Asked by Anonymous
A store owner had 505 kg of fresh honeydews. The honeydews contained 60% water when fresh. One week later, there was only 50% water left in the honeydews. What was the mass of the honeydews one week later?
Answers
Answered by
oobleck
This depends on how you are measuring the water. This is one model:
0.60*505 = 303 kg water to start
If x kg of water was lost, then
(303-x)/(505-x) = 0.50
x = 101
so at the end of one week, the 404kg of melons contained 202 kg of water, and thus 1/3 of the initial water was lost.
If 1/3 of that was lost in the 2nd week, then the remaining water would be
303 * (2/3)^2 = 134.67 kg
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there are, naturally, other models that could be constructed. The best model would have to be determined by actual observation of honeydew populations.
0.60*505 = 303 kg water to start
If x kg of water was lost, then
(303-x)/(505-x) = 0.50
x = 101
so at the end of one week, the 404kg of melons contained 202 kg of water, and thus 1/3 of the initial water was lost.
If 1/3 of that was lost in the 2nd week, then the remaining water would be
303 * (2/3)^2 = 134.67 kg
-----------------------------
there are, naturally, other models that could be constructed. The best model would have to be determined by actual observation of honeydew populations.
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