Carcass wrote:

Country X has three coins in its currency: a duom worth 2 cents, a trippim worth 11 cents, and a megam worth 19 cents. If a man has $3.21 worth of Country X's currency and cannot carry more than 20 coins, what is the least number of trippim he could have?

A. 0

B. 1

C. 2

D. 3

E. It cannot be determined.

Let d, t and m be the number of duom, trippim and megam, respectively. We can create the equation and the inequality:

2d + 11t + 19m = 321 and d + t + m ≤ 20

Since we want to minimize the number of trippim, we want to maximize the number of megam since it is worth the most. Let’s consider 321/19 = 16 R 17. That is, if we have 16 megam, we will have 17 cents left. So we can have 3 duom and 1 trippim to make up the 17 cents. In this case, we see that have only 1 trippim.

Remember we cannot have zero trippim because if we do, the duoms will have a cent value that it even; however 17 is odd.

Therefore, 1 is the least number of trippim we can have under the conditions given.

Answer: B

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Jeffery Miller

Head of GRE Instruction

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