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

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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

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by Brent@GMATPrepNow » Mon Dec 17, 2018 9:45 am
alanforde800Maximus 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 = # of boys in kindergarten
Let G = # of girls in kindergarten
Let J = price (in CENTS) of ONE juice box
Let S = price (in CENTS) of ONE soda

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.
Let scenario 1 = boys buy sodas and girls buy juice boxes
Let scenario 2 = boys buy juice boxes and girls buy sodas
We can start with the following "word equation": (total cost of scenario 1) = (total cost of scenario 2) - 1
So, we get: BS + GJ = BJ + GS - 1

It would be nice if we could factor either side of the equation.
So, let's move some pieces around.
Subtract GS from both sides: BS - GS + GJ = BJ - 1
Subtract GJ from both sides: BS - GS = BJ - GJ - 1
Factor each side: S(B - G) = J(B - G) - 1
Subtract J(B - G) from both sides: S(B - G) - J(B - G) = -1
Simplify: (S - J)(B - G) = -1

IMPORTANT:
Since B and G are both POSITIVE INTEGERS, we know that (B - G) is an INTEGER
Likewise, we know that S and J are both POSITIVE INTEGERS, since the prices cannot be less than one cent.
This means (S - J) is an INTEGER

Since (S - J)(B - G) = -1, we know that one value, (S - J) or (B - G), must be 1 and the other value must be -1

There are more boys than girls in the class
So, B - G must be positive.
This means B - G must equal 1

What is the difference between the number of boys and the number of girls in the class?
B - G = 1

Answer: A

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by [email protected] » Mon Dec 17, 2018 12:11 pm
Hi alanforde800Maximus,

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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alanforde800Maximus 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
$$\left. \matrix{
g\,\,\, = \,\,\# \,\,{\rm{of}}\,\,{\rm{girls}} \hfill \cr
b\,\,\, = \,\,\# \,\,{\rm{of}}\,\,{\rm{boys}}\,\,{\rm{ = }}\,\,\,g + k\,\, \hfill \cr} \right\}\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,?\,\, = \,\,k\,\,\,\,,\,\,\,k \ge 1\,\,\,{\mathop{\rm int}} \,\,\,\,\,\left( {b > g} \right)$$

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

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

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

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


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

Regards,
Fabio.
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by Scott@TargetTestPrep » Sun Mar 10, 2019 6:53 pm
alanforde800Maximus 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 simplify the equation bs + gj = bj + gs - 1:

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. (Note: We must assume that the cost of both soda and juice is in whole cents, not fractions of a cent. If this assumption were not made, then the correct answer would be E.)

Answer: A

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