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

Step 1: Understanding the Question:
The problem provides two conditions relating the initial and final number of boys and girls in a class. We need to translate these conditions into a mathematical system of an equation and an inequality to find a possible total number of students.
Step 2: Key Formula or Approach:
We will set up variables for the number of boys (B) and girls (G). We will then form a linear equation based on the remaining students and a linear inequality based on the initial number of students. Solving this system will give us a constraint on the number of girls, which we can use to find the smallest possible integer solution.
Step 3: Detailed Explanation:
Let 'B' be the initial number of boys and 'G' be the initial number of girls.
Condition 1: Initial State
The number of boys was more than 10 than the number of girls.
\[ B>G + 10 \quad \cdots \text{(Inequality 1)} \] Condition 2: Final State
- 40% of girls left, so 100% - 40% = 60% of girls remain. Number of remaining girls = \(0.6 \times G\).
- 60% of boys left, so 100% - 60% = 40% of boys remain. Number of remaining boys = \(0.4 \times B\).
The remaining number of girls was 8 more than the remaining boys.
\[ 0.6G = 0.4B + 8 \quad \cdots \text{(Equation 2)} \] Now, we solve this system. First, let's simplify Equation 2 by multiplying by 10 and then dividing by 2:
\[ 6G = 4B + 80 \] \[ 3G = 2B + 40 \] From this, we can express B in terms of G:
\[ 2B = 3G - 40 \implies B = \frac{3G - 40}{2} \] Substitute this expression for B into Inequality 1:
\[ \frac{3G - 40}{2}>G + 10 \] \[ 3G - 40>2(G + 10) \] \[ 3G - 40>2G + 20 \] \[ 3G - 2G>20 + 40 \] \[ G>60 \] Since B and G must be integers, B = \(\frac{3G - 40}{2}\) implies that \(3G - 40\) must be an even number. This means \(3G\) must be even, which in turn means G must be an even integer.
So, we need the smallest even integer G that is greater than 60. The smallest such value is \(G = 62\).
Now, we find the corresponding value of B:
\[ B = \frac{3(62) - 40}{2} = \frac{186 - 40}{2} = \frac{146}{2} = 73 \] The initial number of students is \(B + G\).
Total students = \(73 + 62 = 135\).
Step 4: Final Answer:
A possible number of students initially in the class is 135.