Decimal Fraction - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \text{Let } a = 2.39 \text{ and } b = 1.61 \cr & {\text{Given Expression}} \cr & = \frac{{{a^2} - {b^2}}}{{a - b}} \cr & = \frac{{\left( {a + b} \right)\left( {a - b} \right)}}{{ {a - b} }} \cr & = {a + b} \cr & = {2.39 + 1.61} \cr & = 4 \cr} $$

$$\eqalign{ & {\text{Required decimal}} \cr & = \frac{1}{{60 \times 60}} \cr & = \frac{1}{{3600}} \cr & = .00027 \cr} $$

$$\eqalign{ & {\text{Given}}\,{\text{expression}} \cr & = \frac{{{{\left( {0.96} \right)}^3} - {{\left( {0.1} \right)}^3}}}{{{{\left( {0.96} \right)}^2} + \left( {0.96 \times 0.1} \right) + {{\left( {0.1} \right)}^2}}} \cr & = {\frac{{{a^3} - {b^3}}}{{{a^2} + ab + {b^2}}}} \cr & = \frac{\left(a-b\right) \left(a^2 + ab + b^2 \right)}{\left(a^2 + ab + b^2 \right)} \cr & = {a - b} \cr & = {0.96 - 0.1} \cr & = 0.86 \cr} $$

$$\eqalign{ & {\text{Give}}\,{\text{expression}} \cr & = \frac{{{{\left( {0.1} \right)}^3} + {{\left( {0.02} \right)}^3}}}{{{2^3}\left[ {{{\left( {0.1} \right)}^3} + {{\left( {0.02} \right)}^3}} \right]}} \cr & = \frac{1}{8} \cr & = 0.125 \cr} $$

$$\eqalign{ & \frac{{29.94}}{{1.45}} \cr & = \frac{{299.4}}{{14.5}} \cr} $$
  $$ = {\frac{{2994}}{{14.5}} \times \frac{1}{{10}}} $$    [Here, Substitute 172 in the place of $${\frac{{2994}}{{14.5}}}$$ ]
$$\eqalign{ & = \frac{{172}}{{10}} \cr & = 17.2 \cr} $$

$$0.232323... = 0.\overline {23} = \frac{{23}}{{99}}$$

$$\eqalign{ & {\text{Let}}\,\frac{{.009}}{x} = .01; \cr & {\text{Then}}\,x = \frac{{.009}}{{.01}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{.9}}{1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = .9 \cr} $$

Given expression = (11.98)² + (0.02)² + 11.98 × X
For the given expression to be a perfect square,
(a + b)² = a² + b² + 2ab
we must have
11.98 × X = 2 × 11.98 × 0.02
∴ X = 0.04

$$\eqalign{ & {\text{Given}}\,{\text{expression}} \cr & = \frac{{ {0.3333} }}{{ {0.2222} }} \times \frac{{\left( {0.1667} \right)\left( {0.8333} \right)}}{{\left( {0.6667} \right)\left( {0.1250} \right)}} \cr & = \frac{{3333}}{{2222}} \times \frac{{\frac{1}{6} \times \frac{5}{6}}}{{\frac{2}{3} \times \frac{{125}}{{1000}}}} {\text{ [where }} {0.3333} = {\frac{1}{6}}, {0.8333} = {\frac{5}{6}} {\text{ and }} {0.6667} = {\frac{2}{3}} {\text{ ] }} \cr & = {\frac{3}{2} \times \frac{1}{6} \times \frac{5}{6} \times \frac{3}{2} \times 8} \cr & = \frac{5}{2} \cr & = 2.50 \cr} $$

Let 3889 + 12.952 - x = 3854.002.
Then x = (3889 + 12.952) - 3854.002
    = 3901.952 - 3854.002
    = 47.95