Arithmetic

Percentages

Medium

$8\%$ of boys in a class made it past the 90th percentile, while $12\%$ of the girls in the same class made it past that mark. Suppose there are $120$ students in this class: what is the number of boys?

None of these

✅ Drop

Note: For this question, IIM-Kozhikode found a discrepancy and cancelled the question. Full marks were awarded to all candidates.

Let $B =$ number of boys and $G =$ number of girls in the class.

Since there are $120$ students total:

$B + G = 120$

$G = 120 - B$

From the given information:

$8\%$ of boys made it past the $90$ th percentile $= 0.08B$ boys
$12\%$ of girls made it past the $90$ th percentile $= 0.12G$ girls

The $90$ th percentile means the top $10\%$ of all students. The total number of students who made it past the $90$ th percentile is:

$10\% \text{ of } 120 = 12$ students

Students past $90$ th percentile from boys + Students past $90$ th percentile from girls $= 12$

$0.08B + 0.12G = 12$

Substituting $G = 120 - B$ :

$0.08B + 0.12(120 - B) = 12$

$0.08B + 14.4 - 0.12B = 12$

$-0.04B + 14.4 = 12$

$-0.04B = -2.4$

$B = \dfrac{-2.4}{-0.04}$

$B = 60$

With $B = 60$ and $G = 60$ :

Boys past $90$ th percentile $= 8\% \text{ of } 60 = 4.8$ students
Girls past $90$ th percentile $= 12\% \text{ of } 60 = 7.2$ students

The number of students cannot be in decimals, creating an inconsistency in the problem.

Therefore, there are $60$ boys in the class.

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BMSAT Kozhikode Past Year Questions (Topic-Wise)

$8\%$ of boys in a class made it past the 90th percentile, while $12\%$ of the girls in the same class made it past that mark. Suppose there are $120$ students in this class: what is the number of boys?