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Test Code 52, Section 6, Question 7

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Test Code 52, Section 6, Question 7

by irfan_m1973 » Wed Dec 02, 2009 12:24 am
Jaime earned enough money by selling seashells at 25 cents each to buy several used paperback books at 55 cents each. If he spent all of the money he earned selling seashells to buy the books, what is the least number of seashells he could have sold?

(A) 5
(B) 11
(C) 17
(D) 25
(E) 30
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by haydieenano » Wed Dec 02, 2009 12:45 am
i think the problem lacks some information...^^

correct me if im wrong..
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by ace_gre » Wed Dec 02, 2009 12:51 am
Let the number of shells sold be s and number of books bought be b.
Since all the money obtained by the sale of shells was used to buy books 25s=55b.
s=11*5b/(5*5)=11*b/5

To find the least number of shells b must be as small as possible and must be a whole number so b=5==>s=11.

IMO B
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by viju9162 » Wed Dec 02, 2009 2:36 am
Yes, I also agree with ace_gre. earlier, I thought answer is D. But I didnt split 55 into 5 * 11.

Answer is B.
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by papgust » Wed Dec 02, 2009 2:40 am
You can also solve this question by backsolving.

Pick an answer choice, multiply with 0.25 and divide the result by 0.55. The result that is of an integer form will be the right answer.

In this case, it's 11.
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by Stuart@KaplanGMAT » Wed Dec 02, 2009 2:43 am
irfan_m1973 wrote:Jaime earned enough money by selling seashells at 25 cents each to buy several used paperback books at 55 cents each. If he spent all of the money he earned selling seashells to buy the books, what is the least number of seashells he could have sold?

(A) 5
(B) 11
(C) 17
(D) 25
(E) 30
The algebraic solution is correct, but we also could have simply backsolved.

Since we want the least number of seashells, let's start with the smallest answer.

A) 5. 5 * 25 = 125, which isn't a multiple of 55... wrong.
b) 11. 11*25 = 275, which is a multiple of 55... correct.

Done!

Basically, the question is asking us for the lowest common multiple of 25 and 55, so we also could have solved via prime factorization.

25 = 5*5
55 = 5*11

So, the LCM of 25 and 55 will be 5*5*11

Since 25 already has 5*5 in it, we need 11 seashells to reach 5*5*11... choose (B).
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by GMATinsight » Thu Nov 12, 2015 5:16 am
irfan_m1973 wrote:Jaime earned enough money by selling seashells at 25 cents each to buy several used paperback books at 55 cents each. If he spent all of the money he earned selling seashells to buy the books, what is the least number of seashells he could have sold?

(A) 5
(B) 11
(C) 17
(D) 25
(E) 30
His payment is in Multiples of 25 cents

Payment Required is in Multiples of 55 cents

No left over means, Payment Done = Payment Required

i.e. 25x = 55y
i.e. 5x = 11y

i.e. Minimum value of x = 11 (co-efficient of y)
and Minimum value of y = 5 (co-efficient of x)

i.e. Minimum Value of x = 11

Answer: option B
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