Without performing the long division , write down the decimal expansion of
241
/625
Answers
Answer:
Theorem: Let x=
q
p
be a rational number, such that the prime factorisation of q is of the form 2
n
5
m
, where n, m are non-negative integers. Then, x has a decimal expansion which terminates.
(i)
3125
13
Factorise the denominator, we get
3125=5×5×5×5×5=5
5
So, denominator is in form of 5
m
so,
3125
13
is terminating.
(ii)
8
17
Factorise the denominator, we get
8=2×2×2=2
3
So, denominator is in form of 2
n
so,
8
17
is terminating.
(iii)
455
64
Factorise the denominator, we get
455=5×7×13
So, denominator is not in form of 2
n
5
m
so,
455
64
is not terminating.
(iv)
1600
15
Factorise the denominator, we get
1600=2×2×2×2×2×2×5×5=2
6
5
2
So, denominator is in form of 2
n
5
m
so,
1600
15
is terminating.
(v)
343
29
Factorise the denominator, we get
343=7×7×7=7
3
So, denominator is not in form of 2
n
5
m
so,
343
29
is not terminating.
(vi)
2
3
5
2
23
Here, the denominator is in form of 2
n
5
m
so,
2
3
5
2
23
is terminating.
(vii)
2
2
5
7
7
5
129
Here, the denominator is not in form of 2
n
5
m
so,
2
2
5
7
7
5
129
is not terminating.
(viii)
15
6
Divide nominator and denominator both by 3 we get
15
3
So, denominator is in form of 5
m
so,
15
6
is terminating.
(ix)
50
35
Divide nominator and denominator both by 5 we get
10
7
Factorise the denominator, we get
10=2×5
So, denominator is in form of 2
n
5
m
so,
50
35
is terminating.
(x)
210
77
Divide nominator and denominator both by 7 we get
30
11
Factorise the denominator, we get
30=2×3×5
So, denominator is not in the form of 2
n
5
m
so
15
6
is not terminating.
Step-by-step explanation: