The weight of dried cucumbers

Algebra Level 2

Fresh cucumbers contain 99% of water. 100lb of cucumbers are dried so that they contain 98% of water. What's their weight now?

99 90 98 50

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2 solutions

Zico Quintina
Jun 7, 2018
  • Before drying, the non-water components of the cucumbers were 1% of 100 lbs, i.e. the non-water components weighed 1 lb.
  • After drying, the non-water components still weighed 1 lb, but were now 2% of the total weight. Thus the dried cucumbers must weigh 50 lbs.

We are given the ratio of mass of water to total mass of cucumber. The total mass of any amount of cucumber can be expressed as the mass of water plus the mass of everything else present in the cucumber
Let W F W_{F} represent mass of water present in fresh cucumber, W D W_{D} represent mass of water present in dried cucumber, and S S represent mass of non-water components of cucumber, which can ideally be assumed constant through the drying process
We are given W F + S = 100 W_{F} + S = 100 , W F W F + S = 0.99 \frac{W_{F} }{W_{F} + S} = 0.99 and W D W D + S = 0.98 \frac{ W_{D} }{ W_{D} +S } =0.98 . with this information we are asked to find W D + S W_{D} + S , which is the mass of dried cucumber.
subbing W F + S = 100 W_{F} + S = 100 into W F W F + S = 0.99 \frac{W_{F} }{W_{F} + S} = 0.99 and solving gives W F = 99 l b s W_{F} = 99lbs . Subbing that into W F + S = 100 W_{F} + S = 100 gives S = 1 l b S = 1lb
Now solve for W D W_{D} in W D W D + S = 0.98 \frac{ W_{D} }{ W_{D} +S } =0.98 , by multiplying both sides of the equality by W D + S W_{D} +S and subbing S = 1 l b S = 1lb . (also let 0.98 = 49 50 0.98=\frac{49}{50} for convenience)
W D = 49 50 ( W D + 1 W_{D} = \frac{49}{50}(W_{D} + 1 ), giving 1 50 W D = 49 50 \frac{1}{50} W_{D} = \frac{49}{50} and finally W D = 49 l b s W_{D} = 49lbs .
Now, mass of dried cucumber = W D + S = ( 49 l b s ) + ( 1 l b ) = 50lbs \text{mass of dried cucumber} = W_{D} + S = (49lbs) + (1lb) = \textbf{50lbs}


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