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16 Sep 2014, 01:53
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
44% (02:51) correct
56%
(02:12)
wrong
based on 91
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If \(p\), \(q\), \(r\), and \(s\) are consecutive integers, with \(p \lt q \lt r \lt s\), is \(pr \lt qs\)?
(1) \(pq \lt rs\)
(2) \(ps \lt qr\)
(1) \(pq \lt rs\)
(2) \(ps \lt qr\)
16 Sep 2014, 01:53
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Official Solution:
We can solve this problem either by algebra or by number-plugging. Let's use algebra. All four variables can be expressed in terms of just one variable, since they are consecutive integers and we know their order. If we keep \(p\) as the basic variable, then \(q = p + 1\), \(r = p + 2\), and \(s = p + 3\).
Now we can rephrase the question:
Is \(pr \lt qs\)?
Is \(p(p + 2) \lt (p + 1)(p + 3)\)?
Is \(p^2 + 2p \lt p^2 + 4p + 3\)?
Is \(2p \lt 4p + 3\)?
Is \(0 \lt 2p + 3\)?
Is \(-3 \lt 2p\)?
Is \(-\frac{3}{2} \lt p\)?
Since \(p\) is an integer, the question is answered "yes" if \(p = -1\) or greater, and "no" if \(p = -2\) or less.
Statement (1): SUFFICIENT.
We rephrase the statement similarly.
\(pq \lt rs\)
\(p(p + 1) \lt (p + 2)(p + 3)\)
\(p^2 + p \lt p^2 + 5p + 6\)
\(0 \lt 4p + 6\)
\(0 \lt 2p + 3\)
\(-\frac{3}{2} \lt p\)
This is precisely the same condition as asked in the question. Thus, we can answer the question definitively.
Statement (2): INSUFFICIENT.
Again, we rephrase the statement similarly.
\(ps \lt qr\)
\(p(p + 3) \lt (p + 1)(p + 2)\)
\(p^2 + 3p \lt p^2 + 3p + 2\)
\(0 \lt 2\)
Since 0 is always less than 2, no matter the value of \(p\), the statement is always true. Thus, we do not gain any information that would help us answer the question.
Answer: A
We can solve this problem either by algebra or by number-plugging. Let's use algebra. All four variables can be expressed in terms of just one variable, since they are consecutive integers and we know their order. If we keep \(p\) as the basic variable, then \(q = p + 1\), \(r = p + 2\), and \(s = p + 3\).
Now we can rephrase the question:
Is \(pr \lt qs\)?
Is \(p(p + 2) \lt (p + 1)(p + 3)\)?
Is \(p^2 + 2p \lt p^2 + 4p + 3\)?
Is \(2p \lt 4p + 3\)?
Is \(0 \lt 2p + 3\)?
Is \(-3 \lt 2p\)?
Is \(-\frac{3}{2} \lt p\)?
Since \(p\) is an integer, the question is answered "yes" if \(p = -1\) or greater, and "no" if \(p = -2\) or less.
Statement (1): SUFFICIENT.
We rephrase the statement similarly.
\(pq \lt rs\)
\(p(p + 1) \lt (p + 2)(p + 3)\)
\(p^2 + p \lt p^2 + 5p + 6\)
\(0 \lt 4p + 6\)
\(0 \lt 2p + 3\)
\(-\frac{3}{2} \lt p\)
This is precisely the same condition as asked in the question. Thus, we can answer the question definitively.
Statement (2): INSUFFICIENT.
Again, we rephrase the statement similarly.
\(ps \lt qr\)
\(p(p + 3) \lt (p + 1)(p + 2)\)
\(p^2 + 3p \lt p^2 + 3p + 2\)
\(0 \lt 2\)
Since 0 is always less than 2, no matter the value of \(p\), the statement is always true. Thus, we do not gain any information that would help us answer the question.
Answer: A
Yadnesh
Joined: 19 Aug 2014
Last visit: 25 Aug 2015
Posts: 2
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Location: United States
Concentration: Entrepreneurship, General Management
Schools: Schulich '15
GMAT 1: 730 Q49 V41
WE:Other (Other)
01 Oct 2014, 02:41
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We can solve this problem either by algebra or by number-plugging. Let's use algebra. All four variables can be expressed in terms of just one variable, since they are consecutive integers and we know their order. If we keep p as the basic variable,
Bunuel - how do you suggest to solve this by number plugging.
Also im not great at using number plugging to solve questions and generally take the algaebric ways which costs me more time and is very unfeasibe in some questions.
Are you aware of a bank of questions for me to practice number plugging and also some tips and tricks about the same.
Thanks in advance . You are a saviour.
Bunuel - how do you suggest to solve this by number plugging.
Also im not great at using number plugging to solve questions and generally take the algaebric ways which costs me more time and is very unfeasibe in some questions.
Are you aware of a bank of questions for me to practice number plugging and also some tips and tricks about the same.
Thanks in advance . You are a saviour.
