Question

Every morning Aya goes for a $9$ -kilometer-long walk and stops at a coffee shop afterwards. When she walks at a constant speed of $s$ kilometers per hour, the walk takes her $4$ hours, including $t$ minutes spent in the coffee shop. When she walks $s + 2$ kilometers per hour, the walk takes her $2$ hours and $24$ minutes, including $t$ minutes spent in the coffee shop. Suppose Aya walks at $s + \frac{1}{2}$ kilometers per hour. Find the number of minutes the walk takes her, including the $t$ minutes spent in the coffee shop.

Answer

Solution:

Measuring time in hours, the two scenarios say $\frac{9}{s} + \frac{t}{60} = 4 \qquad \text{and} \qquad \frac{9}{s+2} + \frac{t}{60} = \frac{12}{5}.$ Subtracting, $\frac{9}{s} - \frac{9}{s+2} = \frac{8}{5},$ so $\frac{18}{s(s+2)} = \frac{8}{5},$ giving $s(s+2) = \frac{45}{4}.$ The positive root of $s^2 + 2s - \frac{45}{4} = 0$ is $s = \frac{5}{2}.$

Then $\frac{t}{60} = 4 - \frac{9}{5/2} = \frac{2}{5},$ so $t = 24$ minutes. Walking at $s + \frac{1}{2} = 3$ kilometers per hour takes $\frac{9}{3} = 3$ hours, so the total is $180 + 24 = 204$ minutes.

2024 AIME I Problem 1

Solution:

Problem 1 in Other Years