Two amateur wrestling squadsSquad A and Squad Bcompeted against each

Question

Two amateur wrestling squads—Squad A and Squad B—competed against each other. Each squad had the same (nonzero) number of its members who were contestants. In the competition, each contestant in Squads A and B wrestled against exactly one contestant in the other squad, and no non-contestant member of either squad wrestled. Let x be the proportion of Squad A's members who were contestants in the competition and let y be the proportion of Squad B’s members who were contestants in the competition. In terms of x and y only, what is the proportion of the total number of members in the two squads who were contestants in the competition?

chetan2u · Accepted Answer

Question 1
Which one of the following procedures, when applied to the last mathematical expression in the Concise Solution tab, gives a mathematical expression that answers the question asked in the Problem Statement tab?
The equation being talked about is \(\frac{2xa}{a+\frac{x}{y}*a}\)
we want this expression in terms of only x and y. We have an extra in both numerator and denominator => \(\frac{2xa}{a(1+\frac{x}{y})}\)
So, divide numerator and denominator by a to get \(\frac{2x}{1+\frac{x}{y}}\)

Question 2.
A given number is a possible value of [i]a’ if, for each value of x in the Table tab, there is a value of a that is compatible with that value of x and the other information provided. For each of the following numbers, select Yes if, given this condition, it is a possible value of a' Otherwise select No.[/i]
Now x has values 1/3, 1/4 etc, and xa has to be integer as xa is the number of contestants. Thus a has to be a multiple of LCM of denominators of values of x. Thus 12 or multiples of 12.
Hence 9 and 10 are not valid while 12 is.

Question 3.
Assume the information provided is true. For each of the following mathematical statements, select [i]Yes if, according to the information provided, the statement is true for the possible values of x and y. Otherwise select No.[/i]
a'=b'=xa=yb.
Thus, a'=yb is true.
a'=b' is also true.
However, \(a + b < xa + yb\) is not true as xa<a and yb<b, resulting in a+b>xa+yb

KarishmaB · Answer

The video solution of this question is given here: https://www.youtube.com/watch?v=hdCH-U1p8dw GMAT-Club-Forum-a034evp0.png GMAT-Club-Forum-rozergd5.png GMAT-Club-Forum-5f94n313.png GMAT-Club-Forum-jkoxq6ie.png GMAT-Club-Forum-hv5e9917.png

illegallyblonde · Answer

chetan2u  Aren't we talking about a' and not a?

Since a' = xa and we have to select a value for a', it could be anything, right?
1/4 . a = 9 where a = 36, 1/3 . a = 10 where a = 30, etc.

Have I misinterpreted the question? Please let me know. Thanks!

arkaja11 · Answer

I think you're correct. But let's take a' = 9 , for all values of x, it won't return an integer a. Similarly for a'=10.
So, the answer will still be 12.

reeyab · Answer

Wouldnt it be numerators?

Since a'=x*a
=> a= a'*1/x

e.g.: a = 9*3/2 (not possible)

Since a has to be an integer, it would be the reciprocal of x and hence the LCM of the numrator?

nisen20 · Answer

x=\frac{1}{4},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{3}{4}

a=\frac{a'}{x}

a' =9  would fail if the numerator of x were an even number, such as  x=\frac{2}{3}  and  a=\frac{9}{\frac{2}{3}} 
 a' =10  would fail if the numerator of x were an odd number greater than 1, such as  x=\frac{3}{4}  and  a=\frac{10}{\frac{3}{4}}

only  a' =12  can deal with all those possible x listed above.

the numerator of x matters, not the denominator. if three options are 6, 10, and 12, the answer will be 6 and 12.

NEYR0N · Answer

In MSR, do we take the info in each tab as always true? I noticed in your video explanation that you were checking if the math in the concise solution was legit. I am assuming MSR tabs are always true, as in 3rd question it is asking us to validate whether a' = b' (which can only be true if xa = yb) GMAT-Club-Forum-t99f6z3d.png GMAT-Club-Forum-ol3np7r2.png GMAT-Club-Forum-bieuq7bi.png GMAT-Club-Forum-6ttljkjg.png GMAT-Club-Forum-rex5psae.png

anshhh2705 · Answer

For 3.

It's given in the table section that: a' = b' = xa = yb. (YES)
This means that a' = yb is correct, and because b' = yb, a' = b' is also correct. (YES)
However, the statement a + b < xa + yb is incorrect, as we can see, xa < an and yb < b, which actually leads to a + b > xa + yb. (NO)

pulpiestfiction · Answer

but isn't it asking for possible values and it can be true for some of them

Bunuel · Answer

Check the question:

A given number is a possible value of  a’  if,  for each  value of  x  in the Table tab, there is a value of  a  that is compatible with that value of  x  and the other information provided. For each of the following numbers, select  Yes  if, given this condition, it is a possible value of  a'  Otherwise select  No .

	Multiply both the numerator and denominator by a.
	Multiply both the numerator and denominator by y.
	Divide both the numerator and denominator by a.
	Subtract a and then divide the result \(a + \frac{x}{y}a\).
	Divide by \(a + \frac{x}{y}a\). and then subtract a from the result.

Two amateur wrestling squadsSquad A and Squad Bcompeted against each

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Two amateur wrestling squadsSquad A and Squad Bcompeted against each

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