hkm_gmat wrote:

JAI HIND wrote:

21 - 1 alone.

Area of a parallellogram is base x height

So area cannot be greater than 42 as height cannot be greater than either 6 or 7.

Folks I just dont understand how can the area of the llgm be 21.

Please explain.

hkm_gmat, at first I was also confused.

I think that the question asks what are the possible values "among the floowing".

Assuming 7 as the base, if we calculate the largest possible area, it would come out to 7*6 = 42.

Infact the area of the parallelogram would be such that:

0 < Area of ll of gram <= 42

So (III) 84 is out. The other two values 21 & 42 are possible.

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