The ratio of boys to girls in Class A is 3 to 4. The ratio of boys to

gurpreetsingh gives a good algebraic explanation. This question is also easy to solve by using multiples of the original ratios.

We can set up the ratios Ba (boys in A) to Ga (girls in a), and Bb (boys in b) to Gb (girls in B).

Ba/Ga = 3/4
Bb/Gb = 4/5

When combined, we'd like our final ratio ratio to be 17/22. If you write out a few multiples of each of the original ratios,

Ba/Ga = 3/4, 6/8, 9/12...
Bb/Gb = 4/5, 8/10, 12/15...

It's easy to see that by picking 9/12 for A and 8/10 for B will give you a combined ratio of 17/22. Thus, we will have 12 girls in group A.

Answer: E

Hey Bunuel,
While giving test, I have been able to solve it correctly using back solving. However, when I revisited this question, I was unable to solve it at the first go. Request you to help me out..

Let the multiplier be x for Class A . Therefore, number of boys: 3x and number of girls 4x
Let the multiplier be y for Class B. Therefor, number of boys: 4y and number of girls : 5y
It is given that -
3x= 4y+1 ----(A)
and
4x=5y+2 -----(B)
Also
As per question:-
3x+4y/4x+5y = 17/22 (combined boys: combined girls) ---(C)
Putting the value from A and B in Equation C giving me weird results -
y is coming out to be 6/13

I have also gone through the replies posted above and I am not too sure that I would be able to ignore the superfluous statement in the real time exam... so hoping to see the solution utilizing the info.
Thanks
H

Also, if you do use this equation, 3x+4y/4x+5y = 17/22, when you solve it, you get x = 3y/2.
Put x = 3y/2 is one of the equations above, say 3x= 4y+1.
3*3y/2 = 4y + 1
y = 2 (you probably have a calculation error somewhere)
So x = 3
Number of boys and girls in Class A is 3*3 and 4*3.

Mind you, since you have already got two distinct equations (3x= 4y+1 and 4x=5y+2) with two variables, you can easily solve them to get the values of x and y.
This question lets you make 3 equations but there are only 2 variables so you can use any 2 of those equations to get the value of the variables.

Actually speaking, you should not be making any equations. You have the ratio 3:4 and 4:5
You know that the number of boys is class A is 1 more so try and multiply the ratios to figure out where you get a difference of 1. I should multiply 3:4 by a bigger number since the number of boys is more in class A.

(3:4)*3 = 9:12
(4:5)*2 = 8:10

The numbers are very simple so you can arrive at the answer very quickly.

(3x-1)/(4x-2)=4/5

find 4x

sinxe x=3 , 4*3=12 girls in class A

total # of girls = 22k (22, 44, 66 ..)
if A has 2 more girls then # of girls in (a,b) = (12,10), (21,23) ..
at this point, you know that it can only be 12 because 21 is not an option.

The information the ratio of boys to girls in the combined class would be 17 to 22 is superfluous.
Too much info just to try to confuse you. If one counts the unknowns, we have 4: number of boys in class A, girls in A, boys in B and girls in B.
Without the info of the combined rate, we already have in fact four equations:
1) the ratio in class A
2) the ratio in class B
3) one more boy in class A in comparison with class B
4) two more girls in class A in comparison with class B
So, we don't need a fifth equation.

If we denote by B and G the number of boys and girls respectively in class A, we can write the following set of equations:
\(B/G=3/4\)
\((B-1)/(G-2)=4/5\)
Two equations, two unknowns, solve and get G=12.

Or, go with divisibility criteria, G must be a multiple of 4, so choose between A and E.
For A, B=6, G=8 but B-1=5, G-2=6, so \(5/6\neq4/5\).
For E, B=9, G=12 and B-1=8, G-2=10, so 8/10=4/5, OK. Also the additional condition holds, as (9+8)/(12+10)=17/22.

Too many given...

Combination of class equations:
\(3a + 4b = 17\)
\(4a + 5b = 22\)

Another given - Differences of class equations
\(3a = 4b + 1\)
\(4a = 5b + 2\)

Combine \(3a + 4b = 17\) and \(3a = 4b + 1\)

\(17-3a = 3a - 1\)
\(18 = 6a\)
\(a = 3\)

Answer: 4*3 = 12 (E)

Assuming I was short of time on the real test, is it safe to eliminate choices B,C and D on the spot? since 9,10,11 are not divisible by 4?

