Two cyclists start biking on a trail. Cyclist 2 travels at 10 mph and starts 3 hours after cyclist 1 who is traveling at 6 mph. How much time does it take for cyclist 2 to catch cyclist 1?

Answer 1

ANSWER # 1 It looks like cyclist #2 is out to catch cyclist #1. Our #2 cyclist is starting out 3 hours later than cyclist #1, and is traveling at 10 mph versus the 6 mph of cyclist #1. There are a couple of ways to solve this one, so there is no "right way" to do it. Opting for an easy one, let's think of it this way. Think of it like both cyclists are going the same speed, the 6 mph. Yes, our trailing cyclist can't ever catch the lead cyclist, but this is a short cut, okay? If both cyclists ride, the space between them neither increases or decreases. Here's the scoop. It's the space between them that we're gonna look at. That space is 3 hours long. Here's the thinking. Cyclist #2 is going faster than #1 by 4 mph, and that is the "extra" speed that our #2 cyclist is using to close the gap, to make up that 3 hours. If cyclist #2 is going to make up 3 hours at 4 mph (the differential speed), how much ground is he going to have to make up? Well, cyclist #1 is riding at 6 mph for 3 hours to create the gap. That's 6 time 3 or 18 miles that cyclist #2 is going to have to make up. And our #2 guy is going to have to make up the 18 miles at 4 mph (the differential speed). How long is it going to take him? The 18 miles divided by 4 mph equals 4 ½ hours. Presto. It will take our #2 cyclist that 4 ½ hours to catch up to cyclist #1. If we want to check our work, take cyclist #2's 10 mph speed and multiply by 4 ½ hours and we'll see that he has to ride 45 miles to catch cyclist #1. Let's see what happens for cyclist #1. This cyclist will be riding for 4 ½ hours plus the 3 hour head start. That's 7 ½ hours of riding at 6 mph. The 7 ½ times 6 equals 45 miles. Check! We're good here! ANSWER # 2 1st = 6(x +3) ......... 2nd = 10x 10x = 6(x+3) x = 4 (1/2)