Lesson: Number Systems
Exercise 1.1 (21 Multiple choice Questions and
answers)
Question: 1
Every rational number is:
(a) A natural number
(b) An integer
(c) A real number
(d) A whole number
Solution:
c
All natural numbers are part of whole numbers. All
whole numbers are part of integers. All integers are
part of rational numbers and all rational numbers are
part of real numbers.
Question 2
Between two rational numbers:
(a) There is no rational number.
(b) There is exactly one rational number.
(C) There are infinitely many rational numbers.
(d) There are only rational numbers and no irrational
numbers.
Solution:
c
Question 3
Decimal representation of a rational number cannot
be:
(a) Terminating
(b) Non-terminating
(c) Non-terminating repeating
(d) Non-terminating non-repeating
Solution:
d
Decimal representation of a rational number is either
terminating or non-terminating repeating. Decimal
representation of irrational number is non-terminating
and non-repeating.
Question 4
The product of any two irrational numbers is:
(a) Always an irrational number
(b) Always a rational number
(c) Always an integer
(d) Sometimes rational, sometimes irrational
Solution:
d
For example, if
is multiplied by , then the
2
2
product is 2 which is a rational number. If is
2
multiplied by
then the product is which is an
3
6
irrational number. Hence, the product of any two
irrational numbers can sometimes be rational and
sometimes irrational.
Question 5
The decimal expansion of the number is:
2
(a) A finite decimal
(b) 1.41421
(c) non-terminating recurring
(d) non-terminating non-recurring
Solution:
d
The above number is an irrational number. We know
that decimal expansion of an irrational number is non-
terminating and non-recurring.
Question 6
Which of the following is irrational?
(a)
4
9
(b)
12
3
(c)
7
(d)
81
Solution:
c
We find see that
4 2
9 3
12 2 3
2
3 3
81 9
All the above numbers are rational numbers. Only
7
is an irrational number.
Question 7
Which of the following is irrational?
(a) 0.14
(b)
0.1416
(c)
0.1416
(d) 0.4014001400014...
Solution:
d
We can see that 0.14 is a terminating decimal,
0.1416
is a non-terminating repeating decimal, is a
0.1416
non-terminating repeating decimal and
0.4014001400014... is a non-terminating non-
repeating decimal. Therefore, 0.4014001400014... is
an irrational number.
Question 8
A rational number between and is:
2
(a)
2 3
2
(b)
2. 3
2
(c) 1.5
(d) 1.8
Solution:
c
is 1.414…. and is 1.732… Therefore, a
2
rational number between 1.414.. and 1.732… is 1.5.
Question 9
The value of 1.999... in the form , where p and q are
p
q
integers and q 0, is:
(a)
19
10
(b)
1999
1000
(c) 2
(d)
1
9
Solution:
c
1.999...
1.9 (1)
10
10 19.9 (2)
10 19.9 1.9
9 18
2
x
x
x
x x
x
x
Let be equal to
Multiplying both sides by , we get
Subtracting the first equation from the second
equation, we get
Question: 10
is equal to:
2 3 3
(a)
2 6
(b) 6
(c)
3 3
(d)
4 6
Solution:
c
2 3 3
(2 1) 3
3 3
Question: 11
is equal to:
10 15
(a)
6 5
(b)
5 6
(c)
25
(d)
10 5
Solution:
b
10 15
10 15
5 2 3 5
5 2 3
5 6
Question: 12
The number obtained on rationalizing the denominator
of is:
1
7 2
(a)
7 2
3
(b)
7 2
3
(c)
7 2
5
(d)
7 2
45
Solution:
a
1
7 2
1 ( 7 2)
( 7 2)( 7 2)
7 2
7 4
7 2
3
Question: 13
is equal to:
1
9 8
(a)
1
3 2 2
2
(b)
1
3 2 2
(c)
3 2 2
(d)
3 2 2
Solution:
d
1
9 8
9 8
9 8 9 8
9 8
9-8
= 9+ 8
=3+ 2×2×2
=3+2 2
Question: 14
After rationalizing the denominator of , we
7
3 3 2 2
get the denominator as:
(a) 13
(b) 19
(c) 5
(d) 35
Solution:
b
7
3 3 2 2
7 3 3 2 2
3 3 2 2 3 3 2 2
7 3 3 2 2
2 2
3 3 2 2
7 3 3 2 2
27 8
7 3 3 2 2
19
Therefore, the denominator is 19.
