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Difficulty: Very hardPercentages, Percent Change, and Interest

A technology firm monitors its active user base and average monthly server cost per user across four consecutive quarters:

- In Quarter 1, the total server cost was CC.
- In Quarter 2, the active user base increased by 25%25\% relative to Quarter 1, while the average monthly server cost per user decreased by 20%20\% relative to Quarter 1.
- In Quarter 3, the active user base decreased by x%x\% relative to Quarter 2, while the average monthly server cost per user increased by x%x\% relative to Quarter 2, where x>0x > 0.
- In Quarter 4, the active user base increased by 20%20\% relative to Quarter 3, while the average monthly server cost per user decreased by 10%10\% relative to Quarter 3.

If the total server cost in Quarter 4 was 3.68%3.68\% greater than the total server cost in Quarter 1, what is the value of xx?

  1. A
    44
  2. B
    1515
  3. 2020Answer
  4. D
    2525
  5. E
    4040

Answer

The value of xx is 2020.
The correct response is 20. Total cost in Quarter 1 is C1=N1P1=CC_1 = N_1 P_1 = C. In Quarter 2, C2=(1.25N1)(0.80P1)=1.00CC_2 = (1.25 N_1)(0.80 P_1) = 1.00 C. In Quarter 3, the user base and cost per user change by −x%-x\% and +x%+x\%, giving C3=C(1−x100)(1+x100)=C(1−x210,000)C_3 = C \left(1 - \frac{x}{100}\right)\left(1 + \frac{x}{100}\right) = C \left(1 - \frac{x^2}{10,000}\right). In Quarter 4, the overall factor becomes 1.20×0.90=1.081.20 \times 0.90 = 1.08, so C4=1.08C(1−x210,000)C_4 = 1.08 C \left(1 - \frac{x^2}{10,000}\right). Setting 1.08(1−x210,000)=1.03681.08 \left(1 - \frac{x^2}{10,000}\right) = 1.0368 yields 1−x210,000=0.961 - \frac{x^2}{10,000} = 0.96, from which x210,000=0.04\frac{x^2}{10,000} = 0.04, giving x2=400x^2 = 400 and x=20x = 20.

Step-by-Step Solution

1
Define total cost variables for Quarter 1 and Quarter 2.
Quarter 1 cost C1=N1P1=CC_1 = N_1 P_1 = C. Quarter 2 user base N2=1.25N1N_2 = 1.25 N_1 and cost per user P2=0.80P1P_2 = 0.80 P_1, so C2=(1.25N1)(0.80P1)=1.00N1P1=CC_2 = (1.25 N_1)(0.80 P_1) = 1.00 N_1 P_1 = C.
Total cost is the product of user count and cost per user; successive changes in Quarter 2 offset each other completely.
2
Calculate total cost in Quarter 3 in terms of xx.
Quarter 3 user base N3=N2(1−x100)N_3 = N_2\left(1 - \frac{x}{100}\right) and cost per user P3=P2(1+x100)P_3 = P_2\left(1 + \frac{x}{100}\right). Total cost C3=N3P3=C2(1−x100)(1+x100)=C(1−x210,000)C_3 = N_3 P_3 = C_2 \left(1 - \frac{x}{100}\right)\left(1 + \frac{x}{100}\right) = C \left(1 - \frac{x^2}{10,000}\right).
Applying the difference of squares identity (1−u)(1+u)=1−u2(1 - u)(1 + u) = 1 - u^2 for compound percentage change.
3
Calculate total cost in Quarter 4 and set up the equation with Quarter 1 cost.
Quarter 4 user base N4=1.20N3N_4 = 1.20 N_3 and cost per user P4=0.90P3P_4 = 0.90 P_3, giving C4=(1.20)(0.90)C3=1.08C3=1.08C(1−x210,000)C_4 = (1.20)(0.90) C_3 = 1.08 C_3 = 1.08 C \left(1 - \frac{x^2}{10,000}\right). Since C4=1.0368CC_4 = 1.0368 C, we get 1.08(1−x210,000)=1.03681.08 \left(1 - \frac{x^2}{10,000}\right) = 1.0368.
Quarter 4 expenditure compounds the Quarter 3 cost by a factor of 1.20×0.90=1.081.20 \times 0.90 = 1.08.
4
Solve for xx.
1−x210,000=1.03681.08=0.96  ⟹  x210,000=0.04  ⟹  x2=400  ⟹  x=201 - \frac{x^2}{10,000} = \frac{1.0368}{1.08} = 0.96 \implies \frac{x^2}{10,000} = 0.04 \implies x^2 = 400 \implies x = 20.
Dividing 1.03681.0368 by 1.081.08 yields 0.960.96, giving x2=400x^2 = 400 and x=20x = 20 since x>0x > 0.

Key Concept

Compounding successive percent changes across multiple factors and using algebraic identities for net percent change.
Estimated Time:2m 30s
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Topics

Percentages, Percent Change, and Interest

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