2022 AMC 10A Problem 4

P1-10~3 minPrint

Ratios, Percents & Averages Sets, Estimation & Miscellaneous Set up the equation/formula and compute; no special trick needed Test small/specific values or special cases to find or verify the answer

In some countries, automobile fuel efficiency is measured in liters per $100$ kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals $m$ miles, and $1$ gallon equals $l$ liters. Which of the following gives the fuel efficiency in liters per $100$ kilometers for a car that gets $x$ miles per gallon?

Show solution without answering

Key insight

Follow one trip of 100 kilometers: change it to miles, then to gallons of fuel used, then to liters.

Solution

The answer must be the number of liters the car burns while driving $100$ kilometers. So follow one such trip, one unit at a time.

1. Kilometers to miles. One kilometer is $m$ miles, so $100$ kilometers is $100m$ miles.
2. Miles to gallons. The car goes $x$ miles on one gallon. To find how many gallons a trip uses, divide the distance by $x$ . (For example, $200$ miles at $50$ miles per gallon uses $200 \div 50 = 4$ gallons.) So this trip uses $\dfrac{100m}{x}$ gallons.
3. Gallons to liters. Each gallon is $l$ liters, so multiply by $l$ :
$\frac{100m}{x} \cdot l = \frac{100lm}{x} \text{ liters}.$

That is the fuel used per $100$ kilometers.

The answer is $\boxed{\textbf{(E)}\ \frac{100lm}{x}}$ .

Why this works

Changing units is safest one step at a time, with a small example in mind to decide whether to multiply or divide. A quick sense check also helps: a car with a bigger $x$ goes farther on each gallon, so it should use fewer liters. That means $x$ must be on the bottom of the fraction, which already rules out (A) and (B).

Alternative approach

Plug in easy numbers. Say $m = 2$ , $l = 1$ and $x = 50$ . Then $100$ kilometers is $200$ miles, which uses $200 \div 50 = 4$ gallons, which is $4$ liters.

Now put the same numbers into each choice:

Choice	(A)	(B)	(C)	(D)	(E)
Value	$\frac{50}{200} = \frac14$	$\frac{100}{100} = 1$	$\frac{2}{5000}$	$\frac{100}{100} = 1$	$\frac{200}{50} = 4$

Only (E) gives $4$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

We want the number of liters the car burns while driving $100$ kilometers.

Distance: $100$ kilometers is $100m$ miles.

Fuel in gallons: the car covers $x$ miles per gallon, so $100m$ miles use $\dfrac{100m}{x}$ gallons.

Fuel in liters: each gallon is $l$ liters, so the trip uses $\dfrac{100m}{x} \cdot l = \dfrac{100lm}{x}$ liters.

The answer is $\boxed{\textbf{(E)}\ \frac{100lm}{x}}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 6: Ratios and Percents
Intermediate Counting & Probability, Chapter 2: Sets and Logic (§2.2, §2.3)
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)
Prealgebra, Chapter 7: Ratios, Conversions, and Rates (§7.4)