2.Dividing 1 by 9 or number having ones digit 9 end at 1 so

the above division is proceed by multiplying or dividing 1 by 2.

A Multiplication Method:

Step 1 Put the ending digit 1 to the right.
Step 2 Multiply 1 by 2 = 2 put 2 just left to 1 (of step 1)
Step 3 Multiply 2 by 2 = 4 and put 4 just left to 2 (of step 2)
Step 4 Multiply 4 by 2 = 8 and put 8 just left to 4 (of step 3)
Step 5 Multiply 8 by 2 = 16 and put 6 just left to 8 (of step 4), take 1 carry over.
Step 6 Multiply 6 by 2 = 12 plus carry over 1 = 13 put 3 just left to 6 (of step 5) and
take 1 carry over and so on until repetition starts. Do not repeat and put decimal.

11
19 2 0.0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1

186 Fraction and Decimal

B. Division Method: The vedic operator is 1
2
Step 1 Put decimal; divide 1 by 2, quotient (q1) = 0 and remainder (r1) = 1. Put 0 right to
the decimal.

Step 2 Divide the number formed by r1 and q1, i.e 10 by 2 we get q2=5 and r2=0

Step 3 Divide the number formed by r2 and q2 i.e 05 by 2 and so on until repetition starts.

\ 1 1 0.1 0 51 2 6 31 11 51 71 8 91 4 71 31 6 8 4 2 1
19 2

It is interesting that the sum of the first half of the digits and remaining half of the
digits give all 9.

052632578
+ 947368421

999999999 Which further makes the process more simplified.

Example 1: Convert 0...24 into fraction.
Let x = 0.2.4 . ---------- (i)
100x = 24.24 ---------- (ii)

Subtracting (i) from (ii) we get

99x = 24
or, x
= 248
\ 0.24 9933
8
= 33

Example 2: Convert 0.24 into fraction
Solution:
Given recurring decimal 0.2
= 0.24.
Let x = 2.4. (only 4 recurring)
\ 10x ---------- (i)
Again multiplying by 10 both side

100x = 24.4. ----------(ii)
Subtracting i) from ii) we get

90x = 22
2211
x = 90
. . 45

0.24. = 11
45

Prime Mathematics Book - 7 187

Example 3 : Convert 1.212 into fraction
Solution
..
Given recurring decimal 1.212
..
Let x = 1.212
Multiplying by 10. . both side
10x = 12.12 ---------- (i)

Again multiplying .b.oth side by 100 we get

1000x = 1212.12 ---------- (ii)

Subtracting (i) from (ii) we get

990x = 1200 40 .. 40
1200 33 33
x = 990 or x= \ 1.212 =

Exercise 8.3

1. Say whether the following fractions give terminating or non- terminating but
recurring decimal.
1 1 2
a) 1 b) 1 c) 1 d) 16 e) 14 f) 9
3 8 5

g) 1 7 i) 1 j) 3 k) 1 l) 1
25 h) 125 11 13 19 40

2. Convert the following decimals into fractions.

a) 0.25 b) 0.2 c) 0.5 d) 0.124
h) 2.035
e) 0.046875 f) 0. 0016 g)4.35

3. Find the non terminating but recurring decimals of.
1 1
a) 1 b) 6 c) 11 d) 3
3 7

e) 1 f) 1 g) 5 h) 7
14 13 9 30

4. Cao)n0v.e2..r.t the following decbi)m0a.3l..s. into fractions. c) 0.1. . . ..
d) 00..3.16.2
e) 0.3.5 f) 0.4.5 g) 0.1. 24 h)

i) 2.46 j) 1.1 k) 2.9

188 Fraction and Decimal

8.4 Fundamental operations with decimal

Addition and subtraction of decimals

Arrange the numbers to be added vertically so that the digits in same place lie
along the same line.

Add or subtract usually like in the case of addition and subtraction of whole
numbers.

Example 1: Add 236.432 and 384.56
Solution:

2 3 6. 4 3 2
+ 3 8 4. 5 6

6 2 0. 9 9 2
Thus, the sum of 236.432 and 354.56 is 620.992.
Example 2 : Subtract 67.573 from 165.8625
Solution:

1 6 5. 8 6 2 5
- 6 7. 5 7 3

9 8. 2 8 9 5
\165.8625 - 67.573 = 98.2895

Example 3: Simplify 375.47 + 82.56 - 23.27
Solution:

375.47 + 82.56 – 23.27
= 458.03 - 23.27 = 434.76

Example 4: A tree 14.6 m is broken by the wind. If the broken part of the tree is 6.9m,
find the length of the remaining part of the tree.

Solution:
Here, height of the tree = 14.6
\Length of broken part = 6.9m

Length of the remaining part = 14.6m - 6.9m = 7.7 m
\The length of the remaining part is 7.7m

Exercise 8.4

1. Carry out the following additions.
a) 86.421 + 352.56 b) 0.09 + 2.005
c) 345.54 + 23.567 +1745.7 d) 845.034 + 18.304 +18.75

2. Carry out the following subtractions.
a) 1.872-0.359 b) 10.5-4.68
c) 125-4.375 d)13.467-2.009

3. Simplify. b) 0.54-0.72+0.8
a) 0.49+0.87-0.9 d) (5.2+0.08) – (0.8+0.07)
c) 24.37-16.2+6.12

Prime Mathematics Book - 7 189

4. Solve the following word problems.
a) The sum of 5.64 and 12.08 is subtracted from 22.32. What is the result?
b) Sonit secured 39.7 marks in mathematics and failed by 10.3 marks,
what was the pass mark?
c) A rope 36.6 m is made into three pieces. If the length of two pieces
are 9.75 m and 13.6 m, find the length of the remaining piece.
d) Rojesh buys a birthday cake for Rs 575.50, a packet of sweets for
Rs 112.25 and ice cream for Rs 148. If he gives Rs 1000 note to the
shopkeeper, how much does he get as a return?

