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If every boy in a kindergarten class buys a soda and. . . .

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If every boy in a kindergarten class buys a soda and. . . .

by M7MBA » Thu Mar 01, 2018 7:00 am
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

The OA is the option A.

I think the correct answer is the option E. Can any expert help me here, please? I'd be thankful for your help.
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by regor60 » Thu Mar 01, 2018 7:50 am
M7MBA 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

The OA is the option A.

I think the correct answer is the option E. Can any expert help me here, please? I'd be thankful for your help.
Let B and G be the number of boys and girls and S and J be the cost in cents of a soda and a juicebox, respectively

So the total cost for the first situation is: BxS + GxJ in cents

Cost of the second situation is: BxJ + GxS in cents

The problem states that the first costs 1 cent less than the second, so subtract two from one and equate to 1 cent:

BxJ +GxS - BxS - GxJ = 1

This reduces to B(J-S) - G(J-S), which can be further reduced to

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

Now, if (B-G) is > 1, that would mean that (J-S) is less than one, but we can suppose that the products are being priced in whole cents, so (B-G) can't be greater than 1

Similarly, (B-G) can't be less than 1 because we're told B>G and we can safely assume no partial people, so that must mean that

(B-G)=1,A
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by [email protected] » Thu Mar 01, 2018 7:32 pm
Hi M7MBA,

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: E

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by regor60 » Fri Mar 02, 2018 7:55 am
[email protected] wrote:Hi M7MBA,

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: E

GMAT assassins aren't born, they're made,
Rich
Don't understand how this is an issue since you can only deal in whole cents
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juice and soda

by GMATGuruNY » Fri Mar 02, 2018 8:52 am
M7MBA 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
Let:
b = the number of boys
g = the number of girls
s = the number of cents for each soda
j = the number of cents for each juice box
Note:
All of the values above must be POSITIVE INTEGERS.

Case One: If every boy in a kindergarten class buys a soda and every girl in the same class buys a juice box.
In this case, the total amount spent = (number of boys)(number of cents per soda) + (number of girls)(number of cents per juice box) = bs + gj.

Case Two: If every boy in the class buys a juice box and every girl in the class buys a soda.
In this case, the total amount spent = (number of boys)(number of cents per juice box) + (number of girls)(number of cents per soda) = bj + gs.

Since the amount in Case One is 1 cent less than the amount in Case Two, we get:
bs + gj = (bj + gs) - 1
gj - gs + 1 = bj - bs
g(j-s) + 1 = b(j-s)
1 = b(j-s) - g(j-s)
1 = (b-g)(j-s).

All of the values in the resulting equation are POSITIVE INTEGERS.
Since there are more boys than girls -- implying that b-g is positive -- the two factors on the right side must both be equal to 1:
b-g=1 and j-s=1.
Thus, the difference between the number of boys and the number of girls = 1.

The correct answer is A.
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by Scott@TargetTestPrep » Fri Mar 02, 2018 10:01 am
M7MBA 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
We can let b = the number of boys in the class, g = the number of girls in the class, s = price of a soda and j = price of a juice box. From the information given in the problem, we see that:

bs + gj = bj + gs - 1 and b > g

We need to determine the value of b - g.

Let's look at the equation bs + gj = bj + gs - 1 and simplify

bj + gs - bs - gj = 1

bj - bs - gj + gs = 1

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

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

Since b, g, j and s are integers and the only way two integers multiplied together yield a product of 1 is 1 x 1, we see that b - g = 1 and j - s = 1. Thus, we see that b - g = 1.

Answer: A

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by GMATGuruNY » Sat Mar 03, 2018 3:41 am
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.
The case above is not viable.
Since S and J represent the price in cents, they must be positive integers.
It is not possible that the price of a soda is 1/2 cent.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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