In class A, the ratio of boys to girls is 2 : 3. In class B the ratio

Given:
1. In class A, the ratio of boys to girls is 2 : 3.
2. In class B the ratio of boys to girls is 4 : 5.

Asked: If the ratio of boys to girls in both classes put together is 3 : 4, what is the ratio of number of girls in class A to number of girls in class B?

bA:gA = 2:3
bB:gB = 4:5
bA + bB : gA + gB = 3:4

Let bA = 2k; gA = 3k
Let bB = 4m; gB = 5m

(2k + 4m)/(3k+5m) = 3/4
8k + 16m = 9k + 15m
m = k

gA:gB = 3k/5k = 3/5

IMO C

Given:
1. In class A, the ratio of boys to girls is 2 : 3.
2. In class B the ratio of boys to girls is 4 : 5.

Asked: If the ratio of boys to girls in both classes put together is 3 : 4, what is the ratio of number of girls in class A to number of girls in class B?


Class ********* Boys *************Girls **************Total
A ************* 2k *************** 3k************** 5k
B ************* 4m ***************5m **************9m
Combined***** 2k+4m ********* 3k+5m **********5k + 9m

2k + 4m : 3k + 5m = 3:4
8k + 16m = 9k + 15m
k = m

Class ********* Boys *************Girls **************Total
A ************* 2k *************** 3k************** 5k
B ************* 4k ***************5k ****************9k
Combined***** 6k ***** ********* 8k *************14k

The ratio of number of girls in class A to number of girls in class B = 3k : 5k = 3/5

IMO C

Solution

Given

• Ratio of boys to girls in class A = 2:3, in class B = 4:5. Ratio of boys to girls in both classes put together is 3:4.

To find

• Ratio of number of girls in class A to number of girls in class B.

Approach and Working out
The question can be solved using the process skills of ‘Inference’. Let’s see how.

B1, B2 = Boys in classes A and B respectively,
G1 and G2 = Girls in classes A and B respectively.

• B1/G1 = 2/3.
• B1 = 2a, G1 = 3a. (where a is any positive integer)
• B2 / G2 = 4/5
• B2 = 4a, G2 = 5b (where b is any positive integer)
• B1+B2 / G1+G2 = 3/4.

Application of process skill of inference:

• We need to find G1/G2 = 3a/5b. =3/5(a/b).
• Thus if we can find a/b, we can solve the question.

Finding a/b:

• (2a + 4a)/(3b+5b) = 3/4.

• a/b = 1

Therefore (3/5)*1 = 3/5
Correct Answer: Option C

Hello VeritasKarishma, Bunuel, why am I not able to solve the above problem using the teeter totter method ?

It definitely looks like a mixture problem of sorts. When i solved this question using that method, I got the answer as 5/9

Can you guys please explain why this method would not work here ?

I am guessing you mean the weighted average method.

Concentration of girls in A = 3/5
Concentration of girls in B = 5/9
Avg conc = 4/7

No of students in A/No of students in B = (5/9 - 4/7) / (4/7 - 3/5) = 5/9 (Note that this is the ratio of students, not girls)

So if there are 5n students in A, there are 3n girls in A.
If there are 9n students in B, there are 5n girls in B.

Ratio of girls in A: B = 3n/5n = 3/5

Answer (C)

Hello VeritasKarishma,

Yes the teeter totter method is the weighted average method you are talking about. Thank you so much for your input on this question.

Just to confirm, for problems involving the weighted averages here, the weights depict the total mixture ? In the above case you mentioned that the weights 5/9 is not the ratio of weights of the girls from Class A/ Class B but the ratio of the students from the two classes instead.

Could you please explain this line method of weighted avg (or teeter totter method) more clearly. I did not find a lot of information about this on the GMAT Club forum.

I would truly appreciate it !!!!

Solution:

We can let 2x and 3x be the number of boys and girls in class A, respectively. Similarly, we can let 4y and 5y be the number of boys and girls in class B, respectively. We can create the equation:

(2x + 4y) / (3x + 5y) = 3/4

3(3x + 5y) = 4(2x + 4y)

9x + 15y = 8x + 16y

x = y

So there are 3x girls in class A and 5y = 5x girls in class B. Therefore, the ratio of number of girls in class A to number of girls in class B is 3x/5x = 3/5.

Answer: C

Suggest you to review these posts on our blog:
https://anaprep.com/arithmetic-weighted-averages/
https://anaprep.com/arithmetic-mixtures/
and these videos:
https://www.youtube.com/watch?v=_GOAU7moZ2Q
https://www.youtube.com/watch?v=VdBl9Hw0HBg

These discuss the weighted avg method in detail and how to apply it for mixtures. Essentially, this question is a mixtures problem. Class A mixed with Class B to give average. Girls and Boys are ingredients of the two individual solutions.
The concept lies in how average concentration is found (girls as a fraction of total).

Take a simple example.

Say class A has 20% girls and total 50 students (10 girls and 40 boys).
The number of girls = 20% of 50 (not 20% of 10, right?)

Say another class B has 40% girls and 100 students (40 girls and 60 boys).

The avg concentration of girls = (20% of 50 + 40% of 100) / (50 + 100)
This is just your weighted avg formula. So your w1 (50) and w2 (100) are the total number of students of each class.

Got it. So the weights / concentrations will represent the total quantity of the mixture and not parts of it. Understood. Thank you so much VeritasKarishma

Thank you sir!
Really help me to understand the question!! The problem wasn't as easy as I thought.

KarishmaB or ScottTargetTestPrep would we have been able to solve this if a would not have coincidentally been equal to b? Wouldn't it have left us with a variable and this would be unsolvable?

No, it is just a co-incidence that you got the answer as 3:5 and the given Boys:Girls ratios are 2:3 and 4:5.

I will change the rest of the data a little:
In class A, the ratio of boys to girls is 2 : 3. In class B the ratio of boys to girls is 4 : 5. If the ratio of boys to girls in both classes put together is 19:26, what is the ratio of number of girls in class A to number of girls in class B?

The answer has now changed. See what you get and how. Try weighted averages if you are comfortable with it.

karishma I don't understand - I end up with the issue I addressed, which is that 9/5 = a/b. There are two variables.

I did (2a + 4b) / (3a + 5b) = 19/26. Once simplified this gives 9/5 = a/b. In the prior example, given a was equal to b, you could directly use the numbers given in the ratio. In your new example, we don't know what a or b is, just the ratio between them, so how am I supposed to get the girls in each class?

If you insist on taking multiple variables, ensure that you are very clear on what each variable is.

If you say that 2a and 3a is the number of boys and girls in class A (hence total 5a children)
and 4b and 5b is the number of boys and girls in class B (hence total 9b children),
then you get a/b = 9/5 which means IF class A has 5a = 5*9 = 45 children, THEN class B has 9b = 9*5 = 45 children too. (still ratio terms only)
So both classes have equal number of children, whatever the actual number may be (45 or 90 or 135 etc.)

Assume both classes have 45 children each. Then number of girls in class A is 3/5th of 45 = 27
and number of girls in class B is 5/9th of 45 = 25.

So the ratio of girls in class A : girls in class B = 27 : 25
Assume that class A and class B have 90 students each and then check. The ratio will stay the same.

Since the answer required was in terms of ratios only, we could answer it without knowing any actual values.

Since it has got a bit complicated, I like to use weighted averages in which I deal with only the "concentration of girls".