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605-655 (Medium)|   Algebra|      
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gmatretest
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If the total price for n copies of a book is $31.5, what is the per Updated on: 07 Mar 2016, 01:45
Originally posted by gmatretest on 07 Mar 2016, 00:50.
Last edited by Bunuel on 07 Mar 2016, 01:45, edited 1 time in total.
Renamed the topic and edited the question.
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Difficulty:

45% (medium)

Question Stats:

65% (01:55) correct 35% (01:53) wrong based on 105 sessions

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If the total price for n copies of a book is $31.5, what is the per copy of the book?

(1) if twice as many copies were bought for same total price, the price per copy would be $1.75.
(2) if 4 fewer copies were bought for the samt total price, the price per copy would be $2.80 greater.

Happened to work on this question, and I got myself trapped in quadratic equation. The previous related discussion has been locked.

Show SpoilerHere is how I work,
Decipher question: \(P = 31.5\), Find price per copy a.k.a. \(\frac{P}{n}\).

Statement 1: \(\frac{P}{2n} = 1.75\), \(\frac{P}{n} = 1.75*2\).
Cross B, C, E.

Statement 2: \(\frac{P}{n-4} = \frac{P}{n} + 2.8\)
\(\frac{P}{n} = \frac{31.5-2.8n}{n-4}\)
Quadratic. So I thought Statement 2 was insufficient.


Show SpoilerIn the official answer,
Statement 2 is translated into \((n-4)(p+2.8) = 31.5\), where \(p\) is price per copy.
Eventually, by plugging \(pn=31.5\) into equation, we get \((\frac{31.5}{p}-4)(p+2.8) = 31.5\), which is still a non-linear equation.
Where has gone wrong, and how should I prevent myself from falling into a similar quadratic trap in future?
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ENGRTOMBA2018
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If the total price for n copies of a book is $31.5, what is the per 07 Mar 2016, 05:28
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gmatretest
If the total price for n copies of a book is $31.5, what is the per copy of the book?

(1) if twice as many copies were bought for same total price, the price per copy would be $1.75.
(2) if 4 fewer copies were bought for the samt total price, the price per copy would be $2.80 greater.

Happened to work on this question, and I got myself trapped in quadratic equation. The previous related discussion has been locked.

Show SpoilerHere is how I work,
Decipher question: \(P = 31.5\), Find price per copy a.k.a. \(\frac{P}{n}\).

Statement 1: \(\frac{P}{2n} = 1.75\), \(\frac{P}{n} = 1.75*2\).
Cross B, C, E.

Statement 2: \(\frac{P}{n-4} = \frac{P}{n} + 2.8\)
\(\frac{P}{n} = \frac{31.5-2.8n}{n-4}\)
Quadratic. So I thought Statement 2 was insufficient.


Show SpoilerIn the official answer,
Statement 2 is translated into \((n-4)(p+2.8) = 31.5\), where \(p\) is price per copy.
Eventually, by plugging \(pn=31.5\) into equation, we get \((\frac{31.5}{p}-4)(p+2.8) = 31.5\), which is still a non-linear equation.
Where has gone wrong, and how should I prevent myself from falling into a similar quadratic trap in future?
Show more

Good that you figured out the issue with your calculations. DS questions are not difficult from the calculations stand point but are for the very interpretations that we add to the analysis of the statements. You need to do your due diligence in making sure that a 'peuso-quadratic' equation will not get broken down/cancelled to become a linear equation. This question is a classic example of that.

Lets analyze the question:

n= number of the copies C= cost per copy.

Per the question, nC=31.5 and C=? (easy to see here that once you have a unique value of n you should get the value of C)

Per statement 1, set up the equation as: 31.5 / 2n = 1.75 , clearly solvable for n ---> sufficient.

Per statement 2, 31.5 / (n-4) = 31.5 / n + 2.8, although this looks like a quadratic equation, this is in fact a linear equation as when you solve you get 31.5 * n = 31.5 (n-4)+ 2.8 (n-4)

31.5 * n = 31.5 (n-4)+ 2.8 (n-4) ---> you get \(n^2-4n-45 =0\) ---> n = 9 or -5. As number of books can not be <0 ---> n = 9 is the ONLY allowed value.

Thus sufficient.

Make sure to see if you actually get logical values for number of books, number of people, number of oranges or apples or any other fruits, number of pencils, i.e. anything that must be a positive integer.

Hope this helps.
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If the total price for n copies of a book is $31.5, what is the per 08 Mar 2016, 21:14
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gmatretest
If the total price for n copies of a book is $31.5, what is the per copy of the book?

(1) if twice as many copies were bought for same total price, the price per copy would be $1.75.
(2) if 4 fewer copies were bought for the samt total price, the price per copy would be $2.80 greater.

Happened to work on this question, and I got myself trapped in quadratic equation. The previous related discussion has been locked.

