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.
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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