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Difficulty: MediumPercents and Percent Change

In a laboratory experiment, a wet soil sample has a mass of 800800 grams. During a drying process, the sample loses 20%20\% of its mass. After the drying process, a nutrient solution is added, which increases the sample's mass by 15%15\% of its dried mass. What is the final mass, in grams, of the soil sample after the solution is added?

  1. A
    544
  2. B
    655
  3. 736Answer
  4. D
    760

Answer

736
To find the final mass of the soil sample, first calculate the mass after it loses 20%20\% of its initial 800800 grams: 800×(1−0.20)=640800 \times (1 - 0.20) = 640 grams. Next, calculate the final mass after a 15%15\% increase is applied to this intermediate mass: 640×(1+0.15)=736640 \times (1 + 0.15) = 736 grams. This shows that the final mass of the sample is 736736 grams.

Step-by-Step Solution

1
Calculate the mass of the soil sample after losing 20%20\% of its initial mass of 800800 grams.
Dried mass = 800×(1−0.20)=640800 \times (1 - 0.20) = 640 grams.
This establishes the intermediate baseline mass before the nutrient solution is added.
2
Calculate the final mass of the soil sample after a 15%15\% increase is applied to the dried mass of 640640 grams.
Final mass = 640×(1+0.15)=736640 \times (1 + 0.15) = 736 grams.
The 15%15\% increase must be calculated using the dried mass of 640640 grams as the new base.

Key Concept

Multi-step percent change and identifying the correct base value for sequential percentage increases or decreases.

Alternative Method

You can express the sequential percent changes as a single product: 800×0.80×1.15800 \times 0.80 \times 1.15. Multiplying these values directly yields 800×0.92=736800 \times 0.92 = 736 grams.
Estimated Time:1m 15s
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Topics

Percents and Percent Change

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