Show SpoilerHere is how I work,
Decipher question: \(P = 31.5\), Find price per copy a.k.a. \(\frac{P}{n}\).

Statement 1: \(\frac{P}{2n} = 1.75\), \(\frac{P}{n} = 1.75*2\).
Cross B, C, E.

Statement 2: \(\frac{P}{n-4} = \frac{P}{n} + 2.8\)
\(\frac{P}{n} = \frac{31.5-2.8n}{n-4}\)
Quadratic. So I thought Statement 2 was insufficient.


Show SpoilerIn the official answer,
Statement 2 is translated into \((n-4)(p+2.8) = 31.5\), where \(p\) is price per copy.
Eventually, by plugging \(pn=31.5\) into equation, we get \((\frac{31.5}{p}-4)(p+2.8) = 31.5\), which is still a non-linear equation.
Where has gone wrong, and how should I prevent myself from falling into a similar quadratic trap in future?

Good that you figured out the issue with your calculations. DS questions are not difficult from the calculations stand point but are for the very interpretations that we add to the analysis of the statements. You need to do your due diligence in making sure that a 'peuso-quadratic' equation will not get broken down/cancelled to become a linear equation. This question is a classic example of that.

Lets analyze the question:

n= number of the copies C= cost per copy.

Per the question, nC=31.5 and C=? (easy to see here that once you have a unique value of n you should get the value of C)

Per statement 1, set up the equation as: 31.5 / 2n = 1.75 , clearly solvable for n ---> sufficient.

Per statement 2, 31.5 / (n-4) = 31.5 / n + 2.8, although this looks like a quadratic equation, this is in fact a linear equation as when you solve you get 31.5 * n = 31.5 (n-4)+ 2.8 (n-4)

31.5 * n = 31.5 (n-4)+ 2.8 (n-4) ---> you get \(n^2-4n-45 =0\) ---> n = 9 or -5. As number of books can not be <0 ---> n = 9 is the ONLY allowed value.

Thus sufficient.

Make sure to see if you actually get logical values for number of books, number of people, number of oranges or apples or any other fruits, number of pencils, i.e. anything that must be a positive integer.

Hope this helps.
Show more

I do not know if this shortcut is acceptable in determining sufficiency without actually spending time to solve the quadratic equation.
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\(31.5n=(2.8n+31.5)(n-4)\) ≈ \(32n=(3n+32)(n-4)\)
I stopped at \(3n^2+12n-128=0\), assuming \(ax^2+bx-c=0\) yields a positive and a negative factor, therefore sufficient to solve the problem.
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If the total price for n copies of a book is $31.5, what is the per 05 Nov 2018, 19:53
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gmatretest
If the total price for n copies of a book is $31.5, what is the per copy of the book?

(1) if twice as many copies were bought for same total price, the price per copy would be $1.75.
(2) if 4 fewer copies were bought for the same total price, the price per copy would be $2.80 greater.
Show more
\(? = c\,\,\left( {{\rm{dollar}}\,\,{\rm{cost}}\,\,{\rm{per}}\,\,{\rm{copy}}} \right)\)

\(n \cdot c = 31.5\,\,\,\,\left[ \$ \right]\,\,\,\,\,\,\left( * \right)\)


\(\left( 1 \right)\,\,2n \cdot 1.75 = 31.5\,\,\,\, \Rightarrow \,\,\,n\,\,{\rm{unique}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\, ? = c\,\,{\rm{unique}}\,\,\,\, \Rightarrow \,\,\,{\rm{SUFF}}{\rm{.}}\)

\(\left( {\,{\rm{POST - MORTEM}}\,\,:\,\,\,\,c\,\,\mathop = \limits^{\left( * \right)} \,\,{{31.5} \over n} = 2 \cdot 1.75\,} \right)\)


\(\left( 2 \right)\,\,\left( {n - 4} \right)\left( {c + 2.8} \right) = 31.5\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\left( {{{31.5} \over c} - 4} \right)\left( {c + 2.8} \right) = 31.5\)

\(31.5 = 31.5 + {{\left( {31.5} \right)\left( {2.8} \right)} \over c} - 4c - 4 \cdot 2.8\,\,\,\,\, \Rightarrow \,\,\,\,\,4{c^2} + 4 \cdot 2.8 \cdot c - \left( {31.5} \right)\left( {2.8} \right) = 0\)

\({c_1}{c_2} = {{ - \left( {31.5} \right)\left( {2.8} \right)} \over 4} < 0\,\,\,\,\left( {{c_1},{c_2}\,\,{\rm{roots}}} \right)\,\,\,\,\, \Rightarrow \,\,\,c > 0\,\,{\rm{unique}}\,\,\,\,\, \Rightarrow \,\,\,{\rm{SUFF}}.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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