Question

At 10:00 a.m. two boys start out on their bikes to meet each other from towns located 68 miles apart. At 1:00 p.m. they meet. If one boy traveled 3 miles per hour faster than the other, what was the rate of speed of each boy?

Answer 1

To determine the rates of speed of two boys on bikes who started from towns 68 miles apart and met after 3 hours, with one traveling 3 mph faster, we set up an equation based on their combined travel distances equaling 68 miles. The slower boy's speed is approximately 9.83 mph and the faster boy's speed is approximately 12.83 mph.

Step-by-step explanation:

To solve the problem of determining the boys' rates of speed, we use the information that they are 68 miles apart and meet after traveling for 3 hours, and one boy travels 3 mph faster than the other. We let x be the speed of the slower boy, so the speed of the faster boy would be x + 3 mph.

The total distance covered by both boys in 3 hours when they meet will be 68 miles. So, if the slower boy covers 3x miles in 3 hours, the faster boy will cover 3(x + 3) miles. Thus, the equation to represent the situation would be:

3x + 3(x + 3) = 68

Solving for x, we get:
3x + 3x + 9 = 68
6x = 68 - 9
6x = 59
x = 59 / 6
x = 9.83333...

The slower boy's speed is approximately 9.83 mph, and the faster boy's speed is 12.83 mph. Note that we rounded to two decimal places for ease of interpretation.

Answer 2

a = speed of one boy
a + 3 = speed of 2nd boy

d = speed X time ... time is 3 hrs

68 = 3a + 3(a + 3)
6a = 59
a = 59/6 = 9 5/6 miles per hour
a + 3 = 12 5/6 miles per hou