Question: are \(p<q<r<s\) they consecutive?

Statement 1: r-q=1
We can deduce that q and r are consecutive but
No information about p and s hence
NOT SUFFICIENT

Statement 2: \((p-q)(r-s)=1\)
We can deduce that p and q as well as r and s are consecutive but
No information about whether q and r are consecutive or not hence
NOT SUFFICIENT

Combining we get
All are consecutive

Sufficient

Answer: Option C
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Statement 1: r-q=1
We can deduce that q and r are consecutive but
No information about p and s hence
NOT SUFFICIENT.So answer will be B C & E

Statement 2: (p−q)(r−s)=1
From this We conclude that (p-q) must be equal to 1 & (r-s) also must be equal to 1 then only above relation satisfied.
P-q =1 & r-s = 1 means they are consecutive numbers. but
No information about whether q and r are consecutive or not hence
NOT SUFFICIENT.so answer will be C & E

Combining we get
All are consecutive

Sufficient

Answer: Option C

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Beautiful problem, Max. Congrats!


\(p < q < r < s\,\,\,\,{\rm{ints}}\,\,\,\,\,\left( * \right)\)

\({\text{?}}\,\,{\text{:}}\,\,\,{\text{consecutives}}\,\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\boxed{\,\,s - p\,\,\mathop { = \,}\limits^? \,3\,\,}\)


\(\left( 1 \right)\,\,\,\,r = q + 1\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {p,q,r,s} \right) = \left( {0,1,2,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {p,q,r,s} \right) = \left( {0,1,2,4} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)


\(\left( 2 \right)\,\,\left( {p - q} \right)\left( {r - s} \right) = 1\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {p,q,r,s} \right) = \left( {0,1,2,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {p,q,r,s} \right) = \left( {0,1,3,4} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)


\(\left( {1 + 2} \right)\,\,\,\,\left( {p - q} \right)\left( {r - s} \right) = 1\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,p - q\,\,{\rm{and}}\,\,r - s\,\,{\rm{are}}\,\,{\rm{divisors}}\,\,{\rm{of}}\,\,1\,\,\,\,\mathop \Rightarrow \limits^{\left( 2 \right)} \,\,\,\left\{ \matrix{
\,\left( {\rm{I}} \right)\,\,\,\,\left( {p - q,r - s} \right) = \left( {1,1} \right) \hfill \cr
\,\,\,{\rm{OR}} \hfill \cr
\,\left( {{\rm{II}}} \right)\,\,\,\left( {p - q,r - s} \right) = \left( { - 1, - 1} \right) \hfill \cr} \right.\)

\(\left( {\rm{I}} \right)\,\,\,\,\left\{ \matrix{
p - q = 1 \hfill \cr
r - s = 1 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\,\,\,\,\,\,\,\left( {p < q\,\,\,{\rm{and}}\,\,\,r < s} \right)\)

\(\left( {{\rm{II}}} \right)\,\,\,\,\left\{ \matrix{
p - q = - 1 \hfill \cr
r - s = - 1 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\mathop \sim \limits^{\left( 1 \right)} \,\,\,\,\,\,\,\left\{ \matrix{
p - q = - 1 \hfill \cr
q + 1 - s = - 1 \hfill \cr} \right.\,\,\,\,\,\, \sim \,\,\,\,\,\left\{ \matrix{
p - q = - 1 \hfill \cr
q - s = - 2 \hfill \cr} \right.\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\,\,\,\,p - s = - 3\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)


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

Regards,
Fabio.
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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 4 variables (p, q, r and s) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since \(r – q = 1\) or \(r = q + 1\) from condition 1), \(q\) and \(r\) are consecutive integers.
Since we have \((p-q)(r-s) = (q-p)(s-r) =1\), where \(p < q, r < s\) and \(p-q\) and \(r-s\) are integers, we have \(q-p = 1\) and \(s-r = 1\), or \(q = p +1\), and \(s = r +1\).
Thus \(p, q, r\) and \(s\) are consecutive integers and both conditions are sufficient when applied together.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
\(p = 1, q = 2, r = 3\) and \(s = 4\) are consecutive integers satisfying \(r – q = 1\), and the answer is ‘yes’.
\(P = 1, q = 2, r = 3\) and \(s = 5\) satisfy \(r – q = 1\), but are not consecutive integers, and the answer is ‘no’.
Since we don’t have a unique solution, condition 1) is not sufficient.

Condition 2)
\(p = 1, q = 2, r = 3\) and \(s = 4\) are consecutive integers satisfying \((p-q)(r-s)=1,\)
and the answer is ‘yes’.
\(p = 1, q = 2, r = 4\) and \(s = 5\) satisfy \((p-q)(r-s)=1\), but are not consecutive integers, and the answer is ‘no’.
Since we don’t have a unique solution, condition 2) is not sufficient.

Therefore, C is the answer.
Answer: C

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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