Finding class ratio with different mean marks for boys and girls

[Monoxdifly]

In a class, the mean of boys' mark is 7.1 while the mean of girls' mark is 8.1. Their overall mean is 7.5. If there are 40 students in the class, determine:
a. The ratio of boys and girls
b. The number of boys and girls

I got this question from someone I tutored. I got the number of boys as 24 and the number of girls as 16. However, his teacher got vice-versa. Which one was right?

 

[MarkFL]

Let's let $B$ be the number of boys, and so the number of girls is $G=40-B$. Let $S_i$ be the sum of the respective marks. From what we are given, we know:

$$\frac{S_B}{B}=\frac{71}{10}$$

$$\frac{S_G}{G}=\frac{81}{10}$$

$$\frac{S_B+S_G}{40}=\frac{15}{2}$$

Thus, we have the system:

$$B+G=40$$

$$71B=10S_B$$

$$81G=10S_G$$

$$S_B+S_G=300$$

Now, the 2nd and 3rd equations, when added and simplified, give us

$$S_B+S_G=\frac{71B+81G}{10}$$

And so this implies, when considering the 4th equation:

$$71B+81G=3000$$

Then using the 1st equation, and substitution, we obtain:

$$71B+81(40-B)=3000$$

$$71B+3240-81B=3000$$

$$B=24\implies G=16$$

I agree with you. :D

 

[HallsofIvy]

Equivalently, let a be the number of boys and b the number of girls. Then the weighted average of grades is $$\frac{7.1a+ 8.1b}{a+ b}= 7.5$$. Of course, a+ b= 40 so that is $$\frac{7.1a+ 8.1b}{40}= 7.5$$ or 7.1a+ 8.1b= 7.5(40)= 300. We need to solve the simultaneous equations 7.1a+ 8.1b= 300 and a+ b= 40. Multiplying the second equation by 7.1, 7.1a+ 7.1b= 7.1(40)= 284. Subtracting that from the first equation eliminates a leaving b= 16. Then a= 40- 16= 24.

Yet another way: from 7.1 to 8.1 is a difference of 1.0. From 7.1 to 7.5 is a difference of 0.4. 1.0- 0.4= 0.6. The 40 students must be divided into boys and girls the same way 1.0 is divided into 0.6 and 0.4. 0.6(40)= 24, 0.4(40)= 16.

 

[Monoxdifly]

Thank you guys.:D