Question

Two cyclists left simultaneously from cities A and B heading towards each other at constant rates and met in 5 hours. The rate of the cyclist from A was 3 mph less than the rate of the other cyclist. If the cyclist from B had started moving 30 minutes later than the other cyclist, then the two cyclists would have met 31.8 miles away from A. What is the distance between A and B, in miles?

Answer 1

75 miles

Explanation:

Let x mph be the cyclist A rate, then x+3 mph is the cyclist B rate.

1. In 1 hour they both traveled x+x+3=2x+3 miles. In 5 hours they traveled

`5(2x+3)=10x+15\ miles.`

2. Cyclist A spent
`(31.8)/(x)` hours to travel 31.8 miles. If the cyclist from B had started moving 30 minutes (1/2 hour) later than the cyclist A, then he spent
`(31.8)/(x)-(1)/(2)` hours to travel the rest of the distance. In total they both traveled the whole distance 10x+15 miles, thus

`31.8+\left((31.8)/(x)-(1)/(2)\right)\cdot (x+3)=10x+15`

Solve this equation. Multiply it by 2x:

`63.6x+(63.6-x)(x+3)=2x(10x+15)\\ \\63.6x+63.6x+190.8-x^2-3x=20x^2+30x\\ \\-x^2+124.2x+190.8-20x^2-30x=0\\ \\-21x^2+94.2x+190.8=0\\ \\210x^2-942x-1908=0\\ \\35x^2-157x-318=0\\ \\D=(-157)^2-4\cdot 35\cdot (-318)=69169\\ \\x_(1,2)=(-(-157)\pm√(69169))/(2\cdot 35)=(157\pm263)/(70)=-(106)/(70),\ 6`

The rate cannot be negative, thus, x=6 mph.

Hence, the distance between cities A and B is

`10\cdot 6+15=60+15=75\ miles.`