Decimal Fraction - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option B

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

Answer: Option C

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

Answer: Option A

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

Answer: Option B

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

Answer: Option C

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