This was quite a tricky question for me, but I hope that I have been able to crack it! Here's my solution:

Let's assign variables for boys (b), girls (g), sodas (s), juices boxes (j) and the total cents spent as (t). From the information given in the question stem, we also know that boys are greater than girls (b > g). Let's set up some equations:

b*s + g*j = t - 1.... (the total if every boy in the class buys a soda and every girl in the class buys a juice box)

b*j + g*s = t.... (the total if every boy in the class buys a juice box and every girl in the class buys a soda)

We can subtract both equations from each other to cancel out the variable (t). This gives us:

b*j + g*s - (b*s + g*j) = t - (t - 1)..... this gives us: b*j - b*s - g*j + g*s = 1. We can further factorise this equation as b*(j - s) - g*(j - s) = 1........ (b - g)*(j - s) = 1.

From the factorised equation (b - g)*(j - s) = 1, we can say that either (b - g) =1 or (j - s) = 1. This basically means that either the number of boys are one more than the number of girls (b = g + 1) or that the price of a juice box costs 1¢ higher than the price of a soda can (j = s +1). Since we know (from the question stem) that b > g, we could conclude that b = g + 1, and that the prices of both the juices boxes and the sodas are the same!

I think the answer is A.

Please consider giving me KUDOS if you felt this post was helpful and correct! or please enlighten me (in case my answer's incorrect) so that I can learn and improve from my mistakes! Thanks.

I will go with A

Here is my solution:

boys (b) , girls (g) , soda (s) , juice box (j) assuming b,g,s,and j are all integers.
Given that b>g
Q: What is b-g ?

Total amount spent ;
b*s+g*j (boys buy soda , girls buy juice box)
b*j+g*s (boys buy juice box, girls buy soda)

Therefore the equation must be

b*s+g*j+1=b*j+g*s

we should try to solve the equation for (b-g) ;
b*s+g*j - b*j-g*s=-1
b(s-j) - g(s-j)=-1
(b-g)(s-j)= -1
Isolate (b-g) to one side of the equation ;
b-g = -1/(s-j)
We know that both b-g and s-j result integers, and b-g must be positive, therefore s-j can only be -1 which makes b-g=1

CEZZAR89 - Thanks for the kudos man!

Yeah, after re-reading through the question (and some of the other answer posts) again. I think that I have a made a mistake when reasoning my factorised equation of (b - g)*(j - s) = 1. I think that the number of boys are one more than the number of girls (b = g + 1) AND that the price of a juice box costs 1¢ higher than the price of a soda can (j = s +1). I don't think that this is an 'either and or' situation because we are talking about the relationship between boys and girls (b = g + 1) and between the price of a soda can and a juice box (j = s +1) separately.

I still think that that the answer A, because b = g + 1, but I also think that the j = s +1.

I think answer option E is correct.
Reason - We know b and g are integers but we cannot say the same about j and s. Consider an example - what if b=10, g=8, j=1 and s=1/2. We still get (b-g)*(j-s)=1.

Posted from my mobile device

Hi All,

This is a thick, layered question, and would likely take most Test Takers more time than average to solve correctly. The key to solving it is to realize that we don't know the prices of each soda and each juice box - they MIGHT be integers, but they MIGHT NOT. Also, we don't know the relative prices (so one might be more expensive than the other, or vice-versa).

From the given prompt, we have 4 variables:

B = The number of boys
G = The number of girls
S = The price of 1 soda
J = The price of 1 juice box

From the prompt, we can create just 1 equation:

(B)(S) + (G)(J) = (B)(J) + (G)(S) - 1

Here's how we can TEST VALUES to prove that there's more than one answer. Since this IS such a thick question, the key to doing the work quickly is to keep the values SMALL.

We do have a couple of 'restrictions' that we have to work with:
1) B and G are both INTEGERS (since you cannot have a 'fraction' of a boy or girl)
2) We're told that there are MORE boys than girls, so B > G

IF....
B=2
G=1
S=1
J=2
(2)(1) + (1)(2) = (2)(2) + (1)(1) - 1
2 + 2 = 4 + 1 - 1
4 = 4
Here, we have 2 boys and 1 girl, so the difference is 1.

In the above example, both S and J are INTEGERS and S < J. What happens if we make one of those variables a fraction......

IF....
B=3
G=1
S=1/2
J=1
(3)(1/2) + (1)(1) = (3)(1) + (1)(1/2) - 1
1.5 + 1 = 3 + 0.5 - 1
2.5 = 2.5
Here, we have 3 boys and 1 girl, so the difference is 2.

Thus, there's no exact answer....

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

Is the factorization correct?

b-g = 1 ONLY WHEN j-s = 1, similarly

j-s = 1 ONLY WHEN b-g = 1

No where in the problem is given that individual difference is 1
_________________

hi karishma,
this was what i too thought while i answered this Question with whatever knowledge i have on currency. But then there were three points..
1) would the GMAC people test us on knowledge of currency. i may know it but the exam is widely held in different parts of world.
2) if this same question talks of discount of 10% on say both the items, does it mean that .9*juicebox-.9*soda should be an integer. If so, we have to have both these items costing in multiple of 10 cents...
3) if some question talks of some item in terms of dozens, does it mean that it should be costing in multiples of 12.
i know it is logical that the price should be in integer cents, and i dont doubt that for a second.
Ofcourse i know my answer to this question is wrong as it comes from you, the source itself
_________________

VERITAS PREP OFFICIAL SOLUTION:

b = # of boys

g = # of girls

s = cost of a soda, in cents

j = cost of a juice, in cents

Here's how we translate the first sentence into math:

sb + jg + 1 = jb + sg

Now we'll subtract sb and jg from both sides, so we can get all of our integers equal to an integer.

1 = jb - jg + sg - sb

1 = j(b-g) + s(g-b)

1 = j(b-g) + s(-1)*(b-g) we do this we can have a common term of (b - g) that we can then factor out

1 = j(b-g) - s(b-g)

1 = (j - s) * (b - g)

At this point we seem stuck, but notice what we have. We know j, s, b, and g are integers, as they represent prices in cents (which must be integers, e.g. 1 cent, 2 cents, 3 cents, etc.) and people, respectively. We know we have more boys than girls, so (b - g) is positive. If we have a positive integer times another integer = 1, then BOTH integers must be 1. Hence (b - g) = 1 and (s - j) = 1, and the answer is A.
_________________

In that case, since it is physically not available, you would expect the candies to be always sold in multiples of 10 only. The question forces you to think of the real world. Nothing would be priced in non-integer cent terms.
_________________