Question: 15
The value of is equal to:
32 48
8 12
(a)
2
(b) 2
(c) 4
(d) 8
Solution:
b
32 48
8 12
4 2 4 3
2 2 2 3
4 2 3
2 2 3
4
2
2
Question: 16
If 1.4142, then is equal to:
2
2 1
2 1
(a) 2.4142
(b) 5.8282
(c) 0.4142
(d) 0.1718
Solution:
c
2 1
2 1
2 1 2 1
2 1 2 1
2 1 2 2
2 1
3 2 2
1
0.1716
0.4142
Question: 17
equals:
4
3
2
2
(a)
1
6
2
(b)
6
2
(c)
1
6
2
(d)
6
2
Solution:
c
4
3
2
2
1
2
4
3
2
2 1
3 4
2
1
6
2
Question: 18
The product equals:
3 4 12
2. 2. 32
(a)
2
(b) 2
(c)
12
2
(d)
12
32
Solution:
b
3 4 12
2. 2. 32
1 1
1
3 4
5
12
2 . 2
1 1 5
3 4 12
2
4 3 5
12
2
12
12
2
2
Question: 19
Value of is:
2
4
81
(a)
1
9
(b)
1
3
(c) 9
(d)
1
81
Solution:
a
2
4
81
2
2
4
9
2
2
4
9
4
4
9
1
9
1
9
Question: 20
Value of is:
0.16 0.09
256 256
(a) 4
(b) 16
(c) 64
(d) 256.25
Solution:
a
0.16 0.09
256 256
0.25
256
1
2
4
16
1
2
4
16
1
2
16
4
Question: 21
Which of the following is equal to ?
x
(a)
12 5
7 7
x x
(b)
1
12
4
3
x
(c)
2
3
3
x
(d)
12
7
7 12
x x
Solution:
c
2
3 2
3
3
2 3
x x x
Exercise 1.2
Question: 1
Let x and y be rational and irrational numbers,
respectively. Is necessarily an irrational number?
x y
Give an example in support of your answer.
Solution:
As the sum of a rational and an irrational number is
always an irrational number, is necessarily an
x y
irrational number.
For example: is an irrational number.
3 2
Question: 2
Let x be rational and y be irrational. Is xy necessarily
irrational? Justify your answer by an example.
Solution:
No, xy may not be irrational always.
For example:
Let and , then xy = 0 × = 0 which is a
0
x
3
y
rational number.
Question: 3
State whether the following statements are true or
false? Justify your answer.
(i)
is a rational number.
2
3
(ii) There are infinitely many integers between any
two integers.
(iii) Number of rational numbers between 15 and 18 is
finite.
(iv) There are numbers which cannot be written in the
form
both are integers.
,
p
q
0,
q p,q
(v) The square of an irrational number is always
rational.
(vi)
is not a rational number as and
are
12
3
12
not integers.
(vii)
is written in the form ,
and so it is a
15
3
p
q
0
q
rational number.
Solution:
(i) False. The quotient of an irrational number and a
none zero rational number is always irrational.
(ii) False. Between 2 and 3, there is no integer.
(iii) False. Between any two rational numbers we can
find infinitely many rational numbers.
(iv) True. Irrational numbers cannot be written in the
form , where p and q are integers and
p
q
0
q .
(v) False, as
which is not a rational
2
4
2 2
number.
(vi) False, because
which is a rational
12
4 2
3
number.
(vii) False, because , which is not a rational
15
5
3
number.
Question: 4
Classify the following numbers as rational or irrational
with justification:
(i)
196
(ii)
3 18
(iii)
9
27
(iv)
28
343
(v)
0.4
(vi)
12
75
(vii) 0.5918
(viii)
 1 5 4 5
(ix) 10.124124...
(x) 1.010010001...
Solution:
(i) Rational Number.
196 14
14 is a rational number.
(ii) Irrational number
.
3 18 9 2
is the product of the non-zero rational, 9 and the
9 2
irrational number, . The product of a non-zero
2
rational number and an irrational number is an
irrational number.
(iii) Irrational Number.
