Math Revolution DS Expert - Ask Me Anything about GMAT DS

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 ‘Is an integer ‘6’ present in Set S’- where Set S is a set of 11 consecutive integers

Follow the second and the third step: From the original condition, we have 1 variable (n) since n, n+1, n+2, …., n+13. 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 both conditions together.

Condition (1) tells us that the integer −3 is present in the set.

We can have 11 consecutive integers including −3 in the following ways:

=> Set S: { −4, −3, −2, . . . 3, 4, 5, 6}; (‘6’ is present in the set) - YES

OR

=> Set S: {−7, −6, −5, −4, −3 . . . 1, 2, 3}; (‘6’ is present in the set) - NO

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

Condition (2) tells us that the integer 9 is present in the set

We can have 11 consecutive integers including 9 in the following ways:

=> Set S: {−1, . . . 6, 7, 8, 9}; (‘6’ is present in the set) - YES

OR

=> Set S: {9, 10, 11, 12,13 . . . 17, 18, 19}; (‘6’ is not present in the set) - NO

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

Both conditions (1) and (2) tells us that −3 and 9 are present in the set.

We cannot have 11 consecutive integers including −3 and 9.

The answer is not unique, YES and NO, so both conditions (1) and (2) combined are not sufficient according to Common Mistake Type 1 which states that the answer should be a unique YES or a NO.

Both conditions together are not sufficient.

Therefore, E is the correct answer.

Answer: E