Ratio - Aptitude MCQ Questions and Solutions with Explanations

According to the question,
$$\eqalign{ & \frac{{{\text{Story books}}}}{{{\text{Other books}}}} = \frac{7}{2} \cr & {\text{Story books = 1512}} \cr & {\text{7 units}} \to 1{\text{512}} \cr & {\text{1 unit}} \to \frac{{1512}}{7} = 216 \cr & {\text{2 units}} \to 216 \times 2 = 432 \cr & {\text{Other books }} = 432 \cr & {\text{New ratio of}} \cr & \frac{{{\text{Story books}}}}{{{\text{Other books}}}} = \frac{{15}}{4} \cr} $$
As we know that only story books are added
$$\eqalign{ & {\text{4 units}} \to 43{\text{2}} \cr & {\text{1 unit}} \to \frac{{432}}{4} = 108 \cr & {\text{15 units}} \to 108 \times 15 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1620 \cr} $$
New collection of story books = 1620
∴ Number of story books are added = 1620 - 1512 = 108

Let the required number be x
Then,
$$ = \frac{{7 - x}}{{9 - x}} = \frac{{11 - x}}{{15 - x}}$$
⇒ (7 - x)(15 - x) = (11 - x)(9 - x)
⇒ 105 - 22x + x² = 99 - 20x + x²
⇒ 2x = 6
⇒ x = 3

$$\eqalign{ & {\text{Quantity of milk}} \cr & = \left( {80 \times \frac{7}{{10}}} \right){\text{litres}} \cr & = {\text{56 litres}}{\text{.}} \cr & {\text{Quantity of water}} \cr & = \left( {80 - 56} \right){\text{litres}} \cr & = {\text{24 litres}}{\text{.}} \cr} $$
Let the quantity of water to be added be x litres
Then,
$$\eqalign{ & = \frac{{56}}{{24 + x}} = \frac{2}{1} \cr & \Rightarrow 2x = 8 \cr & \Rightarrow x = 4 \cr} $$

Let the numbers be 3x and 5x
Then,
$$ = \frac{{3x - 9}}{{5x - 9}} = \frac{{12}}{{23}}$$
⇒ 23(3x - 9) = 12(5x - 9)
⇒ 69x - 207 = 60x - 108
⇒ 9x = 99
⇒ x = 11
∴ Largest number = 5 × 11 = 55

Initial Ratio of Milk and water = 5 : 1 = 5x : 1x
After adding 5 litres of water in the mixture, Ratio of Milk and water 5 : 2 = 5x : 2x
Water 1 unit ratio increase by adding 5 litres of water. i.e x = 5
The quantity of milk in the mixture = 5 × 5 = 25 litres

Sides of a triangle are in the ratio of 7 : 9 : 12
i.e. 7x : 9x : 12x
Difference between the length of largest and smaller sides = 12x - 7x = 15
⇒ x = 3
∴ largest sides would be 12 × 3 = 36cm

Sum of the opposite angles in cyclic quadrilateral is 180°
∠A + ∠C = 180°
⇒ x + 4x = 180°
∴ x = 36°
Now ∠B = 3x = 3 × 36° = 108°
∠D = 180 - ∠B = 180 - 108 = 72°

Let the number of managers and production - line workers be 5x and 72x respectively
Then,
$$\eqalign{ & = \frac{{5x}}{{72x + 8}} = \frac{5}{{74}} \cr & \Rightarrow 370x = 360x + 40 \cr & \Rightarrow 10x = 40 \cr & \Rightarrow x = 4. \cr} $$
∴ Number of managers = 5 × 4 = 20

Let the number to be added be x
Then,
$$\eqalign{ & = \frac{{3 + x}}{{5 + x}} = \frac{5}{6} \cr & \Rightarrow 6\left( {3 + x} \right) = 5\left( {5 + x} \right) \cr & \Rightarrow 18 + 6x = 25 + 5x \cr & \Rightarrow x = 7 \cr} $$

Let the number of ladies and gents at the party be 3x and 2x respectively.
Then,
$$\eqalign{ & = \frac{{3x}}{{2x + 20}} = \frac{2}{3} \cr & \Rightarrow 9x = 4x + 40 \cr & \Rightarrow 5x = 40 \cr & \Rightarrow x = 8 \cr} $$
∴ Number of ladies = 3 × 8 = 24