9 1
27
3
is the quotient of the non-zero rational, 1 and the
1
3
irrational number,
.3
The quotient of a non-zero rational number and an
irrational number is an irrational number.
(iv) Rational Number
28 2
7
343
is a rational number.
2
7
(v) Irrational
2
0.4
10
is the quotient of the non-zero rational number,
2
10
and the irrational number,
2
.10
The quotient of a non-zero rational number and an
irrational number is an irrational number.
(vi) Rational number
12 2 3 2
5
5 3
75
is a rational number.
2
5
(vii) Rational number
0.5918 is a terminating decimal. Therefore, it is a
rational number.
(viii) Rational number
.
 
1 5 4 5 1 4 5 5
3
is an integer. All integers are rational numbers.
3
(ix) Rational.
10.124124... is a non-terminating and recurring
decimal. Hence, it is a rational number.
(x) Irrational.
1.010010001... is a non-terminating and non-recurring
decimal. All non-terminating and non-recurring
decimals are irrationals.
Exercise 1.3
Question: 1
Find which of the variables x, y, z and u represent
rational numbers and which irrational numbers:
(i)
2
5
x
(ii)
2
9
y
(iii)
2
.04
z
(iv)
17
2
4
u
Solution:
(i) x represents an irrational number.
2
5 5
x x
are irrational numbers.
5
(ii) y represents a rational number.
3
2
y 9 y
are rational numbers.
3
(iii) z represents a rational number.
2
.04 0.2
z z
are rational numbers.
0.2
(iv) u represents an irrational number.
17 17
2
4 2
u u
are irrational numbers.
17
2
Question: 2
Find three rational numbers between:
(i) –1 and –2
(ii) 0.1 and 0.11
(iii)
5 6
7 7
and
(iv)
1 1
4
5
and
Solution:
(i) –1 and –2
Three rational number between –1 and –2 are –1.1, –
1.2 and –1.3.
(ii) 0.1 and 0.11
Three rational numbers between 0.1 and 0.11 are
0.105, 0.106 and 0.107.
(iii)
5 6
7 7
and
5 5 10 50
10
7 7 70
6 6 10 60
10
7 7 70
So, three rational numbers between
are:
5 6
7 7
and
55 56 57
, .
70 70 70
and
(iv)
1 1
4
5
and
LCM of 4 and 5 is 20.
We can write,
1 1 5 5 10 50 1 1 4 4 10 40
4 4 20 10 200 4 20 10 200
5 5 5
and
So, three rational numbers between
1 1
4
5
and are:
41 42 43
,
200 200 200
and
Question: 3
Insert a rational number and an irrational number
between the following:
(i) 2 and 3
(ii) 0 and 0.1
(iii)
1 1
3 2
and
(iv)
2 1
2
5
and
(v) 0.15 and 0.16
(vi)
2 3and
(vii) 2.357 and 3.121
(viii) .0001 and .001
(ix) 3.623623 and 0.484848
(x) 6.375289 and 6.375738
Solution:
(i) A rational number between 2 and 3 is 2.1. An
irrational number between 2 and 3 is 2.040040004...
(ii) A rational number between 0 and 0.1 is 0.03.
An irrational number between 0 and 0.1 is
0.001000100001…..
(iii) A rational number between
is . An irrational
1 2 20 1 3 30
3 2
6 60 6 60
and
25 5
12
60
number between is
1 1
0.333... 0.5
3 2
and
0.414114111...
(iv) A rational number between
2 1
0.4 0.5
2
5
and
is 0.04.
An irrational number between
is
2 1
0.4 0.5
2
5
and
0.001000100001…..
(v) A rational number between 0.15 and 0.16 is 0.155.
An irrational number between 0.15 and 0.16 is
0.155050050005...
(vi) A rational number between
is 1.5.
2 3and
An irrational number between is
2 3and
1.5050050005…
(vii) A rational number between 2.357 and 3.121 is 3.
An irrational number between 2.357 and 3.121 is
3.101101110...
(viii) A rational number between .0001 and .001 is
0.00011. An irrational number between .0001 and .001
is 0.00011313313331...
(ix) A rational number between 3.623623 and
0.484848 is 2.
An irrational number between 3.623623 and 0.484848
is 2.010010001…
(x) A rational number between 6.375289 and
6.375738 is 6.3753. An irrational number between
6.375289 and 6.375738 is 6. 3754141141114...
