Examkin

Question

Difficulty: HardPercent Change and Interest

A commercial real estate developer acquired a property for $P\$P. Over the first year, the market value of the property increased by 25%25\%. During the second year, the property's value decreased by x%x\%. In the third year, the value increased again by 20%20\% relative to its value at the end of the second year. If the final value of the property at the end of the third year was 14%14\% greater than the original acquisition price $P\$P, what is the value of xx?

  1. A
    2121
  2. 2424Answer
  3. C
    3030
  4. D
    3131
  5. E
    7676

Answer

The value of xx is 24.
To find the net effect of successive percentage changes, express each period's change as a multiplier of the value from the preceding period. A 25%25\% increase corresponds to a multiplier of 1.251.25, an x%x\% decrease corresponds to a multiplier of (1−x100)\left(1 - \frac{x}{100}\right), and a 20%20\% increase corresponds to a multiplier of 1.201.20. Combined, these produce a final multiplier of 1.25×1.20×(1−x100)=1.50(1−x100)1.25 \times 1.20 \times \left(1 - \frac{x}{100}\right) = 1.50\left(1 - \frac{x}{100}\right). Setting this equal to the net 14%14\% total increase (1.141.14) yields 1.50(1−x100)=1.141.50\left(1 - \frac{x}{100}\right) = 1.14, which simplifies to 1−x100=0.761 - \frac{x}{100} = 0.76, giving x=24x = 24.

Step-by-Step Solution

1
Express each sequential period's value using multiplier notation.
At Year 1 end: V1=1.25PV_1 = 1.25P. At Year 2 end: V2=1.25P(1−x100)V_2 = 1.25P \left(1 - \frac{x}{100}\right). At Year 3 end: V3=1.25P(1−x100)×1.20V_3 = 1.25P \left(1 - \frac{x}{100}\right) \times 1.20.
Successive percent changes compound on the intermediate values of each period, not on the original base price.
2
Equate the overall combined growth multiplier to the given total net change.
1.25×1.20×(1−x100)=1.14  ⟹  1.50×(1−x100)=1.141.25 \times 1.20 \times \left(1 - \frac{x}{100}\right) = 1.14 \implies 1.50 \times \left(1 - \frac{x}{100}\right) = 1.14
The final value is given as 14%14\% greater than PP, which corresponds to a net multiplier of 1.141.14.
3
Solve the algebraic equation for xx.
1−x100=1.141.50=0.76  ⟹  x100=0.24  ⟹  x=241 - \frac{x}{100} = \frac{1.14}{1.50} = 0.76 \implies \frac{x}{100} = 0.24 \implies x = 24
Subtracting 0.760.76 from 11 yields the fractional decrease of 0.240.24, which equals 24%24\%.

Key Concept

Successive Percent Change and Multipliers
Estimated Time:2m 0s
Rate this question

Topics

Percent Change and Interest

More questions from this topic