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A bookstore that sells used books sells each of its paperback books for a certain price and each of its hardcover books for a certain price. If Joe, Maria, and Paul bought books in this store, how much did Maria pay for 1 paperback book and 1 hardcover book?

(1) Joe bought 2 paperback books and 3 hardcover books for $12.50. (2) Paul bought 4 paperback books and 6 hardcover books for $25.00.

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Re: A bookstore that sells used books sells each of its paperbac [#permalink]

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23 Feb 2014, 07:48

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Expert's post

SOLUTION

A bookstore that sells used books sells each of its paperback books for a certain price and each of its hardcover books for a certain price. If Joe, Maria, and Paul bought books in this store, how much did Maria pay for 1 paperback book and 1 hardcover book?

We should find the value of p+h, where p is the price of one paperback and h is the price of one hard cover book.

(1) Joe bought 2 paperback books and 3 hardcover books for $12.50 --> 2p + 3h = 12.5. Not sufficient.

(2) Paul bought 4 paperback books and 6 hardcover books for $25.00 --> 4p + 6h = 25. Not sufficient.

(1)+(2) We can get 4p + 6h = 25 by multiplying 2p + 3h = 12.5 by 2, thus even when combining the statements we still have only one equation. Not sufficient.

Re: A bookstore that sells used books sells each of its paperbac [#permalink]

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24 Feb 2014, 00:27

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A bookstore that sells used books sells each of its paperback books for a certain price and each of its hardcover books for a certain price. If Joe, Maria, and Paul bought books in this store, how much did Maria pay for 1 paperback book and 1 hardcover book?

(1) Joe bought 2 paperback books and 3 hardcover books for $12.50. (2) Paul bought 4 paperback books and 6 hardcover books for $25.00.

Sol: Let the price of paperbook be "X" and that of Hardcover be "Y". So we need to find X+Y?

St 1: 2X+3Y=12.5. There are multiple values possible so not sufficient. A and D ruled out

St 2: 4X+6Y=25 or 2X+3Y=12.5 which is same information as St 1 So B ruled out

Combining the 2 statements, we get no new information so ans is E.

650 level is okay _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: A bookstore that sells used books sells each of its paperbac [#permalink]

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24 Feb 2014, 04:22

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Let the price of paperbacks be 'x' the price of hard cover book be 'y'

From Statement 1:- Joe bought 2 paperback books and 3 hardcover books for $12.50. Thus 2x+3y = 12.50 This is one equation with 2 unknowns hence it cannot be solved. So statement 1 alone is insufficient

From Statement 2:- Paul bought 4 paperback books and 6 hardcover books for $25.00 Thus 4x+6y = 25 Again it is one equation with 2 unknowns. So statement 2 alone is insufficient.

Combining both the statements we get 2 equations:- 1. 2x+3y = 12.5 2. 4x+6y = 25

Now linear equations with 2 unknowns can only be solved it both the equations are different i.e not the same or multiplies of each other. Here equation 2 is a multiple of equation 1. Equation 2 = (Equation 1)x2

Thus this pair of equations cannot be solved. and hence even after combining both the equations the solution is not possible.

Re: A bookstore that sells used books sells each of its paperbac [#permalink]

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01 Mar 2014, 05:30

Expert's post

SOLUTION

A bookstore that sells used books sells each of its paperback books for a certain price and each of its hardcover books for a certain price. If Joe, Maria, and Paul bought books in this store, how much did Maria pay for 1 paperback book and 1 hardcover book?

We should find the value of p+h, where p is the price of one paperback and h is the price of one hard cover book.

(1) Joe bought 2 paperback books and 3 hardcover books for $12.50 --> 2p + 3h = 12.5. Not sufficient.

(2) Paul bought 4 paperback books and 6 hardcover books for $25.00 --> 4p + 6h = 25. Not sufficient.

(1)+(2) We can get 4p + 6h = 25 by multiplying 2p + 3h = 12.5 by 2, thus even when combining the statements we still have only one equation. Not sufficient.

Re: A bookstore that sells used books sells each of its paperbac [#permalink]

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03 Jun 2014, 22:06

Bunuel wrote:

SOLUTION

A bookstore that sells used books sells each of its paperback books for a certain price and each of its hardcover books for a certain price. If Joe, Maria, and Paul bought books in this store, how much did Maria pay for 1 paperback book and 1 hardcover book?

We should find the value of p+h, where p is the price of one paperback and h is the price of one hard cover book.

(1) Joe bought 2 paperback books and 3 hardcover books for $12.50 --> 2p + 3h = 12.5. Not sufficient.

(2) Paul bought 4 paperback books and 6 hardcover books for $25.00 --> 4p + 6h = 25. Not sufficient.

(1)+(2) We can get 4p + 6h = 25 by multiplying 2p + 3h = 12.5 by 2, thus even when combining the statements we still have only one equation. Not sufficient.

Answer: E.

Hi Bunuel ,

Can't we get Answer as' D' by using trial and error method.

2 PB + 3 HC = 12.50$ The equation satisfies for PB =4 AND HC = 1.5 ;This is the only pair that satisifies the above equation ; Therefore, can't we say that for one PB +one HC = 2+ 1.5 = 3.5 $

Please explain whether my approach is correct (or ) wrong.

Re: A bookstore that sells used books sells each of its paperbac [#permalink]

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04 Jun 2014, 03:38

Expert's post

dheeraj24 wrote:

Bunuel wrote:

SOLUTION

A bookstore that sells used books sells each of its paperback books for a certain price and each of its hardcover books for a certain price. If Joe, Maria, and Paul bought books in this store, how much did Maria pay for 1 paperback book and 1 hardcover book?

We should find the value of p+h, where p is the price of one paperback and h is the price of one hard cover book.

(1) Joe bought 2 paperback books and 3 hardcover books for $12.50 --> 2p + 3h = 12.5. Not sufficient.

(2) Paul bought 4 paperback books and 6 hardcover books for $25.00 --> 4p + 6h = 25. Not sufficient.

(1)+(2) We can get 4p + 6h = 25 by multiplying 2p + 3h = 12.5 by 2, thus even when combining the statements we still have only one equation. Not sufficient.

Answer: E.

Hi Bunuel ,

Can't we get Answer as' D' by using trial and error method.

2 PB + 3 HC = 12.50$ The equation satisfies for PB =4 AND HC = 1.5 ;This is the only pair that satisifies the above equation ; Therefore, can't we say that for one PB +one HC = 2+ 1.5 = 3.5 $

Please explain whether my approach is correct (or ) wrong.

Help is appreciated.

How did you get that p=4 and h=1.5 is the only solution of 2p + 3h = 12.5 ? What about p=4.75 and h=1, p=3.25 and h=2, ... ? _________________

Re: A bookstore that sells used books sells each of its paperbac [#permalink]

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20 Dec 2015, 16:57

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A bookstore that sells used books sells each of its paperbac [#permalink]

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28 Mar 2016, 18:17

Here is a visual that should help.

Observations:

1) Since the cost of all the goods is the same for everyone, it is irrelevant who bought what, so don't bother complicating the question by including this information.

2) Usually, when we have two equations with two variables, this is sufficient...unless the 2nd equation happens to be the same as the first, which is what happened here.

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Screen Shot 2016-03-28 at 6.17.35 PM.png [ 110.42 KiB | Viewed 585 times ]

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A bookstore that sells used books sells each of its paperbac
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28 Mar 2016, 18:17

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