MathRevolution wrote:

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Math Revolution GMAT math practice question]

Is the sum of \(7\) different positive integers greater than or equal to \(48\)?

1) Their median is \(9\)

2) The largest number is \(12\)

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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 many variables (x

1, x

2, …, x

7) 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)

We write the numbers as x

1≤x

2≤x

3≤x

4≤x

5≤x

6≤x

7. Then x

4 is their median.

From condition 1), x

4 = 9 and the smallest possible number is 1 + 2 + 3 + 9 + 10 + 11 + 12 = 48. Therefore, the answer is ‘yes’.

Both conditions 1) & 2) together are sufficient.

Since this question is a statistics 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)

Since we didn’t use condition 2) in the above argument, condition 1) is sufficient on its own.

By condition 1), x

4 = 9 and the smallest possible number is 1 + 2 + 3 + 9 + 10 + 11 + 12 = 48. Therefore, the answer is ‘yes’.

Condition 2)

When the numbers are 6,7,8,9,10,11,12, their sum is 6 + 7 + 8 + 9 + 10 + 11 + 12 = 63 > 48, and the answer is ‘yes’.

When the numbers are 6,7,8,9,10,11,12, their sum is 1 + 2 + 3 + 4 + 5 + 6 + 12 = 33 < 48, and the answer is ‘no’.

Since we don’t obtain a unique answer, condition 2) is not sufficient.

Therefore, A is the answer.

Answer: A

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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