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Gmat club tests M22 Q12 [#permalink]
17 Jan 2010, 17:34

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?

$27 $31 $35 $39 $41

Could someone elaborate on this, I could not understand explanation provided by GMAT club

Re: Gmat club tests M22 Q12 [#permalink]
17 Jan 2010, 18:27

mirzohidjon 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?

$27 $31 $35 $39 $41

Could someone elaborate on this, I could not understand explanation provided by GMAT club

I am sure there is a shorter way but:

15N + 10U = 125 where N = new books sold and U = used books sold 5N + 2U = x where x = total profit

subtract both equations 4U = 125 - 3x

You can substitute each number above and 125-3x must be a multiple of 4 a 125 - 3(27) = 44 b 125 - 3(31) = 32 c 125 - 3(35) = 20 d 8 e 2 (answer not possible)

Re: Gmat club tests M22 Q12 [#permalink]
19 Jan 2010, 03:57

mirzohidjon 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?

$27 $31 $35 $39 $41

Could someone elaborate on this, I could not understand explanation provided by GMAT club

Try this approach: Since Total Sales is $125 (odd), no of new books sold should be odd So following are the four possible scenario 1. 1 New + 11 Old => profit (1*5)+(11*2)=$27 2. 3 New + 8 Old => profit (3*5)+(8*2)=$31 3. 5 New + 5 Old => profit (5*5)+(5*2)=$35 4. 7 New + 2 Old => profit (7*5)+(2*2)=$39

Bunuel, How can I solve this method in two minutes? Any tips? THE OE says "The maximum number of new books that could have been sold is 7. In this case the profit would have been highest: USD. 41 USD is an impossible profit."

If 125 = 15*1 + 11*10=> Maximum number of books = 12 and not 7.

I wrote two equations : 15X + 10y =125; 5X+2Y = (substitute answer choice). However, this took me over 4 minutes.

Bunuel, How can I solve this method in two minutes? Any tips? THE OE says "The maximum number of new books that could have been sold is 7. In this case the profit would have been highest: USD. 41 USD is an impossible profit."

If 125 = 15*1 + 11*10=> Maximum number of books = 12 and not 7.

I wrote two equations : 15X + 10y =125; 5X+2Y = (substitute answer choice). However, this took me over 4 minutes.

Thanks

OE says that "The maximum number of new books that could have been sold is 7" not total number of books. Below is new OE for this question:

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).