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Country X has three coins in its currency: a duom worth
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19 Aug 2017, 08:20
19 Aug 2017, 08:20
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Question Stats:
48% (02:22) correct
51% (02:10) wrong based on 104 sessions
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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.
Re: Country X has three coins in its currency: a duom worth
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20 Feb 2018, 05:49
20 Feb 2018, 05:49
1
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.
So the answer is B.
Re: Country X has three coins in its currency: a duom worth
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05 Mar 2018, 09:14
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.
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.
