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

Step 1: Form the required equation.} Let the initial number of boys be \(B\) and the initial number of girls be \(G\). After some students leave the class, the remaining numbers are: \[ \text{Remaining girls} = (1 - 0.40)G = 0.6G, \] \[ \text{Remaining boys} = (1 - 0.60)B = 0.4B. \] It is given that the remaining number of girls exceeds the remaining number of boys by 8. Hence, \[ 0.6G = 0.4B + 8. \] Multiplying both sides by 10, we get \[ 6G = 4B + 80. \] Dividing throughout by 2, \[ 3G = 2B + 40. \tag{1} \] Step 2: Use the condition of whole numbers Since 40\% of the girls and 60\% of the boys leave, the numbers leaving must be integers. Therefore, \[ 0.4G = \frac{2}{5}G \Rightarrow G \text{ is a multiple of } 5, \] \[ 0.6B = \frac{3}{5}B \Rightarrow B \text{ is a multiple of } 5. \] Let \[ G = 5y \quad \text{and} \quad B = 5x, \] where \(x\) and \(y\) are positive integers. Substituting in equation (1), \[ 3(5y) = 2(5x) + 40, \] \[ 15y = 10x + 40, \] \[ 3y = 2x + 8, \] \[ y = \frac{2x + 8}{3}. \tag{2} \] Step 3: Minimize the total number of students.} The total number of students initially is \[ T = B + G = 5x + 5y = 5(x + y). \] To obtain the minimum value of \(T\), we need the smallest possible value of \(x + y\), subject to: \[ B > 10 \Rightarrow 5x > 10 \Rightarrow x > 2, \] and \(y\) must be an integer from equation (2). Checking successive integer values of \(x\): \[ x = 3 \Rightarrow 2(3) + 8 = 14 \quad (\text{not divisible by } 3), \] \[ x = 4 \Rightarrow 2(4) + 8 = 16 \quad (\text{not divisible by } 3), \] \[ x = 5 \Rightarrow 2(5) + 8 = 18 \Rightarrow y = 6 \quad (\text{valid}). \] Step 4: Find the required numbers. \[ B = 5x = 5 \times 5 = 25 \ (>10), \] \[ G = 5y = 5 \times 6 = 30, \] \[ T = B + G = 25 + 30 = 55. \] Hence, the minimum possible initial number of students in the class is \(55\). 

Answer 2

Step 1: Set up the equation. 
Let the initial number of boys be \(B\) and girls be \(G\). After some students leave: 
\[ \text{Remaining girls} = (1 - 0.40)G = 0.6G, \] \[ \text{Remaining boys} = (1 - 0.60)B = 0.4B. \] Given that the remaining number of girls is 8 more than the remaining number of boys: 
\[ 0.6G = 0.4B + 8. \] Multiply both sides by 10: 
\[ 6G = 4B + 80. \] Divide by 2: 
\[ 3G = 2B + 40. \tag{1} \] 
Step 2: Apply integer constraints. 
Since 40% of girls and 60% of boys leave, the numbers leaving must be integers: 
\[ 0.4G = \frac{2}{5}G \implies G \text{ is a multiple of } 5, \] \[ 0.6B = \frac{3}{5}B \implies B \text{ is a multiple of } 5. \] Let 
\[ G = 5y, \quad B = 5x, \] where \(x\) and \(y\) are positive integers. Substitute into equation (1): 
\[ 3(5y) = 2(5x) + 40 \] \[ 15y = 10x + 40 \] \[ 3y = 2x + 8 \] \[ y = \frac{2x + 8}{3}. \tag{2} \] 
Step 3: Minimize total students. 
Total initial students: 
\[ T = B + G = 5x + 5y = 5(x + y). \] We want to minimize \(T\), i.e., minimize \(x + y\), subject to: 

• \(B > 10 \implies 5x > 10 \implies x > 2\),
• \(y\) must be an integer (from equation (2)).

Check small integer values of \(x\): 

• \(x = 3\): \(2(3) + 8 = 14\) (not divisible by 3),
• \(x = 4\): \(2(4) + 8 = 16\) (not divisible by 3),
• \(x = 5\): \(2(5) + 8 = 18\), so \(y = \frac{18}{3} = 6\) (valid).

Step 4: Compute numbers of boys, girls and total students. 
\[ B = 5x = 5 \times 5 = 25 \quad (>10, \text{valid}), \] \[ G = 5y = 5 \times 6 = 30, \] \[ T = B + G = 25 + 30 = 55. \] Thus, the minimum possible initial number of students in the class is \(55\).