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a bookstore sells new books for $15 each and used books for

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CEO
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a bookstore sells new books for $15 each and used books for [#permalink] New post 22 Oct 2007, 09:15
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 sales were $125, which of the following cannot be the profit on that day?

27
31
35
39
41
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 [#permalink] New post 22 Oct 2007, 10:08
E.

From the given information about sales,
15x + 10y =125.
So, 3x + 2y = 25

Also, from the given information about profits
5x + 2y = 27 or 31 or 25 or 39 or 41.

Lets take the first case : 5x + 2y = 27.
Solve this and the earlier sales eqn, we get x = 1, y = 11. The values of x and y fit well in to both the equations.

Test this for each of the 5 possibilities of value of profit. Its fits for each except for 5x + 2y = 41. If we solve this and sales eqn, we get x = 8. But if we substitute, x = 8 in either of the eqn. y is not an integer. So the answer is E.

Whats the OA ?
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 [#permalink] New post 22 Oct 2007, 10:20
15x + 10y = 125
Max profit is when you sell max of X.
So, max x = 7. 7*15= 105 + 10y = 125, y =2
Thus, max profit is 7*5 + 2*2 = 39.

It can't be higher than 39, so 41 is impossible.
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 [#permalink] New post 22 Oct 2007, 11:04
I get 41 profit is impossible as well.

N = new books sold
U = used books sold
P = profit

15N + 10U = 125
5N + 2U = P

combining equations (mult bottom eq by -3)

4U = 125-3P ----------> 125 - 4U = 3P

By pluggin in numbers, if you try 41 for P you get 125 - 4U = 122, which is impossible.
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 [#permalink] New post 06 Dec 2007, 06:09
ben928 wrote:
15x + 10y = 125
Max profit is when you sell max of X.
So, max x = 7. 7*15= 105 + 10y = 125, y =2
Thus, max profit is 7*5 + 2*2 = 39.

It can't be higher than 39, so 41 is impossible.


Excellent approach

OA is E.
  [#permalink] 06 Dec 2007, 06:09
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