A Students Ratio Problem

[Guest]

I couldn't solve a similar word problem, so I imagined this one in the hope that starting from the solution, I could solve it.

But I can't. So any help is appreciated. Thank you very much.

If you want me to submit the solution, I'll do it. Just ask.

In the classes A and B, the total number of girls was [tex]150%[/tex] more than the total number of boys.

The ratio of boys to girls in class A was [tex]\dfrac{3}{5}[/tex] and the ratio of the boys to girls in class B was [tex]\dfrac{4}{15}[/tex].

There was [tex]10[/tex] more girls in class B than in class A.

Find the number of of boys and girls in each class.

[Guest]

First identify the variables we need to determine and/or are mentioned in the information given in the question.

Let
[tex]B_A[/tex] = number of boys in class A
[tex]G_A[/tex] = number of girls in class A
[tex]B_B[/tex] = number of boys in class B
[tex]G_B[/tex] = number of girls in class B

Next convert the information given in the question into equations that are expressed in terms of the variables.

"the total number of girls was 150% more than the total number of boys."
breaks down in the following way:
"the total number of girls" --> [tex]G_A+G_B[/tex]
"was" --> =
"150% more than" --> [tex](1.5+1)\times[/tex]
"the total number of boys" --> [tex]B_A+B_B[/tex]
which combines to give the equation
[tex]G_A+G_B = 2.5(B_A+B_B)[/tex]

"The ratio of boys to girls in class A was 3/5"
translates to
[tex]B_A/G_A=3/5[/tex]

"the ratio of the boys to girls in class B was 4/15"
becomes
[tex]B_B/G_B = 4/15[/tex]

"There was 10 more girls in class B than in class A"
as an equation is
[tex]G_B = G_A+10[/tex]

You now have 4 equations in 4 unknown variables.

The next step is to try to rearrange the equations and then by using the "elimination method" or the "substitution method" try to reduce the number of variables and equations. There are lots of resources online explaining how these two methods work for the case of simultaneous equations in two variables and then how to apply the methods to larger systems of equations. So I won't go over how the methods work in general here. I will however solve the specific equations given in your question using the substitution method.

We start with
[tex]G_A+G_B = 2.5(B_A+B_B)[/tex]
[tex]B_A/G_A=3/5[/tex]
[tex]B_B/G_B = 4/15[/tex]
[tex]G_B = G_A+10[/tex]

Substitute the 4th equation into the other 3 to get:
[tex]G_A+(G_A+10) = 2.5(B_A+B_B)[/tex]
[tex]B_A/G_A=3/5[/tex] (this equation didn't change as there is no [tex]G_B[/tex] term)
[tex]B_B/(G_A+10) = 4/15[/tex]

The new third equation can be rearranged so that one variable can be written in terms of the others. Specifically
[tex]B_B = 4/15 \times (G_A+10)[/tex]
Substitute this into the other 2 equations to get:
[tex]G_A+(G_A+10) = 2.5(B_A+4(G_A+10)/15)[/tex]
[tex]B_A/G_A=3/5[/tex]

The new second equation can be rearranged so that one variable can be written in terms of the others. Specifically
[tex]B_A = 3/5 \times G_A[/tex]
Substitute this into the other equation to get:
[tex]G_A+(G_A+10) = 2.5(3G_A/5+4(G_A+10)/15)[/tex]
this simplifies to
[tex]2G_A+10 = 5/2\times(3G_A/5+4G_A/15+40/15)[/tex]
continuing to simplify we get
[tex]2G_A+10 = 3G_A/2+2G_A/3+20/3[/tex]
[tex](2-3/2-2/3)G_A = 20/3-10[/tex]
[tex](-1/6)G_A = -10/3[/tex]
[tex]G_A = 20[/tex]

Looking back on the equations we substituted we see that [tex]B_A = 3/5 \times G_A[/tex] and we know [tex]G_A = 20[/tex] which implies [tex]B_A = 3/5\times 20 = 12[/tex].
Similarly we used [tex]B_B = 4/15 \times (G_A+10)[/tex] which implies [tex]B_B = 4/15 \times(20+10) = 8[/tex].
Finally we used [tex]G_B = G_A+10[/tex] which means [tex]G_B = 30[/tex].

