MGMAT QUESTION:
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?

MGMAT ANSWER:
The ratio of boys to girls in Class A is 3:4, so we can start this problem by calling the number of boys in class A 3x and the number of girls 4x. (Remember that ratios always employ a common multiplier to calculate the actual numbers.)

If the number of boys in class A is 3x, then the number of boys in class B is (3x – 1). Similarly, if the number of girls in class A is 4x, then the number of girls in class B is (4x – 2).

We can now use the ratio of boys to girls in class B to solve for x.

3x – 1

4x – 2
=
4

5
5(3x – 1) = 4(4x – 2)
15x – 5 = 16x – 8
3 = x

Now that we know the value of x, we can solve for the number of girls in class A. The number of girls in class A is 4x, or 4(3), which equals 12.

The correct answer is E.

MY QUESTION: is there a different way to solve this that would involve using the information re the 17:22 ratio of boys to girls in the combined class??

Hi,

There is.

Use MGMAT answer to the point where you have the common multiplier and you can express the numbers in terms of x:


Now,
Number of boys in the new class:3x + (3x - 1) = 6x - 1
Number of girls in the new class: 4x + (4x -2) = 8x - 2

The ratio of these two is 17/22. So we have the equation:

(6x - 1)/(8x - 2) = 17/22
=> 22(6x - 1) = 17 (8x - 2)
=> 132x - 22 = 136x - 34
=> 132x - 136x = -34 + 22
=> -4x = -12
=> x = 3

Now that you have x, you can solve the number of boys and girls for any class

Regards

Sunil

PS: Please note that MGMATs approach is still better because it involves easier calculations. In the above, we had to deal with 17 and 22, which would have cost us 20-30 seconds on the test

Thanks Sunil! I agree with your point about the MGMAT strategy being the preferred method.

_________________
Jamie Nelson
ManhattanGMAT Instructor

"If the number of boys in class A is 3x, then the number of boys in class B is (3x – 1). Similarly, if the number of girls in class A is 4x, then the number of girls in class B is (4x – 2)."

Can someone please explain where the (3x-1) and (4x-2) come from?

Sure! We know that the ratio of boys to girls in Class A is 3 to 4.
We don't know how many actual boys and girls there are, but we know that multiplying both the 3 and the 4 by the same number will give us the actual number. We call this multiplication factor "the unknown multiplier" and represent it as "x". Thus, 3x is the actual number of boys in Class A and 4x is the actual number of girls in Class A.

We then are told that Class A has one more boy and two more girls than class B. So if Class A has 3x boys, Class B must have one fewer boy, or 3x-1. If Class A has 4x girls, Class B must have two fewer girls, or 4x-2.

I hope that this helps!

_________________
Jamie Nelson
ManhattanGMAT Instructor

Now this is my understanding in ratio problems to make them simpler:

Class A- remember students in ONE class/group/club will always share a common multiplier eg:(x) BUT this multiplier cannot be used for another group/class/club. SO.

Class B - We use (y) as a multiplier NOW

Given:
Class A - boys/girls = 3x/4x
Class B - boys/girls = 4y/5y

According to question:

4y=3x +1 ---------equation I (given boys of class B are one less than class A hence add one to class A to equate)
Similarly, 5y=4x+2 ------equation II

Solve equations I & II to get x = -3 (the multiplier for students in class A is derivated) (cannot be -ve because we are talking abt students )

Hence Initial number of Girls in class A = 4 * 3 = 12 ANS.

thanks!

_________________
Tim Sanders
Manhattan GMAT Instructor