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Difficulty: HardPercentages, Percent Change, and Interest

A commercial trucking fleet monitored its vehicle fuel efficiency (in miles per gallon) and average fuel price (in dollars per gallon) over a two-year period. In Year 1, vehicle fuel efficiency increased by 25%25\%, while the price of fuel increased by 15%15\%. In Year 2, vehicle fuel efficiency increased by an additional 20%20\% relative to Year 1, while the price of fuel decreased by 10%10\% relative to Year 1. Assuming the total annual miles driven per vehicle remained constant throughout the two-year period, by what percent did the total annual fuel cost per vehicle change from the beginning of Year 1 to the end of Year 2?

  1. 31%31\% decreaseAnswer
  2. B
    69%69\% decrease
  3. C
    40%40\% decrease
  4. D
    28%28\% decrease
  5. E
    45%45\% increase

Answer

A 31%31\% decrease
The total fuel cost is given by Cost=Distance×PriceEfficiency\text{Cost} = \text{Distance} \times \frac{\text{Price}}{\text{Efficiency}}. After Year 1, efficiency becomes 1.25E01.25 E_0 and price becomes 1.15P01.15 P_0. After Year 2, efficiency increases to 1.25E0×1.20=1.50E01.25 E_0 \times 1.20 = 1.50 E_0, while price changes to 1.15P0×0.90=1.035P01.15 P_0 \times 0.90 = 1.035 P_0. The new cost ratio is 1.0351.50=0.69\frac{1.035}{1.50} = 0.69, meaning the new cost is 69%69\% of the original cost. This represents a net decrease of 100%−69%=31%100\% - 69\% = 31\%.

Step-by-Step Solution

1
Formulate the fuel cost relationship in terms of distance, efficiency, and price.
Initial cost C0=D×P0E0C_0 = D \times \frac{P_0}{E_0}, where DD is distance driven, P0P_0 is initial price per gallon, and E0E_0 is initial fuel efficiency in miles per gallon.
Total fuel used is DE0\frac{D}{E_0}, so multiplying by price per gallon gives total cost.
2
Calculate the updated fuel efficiency E2E_2 and price P2P_2 at the end of Year 2 using successive percentage changes.
E2=E0×1.25×1.20=1.50E0E_2 = E_0 \times 1.25 \times 1.20 = 1.50 E_0 and P2=P0×1.15×0.90=1.035P0P_2 = P_0 \times 1.15 \times 0.90 = 1.035 P_0.
Compounding changes sequentially: Year 1 efficiency factor is 1.251.25, Year 2 factor is 1.201.20; Year 1 price factor is 1.151.15, Year 2 factor is 0.900.90.
3
Compute the final fuel cost C2C_2 as a fraction of the initial cost C0C_0.
C2=D×1.035P01.50E0=1.0351.50×C0=0.69C0C_2 = D \times \frac{1.035 P_0}{1.50 E_0} = \frac{1.035}{1.50} \times C_0 = 0.69 C_0.
Substituting the expressions for P2P_2 and E2E_2 into the fuel cost formula.
4
Determine the net percentage change from C0C_0 to C2C_2.
\text{Percent Change} = \frac{0.69 C_0 - C_0}{C_0} \times 100\% = -31\%,representinga, representing a 31\%$ decrease.
Subtracting 1.001.00 from 0.690.69 yields −0.31-0.31, which corresponds to a 31%31\% reduction.

Key Concept

Successive Percentage Changes and Rate Relationships
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Topics

Percentages, Percent Change, and Interest

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