A coin purse contains 13 coins, each worth 1, 5, 10, or 25 cents; the

MANHATTAN GMAT OFFICIAL SOLUTION:

Be very organized with your work.

First, figure out a few parameters for the given information. If the coin purse contains any 1-cent coins, then it must contain either 5 or 10 of them, since all 13 coins add up to 150. The purse can’t contain 10 1-cent coins, because then the remaining 3 coins would have to add up to 140, and this isn’t possible.

What about the other extreme? You do need at least one 25-cent coin; the next-largest denomination is 10 cents, and 13 of those add up to only 130 cents. Great! The purse contains at least one 25-cent coin, so the remaining 12 coins have to add up to 125.

Hmm. Either one additional coin needs to have a 5-cent or a 25-cent value, or the purse must contain 5 1-cent coins. So you’ll start with a total of either two coins (one 25-cent + one 5-cent or one 25-cent) or six coins (one 25-cent + five 1-cent), and the remaining total value needed will end in a 0. (For example, if the purse contains one 25-cent piece and one 5-cent piece, then 11 more coins have to add up to 120 cents.)

In any of those cases, additional 5-or 25-cent pieces would have to be added in pairs or in conjunction with a set of 5 1-cent pieces, so that the total value adds up to 0 in the units place. These coins would get added, then, in combinations of eitehr two coins or six coins.

In other words, the total number of 1-cent, 5-cent, and 25-cent pieces has to be even. There are, then, an odd number of 10-cent pieces—and there is at least one 10-cent piece, since there are 13 coins overall.

Okay, it’s time to bring in the additional info from the statements. Keep in mind that you only need to find two contradictory cases in order to determine that the statement is insufficient.

(1) INSUFFICIENT: If the value of 150 cents can be divided evenly into 5 separate envelopes, then the value in each envelope must be 150 / 5 = 30 cents. The purse must contain at least one 25-cent coin and at least one 10-cent coin.

What coin combinations add to 30 cents?

2 coins (25 + 5 = 30)
3 coins (10 + 10 + 10 = 30)
4 coins (5 + 5 + 10 + 10 = 30)
5 coins (5 + 5 + 5 + 5 + 10 = 30)
6 coins (25 + 1 + 1 + 1 + 1 + 1 = 30)

Start with the givens: one 25 and one 10.

2 coins (25 + 5)
5 coins (5 + 5 + 5 + 5 + 10)

That’s 7 coins total, so you need 6 more coins spread over 3 groups… okay, pick 3 groups of 2 coins each. (Note: you can ignore the total value here because you know each group adds up to 30 cents; pick 5 groups and you will have 150 cents.) Done! You’ve fulfilled the requirements for one 10-cent coin.

Can you create a solution with three or five 10-cent coins?

2 coins (25 + 5)
3 coins (10 + 10 + 10)

That’s 5 coins total. You need to pick 3 more groups that contain a total of 8 coins. A group of 2, 2, and 4 adds up to 8 coins. The total number of 10-cent coins among these groups is 5.

There are at least two different possibilities: one 10-cent coin or five 10-cent coins. Statement 1 is insufficient.

(2) INSUFFICIENT. This set-up is very similar, except the coins are divided into 6 envelopes this time. In this case, the value in each envelope must be 150 / 6 = 25 cents.

These combinations add up to 25 cents:

1 coin (25)
3 coins (10 + 10 + 5 = 25)
4 coins (10 + 5 + 5 + 5 = 25)
5 coins (5 + 5 + 5 + 5 + 5 = 25)

Pick 6 groups that add up to 13 coins (again, ignore the value, because each group equals 25 cents). Start with the givens: need at least one 25-cent coin and one 10-cent coin.

Option 1: Start with a 1 coin group and a 4 coin group. You need 8 more coins across 4 groups: 1 coin, 1 coin, 1 coin, 5 coins. That gives a total of one 10-cent coins.

Next, because on the previous problem you found a solution for five 10-cent coins, try to find the same here. (You’re doing this just in case you ultimately have to test the two statements together—you’ll save yourself some work.)

Option 2: Start with a 1 coin group, a 3 coin group, another 3 coin group, and a 4 coin group; that gives you five 10-cent coins. You need 2 more coins spread over 2 groups, so you need two more 1 coin groups. You have a total of five 10-cent coins.

Statement 2 is insufficient.

(1) AND (2) INSUFFICIENT. First, check the work you did earlier; can you reuse any of it?

For statement 1, you found a total of one 10-cent coins using these groups:

Statement 1:

2 coins (25 + 5)
5 coins (5 + 5 + 5 + 5 + 10)
2 coins (25 + 5)
2 coins (25 + 5)
2 coins (25 + 5)

Statement 2:

1 coin (25)
1 coin (25)
1 coin (25)
1 coin (25)
4 coins (10 + 5 + 5 + 5 = 25)
5 coins (5 + 5 + 5 + 5 + 5 = 25)

Let’s see. There are four 25-cent coins in each group. You’ve also got one 10-cent coin and eight 5-cent coins! It’s a match! You can fulfill all of the requirements for both statements using this same set of coins, resulting in one 10-cent coin.

Next, compare the case where you found a total of five 10-cent coins.

2 coins (25 + 5)
2 coins (25 + 5)
2 coins (25 + 5)
3 coins (10 + 10 + 10)
4 coins (5 + 5 + 10 + 10)

For statement 2, you found a total of five 10-cent coins using these groups:

1 coin (25)
1 coin (25)
1 coin (25)
3 coins (10 + 10 + 5)
3 coins (10 + 10 + 5)
4 coins (10 + 5 + 5 + 5)

Compare the coins. There are three 25-cent coins, five 10-cent coins, and five 5-cent coins in each set. Check! The coin sets are identical, so it is possible to have five 10-cent coins using both statements.

The two statements together still yield at least two possible answers.

The correct answer is (E).