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21 Aug 2010, 22:52
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A bookstore sells new books for $15 each and used books for$10 each. On every new book, the store makes a profit of $5 while on every used book it makes a profit of$2. If on a given day the bookstore's sales amounted to $125, which of the following cannot be the profit made on that day? A.$27
B. $31 C.$35
D. $39 E.$41

m22 q12
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Apr 2014, 13:58, edited 2 times in total.
Edited the question.

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21 Aug 2010, 23:42
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shekar123 wrote:
A bookstore sells new books for $15 each and used books for$10 each. On every new book, the store makes a profit of $5 while on every used book it makes a profit of$2. If on a given day the bookstore's sales amounted to $125, which of the following cannot be the profit made on that day? A.27 B.31 C.35 D.29 E.41 Answer D should read 39, not 29. I can suggest a couple of systematic ways to look at this, though I think it's practical to get the answer within two minutes by guessing-and-checking. First you might notice that the number of new books sold must be odd; otherwise the total sales in dollars would end in 0. You could then take an algebraic approach. If n is the number of new books and u the number of used books, we know that 15n+10u = 125, or dividing by 5, we have: 3n + 2u = 25 We want to know the value of 5n + 2u, which is the profit in dollars. Notice how similar this is to the left side of the equation above: 5n + 2u = 2n + (3n + 2u) = 2n + 25 So we just want to find what values are possible for 2n + 25. Remembering that n must be odd, it's easy enough just to plug in n=1, 3, 5 and 7 to see that every answer choice is possible except for 41. Or, if you know that n is odd, you can replace it with 2k + 1, for some integer k. Then the quantity we're trying to find becomes 2u + 25 = 2(2k+1) + 25 = 4k + 27 = 4k + 24 + 3 = 4(k+6) + 3 So our profit is 3 greater than a multiple of 4, and thus has a remainder of 3 when divided by 4. Thus 41 is impossible (you might, when looking at the answers, see that 41 is a bit suspicious - all of the answer choices give a remainder of 3 when divided by 4 with one exception - 41). That's probably more work than the first approach, though it's perhaps interesting to see why each answer has the same remainder by 4. Actually, the first approach I took when looking at the question was to treat it something like a weighted average. If the store only sells new books, then it makes one third of a dollar profit for each dollar of sales. If it only sells used books, it makes one fifth of a dollar profit for each dollar of sales. So if it sells a combination of new and used books for S dollars, the profit must be somewhere between S/5 and S/3. We know the total sales was$125, so the profit must be between $125/5 and$125/3, or in other words, between $25 and$41.67. Unfortunately that doesn't rule out any answer choices right away, but the answer $41 is suspiciously close to the maximum here (remember we get the max if the store *only* sells new books, and we know the store sold some used books as well since$125 is not a multiple of 15, so $41 seems very unlikely), so I'd be nearly certain$41 was impossible. I'd then find the profit if the store sold as many new books as possible to verify that their max profit was $39, not$41.
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22 Aug 2010, 18:17
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I guess the fastest way could be to calculate the maximum possible profit, which is 7*5+1*2= 37.
Now you know that 41 can't be the answer!
If answer D is actually 39 as IanStewart says, just rule it out...

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22 Aug 2010, 19:40
toshio86 wrote:
I guess the fastest way could be to calculate the maximum possible profit, which is 7*5+1*2= 37.
Now you know that 41 can't be the answer!
If answer D is actually 39 as IanStewart says, just rule it out...

The max profit is 7*5 + 2*2 = 39, and not 37; 37 is actually impossible here.

That's certainly very fast, but I suppose the question is - how do you know to do this instead of finding the minimum profit? And what would you do with different answer choices - say all of the answers were between 27 and 39? It's for that reason that I discussed a few ways to look at the question, but I do think a kind of 'plug in numbers and see what happens' approach is perfectly good here, quite practical to do within 2 minutes.
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22 Aug 2010, 20:23
i couldn't find any combinations that can fit into 29 and 41, so i guess D and E are the answers unless there's a typo for any of them.

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23 Aug 2010, 00:05
Profits can only be 25, 27, 39, 48 dollars!
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29 Aug 2010, 20:55
New books revenues can be either 15,45,75,or 105 and corresponding used book revenues would e 110,80,50 and 20 making profits will be 27,31,35,39. hence answer is E. if D is typo

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30 Aug 2010, 00:41
D should be 39 and not 29....E is the clear answer then

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16 Jun 2011, 01:09
3u + 2n = 25
u(max) = 8
check for u =7,5,3,1

will help in POE
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22 Apr 2014, 13:57
shekar123 wrote:
A bookstore sells new books for $15 each and used books for$10 each. On every new book, the store makes a profit of $5 while on every used book it makes a profit of$2. If on a given day the bookstore's sales amounted to $125, which of the following cannot be the profit made on that day? A.$27
B. $31 C.$35
D. $39 E.$41

Given: $$15n+10u=125$$, where $$n$$ is the number of new books sold and $$u$$ is the number of used books sold.
Question: taking into account above equation which value is not possible for $$5n+2u$$?

Reduce $$15n+10u=125$$ by 5: $$3n+2u=25$$. Notice that this equation to hold true $$n$$ must be odd, since if $$n=even$$ then $$3n+2u=even+even=even$$ so, it cannot equal to odd number 25.

Next, $$5n+2u=2n+(3n+2u)=2n+25$$. Now, you can notice that if $$n$$ is 1, 3, 5, or 7 then options A, B, C and D are possible (else notice that E to be possible $$n$$ must be 8, so even and we know that $$n$$ is odd).

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05 May 2014, 06:12
pretzel wrote:
Bunuel wrote:
Next, $$5n+2u=2n+(3n+2u)=2n+25$$. Now, you can notice that if $$n$$ is 1, 3, 5, or 7 then options A, B, C and D are possible (else notice that E to be possible $$n$$ must be 8, so even and we know that $$n$$ is odd).

Hi Bunnel, why do we have 2n + (3n+2u)?

We break 5n+2u into 2n and (3n+2u) because we know the value of 3n+2u (25), so we can substitute, which further will help to answer the question.
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Re: A bookstore sells new books for $15 each and used books for [#permalink] ### Show Tags 05 May 2014, 22:34 shekar123 wrote: A bookstore sells new books for$15 each and used books for $10 each. On every new book, the store makes a profit of$5 while on every used book it makes a profit of $2. If on a given day the bookstore's sales amounted to$125, which of the following cannot be the profit made on that day?

A. $27 B.$31
C. $35 D.$39
E. $41 m22 q12 Cost of new books is$15 and old is $10. Total sale price is$125. This tells me that number of new books sold was definitely odd since the total sale price ends with a 5. Had number of new books sold been even, the total sale price would have been a multiple of 10. So there must have been 1, 3, 5 or 7 new books and rest old books. So profit would have been an odd multiple of 5 + a multiple of 2.

Number of new and old books could be 1 and 11 (total revenue of 125). Profit 5 + 2*11 = 27
Number of new and old books could be 3 and 8. Profit 15 + 2*8 = 31
Number of new and old books could be 5 and 5. Profit 25 + 5*2 = 35
Number of new and old books could be 7 and 2. Profit 35 + 2*2 = 39

A profit of 41 is not possible.

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