Problem Solving

Set T consists of 100 consecutive odd integers. If k is an integer, which of the following CANNOT equal the median of set T?

A) k² - k - 6
B) k² + 8k + 15
C) 4k² + 4k + 1
D) k³ - 4k² - k
E) 3k³ - 27k²

Answer: C
Source: www.gmatprepnow.com
Estimated difficulty level: 650-700

Cheers,
Brent

Brent@GMATPrepNow wrote:Set T consists of 100 consecutive odd integers. If k is an integer, which of the following CANNOT equal the median of set T?

A) k² - k - 6
B) k² + 8k + 15
C) 4k² + 4k + 1
D) k³ - 4k² - k
E) 3k³ - 27k²

Answer: C
Source: www.gmatprepnow.com
Estimated difficulty level: 650-700

Cheers,
Brent

Set T consists of 100 consecutive odd integers, so the median value will be the average of the 50th and 51st integers.
The average of two consecutive odd integers is the even integer between them.
The only thing we know about the median of Set T is that it is an even integer.
Plug In values for k, looking for the answer that CANNOT be even when k is an integer.

Choice A:
Plug In k = 2, so that k^2 - k - 6 = -4
The number -4 is even, so eliminate choice A

Choice B:
Plug In k = 3, so that k^2 + 8k + 15 = 48
The number 48 is even, so eliminate choice B

Choice C:
Plug In k = 2, so that 4k^2 + 4k + 1 = 25
Plug In k = 3, so that 4k^2 + 4k + 1 = 53
Whether integer k is even or odd, choice C will be odd, so keep choice C.

Choice D:
Plug In k = 2, so that k^3 - 4k^2 - k = -10
The number -10 is even, so eliminate choice D.

Choice E:
Plug In k = 2, so that 3k^3 - 27k^2 = 24 - 108 = -84
The number -84 is even, so eliminate choice E.

The correct answer is choice C.

Brent@GMATPrepNow wrote:Set T consists of 100 consecutive odd integers. If k is an integer, which of the following CANNOT equal the median of set T?

A) k² - k - 6
B) k² + 8k + 15
C) 4k² + 4k + 1
D) k³ - 4k² - k
E) 3k³ - 27k²

Answer: C
Source: www.gmatprepnow.com
Estimated difficulty level: 650-700

Cheers,
Brent

One more approach:

If Set T COULD look like this: { -99, -97 ..... -3, -1, 1, 3......97, 99 }, in which case the median = zero

We can see that answer choices A, B, D and E could equal zero, so we can ELIMINATE them.

Here's what I mean:
A) k² - k - 6 = (k + 2)(k - 3). So, answer choice A could equal zero if k = -2 or k = 3. ELIMINATE A

B) k² + 8k + 15 = (k + 3)(k + 5). So, answer choice B could equal zero if k = -3 or k = -5. ELIMINATE B

D) k³ - 4k² - k. If k = 0, then answer choice D could equal zero . ELIMINATE D

E) 3k³ - 27k². If k = 0, then answer choice E could equal zero. ELIMINATE E


What about answer choice C??
C) 4k² + 4k + 1 = (2k + 1)(2k + 1), so answer choice C could equal zero if k = -1/2. HOWEVER, we're told that k is an integer.
So, 4k² + 4k + 1 CANNOT equal zero.

In fact, if k is an integer, we can see that 4k² + 4k + 1 must be ODD, and as regor60 noted, the median of set T must be EVEN.

Answer: C

Cheers,
Brent