Yes, the ratio is given in its lowest terms 3:4.
So number of boys must be 3a and number of girls must be 4a where a is an integer.

Hi Bunnel,

Can you please point me to the similar ratio problems ? pls

-AJ

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Hope it helps.

The 17/22 ratio is given just for distraction and honestly speaking i fell prey of it . 17 is a prime number so i thought there is some short route and i can solve without using pen and paper but as i did nt have luxury of time , after wasting 20secs i gave up on it and move ahead with simultaneous eq to arrive at answer E.

Actually, you are not wrong. As I mentioned in my post on the previous page, you shouldn't need to make any equations. The only thing is that what will help you arrive at the answer is this "there is one more boy in class A than in class B..."

Here is how you can solve it:

You have the ratios 3:4 and 4:5.
You know that the number of boys in class A is 1 more so try and multiply the ratios to figure out where you get a difference of 1. I should multiply 3:4 by a bigger number since the number of boys is more in class A.

(3:4)*3 = 9:12
(4:5)*2 = 8:10

In this case, the ratio of the total will be 9+8:12+10 = 17:22

The numbers are very simple so you can arrive at the answer very quickly.

The ratio of boys to girls in Class A is 3 to 4. The ratio of boys to girls in Class B is 4 to 5. If the two classes were combined, the ratio of boys to girls in the combined class would be 17 to 22. If Class A has one more boy and two more girls than class B, how many girls are in Class A?

A. 8
B. 9
C. 10
D. 11
E. 12

Hi VeritasPrepKarishma,

If the question did not give the condition of number of Boys 1 more in A & 2 less number of girls in B, and just mentioned an absolute number for boys in Class A lets say 9 (as we figured above). Can we still find out the number of girls in Class B??

I am stuck to solve this using above reasoning.

Thanks in advance.

I'm not Karishma, but I thought this was an interesting question!

The short answer is: it would make the question a little less simple, but you could still answer it. Interestingly, changing the problem like this means that it now requires all of the given information to solve. That makes it a bit more GMAT-like!

If there are 9 boys in Class A, there are 12 girls. There are also 4x boys in Class B, and 5x girls. (You don't know the value of x, but it could be any integer. For instance, there could be 40 boys and 50 girls with x=10, or 16 boys and 20 girls with x=4.)

When you combine the classes together, there are 9 + 4x boys in total, and 12 + 5x girls.

You also know that the ratio, after combining the classes, is 17 to 22.

So, you've got a two-sided equation. The ratio of boys to girls is (9 + 4x) / (12 + 5x). It's also 17/22. One option is to put an equals sign between them, and solve:

(9+4x)/(12+5x) = 17/22
22(9) + 22(4x) = 17(12) + 17(5x)
88x - 85x = 204 - 198
3x = 6
x = 2

8 boys, 10 girls.

The smarter thing to do, would be to check some small values for x, and see whether they give you a 17 to 22 ratio. x = 2 is pretty simple to check, and it works out perfectly: 9 + 4(2) = 17, and 12 + 5(2) = 22.

Yes, it becomes a perfect candidate for weighted average formula in that case.

Class A - Fraction of boys = 3/7
Class B - Fraction of boys = 4/9
Combined - fraction of boys = 17/39

wA/wB = (4/9 - 17/39)/(17/39 - 3/7) = (1/3*39)/(2/7*39) = 7/6

Ratio of students in class A : students in class B = 7:6
Say, number of students in class A is 7n and the number in class B is 6n.
Class A has (3/7)*7n = 9 boys
n = 3
Number of girls in class B = (5/9)*6n = 10

Thank you ccooley & VeritasPrepKarishma. Your explanation helps a lot in understanding the concept better.

I also tried one more thing with the question to check my understanding.

What if the question mentioned that the combined class has 17 boys and we need to determine number of students of Class A.

So, the question stem gives following :

Class A, Boys : Girls = 3: 4
Class B, Boys : Girls = 4:5
Combined Class, Boys : Girls = 17 : 22
Total number of boys in Combined Class = 17
Find the number of students in Class A?

Using the weighted average formula, Ratio of total number of students in Class A : Total in Class B = 7 : 6 (Though the calculation took time, is there an alternative method to find this step? )...(#)

No. of Boys in combined class => \(\frac{3}{7}\)*7n + \(\frac{4}{9}\)*6n = 17
n= 3

From (#), 7n = 21 students in Class A.

Hope this is correct.

Is there any other method to solve this?

Thanks in advance.