[
Math Revolution GMAT math practice question]
If the integers p, q, r, and s satisfy p<q<r<s, are they consecutive?
1) r-q=1
2) (p-q)(r-s)=1
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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
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