Two cyclists, 128 miles apart, start riding toward each other at the same time. One cycles 3 times as fast as the other. If they meet 4 hours later, what is the speed (in \( \mathrm{mi} / \mathrm{h} \) ) of the faster cyclist?

To find the speed of the faster cyclist, let's denote the speed of the slower cyclist as \( x \) miles per hour. Therefore, the speed of the faster cyclist is \( 3x \) miles per hour. The total distance covered by both cyclists when they meet is the sum of the distances they traveled, which equals 128 miles. Since they meet after 4 hours, we can set up the following equation: \[ \text{Distance by slower cyclist} + \text{Distance by faster cyclist} = 128 \text{ miles} \] In 4 hours, the slower cyclist travels \( 4x \) miles, and the faster cyclist travels \( 4(3x) = 12x \) miles. Thus: \[ 4x + 12x = 128 \] Combining like terms gives: \[ 16x = 128 \] Now, solving for \( x \): \[ x = \frac{128}{16} = 8 \text{ miles per hour} \] Thus, the speed of the faster cyclist is: \[ 3x = 3 \times 8 = 24 \text{ miles per hour} \] So, the speed of the faster cyclist is \( 24 \, \mathrm{mi/h} \).