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What is the median of 3 consecutive integers? 1) The product of the in

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What is the median of 3 consecutive integers? 1) The product of the in [#permalink]

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New post 06 Oct 2017, 02:41
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A
B
C
D
E

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  55% (hard)

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60% (00:44) correct 40% (00:43) wrong based on 45 sessions

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[GMAT math practice question]

What is the median of 3 consecutive integers?

1) The product of the integers is 0
2) The sum of the integers and the product of the integers are the same
[Reveal] Spoiler: OA

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Last edited by Bunuel on 06 Oct 2017, 04:28, edited 1 time in total.
Edited the OA.

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Re: What is the median of 3 consecutive integers? 1) The product of the in [#permalink]

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New post 06 Oct 2017, 04:08
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MathRevolution wrote:
[GMAT math practice question]

What is the median of 3 consecutive integers?

1) The product of the integers is 0
2) The sum of the integers and the product of the integers are the same



Answer should be C and not B
Statement 2 can have 2 examples: -1,0,1 or 1,2,3

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Re: What is the median of 3 consecutive integers? 1) The product of the in [#permalink]

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New post 06 Oct 2017, 04:28
KS15 wrote:
MathRevolution wrote:
[GMAT math practice question]

What is the median of 3 consecutive integers?

1) The product of the integers is 0
2) The sum of the integers and the product of the integers are the same



Answer should be C and not B
Statement 2 can have 2 examples: -1,0,1 or 1,2,3


Yes, you are right. The OA is not correct. Will edit.
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What is the median of 3 consecutive integers? 1) The product of the in [#permalink]

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New post 08 Oct 2017, 18:02
=>

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures a solution.
We assume that we have 3 consecutive integers, n, n+1 and n+2.
We have 1 variable and 0 equation from the original condition. Therefore, D is most likely to be the answer.

Condition 1)
n(n+1)(n+2) = 2.
We have n = 0, n = -1 or n = -2.
This is not sufficient.

Condition 2)
n + ( n + 1 ) + ( n + 2 ) = n(n+1)(n+2)
3n + 3 = n(n+1)(n+2)
3(n + 1) = n(n+1)(n+2)
n(n+1)(n+2) - 3(n + 1) = 0
(n+1){ n(n+2) – 3 } = 0
(n+1)(n^2 + 2n – 3) = 0
(n+1)(n-1)(n+3) = 0
n = -1, n = 1 or n = -3
This is not sufficient.
Condition 1) & 2)
From the condition 1), we have n = 0, n = -1 or n = -2.
From the condition 2), we have n = 1, n = -1 or n = -3.
Thus, n = -1.
Both conditions together are sufficient.

Therefore, unlike our expectation, C is the answer.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both con 1) and con 2). Therefore, C has a high chance of being the answer, which is why we attempt to solve the question using con 1) and con 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using con 1) and con 2) together. (It saves us time). Obviously, there may be cases where the answer is A, B, D or E.
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