Simplification - Aptitude MCQ Questions and Solutions with Explanations

According to question,

Factors are :
3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 5
∴ It must be multiplied by
= 2 × 5 × 5
= 50

Clearly, $$\frac{{x + 1}}{x}$$ is the only fraction in which the numerator is greater then the denominator. So, it is greatest fraction.

$$\eqalign{ & \frac{6}{{7/8}} - \frac{{6/7}}{8} \cr & = 6 \times \frac{8}{7} - \frac{6}{7} \times \frac{1}{8} \cr & = \frac{{48}}{7} - \frac{6}{{56}} \cr & = \frac{{384 - 6}}{{56}} \cr & = \frac{{378}}{{56}} \cr & = \frac{{27}}{4} \cr & = 6\frac{3}{4} \cr} $$

Let the value of the estate be Rs. x
$$\eqalign{ & {\text{Then}}\frac{4}{5}{\text{ of }}x = 16800 \cr & \Leftrightarrow x = \left( {\frac{{16800 \times 5}}{4}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 21000 \cr & \Leftrightarrow \frac{3}{7}x = \left( {\frac{3}{7} \times 21000} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = 9000 }} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{3a + 4b}}{{3a - 4b}}{\text{ = }}\frac{{3c + 4d}}{{3c - 4d}} \cr & \Rightarrow \frac{{3a}}{{4b}} = \frac{{3c}}{{4d}} \cr & \Rightarrow ad = bc \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{\frac{1}{3} + \frac{1}{4}\left[ {\frac{2}{5} - \frac{1}{2}} \right]}}{{1\frac{2}{3}\,{\text{of }}\frac{3}{4} - \frac{3}{4}\,{\text{of }}\frac{4}{5}}} \cr & \Rightarrow \frac{{\frac{1}{3} + \frac{1}{4}\left[ {\frac{{4 - 5}}{{10}}} \right]}}{{\frac{5}{3}\, \times \frac{3}{4} - \frac{3}{4}\, \times \frac{4}{5}}} \cr & \Rightarrow \frac{{\frac{1}{3} - \frac{1}{4} \times \frac{1}{{10}}}}{{\frac{5}{4} - \frac{3}{5}}} \cr & \Rightarrow \frac{{\frac{{40 - 3}}{{120}}}}{{\frac{{25 - 12}}{{20}}}} \cr & \Rightarrow \frac{{37}}{{120}} \times \frac{{20}}{{13}} \cr & \Rightarrow \frac{{37}}{{78}} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{0.3555 \times 0.5555 \times 2.025}}{{0.225 \times 1.7775 \times 0.2222}}\, \cr & = \frac{{3555 \times 5555 \times 2025}}{{225 \times 17775 \times 2222}}\, \cr & = 4.5 \cr} $$

$$\eqalign{ & {\text{Green portion}} \cr & = \left[ {1 - \left( {\frac{1}{{10}} + \frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{40}} + \frac{1}{{50}} + \frac{1}{{60}}} \right)} \right] \cr & = \left[ {1 - \frac{1}{{10}}\left( {1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}} \right)} \right] \cr & = 1 - \frac{1}{{10}} \times \frac{{147}}{{60}} \cr & = 1 - \frac{{147}}{{600}} \cr & = \frac{{453}}{{600}} \cr & {\text{let the length of the pole be }}x{\text{ metres}} \cr & {\text{Then, }}\frac{{453}}{{600}}x = 12.08 \cr & \Leftrightarrow x = \left( {\frac{{12.08 \times 600}}{{453}}} \right) = 16 \cr} $$

Let the capacity of the drum be x liters
Then,
$$\eqalign{ & \frac{3}{4}x - \frac{7}{{12}}x = 30 \cr & \Leftrightarrow 9x - 7x = 12 \times 30 \cr & \Leftrightarrow 2x = 360 \cr & \Leftrightarrow x = 180 \cr} $$

Suppose the taller tree grows x feet in 1 year.
Then the shorter tree grows
$$\eqalign{ & = \frac{{3x}}{5}{\text{ft in 1 year}} \cr & \therefore 4x + 4 \times \frac{3}{5}x = 8 \cr & \Leftrightarrow 4x + \frac{{12}}{5}x = 8 \cr & \Leftrightarrow \frac{{32}}{5}x = 8 \cr & \Leftrightarrow x = \frac{{8 \times 5}}{{32}} = \frac{5}{4} \cr & {\text{Growth of shorter tree in}} \cr & {\text{2 years = }}\left( {2 \times \frac{{3x}}{5}} \right){\text{ft}} \cr & {\text{ = }}\left( {2 \times \frac{3}{5} \times \frac{5}{4}} \right){\text{ft}} \cr & {\text{ = }}\frac{3}{2}{\text{ft}} \cr & {\text{ = 1}}\frac{1}{2}{\text{ft}} \cr} $$