Ratio - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{60}}\% \,{\text{of}}\,{\text{A}} = \frac{{\text{3}}}{{\text{4}}}\,{\text{of}}\,{\text{ B}} \cr & \Rightarrow \frac{{60}}{{100}}{\text{A}} = \frac{3}{4}\,{\text{B}} \cr & \Rightarrow \frac{{\text{3}}}{{\text{5}}}{\text{A = }}\frac{{\text{3}}}{{\text{4}}}\,{\text{B}} \cr & \Rightarrow \frac{{\text{A}}}{{\text{B}}}{\text{ = }}\frac{{\text{3}}}{{\text{4}}} \times \frac{{\text{5}}}{{\text{3}}} \cr & \Rightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{5}{4} \cr} $$
⇒ A : B = 5 : 4

A = 8 : a = 8 : b ⇒ 8a = 8b ⇒ a = b.
B = a : 16 = b : 4 ⇒ 4a = 16b ⇒ a = 4b
C = a : 8 = b : 8 ⇒ 8a = 8b ⇒ b = a
D = 32 : a = b : 2 ⇒ ab =64

The total number of members must be a multiple of the sum ratio terms.
3 + 2 = 5
and 25 is a multiple of 5

$$\eqalign{ & m:n = 3:2 \cr & \frac{m}{n} = \frac{3}{2} \cr & \Rightarrow \frac{{4m + 5n}}{{4m - 5n}} \cr & \Rightarrow \frac{{n\left( {4\frac{m}{n} + 5} \right)}}{{n\left( {4\frac{m}{n} - 5} \right)}} \cr & \Rightarrow \frac{{4 \times \frac{3}{2} + 5}}{{4 \times \frac{3}{2} - 5}} \cr & \Rightarrow \frac{{6 + 5}}{{6 - 5}} \cr & \Rightarrow \frac{{11}}{1} \cr & \Rightarrow 11:1 \cr} $$

A + B = 40
A - B = 4
∴ A = 22
B = 18
A : B = 22 : 18
        = 11 : 9

Each exterior angle of n sided polygon is
$${\text{ = }}\left( {\frac{{360}}{n}} \right)$$
And each interior angle of n sided polygon
$$\eqalign{ & {\text{ = }}\frac{{\left( {n - 2} \right) \times 180}}{n} \cr & \therefore \frac{{\frac{{\left( {n - 2} \right) \times 180}}{n}}}{{\frac{{360}}{n}}} = \frac{2}{1} \cr & \Rightarrow \frac{{\left( {n - 2} \right)}}{2} = 2 \cr & \Rightarrow n - 2 = 4 \cr & \Rightarrow n = 6 \cr} $$

$$\eqalign{ & \Rightarrow 7:15 = \frac{7}{{15}} = 0.466 \cr & \Rightarrow 15:23 = \frac{{15}}{{23}} = 0.652 \cr & \Rightarrow 17:25 = \frac{{17}}{{25}} = 0.68 \cr & \Rightarrow 21:39 = \frac{{21}}{{39}} = 0.538 \cr} $$
Clearly, 7 : 15 is the lowest

Let the missing number be x.
Then,
$$\eqalign{ & = \frac{1}{3}:\frac{1}{2}::27y:x \cr & \Rightarrow \frac{3}{4}x = \frac{1}{2} \times {\text{27}}y \cr & \Rightarrow x{\text{ = }}\frac{{27y}}{2} \times \frac{4}{3} \cr & = 18y \cr} $$

$$\eqalign{ & {\text{Harsha = 40}} \cr & {\text{Ritu = 60}} \cr & \frac{{{\text{Harsha}}}}{{{\text{Ritu}}}} = \frac{{40}}{{60}} \cr} $$
∴ Let x years ago their ages ratio was 3 : 5
$$\eqalign{ & \Rightarrow \frac{{40 - x}}{{60 - x}} = \frac{3}{5} \cr & \Rightarrow 200 - 5x = 180 - 3x \cr & \Rightarrow 2x = 20 \cr & \Rightarrow x = 10 \cr} $$

$$\eqalign{ & {\text{A : B}} \cr & \frac{3}{2}:\frac{8}{3} \cr} $$
take L.C.M. of denominator and multiply)
$$ \Rightarrow \frac{3}{2} \times 6:\frac{8}{3} \times 6 = 9x:16x$$
After adding 15 in each we get
$$\eqalign{ & \therefore \frac{{9x + 15}}{{16x + 15}} = \frac{5}{3} \times \frac{2}{5} \cr & \Rightarrow \frac{{9x + 15}}{{16x + 15}} = \frac{2}{3}\left( {{\text{cross multiply}}} \right) \cr & \Rightarrow 27x + 45 = 32x + 30 \cr & \Rightarrow 5x = 15 \cr & \Rightarrow x = 3 \cr & \therefore {\text{Smaller number}} \cr & = {\text{9}} \times {\text{3 = 27}} \cr & {\text{Greater number}} \cr & {\text{ = 16}} \times {\text{3 = 48}} \cr} $$