Question: 4
Represent the following numbers on the number line:
3 12
7,7.2, ,
2
5
Solution:
Question: 5
Locate
and
on the number line.
5, 10
17
Solution:
on the number line:
5
With O as centre and OB as radius, an arc is drawn to
meet the number line at C. The point C represents .
5
on the number line:
10
With O as centre and OB as radius, an arc is drawn to
meet the number line at C. The point C represents
10
on the number line:
17
With O as centre and OB as radius, an arc is drawn to
meet the number line at C. The point C represents .
17
Question: 6
Represent geometrically the following numbers on the
number line:
(i)
4.5
(ii)
5.6
(iii)
8.1
(iv)
2.3
Solution:
(i)
on the number line:
4.5
Draw a number line. Mark point A at 0 and B at 4.5
such that AB = 4.5 units. Mark a point C on the
number line such that BC = 1 unit. Bisect AC such
that O is the mid-point of AC. With O as centre and
OC as radius, draw a semicircle to meet the number
line at C. Draw a line perpendicular to AC at B such
that the line intersects the semi-circle at D. With B as
centre and BD as radius, draw an arc to meet AC at F.
The point F represents
.
4.5
(ii)
on the number line:
5.6
Draw a number line. Mark point A at 0 and B at 5.6
such that AB = 5.6 units. Mark a point C on the
number line such that BC = 1 unit. Bisect AC such
that O is the mid-point of AC. With O as centre and
OC as radius, draw a semicircle to meet the number
line at C. Draw a line perpendicular to AC at B such
that the line intersects the semi-circle at D. With B as
centre and BD as radius, draw an arc to meet AC at F.
The point F represents
5.6.
(iii)
on the number line:
8.1
Draw a number line. Mark point A at 0 and B at 8.1
such that AB = 8.1 units. Mark a point C on the
number line such that BC = 1 unit. Bisect AC such
that O is the mid-point of AC. With O as centre and
OC as radius, draw a semicircle to meet the number
line at C. Draw a line perpendicular to AC at B such
that the line intersects the semi-circle at D. With B as
centre and BD as radius, draw an arc to meet AC at F.
The point F represents
8.1.
(iv)
on the number line:
2.3
Draw a number line. Mark point A at 0 and B at 2.3
such that AB = 4.5 units. Mark a point C on the
number line such that BC = 1 unit. Bisect AC such
that O is the mid-point of AC. With O as centre and
OC as radius, draw a semicircle to meet the number
line at C. Draw a line perpendicular to AC at B such
that the line intersects the semi-circle at D. With B as
centre and BD as radius, draw an arc to meet AC at F.
The point F represents
2.3.
Question: 7
Express the following in the form
where p and q are
p
q
integers and
0:
q
(i) 0.2
(ii) 0.888...
(iii)
5.2
(iv)
0.001
(v) 0.2555...
(vi)
0.134
(vii) .00323232...
(viii) .404040...
Solution:
(i) Now,
2 1
0.2
10
5
0.888 ...
0 .. 1
10 ..
.8
8.8 2
x
x
x
Let be equal to
Subtracting equation (1) from equation (2), we get
8
9
 
10 8.8 0.8
9 8.0
x x
x
x
(iii)
5.2 .............. 1
x
Let be equal to
10 52.2 ................ 2
x
Subtracting equation (1) from equation (2), we get
10 52.2 5.2
9 47
47
9
x x
x
x
(iv)
0.001 ............. 1
x
Let be equal to
1000 1.001 ............. 2
x x
Subtracting equation (1) from equation (2), we get
1000 1.001 0.001
999 1
1
999
x x
x
x
(v)
0.2555 ....
x
Let be equal to
0.25
x
10 2.5 ................. 1
x
100 25.5 ................. 2
x
Subtracting equation (1) from equation (2), we get
100 10 25.5 2.5
90 23
23
90
x x
x
x
(vi)
0.134
x
Let be equal to
10 1.34 ................ 1
x
1000 134.34 ................. 2
x
Subtracting equation (1) from equation (2), we get
1000 10 134.34 1.34
990 133
133
990
x x
x
x
(vii)
0.00323232
x
Let be equal to
0.0032
x
100 0.32 .................. 1
x
10000 32.32 ................ 1
x
Subtracting equation (1) from equation (2), we get
10000 100 32.32 0.32
9900 32
32 8
9900
2475
x x
x
x
(viii) .404040....
