Question

The decimal form of 129/2^2 5^7 7^5
(a)terminating
(b)non-terminating
(c)non-terminating non-repeating
(d)none of the above

Answer 1

your answer is

b

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Answer 2

The decimal form of $\displaystyle \sf{ \frac{129}{ {2}^{2} \times {5}^{7} \times {7}^{5} } }$ is non terminating

Given :

The fraction $\displaystyle \sf{ \frac{129}{ {2}^{2} \times {5}^{7} \times {7}^{5} } }$

To find :

The decimal form of the fraction is

(a) terminating

(b) non-terminating

(c) non-terminating non-repeating

(d) none of the above

Concept :

$\displaystyle\sf{Fraction = \frac{Numerator}{Denominator} }$

A fraction is said to be terminating if prime factorisation of the denominator contains only prime factors 2 and 5

If the denominator is of the form

$\sf{Denominator = {2}^{m} \times {5}^{n} }$

Then the fraction terminates after N decimal places

Where N = max { m , n }

Solution :

Step 1 of 2 :

Write down the given fraction

Here the given fraction is

$\displaystyle \sf{ \frac{129}{ {2}^{2} \times {5}^{7} \times {7}^{5} } }$

Step 2 of 2 :

Check whether terminating or not

$\displaystyle \sf{ Numerator = 129 }$

$\displaystyle \sf{Denominator = {2}^{2} \times {5}^{7} \times {7}^{5} }$

We know that fraction is said to be terminating if prime factorisation of the denominator contains only prime factors 2 and 5

Since prime factorisation of denominator contains prime factor as 7 also

So the decimal expansion of the given fraction is non terminating

Hence the correct option is (b) non-terminating

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