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# M22-12

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Math Expert
Joined: 02 Sep 2009
Posts: 52385

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16 Sep 2014, 00:16
00:00

Difficulty:

55% (hard)

Question Stats:

61% (02:05) correct 39% (02:21) wrong based on 119 sessions

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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

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Joined: 02 Sep 2009
Posts: 52385

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16 Sep 2014, 00:16
5
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).

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Intern
Joined: 17 Mar 2015
Posts: 15

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01 Jul 2016, 23:00
1
How is one supposed to notice such stuff? Is it realistic to notice and solve this in 2 mins?
Manager
Joined: 21 Sep 2015
Posts: 77
Location: India
GMAT 1: 730 Q48 V42
GMAT 2: 750 Q50 V41

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28 Jul 2016, 05:29
11
1
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$_________________ Appreciate any KUDOS given ! Manager Joined: 12 Jun 2016 Posts: 216 Location: India Concentration: Technology, Leadership WE: Sales (Telecommunications) Re: M22-12 [#permalink] ### Show Tags 19 Oct 2017, 08:44 1 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 _________________ My Best is yet to come! Intern Joined: 27 Oct 2017 Posts: 11 Location: Brazil Concentration: Strategy, Marketing Schools: Booth '20 (II) GMAT 1: 600 Q46 V28 GMAT 2: 650 Q46 V34 GPA: 3.32 M22-12 [#permalink] ### Show Tags Updated on: 09 Jan 2018, 09:20 1 My solution is very similar to that of Bunuel, but since the reasoning is slightly different, I decided to post it. Let x be the number of NEW books sold, and y be the number of USED books sold. Then, we'll have: 15x + 10y = 125 Analyzing this equation, you can notice that x MUST be an odd value for the equation to hold true; otherwise, the ending 5 digit of the 125 total sales would disappear. For example, if x=2, then 15x=30, and regardless of the y value, the last digit of total sales value would be 0. Not good. So, our mission is to find the total profit number that will produce an EVEN value for x. Let the profit equation be: 5x + 2y = A (A=profit). 15x + 10y = 125 (divide the whole equation by -5, so we can eliminate y value, since we're not interested in it) Sum the new total sales equation with the profit equation: -3x -2y = -25 5x + 2y = A This is our final equation: 2x = A - 25. Quickly test the alternatives (replacing by A) to see which one yields an EVEN x (EmpowerGMAT suggests that in "which of the following" questions, we start by testing either D or E, since those are more likely to be the correct answer). You will find that alternative E yields x=8, while all others yield ODD values for x. Answer: E It's not a 2-minutes question, but if you have time in the GMAT day maybe it's worth spending a little extra time on it Originally posted by thiagod on 09 Jan 2018, 07:00. Last edited by thiagod on 09 Jan 2018, 09:20, edited 2 times in total. Senior Manager Joined: 08 Jun 2015 Posts: 432 Location: India GMAT 1: 640 Q48 V29 GMAT 2: 700 Q48 V38 GPA: 3.33 Re: M22-12 [#permalink] ### Show Tags 09 Jan 2018, 09:08 1 +1 for option E. Find an answer option that when expressed as a sum of multiples of 2 and 5 and then multiplied by 10 and 15 respectively exceed$125.

41 = 7*5+3*2. The cost is 15*7+10*3. The sum 135>125.

Hence option E
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Intern
Joined: 18 Dec 2017
Posts: 1

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09 Jan 2018, 23:36
Bunuel wrote:
Official Solution:

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).

Why do we know that 5n+2u = 2n+(3n+2u)?
Math Expert
Joined: 02 Sep 2009
Posts: 52385

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09 Jan 2018, 23:39
nluna15 wrote:
Bunuel wrote:
Official Solution:

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).

Why do we know that 5n+2u = 2n+(3n+2u)?

_________________
Re: M22-12 &nbs [#permalink] 09 Jan 2018, 23:39
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# M22-12

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