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Manhattan GMAT Challenge Problem of the Week – 31 Aug 2010

by Manhattan Prep, Aug 31, 2010

Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people enter our challenge, the better the prizes.

As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 mins and go!

Question

Joseph has exactly 185 pesos in 50 peso coins, 25 peso coins, and 1 peso coins only, and he has at least one of each. How many 25 peso coins does he have?

1) He has exactly two 50 peso coins.

2) He has fewer than 40 1 peso coins.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.

Answer

The easiest way to count the number of possibilities for how Joseph could have 185 pesos in 50 peso coins, 25 peso coins, and 1 peso coins is to count via the largest increment, 50. Since 3 is the max number of 50 peso coins he could have, here are all the possibilities:

Table1

1) INSUFFICIENT. Joseph has exactly two 50 peso coins. We now have three options, all with different numbers of 25 peso coins.

Table2

2) INSUFFICIENT. He has fewer than 40 1 peso coins. We now have 5 options, with 5 different numbers for how many 25 peso coins he has.

Table3

1+2) INSUFFICIENT. He has exactly two 50 peso coins and fewer than 40 1 peso coins. Since the chart for statement 1 above is shorter, refer back to it, and eliminate the 1 option that has more than 40 1 peso coins. Two options remain:

Table4

Joseph could have either two or three 25 peso coins.

The correct answer is E.

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