Answer :

(i)

Factorize the denominator we get,

3125 = 5×5×5×5×5 = 5^{5}

The denominator is of the form 5^{m}

Hence, the decimal expansion of is terminating.

(ii)

Factorize the denominator we get,

8 = 2 × 2 ×2 = 2^{3}

The denominator is of the form 2^{m}

Hence, the decimal expansion of is terminating.

(iii)

Factorize the denominator we get,

455 = 5×7×13

Since, the denominator is not in the form of 2^{m} × 5^{n}, and it also contains 7 and 13 as its factors,

Its decimal expansion will be non-terminating repeating.

(iv)

Factorize the denominator we get,

1600 = 2^{6}×5^{2}

The denominator is in the form 2^{m} × 5^{n}

Hence, the decimal expansion of is terminating.

(v)

Factorize the denominator we get,

343 = 7^{3}

Since the denominator is not in the form of 2^{m} × 5^{n}, it has 7 as its factors.

So, the decimal expansion of non-terminating repeating.

(vi)

The denominator is in the form 2^{m}×5^{n}

Hence, the decimal expansion of is terminating.

(vii)

Since, the denominator is not in the form of 2^{m} × 5^{n}, as it has 7 in denominator.

So, the decimal expansion of is non-terminating repeating.

(viii)

The denominator is in the form 5^{n}

Hence, the decimal expansion of is terminating.

(ix)

Factorize the denominator we get,

10 = 2×5

The denominator is in the form 2^{m}×5^{n}

Hence, the decimal expansion of is terminating.

(x)

Factorize the denominator we get,

30 = 2×3×5

Since the denominator is not in the form of 2^{m} × 5^{n}, as it has 3 in denominator.

So, the decimal expansion of non-terminating repeating.

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