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\(x\) is a positive integer less than \(20.\) What is the value of \(x?\)

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by M7MBA » Sat Jul 25, 2020 2:49 pm

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\(x\) is a positive integer less than \(20.\) What is the value of \(x?\)

(1) \(x\) is the sum of two consecutive integers.
(2) \(x\) is the sum of five consecutive integers.

[spoiler]OA=E[/spoiler]

Source: Veritas Prep
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Source: — Data Sufficiency |

M7MBA wrote:
Sat Jul 25, 2020 2:49 pm
\(x\) is a positive integer less than \(20.\) What is the value of \(x?\)

(1) \(x\) is the sum of two consecutive integers.
(2) \(x\) is the sum of five consecutive integers.

[spoiler]OA=E[/spoiler]

Source: Veritas Prep
From the information, we know that \(1 ≤x ≤19\). We have to get the unique value of x.

Let's take each statement one by one.

(1) \(x\) is the sum of two consecutive integers.

Say if the two consecutive integers are 1 and 2, then we have x = 3; however, if the two consecutive integers are 2 and 3, then we have x = 5. No unique answer. Insufficient.

(2) \(x\) is the sum of five consecutive integers.

Say the five consecutive integers are n, (n + 1), (n + 2),(n + 3), and (n + 4).

Thus, x = 5n + 10

=> \(1 ≤ 5n + 10 ≤ 19\)

=> \(1 - 10 ≤ 5n ≤ 19 - 10\)

=> \(-9 ≤ 5n ≤ 9\)

=> \(-9/5 ≤ n ≤ 9/5\)

=> \(-1.8 ≤ n ≤ 1.8\)

=> \(n = -1, 0, or 1\); thus, \(x = 5, 10 or 15\). No unique answer. Insufficient.

(1) and (2) together

Even both statements together are insufficient for 5 or 15 can be the sum of two consecutive integers.

Correct answer: E

Hope this helps!

-Jay
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M7MBA wrote:
Sat Jul 25, 2020 2:49 pm
\(x\) is a positive integer less than \(20.\) What is the value of \(x?\)

(1) \(x\) is the sum of two consecutive integers.
(2) \(x\) is the sum of five consecutive integers.

[spoiler]OA=E[/spoiler]

Source: Veritas Prep
Statement 1:

\(x\) can be \(1, 3, 5, 7, 9, 11, 13, 15, 17, 19\). Insufficient \(\Large{\color{red}\chi}\)

Statement 2:

\(x\) can be \(5, 10, 15\). Insufficient \(\Large{\color{red}\chi}\)

Combining \(1\&2\):

\(x\) can be \(5\) or \(15\). Insufficient \(\Large{\color{red}\chi}\)

Therefore, E
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