You have 100 pounds of potatoes, which are 99 percent water by weight. They dehydrate until they’re 98 percent water. How much do they weigh now?
ive thought about this so many ways and it doesnt seem to work...
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Weight of potatoes at 99% - Weight of potatoes at 98% = x
0.99(100) - 0.98(100 - x) = x
99 - 98 + 0.98x = x
1 = x - 0.98x
1 = 0.02x
50 = x
Therefore the total weight lost is 50lbs. To determine the new eight subtract the lost weight from the total weight:
100-50 = 50%
Therefore, the total final weight of the dehydrated potatoes is 50lbs
By the given context there is 99 pound of water content and other substance of 1 pound in the poatato. Only water dehydrates. The mass of the other substance remains unchachanged in dehydration.
Let the potatos weigh x pounds after dehydration. So the water content in the dehydrated potatos = x-1 pound.
So the percentage of water content in the dehydrated potato = ((x-1)/x)100 % which is equal ro 98% by hypothesis. So,
((x-1/x)100 = 98. Multiply by x.
100x-100 = 98x
100x-98x = 100
2x = 100
x = 50.
So the potatos weigh 50 pounds when their water content is 49 pounds after dehydration of water to 98% of the substance , i.e potatos.
d = Weight of completely dehydrated potatoes
n = Percentage of water in potatoes
p(n) = Weight of potatoes having n percent water.
Then for a given value of n and p(n) the value of d is given by formula:
d = p(n)*(100 - n)/100
And for a given value of d and n weight of potatoes is given by the formula:
Weight of potatoes = 100*d/(100 - n)
Substituting, in above equation, given value of p(n) and n for potatoes with 99 percent water:
d = 100*(100 - 99)/100 = 1 pound
Substituting this value of d and percentage of water as 98, in the equation for weight of potatoes we get:
Weight of potatoes wit 98 percent water = 100*1*/(100 - 98)
= 50 pounds
Please note that in this 50 pounds of potatoes with 98 percent water, the weight of dried potato remains 1 pound, and the weight of water becomes 49 pounds. This amounts to 2 percent dried potatoes and 98 percent water.
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