Ratio - Aptitude MCQ Questions and Solutions with Explanations

Ratio of students in 2012 in institutes A and B = 7 : 15
Let number of students in institute A in 2012 = 700
And Number of students in institutes B in 2012 = 1500
25% increase in the number of students in 2013
Now, number of students in Institute A = 700 + 25% of 700 = 875
Number of students in B in 2013 as 26% students increased in B
= 1500 + 26% of 1500 = 1890
Current Ratio of the students,
$$ = \frac{{875}}{{1890}} = 25:54$$

$$\eqalign{ & \frac{4}{{15}}A = \frac{2}{5}B \cr & \Rightarrow A = \left( {\frac{2}{5} \times \frac{{15}}{4}} \right)B \cr & \Rightarrow A = \frac{3}{2}B \cr & \Rightarrow \frac{A}{B} = \frac{3}{2} \cr & \Rightarrow A:B = 3:2. \cr & \therefore {\text{B's}}\,{\text{share}} = Rs.\left( {1210 \times \frac{2}{5}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,484 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{third}}\,{\text{number}}\,{\text{be}}\,x \cr & {\text{Then,}}\,{\text{first}}\,{\text{number}} \cr & = 120\% \,{\text{of}}\,x = \frac{{120x}}{{100}} = \frac{{6x}}{5} \cr & {\text{Second}}\,{\text{number}} \cr & = 150\% \,{\text{of}}\,x = \frac{{150x}}{{100}} = \frac{{3x}}{2} \cr & \therefore {\text{Ratio}}\,{\text{of}}\,{\text{first}}\,{\text{two}}\,{\text{numbers}} \cr & = {\frac{{6x}}{5}:\frac{{3x}}{2}} = 12x:15x = 4:5 \cr} $$

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively
Then, 4x - 3x = 1000
⇒ x = 1000
∴ B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x)
  $$ = \left( {\frac{{140}}{{100}} \times 5x} \right),$$     $$\left( {\frac{{150}}{{100}} \times 7x} \right)$$   and $$\left( {\frac{{175}}{{100}} \times 8x} \right)$$
$$\eqalign{ & = 7x,\,\,\frac{{21x}}{2}{\text{ and }}14x \cr & \therefore {\text{The required ratio}} \cr & = 7x:\frac{{21x}}{2}:14x \cr & = 14x:21x:28x \cr & = 2:3:4 \cr} $$

Quantity of milk
$$ = \left( {60 \times \frac{2}{3}} \right)\,{\text{litres}} = 40\,{\text{litres}}$$
Quantity of water in it = (60 - 40) litres = 20 litres
New ratio = 1 : 2
Let quantity of water to be added further be x litres
$$\eqalign{ & {\text{Then,}}\,{\text{milk : water}} = {\frac{{40}}{{20 + x}}} \cr & {\text{Now}},\, {\frac{{40}}{{20 + x}}} = \frac{1}{2} \cr & \Rightarrow 20 + x = 80 \cr & \Rightarrow x = 60 \cr & \therefore {\text{Quantity}}\,{\text{of}}\,{\text{water}}\,{\text{to}}\,{\text{be}}\,{\text{added = 60}}\,{\text{litres}}{\text{.}} \cr} $$

Originally, let the number of boys and girls in the college be 7x and 8x respectively
Their increased number is (120% of 7x) and (110% of 8x)
$$\eqalign{ & \Rightarrow \left( {\frac{{120}}{{100}} \times 7x} \right)\,{\text{and}}\,\left( {\frac{{110}}{{100}} \times 8x} \right) \cr & \Rightarrow \frac{{42x}}{5}\,{\text{and}}\,\frac{{44x}}{5} \cr & \therefore {\text{The}}\,{\text{required}}\,{\text{ration}} \cr & = {\frac{{42x}}{5}:\frac{{44x}}{5}} \cr & = 21:22 \cr} $$

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively
$${\text{Then}},\,\frac{{2x + 4000}}{{3x + 4000}} = \frac{{40}}{{57}}$$
⇒ 57(2x + 4000) = 40(3x + 4000)
⇒ 6x = 68,000
⇒ 3x = 34,000
Sumit's present salary
= (3x + 4000)
= Rs. (34000 + 4000)
= Rs. 38,000

$$\eqalign{ & {x \times 5} = {0.75 \times 8} \cr & \Rightarrow x = \frac{6}{5} \cr & \Rightarrow x = 1.2 \cr} $$

Let the three parts be A, B, C. Then,
A : B = 2 : 3 and B : C = 5 : 8
$$\eqalign{ & = \left( {5 \times \frac{3}{5}} \right):\left( {8 \times \frac{3}{5}} \right) = 3:\frac{{24}}{5} \cr & \Rightarrow A:B:C = 2:3:\frac{{24}}{5} = 10:15:24 \cr & \Rightarrow B = {98 \times \frac{{15}}{{49}}} = 30 \cr} $$