Decimal Fraction - Aptitude MCQ Questions and Solutions with Explanations

Given expression :
$$\eqalign{ & \left[ {\frac{{8{{\left( {3.75} \right)}^3} + 1}}{{{{\left( {7.5} \right)}^2} - 6.5}}} \right] \cr & = \frac{{{{\left( {2 \times 3.75} \right)}^3} + {1^3}}}{{{{\left( {7.5} \right)}^2} - \left( {7.5 \times 1} \right) + {1^2}}} \cr & = \frac{{{{\left( {7.5} \right)}^3} + {1^3}}}{{{{\left( {7.5} \right)}^2} - \left( {7.5 \times 1} \right) + {1^2}}} \cr & = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right) \cr & = \frac{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}{{\left( {{a^2} - ab + {b^2}} \right)}} \cr & = \left( {a + b} \right) \cr & = \left( {7.5 + 1} \right) \cr & = 8.5 \cr} $$

The given expression can be written in this form also
N = $${\text{ 0}}{\text{.39}}\overline {{\text{39}}} $$ ..... (i)
Multiply equation (i) with 100 on both sides.
100N = $${\text{39}}{\text{.}}\overline {{\text{39}}} $$ ..... (ii)
Subtracting equation (i) from (ii) we get,
⇒ 100N - N = $${\text{39}}\overline {{\text{.39}}} $$  - $${\text{0}}\overline {{\text{.39}}} $$
⇒ 99N = 39
⇒ N = $$\frac{39}{99}$$ = $$\frac{13}{33}$$

Given expression :
$$ = \frac{{{{\left( {0.051} \right)}^3} + {{\left( {0.041} \right)}^3}}}{{{{\left( {0.051} \right)}^2} - \left( {0.051 \times 0.041} \right) + {{\left( {0.041} \right)}^2}}}$$
Let 0.051 = $$a$$ and 0.041 = $$b$$
$$\eqalign{ & = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right) \cr & = (a + b) \cr & = \left( {0.051 + 0.041} \right) \cr & = 0.092 \cr} $$

Given expression :
$$\eqalign{ & = \frac{{{{\left( {10.3} \right)}^3} + {1^3}}}{{{{\left( {10.3} \right)}^2} - \left( {10.3 \times 1} \right) + {1^2}}} \cr & = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right) \cr & = (a + b) \cr & = \left( {10.3 + 1} \right) \cr & = 11.3 \cr} $$

Given expression :
$$ = \frac{{{{\left( {4.53 - 3.07} \right)}^3} + {{\left( {3.07 - 2.15} \right)}^3} + {{\left( {2.15 - 4.53} \right)}^3}}}{{\left( {4.53 - 3.07} \right)\left( {3.07 - 2.15} \right)\left( {2.15 - 4.53} \right)}}$$
Let (4.53 - 3.07) = $$a$$, (3.07 - 2.15) = $$b$$ and (2.15 - 4.53) = $$c$$
$$\eqalign{ & = \frac{{{a^3} + {b^3} + {c^3}}}{{abc}} \cr & = \frac{{3abc}}{{abc}} \cr & = 3 \cr} $$
[∵ If a + b + c = 0, a ³ + b³ + c³ = 3abc]

Given expression :
$$\eqalign{ & \frac{{0.125 + 0.027}}{{0.5 \times 0.5 + 0.09 - 0.15}} \cr & = \frac{{{{\left( {0.5} \right)}^3} + {{\left( {0.3} \right)}^3}}}{{{{\left( {0.5} \right)}^2} + {{\left( {0.3} \right)}^2} - \left( {0.5 \times 0.3} \right)}} \cr & = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} + {b^2} - ab}}} \right) \cr & = \frac{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}{{\left( {{a^2} - ab + {b^2}} \right)}} \cr & = (a + b) \cr & = \left( {0.5 + 0.3} \right) \cr & = 0.8 \cr} $$

$$\frac{2}{5}$$%
= $$\frac{2}{5}$$ × $$\frac{1}{100}$$
= $$\frac{1}{250}$$

Given expression :
$$\eqalign{ & = 35.7 - \left( {3 + \frac{1}{{\frac{{10}}{3}}}} \right) - \left( {2 + \frac{1}{{\frac{5}{2}}}} \right) \cr & = 35.7 - \left( {3 + \frac{3}{{10}}} \right) - \left( {2 + \frac{2}{5}} \right) \cr & = 35.7 - \frac{{33}}{{10}} - \frac{{12}}{5} \cr & = 35.7 - \left( {\frac{{33}}{{10}} + \frac{{12}}{5}} \right) \cr & = 35.7 - \frac{{57}}{{10}} \cr & = 35.7 - 5.7 \cr & = 30 \cr} $$

Given expression :
$$ = \frac{{{a^2} + {b^2} + {c^2}}}{{{{\left( {\frac{a}{{10}}} \right)}^2} + {{\left( {\frac{b}{{10}}} \right)}^2} + {{\left( {\frac{c}{{10}}} \right)}^2}}}$$
Where a = 0.06, b = 0.47 and c = 0.079
$$\eqalign{ & = \frac{{100\left( {{a^2} + {b^2} + {c^2}} \right)}}{{\left( {{a^2} + {b^2} + {c^2}} \right)}} \cr & = 100 \cr} $$