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M22-12

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M22-12 [#permalink]

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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
[Reveal] Spoiler: OA

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Re M22-12 [#permalink]

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New post 16 Sep 2014, 01:16
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Official Solution:

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: M22-12 [#permalink]

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New post 02 Jul 2016, 00:00
How is one supposed to notice such stuff? Is it realistic to notice and solve this in 2 mins?

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Re: M22-12 [#permalink]

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New post 28 Jul 2016, 06:29
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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

Alternate solution :

Try to find the max and min profit

Min profit will happen when the store sells maximum used books i.e 11 used books and 1 new book
Min profit = 22+5 =27$

Max profit will happen when the store sells maximum new books i.e 7 new books and 2 old books
Max profit = 35 + 4 = 39$

Therefore it is not possible to make a profit of 41$
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Re: M22-12 [#permalink]

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New post 19 Oct 2017, 09:44
Here is a (long) Algebraic approach.

Let \(n\) be the number of new books and \(u\) be the number of used books.

Given: \(15n+10u=125 => 3n+2u = 25\)

\(=>u = \frac{(25-3n)}{2}\)

Since \(n\) can only take integral Values, the Possible values of \(n\) = 1, 3, 5, and 7. Corresponding values of \(u\) = 11, 8, 5, and 2

So the possible values of \((n, u)\) are - { (1, 11) , (3,8) , (5,5), and (7,2) }

To find the possible profits Substitute \((n, u)\) in \(5n+2u\). we find apart from E all answer choices hold true.

Final Answer = E
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Re: M22-12   [#permalink] 19 Oct 2017, 09:44
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