Decimal Fraction - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Given,}} \cr & \frac{{52416}}{{312}} = 168 \cr & \Leftrightarrow \frac{{52416}}{{168}} = 312 \cr & {\text{Now,}} \cr & \frac{{52.416}}{{0.0168}} \cr & = \frac{{524160}}{{168}} \cr & = \frac{{52416}}{{168}} \times 10 \cr & = 312 \times 10 \cr & = 3120 \cr} $$

Given expression :
(1.25)³ - (0.75)³ - 3 × (1.25)² × 0.75 + 3 × 1.25 × (0.75)²
= (1.25 - 0.75)³
= (0.5)³
= $${\left( {\frac{1}{2}} \right)^3}$$
= $$\frac{1}{8}$$
[∵ (a - b)³ = a³ - b³ - 3a² b + 3ab²]

Let the missing number be x
$$\eqalign{ & \frac{{21.5}}{5} + \frac{{21}}{6} - \frac{{13.5}}{{15}} = \frac{{{{\left( x \right)}^{\frac{1}{3}}}}}{4} + \frac{{17}}{{30}} \cr & \frac{{21.5}}{5} + \frac{{21}}{6} - \frac{{13.5}}{{15}} - \frac{{17}}{{30}} = \frac{{{{\left( x \right)}^{\frac{1}{3}}}}}{4} \cr} $$
L.C.M of 5, 6, 15 and 30 is 30
$$\eqalign{ & \frac{{129 + 105 - 27 - 17}}{{30}} = \frac{{{{\left( x \right)}^{\frac{1}{3}}}}}{4} \cr & \root 3 \of x = \frac{{190 \times 4}}{{30}} \cr & \root 3 \of x = 25.33 \approx 25 \cr & x = {25^3} \cr & x = 15625 \cr} $$
Hence, the numbers 15625

Converting each of the given fractions into decimal form, we get:
$$\frac{9}{31}$$ = 0.29
$$\frac{3}{17}$$ = 0.176
$$\frac{6}{23}$$ = 0.26
$$\frac{4}{11}$$ = 0.363
$$\frac{7}{25}$$ = 0.28
Clearly, 0.363 > 0.29 > 0.28 > 0.26 > 0.176
So, $$\frac{4}{11}$$ > $$\frac{9}{31}$$ > $$\frac{7}{25}$$ > $$\frac{6}{23}$$ > $$\frac{3}{17}$$

$$\eqalign{ & 0.\overline {142857} \div 0.\overline {285714} \cr & = \frac{{142857}}{{999999}} \div \frac{{285714}}{{999999}} \cr & = \left( {\frac{{142857}}{{999999}} \times \frac{{999999}}{{285714}}} \right) \cr & = \frac{1}{2} \cr} $$

Given expression :
58.621 - (13.829 + 7.302 + 1.214)
= 58.621 - 22.345
= 36.276

Let original fraction be $$\frac{a}{b}$$
Now, according to the question,
$$\eqalign{ & \Leftrightarrow \frac{{a - a \times \frac{{25}}{{100}}}}{{b + b \times \frac{{250}}{{100}}}} = \frac{6}{5} \cr & \Rightarrow \frac{{0.75a}}{{3.50b}} = \frac{6}{5} \cr & \Rightarrow \frac{a}{b} = \frac{6}{5} \times \frac{{3.50}}{{0.75}} \cr & \Rightarrow \frac{a}{b} = \frac{{6 \times 350 \times 100}}{{5 \times 75 \times 100}} \cr & \Rightarrow \frac{a}{b} = \frac{{28}}{5} \cr} $$

$$\frac{- 7}{10}$$ = - 0.7
$$\frac{5}{- 8}$$ = - $$\frac{5}{8}$$ = - 0.625,
$$\frac{2}{- 3}$$ = - $$\frac{2}{3}$$ = - 0.66
Since, -0.7 < -0.66 < -0.625
So, $$\frac{- 7}{10}$$ < $$\frac{2}{- 3}$$ < $$\frac{5}{- 8}$$

Given expression :
$$ = \frac{8 - 2.8}{1.3}$$
$$ = \frac{5.2}{1.3}$$
$$ = \frac{52}{13}$$
= 4