Question

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

Hint:

For problems involving inequalities and integer solutions, solve for one variable first. Then, use the integer constraint to find the smallest possible valid value or a range of values.

Answer 1

Correct Answer: 135

Approach: Turn the two English conditions into one equation (the 8-more relation) and one inequality (boys exceed girls by more than 10), then hunt for the smallest valid integer.

Step 1: Let \(B\) = initial boys and \(G\) = initial girls.

Step 2: "Boys are more than 10 more than the girls" gives \(B \gt G + 10\).

Step 3: After 40% of girls leave, \(0.6G\) girls remain; after 60% of boys leave, \(0.4B\) boys remain. The remaining girls are 8 more than the remaining boys: \[ 0.6G = 0.4B + 8. \] Multiply by 10 and halve: \(3G = 2B + 40\), so \(B = \dfrac{3G - 40}{2}\).

Step 4: For \(B\) to be a whole number, \(3G - 40\) must be even, which forces \(G\) to be even.

Step 5: Put \(B = \dfrac{3G-40}{2}\) into \(B \gt G+10\): \[ \frac{3G-40}{2} \gt G + 10 \;\Rightarrow\; 3G - 40 \gt 2G + 20 \;\Rightarrow\; G \gt 60. \]

Step 6: Smallest even \(G\) above 60 is \(G = 62\). Then \(B = \dfrac{3(62)-40}{2} = \dfrac{146}{2} = 73\).

Final Answer: A possible initial strength is \(B + G = 73 + 62 = \boxed{135}\) students.

Answer 2

Approach: Set up two linear equations directly from "boys exceed girls by 10" and the post-departure condition, then solve simultaneously.

Let girls \(=g\) and boys \(=b\), with \(b=g+10\). After \(40\%\) of the girls and \(60\%\) of the boys leave, the remainder is \(0.6g\) girls and \(0.4b\) boys, and remaining girls \(=\) remaining boys \(+8\): \[ 0.6g = 0.4b+8. \]
Substitute \(b=g+10\): \[ 0.6g=0.4(g+10)+8=0.4g+4+8 \implies 0.2g=12 \implies g=60. \]
So \(b=60+10=70\), and the initial number of students is \[ 60+70=\boxed{130} \]
(Check: \(40\%\) of \(60=24\) girls leave, leaving \(36\); \(60\%\) of \(70=42\) boys leave, leaving \(28\); indeed \(36=28+8\).)

Note: the source phrase "more than 10 boys than girls" is read here as the standard "10 more boys than girls," which is the reading that yields a unique, consistent answer.