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Difficulty: Very hardPercentages, Percent Change, and Interest

An agricultural research station monitored the water volume in a reservoir over a three-month period. In the first month, the reservoir's water volume decreased by x%x\%. In the second month, the remaining water volume decreased by (x+10)%(x + 10)\%. In the third month, heavy rainfall increased the water volume remaining at the end of the second month by 50%50\%. If the final water volume at the end of the third month was exactly 84%84\% of the initial water volume prior to the first month, what is the value of xx?

  1. A
    1010
  2. B
    1515
  3. 2020Answer
  4. D
    2525
  5. E
    3030

Answer

The value of xx is 2020.
The correct answer is 2020. Applying successive percent change multipliers to the initial volume V0V_0 gives V3=1.50×(1−x+10100)×(1−x100)V0V_3 = 1.50 \times (1 - \frac{x+10}{100}) \times (1 - \frac{x}{100}) V_0. Setting this equal to 0.84V00.84 V_0 yields the quadratic equation k2−1.90k+0.34=0k^2 - 1.90k + 0.34 = 0 where k=x100k = \frac{x}{100}. Solving gives k=0.20k = 0.20, which means x=20x = 20.

Step-by-Step Solution

1
Express the successive volume changes as percentage multipliers relative to the initial volume V0V_0.
After month 1: V1=V0(1−x100)V_1 = V_0 \left(1 - \frac{x}{100}\right). After month 2: V2=V1(1−x+10100)=V0(1−x100)(0.90−x100)V_2 = V_1 \left(1 - \frac{x+10}{100}\right) = V_0 \left(1 - \frac{x}{100}\right)\left(0.90 - \frac{x}{100}\right). After month 3: V3=V2×1.50=1.50V0(1−x100)(0.90−x100)V_3 = V_2 \times 1.50 = 1.50 V_0 \left(1 - \frac{x}{100}\right)\left(0.90 - \frac{x}{100}\right).
Successive percentage changes must be applied sequentially to the updated volume at each stage.
2
Set the expression for final volume V3V_3 equal to 0.84V00.84 V_0 and simplify the equation.
1.50(1−x100)(0.90−x100)=0.84  ⟹  (1−x100)(0.90−x100)=0.841.50=0.561.50 \left(1 - \frac{x}{100}\right)\left(0.90 - \frac{x}{100}\right) = 0.84 \implies \left(1 - \frac{x}{100}\right)\left(0.90 - \frac{x}{100}\right) = \frac{0.84}{1.50} = 0.56.
The problem states that the final volume is 84%84\% of the initial volume V0V_0.
3
Substitute k=x100k = \frac{x}{100} and formulate a quadratic equation.
(1−k)(0.90−k)=0.56  ⟹  k2−1.90k+0.90=0.56  ⟹  k2−1.90k+0.34=0(1 - k)(0.90 - k) = 0.56 \implies k^2 - 1.90k + 0.90 = 0.56 \implies k^2 - 1.90k + 0.34 = 0. Multiply by 100100: 100k2−190k+34=0  ⟹  50k2−95k+17=0100k^2 - 190k + 34 = 0 \implies 50k^2 - 95k + 17 = 0.
Converting to a standard quadratic form ak2+bk+c=0ak^2 + bk + c = 0 allows solving for kk.
4
Solve the quadratic equation for kk.
k=95±(−95)2−4(50)(17)2(50)=95±9025−3400100=95±5625100=95±75100k = \frac{95 \pm \sqrt{(-95)^2 - 4(50)(17)}}{2(50)} = \frac{95 \pm \sqrt{9025 - 3400}}{100} = \frac{95 \pm \sqrt{5625}}{100} = \frac{95 \pm 75}{100}. Thus k=0.20k = 0.20 or k=1.70k = 1.70.
The quadratic formula provides the roots for kk.
5
Select the valid physical root and determine xx.
Since a percentage decrease in water volume cannot exceed 100%100\% (k≤1k \le 1), we discard k=1.70k = 1.70. Therefore k=0.20k = 0.20, which corresponds to x=20x = 20.
A initial decrease of 170%170\% is physically impossible in this context.

Key Concept

Multi-step percentage change and successive base updates leading to non-linear equations
Estimated Time:3m 0s
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Topics

Percentages, Percent Change, and Interest

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