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medians

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Source: — Data Sufficiency |

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by Anju@Gurome » Thu Apr 25, 2013 3:37 am
sivanhas wrote:D is not equal to 0. Is D > 0?

i. D is the median of D, 1/D, -D
ii. D^3 is the median of D, D^2, D^3
Whenever comparison of powers of any variable is involved, always try to follow the steps as applicable... (move to the next step only if you can't found a contradiction in previous step)
  • #1. Plug 1 and -1 and check.
    #2. If the variable is positive, plug values greater than 1 as well as less than 1.
    #3. If the variable is negative, plug values less than -1 as well as greater than -1.
In this case, let us plug D = 1 and D = -1,
  • D = 1 ---> median of 1, 1/1, and -1 is 1 ---> median of 1, 1², and 1³ is 1³
    D = -1 ---> median of -1, -1/1, and -(-1) is -1 ---> median of -1, (-1)², and (-1)³ is (-1³)
So, both the cases satisfy all the given conditions but in the first case D > 0, but in the second D < 0.

The correct answer is E.
Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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by srcc25anu » Thu Apr 25, 2013 5:48 am
Ans E
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