Uncovering the Number of Pupils in Class 6A

[wailingkoh]

Hi all
I need help

In the classes 6A and 6B, the total number of girls was 100% more than the total number of boys. The ratio of boys to girls in class 6A was 3 : 4 and the ratio of the boys to girls in class 6B was 1 : 6. If there were 8 more girls in class 6A than class 6B, find the number of pupils in class 6A.

Tks :)

 

[MarkFL]

Hello wailingkoh! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far in your latest 2 threads?

Also, our "Other Topics" forum in the "Pre-University" forum group is meant for "Algebra-based Physics, Chemistry, Biology" questions. The questions you have been posting are best posted in our "Pre-Algebra and Algebra" forum, and so that's why I have been moving them.

 

[MarkFL]

Hi all
I need help

In the classes 6A and 6B, the total number of girls was 100% more than the total number of boys. The ratio of boys to girls in class 6A was 3 : 4 and the ratio of the boys to girls in class 6B was 1 : 6. If there were 8 more girls in class 6A than class 6B, find the number of pupils in class 6A.

Tks :)

Let $B_A$ be the number of boys in class 6A, $G_A$ be the number of girls in class 6A, $B_B$ be the number of boys in class 6B, and $G_B$ be the number of girls in class 6B.

Given that "the total number of girls was 100% more than the total number of boys," we may write:

\(\displaystyle 2\left(B_A+B_B\right)=G_A+G_B\)

Given that "The ratio of boys to girls in class 6A was 3 : 4," we may also write:

\(\displaystyle 4B_A=3G_A\)

Given that "the ratio of the boys to girls in class 6B was 1 : 6," we may also write:

\(\displaystyle 6B_B=G_B\)

And given that "there were 8 more girls in class 6A than class 6B," we may also write:

\(\displaystyle G_A=G_B+8\)

Let's begin with the first equation:

\(\displaystyle 2\left(B_A+B_B\right)=G_A+G_B\)

The 4th equation implies \(\displaystyle G_B=G_A-8\) and so we now have:

\(\displaystyle 2\left(B_A+B_B\right)=G_A+G_A-8\)

\(\displaystyle 2\left(B_A+B_B\right)=2G_A-8\)

\(\displaystyle B_A+B_B=G_A-4\)

Using the 3rd and 4th equations, we find:

\(\displaystyle B_B=\frac{G_B}{6}=\frac{G_A-8}{6}\)

And so we now have:

\(\displaystyle B_A+\frac{G_A-8}{6}=G_A-4\)

\(\displaystyle 6B_A+G_A-8=6G_A-24\)

\(\displaystyle 6B_A+16=5G_A\)

Using the 2nd equation, there results:

\(\displaystyle 3G_A+2B_A+16=5G_A\)

\(\displaystyle B_A+8=G_A\)

\(\displaystyle 7B_A+56=7G_A\)

\(\displaystyle 3B_A+3G_A+56=4B_A+4GA\)

\(\displaystyle B_A+G_A=56\)

Thus, we conclude there are 56 pupils in class 6A.