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Difficulty: MediumRatios and Proportions

A tactical transport vehicle's fuel efficiency is evaluated under two conditions. On paved roads, the ratio of miles driven to gallons of fuel consumed is 15:115:1. In off-road conditions, the vehicle's fuel efficiency decreases by 20%20\%. How many gallons of fuel are required to drive 60 miles60\text{ miles} in off-road conditions?

  1. A
    4.84.8
  2. B
    3.23.2
  3. 55Answer
  4. D
    0.20.2

Answer

The correct amount of fuel required is 55 gallons.
The original fuel efficiency is 15 miles per gallon. A 20% decrease reduces this rate by 3 mpg (0.20 × 15), resulting in a new off-road efficiency of 12 mpg. To find the required fuel for a 60-mile trip, divide the total distance by the new unit rate: 60 / 12 = 5 gallons.

Step-by-Step Solution

1
Identify the original fuel efficiency from the given ratio.
The ratio 15:115:1 means the vehicle gets 15 miles per gallon15\text{ miles per gallon} (mpg) on paved roads.
Converting the ratio into a unit rate makes it easier to apply the percentage change.
2
Calculate the off-road fuel efficiency by applying the 20%20\% decrease.
15−(0.20×15)=15−3=12 mpg15 - (0.20 \times 15) = 15 - 3 = 12\text{ mpg}.
The problem states the efficiency decreases by 20%20\% in off-road conditions.
3
Set up a proportion to find the gallons required for the 60-mile60\text{-mile} trip.
12 miles1 gallon=60 milesx gallons\frac{12\text{ miles}}{1\text{ gallon}} = \frac{60\text{ miles}}{x\text{ gallons}}
Fuel consumption is a direct proportion between miles driven and gallons used at a constant rate.
4
Solve for xx.
12x=60  ⟹  x=5 gallons12x = 60 \implies x = 5\text{ gallons}.
Isolating the variable gives the final required amount of fuel.

Key Concept

Applying percentage changes to unit rates and setting up proportions to solve for an unknown quantity.

Alternative Method

Calculate the fuel needed for paved roads first (60/15=4 gallons60 / 15 = 4\text{ gallons}). Recognize that since the new efficiency is 80%80\% (or 45\frac{4}{5}) of the original, the fuel needed will be the inverse ratio (54\frac{5}{4}) of the original fuel amount. Multiply 4 gallons×54=5 gallons4\text{ gallons} \times \frac{5}{4} = 5\text{ gallons}.
Estimated Time:1m 0s
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Topics

Ratios and Proportions

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