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2017 AMC 10A Problem 9

2017 AMC 10A Problem 9AMC 10 Step by Step

Name: Date: October 28, 2026

Linear Equations & Word Problems · difficulty 2 of 5 · about 3 min · position P1-10

Minnie rides on a flat road at $20$ kilometers per hour (kph), downhill at $30$ kph, and uphill at $5$ kph. Penny rides on a flat road at $30$ kph, downhill at $40$ kph, and uphill at $10$ kph. Minnie goes from town $A$ to town $B$ , a distance of $10$ km all uphill, then from town $B$ to town $C$ , a distance of $15$ km all downhill, and then back to town $A$ , a distance of $20$ km on the flat. Penny goes the other way around using the same route. How many more minutes does it take Minnie to complete the $45$ -km ride than it takes Penny?

A) $45$
B) $60$
C) $65$
D) $90$
E) $95$

Answer

Key insight

Time each stretch as distance divided by speed. Penny rides the loop backward, so Minnie's uphill stretch is Penny's downhill one, and the other way round.

Solution

Each rider changes speed on every stretch, so time each stretch on its own with $\text{time} = \text{distance} \div \text{speed}$ .

Minnie rides $A \to B \to C \to A$ .

Stretch	Distance	Type	Speed	Time
$A \to B$	$10$ km	uphill	$5$ kph	$2$ h $= 120$ min
$B \to C$	$15$ km	downhill	$30$ kph	$\tfrac12$ h $= 30$ min
$C \to A$	$20$ km	flat	$20$ kph	$1$ h $= 60$ min

Minnie's total: $120 + 30 + 60 = 210$ minutes.

Penny rides the loop the other way, $A \to C \to B \to A$ . A road that goes downhill from $B$ to $C$ goes uphill from $C$ to $B$ , so the hills swap.

Stretch	Distance	Type	Speed	Time
$A \to C$	$20$ km	flat	$30$ kph	$\tfrac23$ h $= 40$ min
$C \to B$	$15$ km	uphill	$10$ kph	$1\tfrac12$ h $= 90$ min
$B \to A$	$10$ km	downhill	$40$ kph	$\tfrac14$ h $= 15$ min

Penny's total: $40 + 90 + 15 = 145$ minutes.

Difference: $210 - 145 = 65$ minutes.

The answer is $\boxed{\textbf{(C)}\ 65}$ .

Why this works

When speed changes along a trip, there is no shortcut: find the time for each part and add. A table keeps the numbers straight. The one idea to watch is that a hill is uphill in one direction and downhill in the other, so riding the loop backward changes which stretch is slow.

The trap

Giving Penny the same uphill and downhill stretches as Minnie. Going the opposite direction turns uphill into downhill.

Concise solution

Compute each rider's time leg by leg with $\text{time} = \text{distance}/\text{speed}$ .

Minnie: $A \to B$ is $10$ km uphill at $5$ kph, $2$ hours. $B \to C$ is $15$ km downhill at $30$ kph, $\tfrac12$ hour. $C \to A$ is $20$ km flat at $20$ kph, $1$ hour. Total $3\tfrac12$ hours $= 210$ minutes.

Penny rides $A \to C \to B \to A$ , so the slopes are reversed. $A \to C$ is $20$ km flat at $30$ kph, $\tfrac23$ hour $= 40$ minutes. $C \to B$ is $15$ km, now uphill, at $10$ kph, $1\tfrac12$ hours $= 90$ minutes. $B \to A$ is $10$ km, now downhill, at $40$ kph, $\tfrac14$ hour $= 15$ minutes. Total $145$ minutes.

Difference: $210 - 145 = 65$ minutes.

The answer is $\boxed{\textbf{(C)}\ 65}$ .

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)

Techniques

Set up the equation/formula and compute; no special trick needed