Que: Is ‘a’ an integer?

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Que: Is ‘a’ an integer?

by Max@Math Revolution » Tue Aug 10, 2021 9:24 pm

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Que: Is ‘a’ an integer?

(1) \(a^3\) is an integer.
(2) \(6a\) is an integer.

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Re: Que: Is ‘a’ an integer?

by Max@Math Revolution » Wed Aug 11, 2021 8:56 pm

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Solution: To save time and improve accuracy on DS questions in GMAT, learn and apply the Variable Approach.

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.

Visit https://www.mathrevolution.com/gmat/lesson for details.

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 find whether ‘a’ is an integer.

Follow the second and the third step: From the original condition, we have 1 variable (a). 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, D would most likely be the answer.

Recall 3- Principles and Choose D as the most likely answer. Let’s look at each condition separately.

Condition (1) tells us that the \(a^3\) is an integer.

=> If \(a^3=8\), then a = 2. YES

=> But if \(a^3=16\), then a is not an integer. NO

The answer is not a unique YES or a NO, so Condition (1) alone is not sufficient according to Common Mistake Type 1 which states that the answer should be a unique value.

Condition (2) tells us that the 6a is an integer.

=> If a = 10, then 6a = 10 * 6 = 60.

=> But if a = 7.5, then 6a = 6 * 7.5 = 45.

The answer is not a unique YES or a NO, so Condition (2) alone is not sufficient according to Common Mistake Type 1 which states that the answer should be a unique value.


Let’s look at both conditions combined:

=> If \(a^3=8\), then a = 2 and then 6a = 6 * 2 = 12 is an integer – YES

=> If \(a^3=1\), then a = 1 and then 6a = 6 * 1 = 6 is an integer – YES

=> But if, \(a^3=16\), then a = 2.52 and then 4a = 6 * 2.52 = is not an integer

Only if ‘a’ is an integer then, \(a^3\) and 6a will be an integer.

The answer is a unique value YES; both conditions combined are sufficient according to Common Mistake Type 1 which states that the answer should be a unique YES or a NO.

Both conditions together are sufficient.

Therefore, C is the correct answer.

Answer: C