Question

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?

Hint:

When percentages of people leave, ensure that those percentages give integer counts. Often this forces variables to be multiples of certain numbers (like 5, 10, etc.), which helps in solving and minimizing or maximizing totals.

Answer 1

Correct Answer: 55

Approach: Turn the words into one linear equation, then remember the hidden integer rules — 40% of girls and 60% of boys must be whole numbers, so both counts are multiples of 5. Then push for the smallest total.

Step 1: Frame the equation. Let boys \(= B\), girls \(= G\). After 40% of girls and 60% of boys leave, remaining girls \(= 0.6G\), remaining boys \(= 0.4B\). Given remaining girls are 8 more than remaining boys:
\[ 0.6G = 0.4B + 8 \;\Rightarrow\; 3G = 2B + 40. \tag{1} \]

Step 2: Use the "whole-number-leaving" condition. Since exactly 40% of girls leave, \(0.4G = \tfrac{2}{5}G\) is an integer, so \(G\) is a multiple of 5. Similarly \(0.6B = \tfrac{3}{5}B\) integer makes \(B\) a multiple of 5. Write \(B = 5x,\ G = 5y\).

Step 3: Reduce. Putting these in (1): \[ 15y = 10x + 40 \;\Rightarrow\; 3y = 2x + 8 \;\Rightarrow\; y = \frac{2x+8}{3}. \]

Intuition: Total \(= 5(x+y)\), so to minimise it we hunt for the smallest \(x\) (with \(B>10\), i.e. \(x\ge 3\)) that makes \(2x+8\) divisible by 3.

Step 4: Test values. \(x=3:\ 14\) (no), \(x=4:\ 16\) (no), \(x=5:\ 18 \Rightarrow y=6\) (yes).
So \(B = 25\ (>10)\), \(G = 30\). Check: remaining \(0.6\times30 = 18\), \(0.4\times25 = 10\), and \(18-10 = 8\). ✓

Step 5: Answer. \[ T = B + G = 25 + 30 = \boxed{55}. \]

Answer 2

Approach: Work directly with the original boy and girl counts, using divisibility to pin down which values are even possible, then search upward.

Step 1: Set up the equation. Let \(B\) = initial boys, \(G\) = initial girls. Remaining boys \(=0.4B\), remaining girls \(=0.6G\), and \(0.6G = 0.4B+8\), i.e. \(3G-2B=40\).

Step 2: Apply divisibility. Since \(0.4B\) and \(0.6G\) must be whole numbers of students, \(B\) must be a multiple of 5 (so \(2B/5\) is an integer) and \(G\) must be a multiple of 5 too. Also \(B>10\), so the smallest candidate is \(B=15\).

Step 3: Search upward. Test \(B=15,20,25,\ldots\) in \(3G=40+2B\): \(B=15\Rightarrow 3G=70\) (not divisible by 3); \(B=20\Rightarrow 3G=80\) (not divisible); \(B=25\Rightarrow 3G=90\Rightarrow G=30\) (works).

Step 4: Total. \(B+G=25+30=55\), and no smaller valid multiple-of-5 pair exists.

\[ \boxed{55} \]