Bunuel wrote:
Tough and Tricky questions: Inequalities.
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) \(ps \lt qr\)
Kudos for a correct solution. Official Solution:If \(p\), \(q\), \(r\), and \(s\) are consecutive integers, with \(p \lt q \lt r \lt s\), is \(pr \lt qs\)? 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.
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