0.404040
x
Let be equal to
0.40 ............... 1
x
Multiplying both sides of the equation (1) by 100, we
get
100 40.40 .................. 2
x
Subtracting equation (1) from equation (2), we get
100 40.40 0.40
40
99 40
99
x x
x x
Question: 8
Show that 0.142857142857...
.
Solution:
0.142857142857
x
Let be equal to
0.142857 ................... 1
x
Multiplying both sides of the equation (1) by 1000000,
we get
1000000 142857.142857 ................... 2
x
Subtracting equation (1) from equation (2), we get
1000000 142857.142857 0.142857
999999 142857
142857 1
999999
7
x x
x
x
Question: 9
Simplify the following:
(i)
45 3 20 4 5
(ii)
24 54
8 9
(iii)
4 12 7 6
(iv)
4 28 3 7
(v)
7
3 3 2 27
3
(vi)
2
3 2
(vii)
3
4 5
81 8 216 15 32 225
(viii)
3 1
8 2
(ix)
2 3 3
3
6
Solution:
(i)
45 3 20 4 5
3 3 5 3 2 2 5 4 5
3 5 6 5 4 5
5
(ii)
24 54
8 9
2 2 2 3 2 3 3 3
8 9
2 6 3 6
8 9
6 6 3 6 4 6 7 6
4 3 12 12
(iii)
4 12 7 6
4 2 2 3 7 2 3
8 3 7 2 3
56 3 2 3
168 2
(iv)
4 28 3 7
4 28 4 2 2 7 8 7 8
3
3 7 3 7 3 7
(v)
7
3 3 2 27
3
7 3
3 3 2 3 3 3
3 3
7 3
3 3 6 3
3
7
3+6 3
3
34
3
3
(vi)
2
3 2
2 2
3 2 2 3 2
3 2 2 6
5 2 6
(vii)
3
4 5
81 8 216 15 32 225
3
4 5
3 3 3 3 8 6 6 6 15 2 2 2 2 2 15 15
3 8 6 15 2 15
3 48 30 15 48 48 0
(viii)
3 1
8 2
3 1
2 2 2 2
3 1
2 2 2
3 2 5 2 5 2
4
2 2 2 2 2
(ix)
2 3 3
3
6
4 3 3
6
3 3 3
2
6
Question: 10
Rationalise the denominator of the following:
(i)
2
3 3
(ii)
40
3
(iii)
3 2
4 2
(iv)
16
41 5
(v)
2 3
2 3
(vi)
6
2 3
(vii)
3 2
3 2
(viii)
3 5 3
5 3
(ix)
4 3 5 2
48 18
Solution:
(i)
2
3 3
2 3 2 3
9
3 3 3
(ii)
40
3
40 3 120 2 2 30 2 30
3 3 3
3 3
(iii)
3 2
4 2
3 2 2 3 2 2
8
4 2 2
(iv)
16
41 5
16 41 5 16 41 5
16 41 5
41 5
41 25 16
41 5 41 5
(v)
2 3
2 3
 
2
2 3
2
2 3 2 3
2 3
4 3
2 3 2 3
4 3 4 3 7 4 3
(vi)
6
2 3
6 2 3 12 18
2 3
2 3 2 3
2 2 3 2 3 3 2 3 3 2
3 2 2 3
1 1
(vii)
3 2
3 2
2
3 2
3 2 3 2
3 2
3 2 3 2
2
3 2 3 2 2 6 5 2 6
(viii)
3 5 3
5 3
3 5 3 5 3 15 3 15 15 3 18 4 15
2
5 3
5 3 5 3
9 2 15
(ix)
4 3 5 2
48 18
4 3 5 2 4 3 3 2 16 3 12 6 20 6 15 2
16 3 9 2
4 3 3 2 4 3 3 2
48 8 6 30 18 8 6 9 4 6
48 18 30
15
Question: 11
Find the values of a and b in each of the following:
(i)
5 2 3
6 3
7 4 3
a
(ii)
3 5 19
5
11
3 2 5
a
(iii)
2 3
2 6
3 2 2 3
b
(iv)
7 5 7 5 7
5
11
7 5 7 5
a b
Solution:
(i)
5 2 3
6 3
7 4 3
a
5 2 3 7 4 3
6 3
