Lesson Problems on percentage that lead to unexpected results

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Problems on percentage that lead to unexpected results

The percentage problems are usually considered as boring and tedious.
In this lesson some percentage problems are presented what lead to unexpected and even shocking results.

Problem 1

Water makes up 96% of the mass of an apricot. When apricots were left out in the sun, some water evaporated,
and water then makes up 95% of the mass of the sun dried apricots.
What percentage of its original mass do the apricots lose when the water evaporates?

Solution

Let us assume that the original mass of the apricotss was 100 kg.

It means that the content of water was 96 kilograms originally. Respectively, the mass of the solid part (of "the fiber") was 4 kilograms.

After drying, when the part of the water went out, we have the same mass of the fiber of 4 kg, and some new mass of remaining water,
let say w kilograms. According to the condition, the equation for the remaining mass of the water after drying w is

w%2F%284%2Bw%29

= 0.95.

Let us solve it. For it, multiply both sides by (4+w). You will get

w = 4*0.95 + 0.95w, or

w*(1-0.95) = 4*0.95, or

0.05*w = 4*0.95.

Hence, w = %284%2A0.95%29%2F0.05

= 4*19 = 76 kg.

OK, very good. Then new mass of the apricots after drying is 76 + 4 = 80 kilograms.
Thus the fruits do lose 20% of theirs original mass. Indeed, %28%28100-80%29%2F100%29%2A100

= 20%.

It is an unexpected result, isn't it? But this is a well known fact.

Problem 2

A raisin is a dried grape. Originally, grapes contain 98% of water by mass.
100 kilograms of grapes were dried and water content was decreased till 96%. Find the mass of the raisins obtained.

Solution

According to the condition, 100 kilograms of grapes contain 98% of water by mass, i.e. 98 kilograms of water.
Respectively, the mass of the solid part (of the fiber plus cristallised sugar plus protein) is 2 kilograms.

After drying, when the part of the water went out, we have the same mass of the fiber of 2 kg, and some new mass of remaining water,
let say w kilograms. According to the condition, the equation for the remaining mass of the water after drying w is

w%2F%282%2Bw%29

= 0.96.

Let us solve it. For it, multiply both sides by (2+w). You will get

w = 2*0.96 + 0.96w, or

w*(1-0.96) = 2*0.96, or

0.04*w = 2*0.96.

Hence, w = %282%2A0.96%29%2F0.04

= 2*24 = 48 kg.

Thus the obtained mass of raisins after drying is 48 + 2 = 50 kilograms.
Thus the fruit does lose 50% of its original mass after drying.

My other lessons on percentage problems in this site are
    - Percentage problems
    - Percentage word problems (Type 1 problems, Finding the Part)
    - Percentage word problems (Type 2 problems, Finding the Rate)
    - Percentage word problems (Type 3 problems, Finding the Base)
    - Simple percentage problems
    - More complicated percentage problems
    - Advanced problems on percentage
    - Percentage problems on chains of discounts
    - Simple interest percentage problems
    - Compounded interest percentage problems
    - Buying price, selling price and profit percentage problems
    - Business-related entertainment problem on percentage
    - OVERVIEW of lessons on percentage problems

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