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Difficulty: Very hardPercents and Percent Change

A researcher is drying a sample of wet soil in a laboratory. Initially, water makes up 40%40\% of the total weight of the sample. During the first stage of drying, 25%25\% of the water in the sample evaporates. During the second stage of drying, an additional quantity of water evaporates such that the water in the sample now makes up 20%20\% of the sample's total weight. What percent of the water that remained after the first stage evaporated during the second stage?

  1. A
    20%20\%
  2. B
    37.5%37.5\%
  3. 50%50\%Answer
  4. D
    33.3%33.3\%

Answer

50%50\%
The correct answer is the option stating 50%50\%. After the first stage of drying, the sample contains 3030 grams of water and has a total weight of 9090 grams. In the second stage, 1515 grams of water evaporates, leaving 1515 grams of water and a total weight of 7575 grams. Since 1515 is exactly 20%20\% of 7575, the final condition is met. The 1515 grams of water that evaporated in the second stage represents exactly 50%50\% of the 3030 grams of water that remained after the first stage.

Step-by-Step Solution

1
Define the initial weights using a convenient total weight.
Let the initial total weight of the wet soil sample be 100100 grams. Since water makes up 40%40\% of this weight, the initial water weight is 4040 grams and the dry soil weight is 6060 grams.
Choosing a concrete value of 100100 grams simplifies the percent calculations.
2
Calculate the water remaining after the first stage of drying.
In the first stage, 25%25\% of the water evaporates. The evaporated water is 0.25×40=100.25 \times 40 = 10 grams. The remaining water weight is 40−10=3040 - 10 = 30 grams. The total weight of the sample is now 60 (dry soil)+30 (water)=9060 \text{ (dry soil)} + 30 \text{ (water)} = 90 grams.
We must establish the new baseline amount of water and total sample weight before the second drying stage.
3
Set up an equation for the second stage of drying.
Let xx be the weight of water in grams that evaporates during the second stage. The final water weight is 30−x30 - x grams, and the final total weight of the sample is 90−x90 - x grams. Since the final water weight is 20%20\% of the final total weight, we write: 30−x90−x=0.20\frac{30 - x}{90 - x} = 0.20.
This relates the unknown quantity of water evaporated to the given final water percentage.
4
Solve the equation for the evaporated water weight xx.
Multiplying both sides by 90−x90 - x gives 30−x=0.20(90−x)⇒30−x=18−0.20x30 - x = 0.20(90 - x) \Rightarrow 30 - x = 18 - 0.20x. Rearranging terms yields 12=0.80x12 = 0.80x, which gives x=15x = 15 grams.
Solving the algebraic equation finds the exact amount of water that evaporated during the second stage.
5
Calculate the percent of the remaining water that evaporated.
Divide the water evaporated in the second stage (1515 grams) by the water remaining after the first stage (3030 grams): 1530×100%=50%\frac{15}{30} \times 100\% = 50\%.
This answers the specific question asked using the correct base weight.

Key Concept

Multi-step percent change with a shifting base

Alternative Method

Instead of assuming a weight of 100100 grams, the problem can be solved using variables. Let the initial total weight of the wet soil sample be WW. The initial water weight is 0.40W0.40W and the dry soil is 0.60W0.60W. After the first stage, remaining water is 0.75×0.40W=0.30W0.75 \times 0.40W = 0.30W, and total weight is 0.90W0.90W. Let EE be the water evaporated in the second stage. The final water content is 0.30W−E0.30W - E, and the final total weight is 0.90W−E0.90W - E. The equation is 0.30W−E=0.20(0.90W−E)0.30W - E = 0.20(0.90W - E), which solves to E=0.15WE = 0.15W. The ratio of evaporated water to remaining water is 0.15W0.30W=0.50\frac{0.15W}{0.30W} = 0.50, or 50%50\%.
Estimated Time:3m 0s
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Topics

Percents and Percent Change

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