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. If k is an i
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Since there are an even number(100) of odd numbers, the median will be the average of the two "middle numbers", which means that it will be an even number.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
Examining C., 4k^2+4K+1 reduces to 4K(K+1) +1. Since anything multiplied by 4 is even, adding 1 to this will result in an odd number, C
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Set T consists of 100 consecutive odd integers, so the median value will be the average of the 50th and 51st integers.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
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.
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One more approach: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
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