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24 Jun 2017, 08:24
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
92% (00:53) correct
8%
(01:22)
wrong
based on 116
sessions
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If x is the median of three consecutive integers, what is the product of these three integers in terms of x?
A. \(x^3 + 6x^2 + 8x\)
B. \(x^3 – 3x^2 + 2x\)
C. \(x^3 – 3x^2 – 2x\)
D. \(x^3 – x^2\)
E. \(x^3 – x\)
A. \(x^3 + 6x^2 + 8x\)
B. \(x^3 – 3x^2 + 2x\)
C. \(x^3 – 3x^2 – 2x\)
D. \(x^3 – x^2\)
E. \(x^3 – x\)
sashiim20
Joined: 04 Dec 2015
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Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
Posts: 620
24 Jun 2017, 08:31
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Bunuel
If x is the median of three consecutive integers, what is the product of these three integers in terms of x?
A. \(x^3 + 6x^2 + 8x\)
B. \(x^3 – 3x^2 + 2x\)
C. \(x^3 – 3x^2 – 2x\)
D. \(x^3 – x^2\)
E. \(x^3 – x\)
A. \(x^3 + 6x^2 + 8x\)
B. \(x^3 – 3x^2 + 2x\)
C. \(x^3 – 3x^2 – 2x\)
D. \(x^3 – x^2\)
E. \(x^3 – x\)
Given x is the median of 3 consecutive integers. Therefore the 3 integers would be \((x-1), x , (x + 1)\)
Product of these 3 consecutive integers;
\((x-1) *x * (x + 1)\)
\((x - 1)(x + 1) x\)
\((x^2-1)x\)
\(x^3 - x\) . Answer (E) ...
24 Jun 2017, 08:32
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Bunuel
If x is the median of three consecutive integers, what is the product of these three integers in terms of x?
A. \(x^3 + 6x^2 + 8x\)
B. \(x^3 – 3x^2 + 2x\)
C. \(x^3 – 3x^2 – 2x\)
D. \(x^3 – x^2\)
E. \(x^3 – x\)
A. \(x^3 + 6x^2 + 8x\)
B. \(x^3 – 3x^2 + 2x\)
C. \(x^3 – 3x^2 – 2x\)
D. \(x^3 – x^2\)
E. \(x^3 – x\)
the three consecutive numbers are (x-1), x & (x+1)
multiplying them = (x-1)*x*(x+1) = \(x^3 – x\)
Hence E
Another way could be to work through options
Assuming E to be the correct choice -
x^3-x = x(x^2-1) = x(x-1)(x+1).......which is the product of three consecutive integers with "x" being the median