7 4 3 7 4 3
35 20 3 14 3 8 3
6 3
2
2
7 4 3
11 6 3
6 3
49 48
11 6 3 6 3
11
a
a
a
a
a
(ii)
3 5 19
5
11
3 2 5
a
3 5 3 2 5 19
5
11
3 2 5 3 2 5
9 6 5 3 5 2 5 19
5
2 11
2
3 2 5
19 9 5 19
5
9 20 11
19 9 5 19
5
11 11
9 5 19 19
5
11 11
9 19 19
5 5
11 11 11
9
11
a
a
a
a
a
a
a
(iii)
2 3
2 6
3 2 2 3
2 3 3 2 2 3
2 6
3 2 2 3 3 2 2 3
3 2 2 6 3 6 2 3
2 6
2 2
3 2 2 3
12 5 6
2 6
18 12
12 5 6
2 6
6
5
2 6 2 6
6
5
6
b
b
b
b
b
b
b
(iv)
7 5 7 5 7
5
11
7 5 7 5
a b
2 2
7 5 7 5
7
5
2 11
2
7 5
49 5 14 5 49 5 14 5
7
5
11
49 5
28 5 7
5
44 11
7 7
0 5 5
11 11
0, 1
a b
a b
a b
a b
a b
Question: 12
If then find the value of
, 2 3
a
1
.
a
a
Solution:
Given that , therefore, we have
2 3
a
 
1 1
2 3
1 1 2 3
2 3 2 3
1 2 3
2
2
2 3
1 2 3
4 3
1
2 3
1
2 3 2 3
2 3
a
a
a
a
a
a
a
Now
Question: 13
Rationalise the denominator in each of the following
and hence evaluate by taking
2 1.414, 3 1.732
and
up to three places of decimal.
5 2.236,
(i)
4
3
(ii)
6
6
(iii)
10 5
2
(iv)
2
2 2
(v)
1
3 2
Solution:
(i)
4
3
4 3 4 3 4 1.732 6.928
2.3093 2.309
3 3 3
3 3
(ii)
6
6
6 6 6 6
6 2 3 1.414 1.732
6
6 6
2.4490 2.449
(iii)
10 5
2
5 2 1
2.236 1.414 1
1.118 0.414
2 2
0.4628 0.462
(iv)
2
2 2
2 2 1
2 2 2 2 2 2
2 4 2
2 2 2 2
2
2 2
2 1.414 1
0.414
2
(v)
1
3 2
1 3 2 3 2 1.732 1.414
0.318
2 2 3 2
3 2 3 2
3 2
Question: 14
Simplify:
(i)
1
3 3 3
2
1 2 3
(ii)
4 12 6
3 8 32
5 5 5
(iii)
2
1
3
27
(iv)
2
1
4
1
625
2
(v)
1 1
3 2
9 27
1
2
6 3
3 3
(vi)
1 1 2
3 3 3
64 64 64
(vii)
1 1
3 3
8 16
1
3
32
Solution:
(i)
1
3 3 3
2
1 2 3
1
1 8 27
2
1
36
2
1
2
2
6
6
(ii)
4 12 6
3 8 32
5 5 5
4 12 6
3 8 32
4 12
6
5 5
5
4 12 6
3 8 32
4 12 6
5 5 5
6
12
4 3 5
3 2 2
4 12 6
5
4 36 30
3 2 2
2
5
4 2
3 5
6
2
81 25
64
2025
64
(iv)
2
1
4
1
625
2
2
1
1
4
2
2
25
2
1
1
4
25
2
1
4
25
1
2
25
1
2
2
5
5
(v)
1 1
3 2
9 27
1
2
6 3
3 3
4 9 5
1 1 2 3 2 3
2 3
3 2 3 2 3 2 6 6
3 3 3 3 3 3 3
1 1 2 1 2 1 4 3
2
3 3
6 3 6 6 6 6
3 3 3 3 3 3
5 3 2
1
6 6 6 3
3 3 3
(vi)
1 1 2
3 3 3
64 64 64
1 1 2
3 3 3
3 3 3
4 4 4
1 2
4 4 4
4 16
4
12
3
4
(vii)
1 1
3 3
8 16
1
3
32
1 1
3 4
3 3
2 2
1
5
3
2
3 4
3 3
2 2
5
3
2
3 4 5
3 3 3
2
12
3
2
4
2
16
Exercise 1.4
Question: 1
Express in the ,
.
form wherepandqare
integers and
p
0.6 0.7 0.47
q
0
q
Solution:
 
0.6
6
10
3
or
5
. 10.7
x
x
x
y
Let be equal to
Let be equal to
 
10,
..10y 7.7
10 = 7.7 0.
