Ratio AND Proportion Questions FOR Bank Exams with Detailed Solutions

Given
A : M : D = 4 : 5 : 3
Extra = 360 (equally)
Dev Final = 600

Formula Used
Total = Original + Share

Solution
Share each = 360 / 3 = 120
Original Dev = 3x
3x + 120 = 600 => 3x = 480 => x = 160
Arjun Original = 4x = 4 × 160 = 640

Given
Income ratio (A:K) = 4 : 7
Spending ratio (A:K) = 6 : 11
A's saving = 1/3 of income

Formula Used
Saving = Income - Spending

Solution
Let incomes be 4x, 7x and spendings be 6y, 11y.
A's saving = 4x - 6y
Also, 4x - 6y = $$\frac{1}{3}(4x)$$​​
12x - 18y = 4x
8x = 18y => x =$$\frac{9}{4}y$$​​
A's saving =$$\frac{1}{3} \times 4(\frac{9}{4}y) = 3y$$​​
K's income =$$7(\frac{9}{4}y) = \frac{63}{4}y$$​

K's saving =$$\frac{63}{4}y - 11y = \frac{63y - 44y}{4} = \frac{19}{4}y$$​​
Ratio A : K = 3y :$$\frac{19}{4}y$$​= 12 : 19

Final Answer
So the correct answer is (d)

​$$\textbf{Given:}\\\frac{3}{4} \text{ of one number is } 24 \text{ less than } \frac{3}{5} \text{ of the other number}\\\text{Sum of both numbers} = 292\\[6pt]\textbf{Formula Used:}\\\text{Linear equations method}\\[6pt]\textbf{Solution:}\\\text{Let the first number} = x\\\text{Let the second number} = y\\[6pt]\frac{3}{4}x = \frac{3}{5}y - 24\\\frac{3}{4}x - \frac{3}{5}y = -24\\[6pt]\text{Multiply by } 20 \text{ to remove denominators}\\15x - 12y = -480\\[6pt]\text{Also given:}\\x + y = 292\\[6pt]\text{Multiply by } 12\\12x + 12y = 3504\\[6pt]\text{Add both equations:}\\27x = 3024\\x = 112\\[6pt]y = 292 - 112 = 180\\[6pt]\textbf{Final Answer:}\\112$$

Exam Hall Method:

let total employees in the company be 100x

Exam Hall Approach

Given
Income ratio (A:B:C) = 12 : 9 : 7
Spending ratio (A:B:C) = 15 : 9 : 8
A saves 25% of his income.

Formula Used
Saving = Income - Spending

Solution
Let incomes be 12x, 9x, 7x and spendings be 15y, 9y, 8y.
A's saving = 12x - 15y
Also, A's saving = 25% of 12x = 3x
12x - 15y = 3x => 9x = 15y => x =$$\frac{5}{3}y$$​​
Income A = $$12(\frac{5}{3}y)$$​ = 20y; Spending A = 15y; Saving A = 5y
Income B =$$9(\frac{5}{3}y)$$​= 15y; Spending B = 9y; Saving B = 6y
Income C =$$7(\frac{5}{3}y) = \frac{35}{3}y$$​; Spending C = 8y; Saving C =$$\frac{35}{3}y - 8y = \frac{11}{3}y$$​​
Ratio A:B:C = 5y : 6y :$$\frac{11}{3}y = 15 : 18 : 11$$​​

Final Answer
So the correct answer is (a)

Given
Ratio of incomes (S : R : K) = 84 : 76 : 89
R's annual income = Rs. 4,56,000

Formula Used
Individual value = (Total Value / Total Ratio) × Individual Ratio

Solution
Let the annual incomes be 84x, 76x, and 89x.
76x = 4,56,000
x = $$\frac{456000}{76} = 6000$$​​
Sum of S and K annual incomes = 84x + 89x = 173x
Sum = 173 × 6000 = 1,038,000

Final Answer
So the correct answer is (d)

Given
Ratio Rs. 1 : Rs. 2 = 5 : 7
Ratio Rs. 2 : Rs. 5 = 7 : 6
Value of Rs. 1 coins = Rs. 15

Formula Used
Number of coins = Total Value / Value of one coin

Solution
Combined ratio of number of coins (Rs. 1 : Rs. 2 : Rs. 5) = 5 : 7 : 6
Number of Rs. 1 coins = $$\frac{15}{1} = 15$$​​
Since 5 units in ratio = 15 coins,
1 unit = 3
Number of Rs. 5 coins = 6 units = 6 × 3 = 18 coins
Total value of Rs. 5 coins = 18 × 5 = Rs. 90

Final Answer
So the correct answer is (b)

​$$\textbf{Given:} \\ \space \, \\\text{Ratio of boys to girls in class X} = 6:5 \\ \space \, \\\text{Total students in both classes} = 242 \\ \space \, \\\textbf{Formula Used:} \\ \space \, \\\text{Total students} = \text{Boys} + \text{Girls} \\ \space \, \\\textbf{Solution:} \\ \space \, \\\text{Let boys and girls in class X be } 6x \text{ and } 5x \\ \space \, \\\text{Boys in class Y} = 6x + 12 \\ \space \, \\\text{Girls in class Y} = 5x + 20\% \text{ of } 5x = 6x \\ \space \, \\\text{Total students} = 11x + (12x + 12) = 23x + 12 \\ \space \, \\23x + 12 = 242 \\ \space \, \\23x = 230 \\ \space \, \\x = 10 \\ \space \, \\\text{Total boys} = 6x + (6x + 12) = 132 \\ \space \, \\\textbf{Final Answer:} \\ \space \, \\132$$​​

Information Given:
The initial ratio of boys to girls in the school is 5:3.
After 100 boys leave and 50 girls join, the new ratio of boys to girls becomes 10:11.
Asked: We need to find the difference between the initial number of boys and girls.
Explanation:
Let the initial number of boys be 5x and the initial number of girls be 3x,
After 100 boys leave the school and 50 girls join,
the number of boys = 5x – 100
And the number of girls becomes 3x + 50.
ATQ, (5x - 100)/(3x + 50) = 10/11
(5x - 100) ×11 = (3x + 50) ×10
55x - 1100 = 30x + 500
55x - 30x = 500 + 1100
25x = 1600
x = 1600/25
x = 64
Initial number of boys = 5x = 5×64 = 320
Initial number of girls = 3x = 3×64 = 192
Required difference = 320 - 192 = 128