Last Week's Problem: "How Many Coins Are In Your Purse?"

[shaji]

A coin purse contains 13 coins, each worth 1, 5, 10, or 25 cents; the total value of the coins is 150 cents. How many 10-cent coins are in the purse?

(1) The 13 coins can be divided among five separate envelopes so that each envelope contains the same total monetary value.

(2) The 13 coins can be divided among six separate envelopes so that each envelope contains the same total monetary value.
QUICKER APPROACH:
Statement1: Let the 5 envelopes be A,B,C,D & E & the coins as follows:
A,B=(25,5)=4 coins
C, D & E=(10,10,10)=9 coins.
Another manner:
A,B,C,D=(25,5)=8 coins
E=(5,5,5,5,10)=5 coins.
Statement1 is not sufficient.

Statement2:Let the 6 envelopes be A,B,C,D , E & F and the coins as follows:
A,B,C=(25)=3 coins
D & E=(10,10,5)=6 coins
F=(10,5,5,5)=4 coins
Another manner:
A,B,C,D=(25)=4 coins
E=(5,5,5,10)=4 coins
F=(5,5,5,5.5)=5 coins

The correct answer is E.

[RonPurewal]

QUICKER APPROACH:

... well, this isn't really a "quicker approach"; you're just showing the results, without showing the reasoning that got you there. if we did the same thing, then our approach would look just as "quick".

when we write the solutions, we can't just hand you a list of possibilities and say, "there! these things can happen!", as though we had pulled them out of thin air by pure magic.
we're trying to train readers so that they can actually solve these problems themselves. so, we can't do what you're doing here; we have to show how we get what we get.

[RonPurewal]

also, on top of that, your analysis doesn't actually prove that the answer is (e).
to prove that the answer is (e), you actually need to find two sets of coins with different numbers of 10-cent pieces -- with EACH of those sets satisfying BOTH statements. you haven't done that here.

* in your analysis of statement 1, you presented ...
... a set of 2 quarters, 2 nickels, and 9 dimes, and
... a set of 4 quarters, 8 nickels, and 1 dime.

* in your analysis of statement 2, you presented ...
... a set of 3 quarters, 5 nickels, and 5 dimes, and
... a set of 4 quarters, 8 nickels, and 1 dime.

the problem is that the only solution in common -- at least among the examples you've presented here -- is "4 quarters, 8 nickels, 1 dime".
i.e., if the only possibilities were the ones you've mentioned here, then the answer would actually be (c), not (e).

in order to prove that the answer is (e), you actually have to find the other case that's listed in our solution.
the two cases we listed are the only two cases that satisfy both statements, so, regardless of your solution method, you're going to have to come up with the same two sets of coins.