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Answer the question based on the following information.
There are blue vessels with known volumes $v_1 , v_2 ..., v_m$, arranged in ascending order of volume, $v_1 > 0.5$ litre, and $v_m < 1$ litre. Each of these is full of water initially. The water from each of these is emptied into  a minimum number of empty white vessels, each having volume $1$ litre. The water from a blue vessel is not  emptied into a white vessel unless the white vessel has enough empty volume to hold all the water of the  blue vessel. The number of white vessels required to empty all the blue vessels according to the above rules was n.
Among the four values given below, which is the least upper bound on e, where e is the total empty volume in the white vessels at the end of the above process?

  1. $mv_m$
  2. $m(1-v_m)$
  3. $mv_1$
  4. $m(1− v_1)$
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B. m(1-vₘ) should be the answer. Since the blue vessels are arranged in increasing order of volumes, these when poured into white vessels would result in decreasing order of empty space. Therefore, if there are m blue vessels, (1-vₘ) would be the smallest empty volume of the lot. Given that the white vessels are of capacity one litre and each of the blue vessels have volumes > 0.5, no white vessel would be able to hold water from more than one blue vessel. Therefore, the smallest upper bound for the total empty space in white vessels would be m(1-vₘ). And I think D. m(1-v₁) would be the greatest upper bound.

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