2
1 2
9 = 7.0
.
7
7
9
10.47
y y
y
y
Multiplying both sides by we get
Subtracting equation from equation , we get
Let z be equal to
Multiplying both sides of equation by , we get
. 10 4.7
1 10
2
z
Multiplying both sides of equation by , we get
. 100 47 3.7
2 10
z
Subtracting equation (2) from equation (3), we get
Now
  
100 - 10 = 47.7- 4.7
90 43
43
90
0.6 0.7 0.47
3 7 43
5 9 90
54 70 43 167
90 90
z z
z
z
x zy
Question 2
Simplify :
7 3 2 5 3 2
- -
10 + 3 6 + 5 15 +3 2
Solution:
2 2 2 2 2 2
7 3 2 5 3 2
10 3 6 5 15 3 2
7 3 10 - 3 2 5
10 3 10 - 3 6 5
6 5 3 2 15 3 2
6 5 15 3 2 15 3 2
7 3 10 3 2 5 6 5 3 2 15 3 2
10 3 6 5 15 3 2
7 30 7 3 2 30 2 5 3 30 9 2
= -
10 3 6 5 15 18
7 30 3 3 30 6
2 30 10
=
7 1 3
= 30 3 2 30 10 30 6
= 30 3 2 30 10 30 6=1
Question 3
2 1.414, 3 =1.732
4 3
3 3 2 2 3 3 2 2
If then find the value of,
.
Solution:
4 3
3 3 2 2 3 3 2 2
2 2
4 3 3 2 2 3 3 3 2 2
=
3 3 2 2 3 3 2 2
12 3 8 2 9 3 6 2
=
3 3 2 2
21 3 2 2
=
27 8
21 1.732 2 1.414
=
19
36.372 2.828
=
19
39.2
= = 2.063
19
Question 4
If then find the value of ,
2
2
3+ 5 1
+ .
2
a a
a
Solution:
Given :
×
2
2
2
3 5
=
2
3 5
=
2
9 5 6 5
4
14 6 5
4
7 3 5
2
1 2
7+3 5
2 7 3 5
7 3 5 7 3 5
14 6 5
=
49- 45
14 6 5
=
4
7 3 5
=
2
a
a
a
So
2
2
,
1 7 3 5 7 3 5 14
= 7
2 2 2
a
a
Question 5
2 2
If and then find the vy , alue
of .
3 2 3 2
3 2 3 2
x
x y
Solution:
2
2
3 2
=
3 2
3 2 3 2
= ×
3 2 3 2
3 2
=
3- 2
= 3 2
= 3 2 2 6
= 5 2 6
x
So,
2
2
2 2
2 2
3 2
=
3 2
3 2 3 2
= ×
3 2 3 2
3 2
3 2
3 2
3 2 2 6
5 2 6
5 2 6 5 2 6
25 24 20 6 25 24 20 6 = 98
y
x y
Question 6
Simplify :
-3
2
4
256
Solution:
3
2
3
2
3
2
2
3
4
1
4
1
2
1 1
2 8
1
1
8
8 1
8
8
256
= 256
= 256
= 256 = 256
1
= 2 = 2 = 2 =
2
Question: 7
Find the value of
2 3 1
3 4 5
4 1 2
216 256 243
Solution:
Given that :
2 3 1
3 4 5
2 3 1
3 4 5
2 3 1
3 4 5
3 4 5
4 1 2
216 256 243
4 1 2
216 256 243
4 1 2
6 4 3
2 3 1
3× 4× 5×
3 4 5
2 3 1
2 3
4 1 2
= + +
6 4 3
4 1 2
= + +
6 4 3
= 6 +1×4 +2×3
= 36+1×64+6
=144+64+6
= 214