In a class, there were more than 10 boys and a certain number of girls. After 40% of the girls and 60% of the boys left the class, the remaining number of girls was 8 more than the remaining number of boys. Then, the minimum possible number of students initially in the class was

Entered answer:

✅ Correct Answer: 55

Let $B$ = number of boys initially and $G$ = number of girls initially.

After the students leave:

Boys remaining $= 40\%$ of $B = 0.4B$ (since $60\%$ left)

Girls remaining $= 60\%$ of $G = 0.6G$ (since $40\%$ left)

Since remaining girls is 8 more than remaining boys:

$0.6G = 0.4B + 8$

$6G = 4B + 80$

$3G = 2B + 40$

$G = \dfrac{2B + 40}{3}$

Since the number of students must be whole numbers, and a "fraction of a student" cannot leave:

$60\%$ of $B$ must be a whole number, so $B$ must be a multiple of $5$

$40\%$ of $G$ must be a whole number, so $G$ must be a multiple of $5$

Also, $B > 10$ (given).

So possible values of $B$ : $15, 20, 25, 30, ...$

To find the minimum total, start with the smallest $B$ :

For $B = 15$ : $G = \dfrac{2(15) + 40}{3} = \dfrac{70}{3} = 23.33$ — not a whole number

For $B = 20$ : $G = \dfrac{2(20) + 40}{3} = \dfrac{80}{3} = 26.67$ — not a whole number

For $B = 25$ : $G = \dfrac{2(25) + 40}{3} = \dfrac{90}{3} = 30$ — whole number and a multiple of $5$

Total students $= B + G$

$= 25 + 30$

$= 55$

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In a class, there were more than 10 boys and a certain number of girls. After 40% of the girls and 60% of the boys left the class, the remaining number of girls was 8 more than the remaining number of boys. Then, the minimum possible number of students initially in the class was

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