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.
The correct answer is E.