Q1
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:
(i) 13/3125
(ii) 17/8
(iii) 64/455
(iv) 15/1600
(v) 29/343
(vi) 23/2³×5²
(vii) 129/2²×5⁷×7⁵
(viii) 6/15
(ix) 35/50
(i) $\frac{13}{3125}$
Prime factorization of denominator:
$$3125 = 5^5$$
The denominator is of the form $5^m$.
✔️ Terminating decimal expansion
(ii) $\frac{17}{8}$
Prime factorization of denominator:
$$8 = 2^3$$
The denominator is of the form $2^m$.
✔️ Terminating decimal expansion
(iii) $\frac{64}{455}$
Prime factorization of denominator:
$$455 = 5 \times 7 \times 13$$
Since the denominator is not in the form $2^m \times 5^n$, and it also contains 7 and 13 as its factors:
✖️ Non-terminating repeating decimal expansion
(iv) $\frac{15}{1600}$
Simplify the fraction first:
$$\frac{15}{1600} = \frac{3}{320}$$
Prime factorization of denominator:
$$320 = 2^6 \times 5$$
$$1600 = 2^6 \times 5^2$$
The denominator is of the form $2^m \times 5^n$.
✔️ Terminating decimal expansion
(v) $\frac{29}{343}$
Prime factorization of denominator:
$$343 = 7^3$$
Since the denominator is not in the form $2^m \times 5^n$, and it has 7 as its factor:
✖️ Non-terminating repeating decimal expansion
(vi) $\frac{23}{2^3 \times 5^2}$
The denominator is already in the form $2^m \times 5^n$.
✔️ Terminating decimal expansion
(vii) $\frac{129}{2^2 \times 5^7 \times 7^5}$
Since the denominator is not of the form $2^m \times 5^n$, and it also has 7 as its factor:
✖️ Non-terminating repeating decimal expansion
(viii) $\frac{6}{15}$
Simplify the fraction first:
$$\frac{6}{15} = \frac{2}{5}$$
The denominator is of the form $5^n$.
✔️ Terminating decimal expansion
(ix) $\frac{35}{50}$
Simplify the fraction first:
$$\frac{35}{50} = \frac{7}{10}$$
Prime factorization of denominator:
$$10 = 2 \times 5$$
The denominator is of the form $2^m \times 5^n$.
✔️ Terminating decimal expansion