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Re: Country X has three coins in its currency: a duom worth [#permalink]
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Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



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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Re: Country X has three coins in its currency: a duom worth [#permalink]
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Let number of duom coins be x and trippim y and megam z. Since maximum number of coins is 20

x + y + z = 20

2x + 11y + 19z = 321
2x + 2y + 2z = 40

Subtracting the two equations to eliminate x gives
9y + 17z = 281

To get minimum value of y we need to maximize z

9y = 281 - 272

9y = 9

y = 1

Answer B

Adewale Fasipe, GRE, GMAT quant instructor from Lagos, Nigeria.
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