So the solution is:
[tex]B_A = 12[/tex]
[tex]G_A = 20[/tex]
[tex]B_B = 8[/tex]
[tex]G_B = 30[/tex]

Hope this helped,

R. Baber.

[Guest]

Wow! I could never reach the solution by my own. Thank you sooo much, and I really wish you the best!
Your help is great, R. Baber.
Just for the record, this is how I designed The problem;

*********Boys*****Girls*****Ratio
Class A***12********20*******3/5
Class B****8********30******4/15

(Total number of girls)-(Total number of boys)
-------------------------------------------------------------- x 100 =150%
(Total number of boys)

Again, thank you very much.

[Guest]

If you prefer you could have used your formula to get the first equation
("Total number of girls"-"Total number of boys")/"Total number of boys" [tex]\times[/tex] 100 = 150
becomes
[tex]((G_A+G_B)-(B_A+B_B))/(B_A+B_B)\times 100 = 150[/tex]
this rearranges in the following way:
[tex]((G_A+G_B)-(B_A+B_B))/(B_A+B_B) = 150/100[/tex]
[tex](G_A+G_B)-(B_A+B_B) = 1.5(B_A+B_B)[/tex]
[tex]G_A+G_B = 1.5(B_A+B_B)+(B_A+B_B)[/tex]
[tex]G_A+G_B = 2.5(B_A+B_B)[/tex]
which is the same equation I gave in my post.

Hope this helped,

R. Baber.

[Guest]

If you prefer you could have used your formula to get the first equation
("Total number of girls"-"Total number of boys")/"Total number of boys" [tex]\times[/tex] 100 = 150
becomes
[tex]((G_A+G_B)-(B_A+B_B))/(B_A+B_B)\times 100 = 150[/tex]
this rearranges in the following way:
[tex]((G_A+G_B)-(B_A+B_B))/(B_A+B_B) = 150/100[/tex]
[tex](G_A+G_B)-(B_A+B_B) = 1.5(B_A+B_B)[/tex]
[tex]G_A+G_B = 1.5(B_A+B_B)+(B_A+B_B)[/tex]
[tex]G_A+G_B = 2.5(B_A+B_B)[/tex]
which is the same equation I gave in my post.

Hope this helped,

R. Baber.

Your first answer is nice and crystal clear. And believe me, I couldn't have solved the problem without assistance, without your precious assistance, in fact.
I learnt a great deal from your answer: I learnt how to translate the problem to equations, and I'm still reading it to learn again how to elaborate a very good answer.
My mistake was that I believed without reason that the problem will translate to two equations, not more. Besides I lack experience. Such experience will probably come with practice and time.

[Guest]

One handy rule is that (in general) you should have the same number of equations as you have unknown variables that you are trying to determine.
This rule doesn't always work, but the vast vast majority of questions you will come across will have been specifically designed with this in mind.

In your question you have 4 unknowns, and so you should expect 4 bits of information from the question which can be translated into 4 equations.
As you saw substituting one equation into the others got rid of a variable, turning the problem into one of 3 variables with 3 equations. Repeating the procedure got us down to 2 variables and 2 equations, and repeating a final time got us down to 1 variable and 1 equation (which is significantly easier to solve than our original problem). So this gives you some idea why ideally we should have the same number of equations as variables.

You are absolutely right that with experience things will get easier. My explanation was written to try to explain what was going on as well as answering the question. When answering such questions for exams and homework you don't need to go into as much detail. It is usually enough to state the equations (without quoting which part of the question it was translated from) and then to start manipulating the equations giving some indication of generally what you are doing, so that the examiner can follow your argument (even if you make an accidental mistake).

Hope this helped,

R. Baber.

[Guest]

Let class A be 3:5 (boys:girls), meaning that BA = 3K.
GA = 5k
  If class B is 4:15, then BB = 4m.
GB= 15m.
"10 more girls in B than A"
15m = 5k + 10−k = 3m−2.
"Total girls 150 more than total boys"
(5k+15m)−(3k+4m) = 150 ×2k+11m = 150.

Replace k=3m−2 .
2(3m−2)+11m = 150 
17m = 154
 m = 154/17 

not a whole number. Therefore, the data (a ratio or one of the integers) must be misstated; there is no integer answer.