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A bookstore sells new books for $15 each and used books for [#permalink]
21 Aug 2010, 21:52

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E

Difficulty:

45% (medium)

Question Stats:

57% (03:17) correct
43% (01:45) wrong based on 86 sessions

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?

Re: 700 level question [#permalink]
21 Aug 2010, 22:42

1

This post received KUDOS

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

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. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: 700 level question [#permalink]
22 Aug 2010, 17:17

1

This post received KUDOS

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...

Re: 700 level question [#permalink]
22 Aug 2010, 18:40

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. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: 700 level question [#permalink]
29 Aug 2010, 19:55

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

Re: A bookstore sells new books for $15 each and used books for [#permalink]
22 Apr 2014, 12:57

Expert's post

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).

Re: A bookstore sells new books for $15 each and used books for [#permalink]
05 May 2014, 04:47

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).

Re: A bookstore sells new books for $15 each and used books for [#permalink]
05 May 2014, 05:12

Expert's post

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. _________________

Re: A bookstore sells new books for $15 each and used books for [#permalink]
05 May 2014, 21:34

Expert's post

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