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A bookstore sells new books for $15 each and used books for
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Updated on: 22 Apr 2014, 13:58
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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
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Originally posted by shekar123 on 21 Aug 2010, 22:52.
Last edited by Bunuel on 22 Apr 2014, 13:58, edited 2 times in total.
Edited the question.



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Re: 700 level question
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21 Aug 2010, 23:42
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 guessingandchecking. 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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Re: 700 level question
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22 Aug 2010, 18:17
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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Re: 700 level question
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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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Re: 700 level question
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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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Re: 700 level question
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23 Aug 2010, 00:05
Profits can only be 25, 27, 39, 48 dollars!
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Re: 700 level question
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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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Re: A bookstore sells new books for $15 each and used books for
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22 Apr 2014, 12:58
My way OK so we have that: 15N + 10U = 125 3N + 2U = 25 (1) Now we are trying to find 5N+2U Therefore, we will have 25 + 2N, which one CANNOT be expressed in this form? Well if 25 + 2N = 41 then 2N=16, N=8 Therefore replacing in (1) , 3N = 24 and 2U = 1, U=1/2 which CAN'T be correct because we can't sell 1/2 a book Thus E stands Hope this helps Cheers! J



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Re: A bookstore sells new books for $15 each and used books for
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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). Answer: E.
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Re: A bookstore sells new books for $15 each and used books for
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05 May 2014, 05:47
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).
Answer: E. Hi Bunnel, why do we have 2n + (3n+2u)?



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Re: A bookstore sells new books for $15 each and used books for
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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).
Answer: E. 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
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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. Answer (E)
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