Class Act : MGMAT problem, too many equations. Please help.

[singhpk2]

This is the question from one of the MGMAT online test (Question Code= Class Act):
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 the number of boys in Class B is one less than the number of boys in Class A, and if the number of girls in Class B is two less than the number of girls in Class A, how many girls are in Class A?

My approach:
Bys Gls
A 3x 4x (from:The ratio of boys to girls in Class A is 3 to 4)
B 4y 5y (from:The ratio of boys to girls in Class B is 4 to 5)
----------------------------
Total Bys=3x+4y
Gls=4x+5y

From: If the two classes were combined, the ratio of boys to girls in the combined class would be 17 to 22
(3x+4y)/(4x+5y)=17/22

From: If the number of boys in Class B is one less than the number of boys in Class A
4y=3x-1

From: if the number of girls in Class B is two less than the number of girls in Class A
5y=4x-2

THE ISSUE:
We have 3 equations for 2 unknowns so numerous solutions can be found.

The options in this MGMAT questions are 8,9,10,11,12

Please help.

[singhpk2]

here is the image of it

[sanjeev]

hi,

I think you are going well. But your assumption is wrong.
Number of equations doesnt determine the total number of solution. Its the degree of equation which determine the number of solutions.

Example of degree 2 equation, x^2 - 4x + 5 = 0, Here x can have 2 values.
Example of degree 3 equation x^3 - 4x^2 +5 =0, Here we have 3 values.

In all the 3 equations you have (3x + 4y)/ (4x + 5y) = 17/22 -(1)
3x - 1 = 4y -(2)
4x -2 = 5y -(3)
the degree is 1. So you will have only 1 values for x and y.

You were very near to solution. Just substitute the values of 4y and 5y from (2) and (3) in (1)

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

So total number of girls in section A is 4x = 12

Thanks

[singhpk2]

Thanks Sanjeev.

Yes, I am able to solve it now.
However, I feel the third condition is unnecessary (may be it is there to confuse us :evil: ).
I was able to get the same answer using eqn(1) and eqn(2).

[DCE]

I tried doing it the hard way:

I assumed the ratio of number of students in Class A : Class B = k:1

Now the ratio of number of boys : total number of students before and after is as follows

17/39 = (3/7 K + 4/9 ) /(k+1)

k = 7/6

equating this to 7x/9y = 7/6; we get 2x = 3y.

and using the second equation we get perfectly get the answer.

But now if I try something more extraordinary

ratio of number of boys / ratio of number of girls ; we get

3/4 K + 4/5 = (17/22)(k+1)

This gives k = 6/5 which is incorrect; where am I going wrong here.

Thanks
DCE

[Guest]

Ans : 12

We do not need this eq at all " the ratio of boys to girls in the combined class would be 17 to 22" to answer this . Rest of the information is sufficient to answer

[RonPurewal]

But now if I try something more extraordinary

ratio of number of boys / ratio of number of girls ; we get

3/4 K + 4/5 = (17/22)(k+1)

This gives k = 6/5 which is incorrect; where am I going wrong here.

Thanks
DCE

you can't multiply 3/4 by k, because nothing is 3/4 of the total population of students.

your first approach works, because the number of boys in the first class is actually 3/7 of the total. therefore, boys = (3/7)(total), an expression that reduces to 3k/7 when you put it into a ratio.
on the other hand, nothing is (3/4)(total), so nothing will reduce to 3k/4 in the ratio.

it also doesn't make sense to multiply 17/22 by (k + 1), for pretty much exactly the same reason.

[manjk7]

This is a very simple question no need to solve equations

the number of girls or boys = an integer , and the answers are also given (I think this is not a GMAT type question its too easy )

we know the ratio of B to G is class A (3:4) and in B (4:5) and in Combined (17:22)

so number of girls in class A must be a multiple of 4 so 9,10,11 ......out

that leaves 8 and 12 but it cannot be 8 because of the given conditions

so its 12 ..................it will take no more than 20s secs to answer such questions

oops not even 10 secs

[JonathanSchneider]

Yes, nice reasoning. I think this still could appear as a real GMAT question. But it would certainly make it a bit easier!

[ctrajaram]

Nice approaches.

BA-> Boys in class A
GA-> Girls in class A
BB-> Boys in class B
GB-> Girls in class B

The other traditional approach could be solve straight for GA just so u can get to what's asked for .

Given:
BA/GA = 3/4 (1)
BB/GB=4/5 (2)
BA+BB/GA+GB = 17/22 (3)
BB = BA-1 (4)
GB=GA-2 (5)

To find: GA

From (3):

22(BA+BB) = 17(GA+GB)

22 [BA+(BA-1)] = 17[GA+(GA-2)]

22 (2BA-1) = 17(2GA-2)

22 (2 (3GA/4) - 1 ) = 17(2GA-2)

11(3GA) - 22 = 34GA - 34

GA= 12

[esledge]

Nice variety of approaches! And, I had to verify that this was correct:

Ans : 12

We do not need this eq at all " the ratio of boys to girls in the combined class would be 17 to 22" to answer this . Rest of the information is sufficient to answer

Following an algebraic approach as proposed by ctrajaram above, I was in fact able to solve with just given equations (1), (2), (4), and (5). It's very unusual for the GMAT to provide extra information on a problem solving question, but not completely impossible.