AAPL wrote:Veritas Prep

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

Although some "common-sense" can avoid math... why avoid math (it will be fast using it, by the way) ?!
$$\left. \matrix{
g\,\,\, = \,\,\# \,\,{\rm{of}}\,\,{\rm{girls}} \hfill \cr
b\,\,\, = \,\,\# \,\,{\rm{of}}\,\,{\rm{boys}}\,\,{\rm{ = }}\,\,\,g + k\,\, \hfill \cr} \right\}\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,?\,\, = \,\,k\,\,\,\,,\,\,\,k \ge 1\,\,\,{\mathop{\rm int}} \,\,\,\,\,\left( {b > g} \right)$$
$$\left. \matrix{
s\,\, = \,\,\,{\rm{one}}\,\,{\rm{soda}} \hfill \cr
j\,\, = \,\,\,{\rm{one}}\,\,{\rm{juce}}\,\, \hfill \cr} \right\}\,\,\,\,\,{\rm{cost}}\,\,\,\left( {{\rm{in}}\,\,{\rm{cents}}} \right)$$
$$g,j,k,s\,\,\,\, \ge \,\,\,1\,\,\,{\rm{ints}}\,\,\,\,\left( * \right)$$
$$\left[ {\left( {g + k} \right)\,j\,\, + \,g\,s} \right]\,\, - \,\,\,\left[ {\left( {g + k} \right)\,s\,\, + \,g\,j} \right]\,\,\, = 1\,\,\,\,\,\,\,\,\,\left[ {\,{\rm{cents}}\,} \right]$$
$$k\left( {j - s} \right) = 1\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,k\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{positive}}\,\,{\rm{divisor}}\,\,{\rm{of}}\,\,\,1\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = k = 1$$

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

Regards,
Fabio.

AAPL wrote:Veritas Prep

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.