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The Quest for 700: Weekly GMAT Challenge (Answer)

Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the 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:

 50 peso 25 peso 1 peso 3 1 10 2 3 10 2 2 35 2 1 60 1 5 10 1 4 35 1 3 60 1 2 85 1 1 110

1) INSUFFICIENT. Joseph has exactly 2 50 peso coins. We now have three options, all with different numbers of 25 peso coins.

 50 peso 25 peso 1 peso 2 3 10 2 2 35 2 1 60

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.

 50 peso 25 peso 1 peso 3 1 10 2 3 10 2 2 35 1 5 10 1 4 35

1+2) INSUFFICIENT. He has exactly 2 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:

 50 peso 25 peso 1 peso 2 3 10 2 2 35

Joseph could have either 2 or 3 25 peso coins.

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