Question 36·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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A car's fuel efficiency is 32 miles per gallon. Using the conversions $1$ mile $= 1.61$ kilometers and $1$ gallon $= 3.79$ liters, which choice is closest to the car's fuel consumption in liters per 100 kilometers?

Fuel efficiency is a rate in distance per fuel, but fuel consumption asks for fuel per distance. Reverse the rate before converting its units. Then use dimensional analysis: arrange the conversion fractions so the old units cancel, and scale the result to the requested distance. Multiplying by the kilometers-per-mile factor in the wrong direction gives a rate with the wrong units.

Hints

Hint 1
Fuel efficiency counts distance per unit of fuel. Consumption counts fuel per distance, so take the reciprocal, meaning flip the top and bottom of the given rate. How many gallons does the car use for one mile?
Hint 2
A conversion factor is a fraction of two equal amounts. Put gallons on the bottom of the liters factor and miles on top of the kilometers factor, so both old units cancel.
Hint 3
After converting, the rate describes liters for one kilometer. The question asks about a longer distance. Multiply the rate by that distance in kilometers, then check that only liters remain.

Step-by-step

Reverse the rate, then convert the units
Step 1
Find the fuel used per mile
The given $32$ miles per gallon measures distance per fuel, but the question asks for fuel per distance. Take the reciprocal, meaning flip the top and bottom, to get gallons per mile: $\frac{1\text{ gallon}}{32\text{ miles}}$ .
Step 2
Convert gallons per mile to liters per kilometer
A conversion factor compares equal amounts, so it changes the units without changing the amount of fuel used. Put gallons below liters and miles above kilometers so both old units cancel: $\frac{1\text{ gallon}}{32\text{ miles}}\left(\frac{3.79\text{ liters}}{1\text{ gallon}}\right)\left(\frac{1\text{ mile}}{1.61\text{ kilometers}}\right)=\frac{3.79}{32(1.61)}\frac{\text{liters}}{\text{kilometer}}$ .
Step 3
Scale the rate to the requested distance
The target is 100 kilometers, not 100 miles. Multiply the liters-per-kilometer rate by $100$ kilometers; the kilometers cancel: $\frac{3.79}{32(1.61)}\frac{\text{liters}}{\text{kilometer}}(100\text{ kilometers})=\frac{100(3.79)}{32(1.61)}\text{ liters in 100 kilometers}$ .
Step 4
Calculate and choose the closest value
Type $\frac{100(3.79)}{32(1.61)}$ into Desmos. It shows about $7.36$ , so the car uses about $7.36$ liters in $100$ kilometers. The closest choice is $7.4$ . Choice C.

Question 36·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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