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Bunuel
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Bunuel
Grass on a lawn grows everyday by a fixed percent of the height on the previous day. However, cows graze and reduce the height of the grass by 1 mm every day. What was the height of the grass on the first day before the cows started grazing? Assume that the cows graze at the end of the day, after which the grass grows in the beginning of the next day.

(1) The grass grows every day by 2 percent of the height on the previous day.
(2) The grass is completely grazed in 3 days.

Assume growth everyday is x% and initial height is H

(1) X = 2. But this statement does not say anything about the height on any particular day. NOT SUFFICIENT

(2) H(1+x)^3 - 3(1) =0. NOT SUFFICIENT as no information on H and x provided

(1) + (2) . x = 2. Hence, H(1+2%)^3 - 3(1) =0. Solvable and Sufficient

Hence C
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Bunuel
Grass on a lawn grows everyday by a fixed percent of the height on the previous day. However, cows graze and reduce the height of the grass by 1 mm every day. What was the height of the grass on the first day before the cows started grazing? Assume that the cows graze at the end of the day, after which the grass grows in the beginning of the next day.

(1) The grass grows every day by 2 percent of the height on the previous day.
(2) The grass is completely grazed in 3 days.

Let initial height be h
Acc to 1===> then at the end of third day after grazing the height would be H(final) = [[[1.2h -1]*1.2 - 1]*1.2 - 1], but we do not know the final height H(final) so we cannot determine h, insufficient
Acc to 2====> We do not know how much does the percent the height increases daily.

Combined we know H(final) =0 ==> [[[1.2h -1]*1.2 - 1]*1.2 - 1] =0, and we can determine h.....sufficient
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