Question

Two cyclists, 80 miles apart, start riding toward each other at the same time. One cycles 3 times as fast as the other. If they meet 2 hours later, what is the speed (in mi/h) of the faster cyclist?

Answer 1

The speed of the faster cyclist will be 30 miles/hour.

Explanation

One cyclist is 3 times as fast as the other.

Lets assume that the speed of the slower cyclist is  $x$ miles/hour.

So, the speed of the faster cyclist will be:  $3x$ miles/hour.

They start riding toward each other at the same time and meet 2 hours later.

We know that,  $Distance= speed*time$

So, the distance traveled by slower cyclist in 2 hours $= (x*2)miles= 2x$ miles and the distance traveled by faster cyclist $=(3x*2) miles = 6x$ miles.

Given that, they were 80 miles apart at the starting. So, the equation will be.....

$2x+6x=80\\ \\ 8x=80\\ \\ x=10$

So, the speed of the faster cyclist will be: (3×10)= 30 miles/hour.

Answer 2

Distance between two cyclists = 80 miles

Let speed of slower cyclist = s

We know that one cycles 3 times as fast as the other

⇒ Speed of faster cyclist = 3s

We also know that they meet 2 hours later

Total Distance = Distance covered by slower cyclist + Distance covered by faster cyclist

⇒ Total Distance = Speed of slower cyclist × Time taken by slower cyclist + Speed of faster cyclist × Time taken by faster cyclist

⇒ 80 = s × 2 + 3s × 2

⇒ 80 = 8s

⇒ s = 10

and 4s = 30

Hence, speed of the faster cyclist is 30 miles per hour.