8.5 Multiplication and division of decimals.

A. Multiplication of a decimal by a whole number :

Consider the product of 0.25 on a number line

We observe 0.25 0.25
0.25 × 5 = 1.25 ×5 ×10
1.25 00 0.25
×15
0.25 × 10 = 2.50 +25 125
2.50 +25
3.75
0.25 × 15 = 3.75

Multiply usually as in whole numbers.

Number of digits after decimals remain same.

B. Multiplication of decimal number by other decimals.

Consider the multiplication 0.75 × 0.25
75 25 1875

Here, 0.75 × 0. 25 = 100 × 100 = 10000 = 0.1875

Multiply usually as in whole numbers.
No. of digits after decimal in the product is the sum of the digits after decimals
in the multiplicants.

Example 1: Carry out the multiplication 17.43 × 3.5
Solution:
17.43 × 3.5 = 17.43
× 3.5 Multiply 17.43 × 3.5 usually irrespective of decimal point
8715 No of digits after decimal in the product = 2 + 1 = 3

5229 Put the decimal after 3 digits counting from right to left

61.005 \17.43 × 3.5 = 61.005

190 Fraction and Decimal

Example 2: The length of a rectangular field is 32.75 m and breadth is 13.25 m. Find

the area of the field. 32.75
Solution: × 13.25
Here, length (l) = 32.75m, breadth (b) = 13.25 m
16375
\ Area of the rectangular field= l × b 6550
9825
= 32.75m × 13.25 m = 433.9375 m2 +3275

\ The area of the field is 433.9375m2

Division of Decimals 433.9375

Division involved with decimals.

Consider the division 13 + 8

8) 13 ( 1 whole number, 6 tenths, 2 hundredths, 5 thousandths = 1.625
-8
50 → 5 ones = 50 tenths
- 48
20 → 2 tenths = 20 hundredths.
-16
40 → 4 hundredths = 40 thousandths
- 40
0

Division of a decimal by a whole number.
Consider the division 197.52 + 16

16) 197.52 (12 whole number, 3 tenths, 4 hundredths, 5 thousandths = 12.345
- 16
37
- 32

55 → 5 ones = 50 tenths + 5 tenths = 55 tenths.

-48

72 → 7 tenths = 70 hundreds + 2 hundreds = 72 hundreds

- 64

80 → 8 hundreds = 80thousands.

- 80

0 \ 197.52 + 16 = 12.345

Division of a decimal by a decimal.
Consider the division 10.5765 + 0.25
100
Here, 10.5765 = 10.5765 × 100
0.25 0.25

= 1057.65211.53 = 211.53 = 42.306
25 5 5

Prime Mathematics Book - 7 191

Example 1. Carry out the division 34 + 16 Since decimals in the dividend
and the divisor can be converted
Solution to whole numbers, divisor of a
decimal by a decimal is same
Here 34 + 16 as division of decimal by whole
number or division of whole
16) 34 ( 2.125 number by a whole number.
-32
20 tenths \ 34 +16 = 2.125
-16
40 hundreds
-32
80 thousands
-80
0

Example 2. Divide 54.32 by 1.4

Solution: Now,
7) 271.6 (38.8
54.32 + 1.4
54.32 × 10 -21
= 1.4 × 10 61
-56
= 543.2 56
14 -56
271.6 0
= 7

\ 54.32 + 1.4 = 38.8

Example 3. Carry out the division 180.33 + 2.5.

Solution: Now,

Here, 180.33 + 2.5

= 180.33 × 10 25) 1803.3 ( 72.132
2.5 × 10 -175
1803.3 53
= 25 -50
33
-25
80
-75
50
-50

0 \ 180.35 + 2.5 = 72.132

192 Fraction and Decimal

Example 4. Simplify: 9.26 × 74.375 + 3.4

Solution: 74.375 + 3.4 = 743.75
Here, 34
34) 743.375 (21.875 21.875
9.26 × 74.375 + 3.4 × 9.26
-68 255 131250
= 9.26 × 21.875 63 - 238 43750
= 202.5625 -34 +196875
297 170 202.56250
-272 - 170

0

Example 5. Find the perimeter of the rectangle with length 3.42cm and breadth 2.25 cm.
Solution:
Here length (l) = 3.42 cm, breadth (b) = 2.25 cm

We have perimeter of rectangle = 2(l+b)
= 2(3.42cm + 2.25cm)
= 2 × 5.67 cm

\ The perimeter of the rectangle is 11.34cm.

Exercise 8.5

1. Carry out the following multiplication
(a) 6.78 × 0.37 (b) 46.5 × 1.28

(c) 9.425 × 7.26 (e) 3.46 × 2.8

2. Carry out the following divisions.

(a) 28.48 + 8 (b) 23.04 + 2.4

(c) 0.2184 + 0.28 (d) 65.326 + 7.34

3. Simplify: (b) 3.08 × 1.2 + 0.8 × 0.5
(a) 4.08 - 2.5 × 0.74

(c) 7.2 + 0.24 × 0.35 (d) (216.35 + 111.25) + 18.2

(e) (0.85 + 0.59) + (3.59 - 3.47)

4 Solve the following problem.
(a) If 1 kg of rice costs Rs 52.25, what is the cost of 12.4kg of the rice?
(b) The length of a rope is 75m, how many pieces of ropes 1.5m long can be cut from
the rope?
(c) If the area of a rectangular field is 328.018m2 and its length is 20.45m, find the
breadth.

Prime Mathematics Book - 7 193