Country X has three coins in its currency: a duom worth 2 cents, a tri

Question

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.

A. 0

B. 1

C. 2

D. 3

E. It cannot be determined.

Grewal4GMAT · Answer

If a man is carrying at least 20 coins, then

.19*16 coins =3.04
.11*1 coin = .11
.2*3 coins= .6

Total = 3.21

Answer is B.

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EgmatQuantExpert · Answer

Solution

Given:  
Country X has three coins:
• A duom, worth 2 cents; A trippim, worth 11 cents; A megam, worth 19 cents
• A man has $3.21 or 321 cents worth country X’s currency
• He cannot carry more than 20 coins

To find:  
• The least number of trippim the man can have

Approach and Working:   
If we are trying to find out the least number of trippim, we need to maximise the other coin types

In 321 cents,
• Maximum megam possible = 16 (As 16 * 19 = 304 and 323 < 304 < 321)
• Remaining amount = 321 – 304 = 17
• As a duom is worth 2 cents, we can’t have all duoms in 17 cents
 o If we take 1 trippim, then amount with trippim = 1 * 11 = 11 
• Remaining amount = 17 – 11 = 6 cents, which can be paid by 3 duoms
• And also the total number of coins = 16 + 1 + 3 = 20

Hence, the correct answer is option B.

Answer: B

Just a small rectification: in the question it is given that the man   cannot carry more than 20 coins  , whereas you have mentioned that the man   is carrying at least 20 coins  . Those two expressions mean different.

wishalsp333 · Answer

I have a query regarding the answer. Here it goes,

In the question, it says it cannot carry more than 20 coins,

if $3.21 dollars in converted into cents it is 321 cents. So dividing 321/19 = 16 (trippams). Here we don't need other coins at all.

So shouldn't the answer be 0?

EgmatQuantExpert · Answer

If you are paying by megam (worth 19 cents) only, then as you mentioned, you can pay only 16 * 19 = 304 cents.
But in the question it is mentioned that the man has 321 cents, not 304 cents.

For the remaining 321 - 304 = 17 cents, you must have currencies of other denominations.

Hero8888 · Answer

16*19=304, not 321, so you have to add 19 cents with max 4 coins, so you add 3 coins*2 + 1 coin*11 = 19 cents

JeffTargetTestPrep · Answer

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

akadiyan · Answer

Critical information:
Total: $3.21
Total number of coins cannot be more than 20.

Problem: What is  least  number of trippim man could have.

Least is the key word here. We need to make up as much as possible with other 2 currency so that we could have least number of trippim.

Step 1: So lets start with highest currency megam: highest number we could have is 16 coins - 16*19 = 304.

We have 4 coins left to reach 321. we could not do that with all duom  . So we need  atleast 1 trippim .

Step 2: So lets use 1 trippim - 304 + 11 = 315

Step 3: We have 6 cents remaining to reach total and exactly 3 coins left , we could use 3 duom. 3 duom*2 cents = 6 cents.

Total: 3.15+0.06 = 3.21

Ans:   1

bumpbot · Answer

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Country X has three coins in its currency: a duom worth 2 cents, a tri

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