Question 1 (iii)
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
$\frac{64}{455}$

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Solution

$\frac{64}{455}$
Factorize the denominator; we get,
$455 = 5 \times 7 \times 13$
There are 7 and 13 also in denominator so denominator is not in the form of $2^{m} \times 5^{n}$ .
So, it is not terminating.

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Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating decimal expansion : $\frac{64}{455}$ [1 MARK]

Question 1 (iii)
Without actually performing the long division, state whether $\frac{64}{455}$ will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Question 1 (iii)
Without actually performing the long division, state whether $\frac{64}{455}$ will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Q. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

129÷2²5⁷7⁵

Without actually performing the long division, state whether the following rational number will have a terminating or a non-terminating repeating decimal expansion.

\frac{64}{455}

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Question 1 (iii) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 64455

64455 Factorize the denominator; we get, 455=5×7×13 There are 7 and 13 also in denominator so denominator is not in the form of 2m×5n​​​​​​. So, it is not terminating.

Question 1 (iii)
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
$\frac{64}{455}$

$\frac{64}{455}$
Factorize the denominator; we get,
$455 = 5 \times 7 \times 13$
There are 7 and 13 also in denominator so denominator is not in the form of $2^{m} \times 5^{n}$ .
So, it is not terminating.