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Algebraic solution - coins

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by AleksandrM » Sat Jun 14, 2008 1:12 pm
I AM LOOKING FOR AN ALGEBRAIC SOLUTION TO THE FOLLOWING:

Billy has an unlimited supply of the following coins: pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), and half-dollars (50¢). On Monday, Billy bought one candy for less than a dollar and paid for it with exactly four coins (i.e., he received no change). On Tuesday, he bought two of the same candy and again paid with exactly four coins. On Wednesday, he bought three of the candies, on Thursday four of the candies, and on Friday five of the candies; each day he was able to pay with exactly four coins. Which of the following could be the price of one candy in cents?

A) 8c
B) 13c
C) 40c
D) 53c
E) 66c

Thanks!
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by anshul265 » Sat Jun 14, 2008 2:22 pm
Answer: C

I don't think there can be a pure algebraic solution to this problem. Firstly there are too many variables and few equations. Secondly, this condition can be fulfilled by more than one solution (atleast for both 20 and 40).

I have used elimination method with one algebraic support. If you want i can explain that to you.. but probably you have that figured out already!!

And yes, I think for a problem like this, elimination takes much lesser time than pure algebraic methods.
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by AleksandrM » Sat Jun 14, 2008 3:21 pm
Based on what I think and what you said, I believe that this kind of a problem would not appear on the GMAT. From what I have observed, all problems on the GMAT - at least those from OG and GMATPrep - can be solved using algebra. The only time I have seen it necessary to use answer choices is when the answers are incomplete and have to be solved to get the answer. As for word problems, they have all had an algebraic solution.

Thanks for the response.
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