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The scores of Amal and Bimal in an examination are in the ratio 11:1411:14. After an appeal, their scores increase by the same amount and their new scores are in the ratio 47:5647 : 56. The ratio of Bimal's new score to that of his original score is

Solution

✅ Correct Option: 4

Let's assign variables to make this problem manageable. Since Amal and Bimal's original scores are in the ratio 11:14, we can write:

Amal's original score = 11x11x

Bimal's original score = 14x14x

When we have a ratio like 11:14, we can think of it as 11 parts to 14 parts. By using 11x11x and 14x14x, we maintain this ratio while keeping our work algebraic.


The problem states that both scores increase by the same amount. Let's call this increase dd.

After the increase:

Amal's new score = 11x+d11x + d

Bimal's new score = 14x+d14x + d


We're told the new scores are in the ratio 47:56. This means:

Amal’s new scoreBimal’s new score=4756\tfrac{\text{Amal's new score}}{\text{Bimal's new score}} = \tfrac{47}{56}

Substituting our expressions:

11x+d14x+d=4756\tfrac{11x + d}{14x + d} = \tfrac{47}{56}


Cross-multiplying gives us:

56(11x+d)=47(14x+d)56(11x + d) = 47(14x + d)

616x+56d=658x+47d616x + 56d = 658x + 47d


Collecting like terms:

56d−47d=658x−616x56d - 47d = 658x - 616x

9d=42x9d = 42x

d=42x9=14x3d = \tfrac{42x}{9} = \tfrac{14x}{3}

The increase for both students is 14x3\tfrac{14x}{3}.


Now we can find Bimal's actual scores:

Original score: 14x14x

New score: 14x+d=14x+14x3=14x(1+13)=14x×43=56x314x + d = 14x + \tfrac{14x}{3} = 14x\left(1 + \tfrac{1}{3}\right) = 14x \times \tfrac{4}{3} = \tfrac{56x}{3}


The ratio of Bimal's new score to his original score is:

New scoreOriginal score=56x314x=56x3×114x=5642=43\tfrac{\text{New score}}{\text{Original score}} = \tfrac{\tfrac{56x}{3}}{14x} = \tfrac{56x}{3} \times \tfrac{1}{14x} = \tfrac{56}{42} = \tfrac{4}{3}

Therefore, the ratio is 4:3.

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