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Standard VII

Mathematics

Writing Rational Numbers as Decimals

Rational numb...

Question

Rational number $\frac{p}{q}, q \neq 0$ will be terminating decimal if the prime factorisation of q is of the form (m and n are non negative integers) :

2^{m} \times 3^{n}

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2^{m} \times 5^{n}

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3^{m} \times 5^{n}

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3^{m} \times 7^{n}

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Solution

The correct option is B $2^{m} \times 5^{n}$
If numerator and denominator are co-prime then fraction will terminate only if denominator will be of type $2^{m} 5^{n}$ .
So the correct answer will be option B.

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Similar questions

Rational number p/q, q ≠ 0 will be terminating decimal if the prime factorisation of q is of the form. (m and n are non negative integers):

Q. In a rational number of the form

\frac{p}{q}

, the prime factorisation of

q

is of the form

2^{m} \times 5^{n}

, where

m

and

n

are positive integers. The decimal expansion terminates after ____ decimal places if

m > n

Any rational number x in the form p/q (where p and q are co primes), whose decimal expression terminates. Which of the following represents the correct form of prime factorisation of q?

Q. A rational number whose decimal expansion is non-terminating can be expressed in the form

\frac{p}{q}

, where

p

and

q

are coprime, and the prime factorisation of

q

is NOT of the form _____ , where

n

and

m

are non-negative integers.

Q. Let

x = \frac{p}{q}

be a rational number, such that the prime factorisation of q is not of the form

2^{m} 5^{n}

, where n, m are non-negative integers. Then, x has a decimal expansion which is:

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Rational number pq, q≠0 will be terminating decimal if the prime factorisation of q is of the form (m and n are non negative integers) :

The correct option is B 2m×5nIf numerator and denominator are co-prime then fraction will terminate only if denominator will be of type 2m5n .So the correct answer will be option B.

Rational number $\frac{p}{q}, q \neq 0$ will be terminating decimal if the prime factorisation of q is of the form (m and n are non negative integers) :

The correct option is B $2^{m} \times 5^{n}$
If numerator and denominator are co-prime then fraction will terminate only if denominator will be of type $2^{m} 5^{n}$ .
So the correct answer will be option B.