[GMAT math practice question]
(Number Properties) p, q, and r are positive integers. What is the value of p + q + r?
1) p, q, and r are prime numbers.
2) The product of p, q, and r is 5 times the sum of p, q, and r.
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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.
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Now we will solve this DS question using the Variable Approach.
Let’s apply the 3 steps suggested previously.
Follow the first step of the Variable Approach by modifying and rechecking the original condition and the question.
We have to determine the value of p + q + r.
Follow the second and the third step: From the original condition, we have 3 variables (p, q, and r). To match the number of variables with the number of equations, we need 1 equation. Since conditions (1) and (2) will provide 1 equation each, E would most likely be the answer.
Recall 3 Principles and choose E as the most likely answer. Let’s look at both conditions 1) & 2) together.
Since we have 3 variables (p, q, and r) and 0 equations, E is most likely 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) together give us that:
Since we have pqr = 5(p + q + r), we assume p = 5.
Then we have 5qr = 5(5 + q + r) or qr = q + r + 5.
Since we have qr – q – r = 5, we have qr – q – r + 1 = 6 (by adding 1 to both sides so that it can easily be factored). Factoring gives us (q - 1)(r - 1) = 6.
Then we have the possible pairs of q - 1 and r - 1 are (1, 6), (2, 3), (3, 2), and (6, 1).
If we have q – 1 = 1 and r – 1 = 6, we have q = 2 and r = 7.
If we have q – 1 = 2 and r – 1 = 3, we have q = 3 and r = 4.
If we have q – 1 = 3 and r – 1 = 2, we have q = 4 and r = 3.
If we have q – 1 = 6 and r – 1 = 1, we have q = 7 and r = 2.
However, p, q, and r are prime numbers from condition 1), and their possible numbers are 2, 5, and 7.
Thus, we have p + q + r = 2 + 5 + 7 = 14.
The answer is unique, so both conditions together are sufficient, according to Common Mistake Type 2, which states that the number of answers must be only one. So, C seems to be the answer.
However, since this question is an integer question, which is also one of the key questions, we should apply CMT 4(A), which states that if an answer C is found too easily, either A or B should be considered as the answer. Let’s look at each condition separately.
Condition 1) tells us that we don’t have a unique solution, obviously.
If p = 2, q = 3, and r = 5, we have p + q + r = 2 + 3 + 5 = 10.
If p = 2, q = 5, and r = 7, we have p + q + r = 2 + 5 + 7 = 14.
The answer is not unique, and the condition is not sufficient, according to Common Mistake Type 2, which states that the number of answers must be only one.
Let’s look at the condition 2). It tells us that we don’t have a unique solution.
Assume p = 5 and since we have qr – q – r = 5, we have qr – q – r + 1 = 6 or (q - 1)(r - 1) = 6.
Then the possible pairs of q - 1 and r - 1 are (1, 6), (2, 3), (3, 2), and (6, 1).
If we have q – 1 = 1 and r – 1 = 6, we have q = 2, r = 7 and p + q + r = 5 + 2 + 7 = 14.
If we have q – 1 = 2 and r – 1 = 3, we have q = 3, r = 4 and p + q + r = 5 + 3 + 4 = 12.
The answer is not unique, and the condition is not sufficient, according to Common Mistake Type 2, which states that the number of answers must be only one.
Thus, really, both conditions 1) & 2) together are sufficient.
Therefore, C is the correct answer.
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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