2026 AMC 10A Problem 8

P1-10~3 minPrint

Linear Equations & Word Problems Bound the quantity above/below or estimate to pin it down Consider the largest/smallest element or boundary case

A driver sees an electronic highway sign stating that the next exit is $7$ miles away and it will take $5$ minutes to reach it. Assume both numbers have been rounded to the nearest integer. Among the following options, what is the lowest speed limit (in miles per hour) that would possibly allow the driver to legally reach the exit within the given time frame?

Show solution without answering

Key insight

To make the needed speed as small as possible, take the shortest distance that rounds to 7 miles and the longest time that rounds to 5 minutes.

Solution

The sign shows rounded numbers, so the true distance and the true time are not exactly $7$ and $5$ . First find what they could really be.

What the sign could mean. A number rounds to $7$ when it is at least $6.5$ and less than $7.5$ . A number rounds to $5$ when it is at least $4.5$ and less than $5.5$ .

Sign says	True value could be
$7$ miles	from $6.5$ miles up to just under $7.5$ miles
$5$ minutes	from $4.5$ minutes up to just under $5.5$ minutes

Pick the easiest situation. We want the lowest speed limit that could work, so look at the situation needing the least speed: the shortest possible distance and the longest possible time. That is $6.5$ miles in (almost) $5.5$ minutes.

3. Find that speed. $5.5$ minutes is $\tfrac{5.5}{60}$ of an hour, so the speed is
$6.5 \div \frac{5.5}{60} = \frac{6.5 \times 60}{5.5} = \frac{390}{5.5} \approx 70.9 \text{ miles per hour}.$
Even this best case needs a little more than $70$ miles per hour. This is a "least possible" question, so check both halves.

4. A limit of $70$ is not enough. At $70$ miles per hour a car goes $\tfrac{70}{60}$ of a mile each minute. In $5.5$ minutes that is
$\frac{70}{60} \times 5.5 = \frac{385}{60} \approx 6.42 \text{ miles},$
and the true time is less than $5.5$ minutes, so the car covers even less. But the exit is at least $6.5$ miles away. So $70$ cannot work, and $65$ is slower still.

5. A limit of $75$ can work. At $75$ miles per hour a car goes $\tfrac{75}{60} = 1.25$ miles each minute. Suppose the exit is really $6.5$ miles away (which the sign rounds to $7$ ). Then the trip takes
$6.5 \div 1.25 = 5.2 \text{ minutes},$
which the sign rounds to $5$ . So everything on the sign is consistent with a driver going exactly $75$ miles per hour.

The lowest choice that could work is $75$ .

The answer is $\boxed{\textbf{(C)}\ 75}$ .

Why this works

A rounded number really stands for a whole range of values. When a question asks what is possibly enough, choose the values in those ranges that make the job easiest; here that means the shortest distance and the longest time. Then confirm both sides: show the next smaller choice fails even in the best case, and show your choice succeeds in at least one case.

Concise solution

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

The true distance $d$ satisfies $6.5 \le d < 7.5$ miles and the true time $t$ satisfies $4.5 \le t < 5.5$ minutes. The required speed $d/t$ is smallest when $d$ is smallest and $t$ largest, so it always exceeds
$\frac{6.5}{5.5/60} = \frac{390}{5.5} \approx 70.9 \text{ mph}.$
Hence a limit of $70$ (or $65$ ) never suffices. A limit of $75$ does: at $1.25$ miles per minute, a $6.5$ -mile trip takes $5.2$ minutes, which displays as $7$ miles and $5$ minutes.

The answer is $\boxed{\textbf{(C)}\ 75}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 7: Proportion (§7.4)