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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
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GMATinsight wrote:
Bunuel wrote:
If every boy in a kindergarten class buys a soda and every girl in the same class buys a juice box, the class will spend 1¢ less in total than it would if every boy in the class buys a juice box and every girl in the class buys a soda. If there are more boys than girls in the class, what is the difference between the number of boys and the number of girls in the class?

A. 1
B. 3
C. 4
D. 12
E. Cannot be uniquely determined

Kudos for a correct solution.


Bunuel : The OA needs correction. Correct answer is Option E

Answer: Option E

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Bunuel: The answer to this question is (A), not (E).

The difference in total price is given as 1 cent. A cent is the smallest measure of currency and hence cost of soda (S) and cost of juice box (J) in cents must be integers. We cannot have a difference of 0.25 cents in the two costs.
Test takers are expected to understand how currencies work.
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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
I will go with answer E. We need to know the relation between Soda and Juice.

Bunuel wrote:
If every boy in a kindergarten class buys a soda and every girl in the same class buys a juice box, the class will spend 1¢ less in total than it would if every boy in the class buys a juice box and every girl in the class buys a soda. If there are more boys than girls in the class, what is the difference between the number of boys and the number of girls in the class?

A. 1
B. 3
C. 4
D. 12
E. Cannot be uniquely determined

Kudos for a correct solution.
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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
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HI Bunuel,

I will note B the number of boys, G the number of girls, S the number of soda cans and J the number of juice boxes.

We know that B > G and the Q is What is B-G ? ( we don't need to find separately each unknown)

Total amount spent, I note $ the TOTAL money spent

B x S + G x J = $-1 => B x S + G x J + 1 = $
B x J + G x S = $

=> B x S + G x J + 1 = B x J + G x S
=> B X S - B X J + G X J - G X S = -1
=> B(S-J) - G (S-J) = -1
=> (B-G)(S-J) = -1
=> B-G = (-1)/(S-J)

As B-G must be a positive integer, then s-j can only be -1 so that -1/-1 = 1 ( any other number would result in a negative / fraction result)

CORRECT ANSWER 1

Bunuel wrote:
If every boy in a kindergarten class buys a soda and every girl in the same class buys a juice box, the class will spend 1¢ less in total than it would if every boy in the class buys a juice box and every girl in the class buys a soda. If there are more boys than girls in the class, what is the difference between the number of boys and the number of girls in the class?

A. 1
B. 3
C. 4
D. 12
E. Cannot be uniquely determined

Kudos for a correct solution.
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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
kdatt1991 wrote:
Bunuel wrote:
If every boy in a kindergarten class buys a soda and every girl in the same class buys a juice box, the class will spend 1¢ less in total than it would if every boy in the class buys a juice box and every girl in the class buys a soda. If there are more boys than girls in the class, what is the difference between the number of boys and the number of girls in the class?

A. 1
B. 3
C. 4
D. 12
E. Cannot be uniquely determined

Kudos for a correct solution.


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. :)


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.
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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
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Let, B(no of boys), G(no og girls), S(cost of soda) and J(cost of juice).

Then from the given condition we can safely say -
BS + GJ + 1 = BJ + GS
or B(J-S) - G(J-S) = 1
or (B-G)(J-S) = 1

Here, we know that B and G will be integers (since they represent no of boys and girls in a class).
Hence, in the above equation B-G will be integer.

Now how (B-G) (J-S) will be equal to 1, when B-G is an integer.
This means both the multipliers can be 1 or -1. They cant be -1 since it is given B>G
So we are left with B-G =1 (This is what has been asked)!!

Ans- A
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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
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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.

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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
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CEZZAR89 wrote:
(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


hi CEZZAR, kdatt and many others who have given answer as A ie 1..

You all are forgetting that the price of soda and juice box need not be an integer.

Although, you all have correctly come down to the final equation, which is(b-g)(s-j)= 1... but hereon we all are going wrong
b- g can take infinite values depending on the value of (s-j), which can be 1, 1/2,1/3,1/100 and so on..
so ans should be E, depending on values of juice and soda....

Let the price of soda be s and that of juice be j
1) In the first instance, b boys buy b sodas and g girls buy g juice box => bs+gj
2) In the second instance, b boys buy b juices and g girls buy g sodas => bj+gs

\(bs+gj=bj+gs-1.......bj-bs+gs-gj=1.......b(j-s)-g(j-s)=1......(b-g)(j-s)=1\)

If the prices have a difference of 1, that is j-s=1, then b-g=1
If the prices have a difference of \(\frac{1}{2}\), that is \(j-s=\frac{1}{2}\), then \(b-g=2\)

E
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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
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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:

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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
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Answer = E. Cannot be uniquely determined

b = The number of boys
g = The number of girls
s = The price of 1 soda
j = The price of 1 juice box

We just reach the point below:

\(b-g = \frac{1}{j-s}\)

To satisfy this equation, j-s has to be 1, means cost of juice has to be greater than cost of soda by 1, however nowhere in the question, this information is given.

Hence Answer = E
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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
kdatt1991 wrote:
Bunuel wrote:
If every boy in a kindergarten class buys a soda and every girl in the same class buys a juice box, the class will spend 1¢ less in total than it would if every boy in the class buys a juice box and every girl in the class buys a soda. If there are more boys than girls in the class, what is the difference between the number of boys and the number of girls in the class?

A. 1
B. 3
C. 4
D. 12
E. Cannot be uniquely determined

Kudos for a correct solution.


