Question 1016911
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Fresh cranberries have 99% of water. Partially dried cranberries have 98% of water. Jack picked 100 lbs of cranberries. 
How much will they {{{highlight(cross(weigh))}}} weight after drying?
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100 lbs of cranberries contain 99% of water, i.e. 99 lbs of water, according to the first part of the condition. 
And only 1 lbs of dry fiber.

At the drying process, part of water evaporates and goes out.
The other part of water still remains in the partly dried cranberries. 
Together with 1 lbs of the dry fiber, of course.

Let w be the weight of the water remained in the 98% partly dried cranberries.

Then you have this proportion

{{{w/(1+w)}}} = 0.98

to find the amount of water in the 98% partly dried cranberries.

Let us solve it. For it, multiply both sides of the equation by (1+w) to get off the denominator. You will get

w = 0.98*(1 + w)   --->   w = 0.98 + 0.98w  --->  w - 0.98w =0.98  --->  0.02w = 0.98  --->  w = {{{0.98/0.02}}} = 49.

So, after drying, the water content is 49 lbs in the cranberries that contain 98% water.

Add 1 lbs of the dry fiber, and you will have the total weight of 50 lbs partly dried cranberries.


Unexpected result ?


O, yes. Nevertheless, it is true.

And it is very well known fact.

As well as very old well known problem, which produces such unexpected result.

See the lesson 
<A HREF=Problems on percentage that lead to unexpected results>http://www.algebra.com/algebra/homework/Percentage-and-ratio-word-problems/Problems-on-percentage-that-lead-to-unexpected-results.lesson</A> 
in this site, where you will find other similar problems.
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