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Country X has three coins in its currency: a duom worth

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Country X has three coins in its currency: a duom worth [#permalink] New post 19 Aug 2017, 08:20
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Question Stats:

51% (02:07) correct 48% (01:52) wrong based on 60 sessions


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.
[Reveal] Spoiler: OA

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Re: Country X has three coins in its currency: a duom worth [#permalink] New post 20 Feb 2018, 05:49
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A duom worth 2 cents, a trippim worth 11 cents, and a megam worth 19 cents. So in order to minimize the number of trip­pim we have to maximize the number of other coins. Since we can use max 20 coins, starting with option A, we can maximize the megam to 16 and remain 17 cents which cannot be covered with duoms. Considering option B, we can maximize the megam to 16 and remain 6 cents which can be covered with duoms.
So the answer is B.
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Re: Country X has three coins in its currency: a duom worth [#permalink] New post 05 Mar 2018, 09:14
YMAkib wrote:
A duom worth 2 cents, a trippim worth 11 cents, and a megam worth 19 cents. So in order to minimize the number of trip­pim we have to maximize the number of other coins. Since we can use max 20 coins, starting with option A, we can maximize the megam to 16 and remain 17 cents which cannot be covered with duoms. Considering option B, we can maximize the megam to 16 and remain 6 cents which can be covered with duoms.
So the answer is B.

I also applied the same logic to this question, and concluded B as the answer, but I was not confident, as I thought perhaps there is a trick in the question I did not notice. Is there other, algebraic solution other than using logic.
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Re: Country X has three coins in its currency: a duom worth [#permalink] New post 21 May 2018, 10:31
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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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Re: Country X has three coins in its currency: a duom worth   [#permalink] 21 May 2018, 10:31
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