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. :)


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
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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
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chetan2u wrote:
CEZZAR89 wrote:
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


hi CEZZAR, kdatt and many others who have given answer as A ie 1..
just one point to ponder..
you all have correctly come down to the final equation, which is
(b-g)(s-j)= 1... but hereon we all are going wrong
b- g can take infinite values depending on the value of (s-j), which can be 1, 1/2,1/3,1/100 and so on..
so ans should be E, depending on values of juice and soda....


Thanks for recapping Chetan, but here is the problem - Everyone agrees that b and g will take integer values but forget that s and j must take integer values too. Note that the cost is given in cents, the smallest monetary unit. The difference between the cost of soda and juice box cannot be less than 1 cent.

When you get the equation (b-g)(s-j)= 1,
b-g can be 4 only if (s-j) is 1/4. But note that prices are defined only up to a cent.
Hence b-g and s-j both must be 1.

Therefore, the difference between the number of boys and number of girls MUST be 1.

Answer (A)
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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
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VeritasPrepKarishma wrote:
chetan2u wrote:
CEZZAR89 wrote:
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


hi CEZZAR, kdatt and many others who have given answer as A ie 1..
just one point to ponder..
you all have correctly come down to the final equation, which is
(b-g)(s-j)= 1... but hereon we all are going wrong
b- g can take infinite values depending on the value of (s-j), which can be 1, 1/2,1/3,1/100 and so on..
so ans should be E, depending on values of juice and soda....


Thanks for recapping Chetan, but here is the problem - Everyone agrees that b and g will take integer values but forget that s and j must take integer values too. Note that the cost is given in cents, the smallest monetary unit. The difference between the cost of soda and juice box cannot be less than 1 cent.

When you get the equation (b-g)(s-j)= 1,
b-g can be 4 only if (s-j) is 1/4. But note that prices are defined only up to a cent.
Hence b-g and s-j both must be 1.

Therefore, the difference between the number of boys and number of girls MUST be 1.

Answer (A)


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
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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
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chetan2u wrote:

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


1) Perhaps not. Note that this question is an experimental question in our question bank. Through it, we are trying to bring across the importance of logic and reasoning in this test. We are forcing you to imagine the situation in the real world. We expect that most people who come across this question would know that 1 cent is the smallest currency denomination. In case we find that it is not so, we may edit the question a bit in the future.

If the discount given is exactly 10%, then it is logical that the cost would be a multiple of 10 cents. As for a dozen items, it depends on how the question is framed. If the items are available only as a closed pack with dozen items, it can cost anything. If items can be sold loose out of the pack, the cost of each item would be specified.

Recall that you do use such logic in Quant questions such as when you deal with integer solutions of equations in 2 variables etc. The number of pens should be an integer etc.
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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
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Bunuel wrote:
If every boy in a kindergarten class buys a soda and every girl in the same class buys a juice box, the class will spend 1¢ less in total than it would if every boy in the class buys a juice box and every girl in the class buys a soda. If there are more boys than girls in the class, what is the difference between the number of boys and the number of girls in the class?

A. 1
B. 3
C. 4
D. 12
E. Cannot be uniquely determined

Kudos for a correct solution.


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.
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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
If the price of 10 candies is 1 cent, price per candy is 1/10th of a cent. 0.1 cent is a perfect denomination, although physically not available.

What makes us think j-s is an integer?
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Re: If every boy in a kindergarten class buys a soda and every girl in the [#permalink]
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manhasnoname wrote:
If the price of 10 candies is 1 cent, price per candy is 1/10th of a cent. 0.1 cent is a perfect denomination, although physically not available.

What makes us think j-s is an integer?


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.
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Bunuel wrote:
If every boy in a kindergarten class buys a soda and every girl in the same class buys a juice box, the class will spend 1¢ less in total than it would if every boy in the class buys a juice box and every girl in the class buys a soda. If there are more boys than girls in the class, what is the difference between the number of boys and the number of girls in the class?

A. 1
B. 3
C. 4
D. 12
E. Cannot be uniquely determined

\(\left. \matrix{\\
g\,\,\, = \,\,\# \,\,{\rm{of}}\,\,{\rm{girls}} \hfill \cr \\
b\,\,\, = \,\,\# \,\,{\rm{of}}\,\,{\rm{boys}}\,\,{\rm{ = }}\,\,\,g + k\,\, \hfill \cr} \right\}\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,?\,\, = \,\,k\,\,\,\,,\,\,\,k \ge 1\,\,\,{\mathop{\rm int}} \,\,\,\,\,\left( {b > g} \right)\)

\(\left. \matrix{\\
s\,\, = \,\,\,{\rm{one}}\,\,{\rm{soda}} \hfill \cr \\
j\,\, = \,\,\,{\rm{one}}\,\,{\rm{juice}}\,\, \hfill \cr} \right\}\,\,\,\,\,{\rm{cost}}\,\,\,\left( {{\rm{in}}\,\,{\rm{cents}}} \right)\)

\(g,j,k,s\,\,\,\, \ge \,\,\,1\,\,\,{\rm{ints}}\,\,\,\,\left( * \right)\)

\(\left[ {\left( {g + k} \right)\,j\,\, + \,g\,s} \right]\,\, - \,\,\,\left[ {\left( {g + k} \right)\,s\,\, + \,g\,j} \right]\,\,\, = 1\,\,\,\,\,\,\,\,\,\left[ {\,{\rm{cents}}\,} \right]\)

\(k\left( {j - s} \right) = 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,k\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{divisor}}\,\,{\rm{of}}\,\,\,1\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = k = 1\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
GMAT Club Bot
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