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

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

Answer 2

The rate of each boys are as follows:

The boy at slower speed = 9.8 mph (slow speed)

The boy at faster speed= 12.8 mph (faster speed)

Calculation of speed

Speed can be defined as the distance covered per unit time. That is, speed = distance/time.

Since the faster boy travels 3 miles per hour faster than the other, therefore

slower speed= x and

faster speed= x + 3

The total time for each boy = 3 hours

To find out the distance for each boy, you must multiply the 3 hours (time) and speed together.

so distance = time*speed

Therefore,

slower distance = 3x

faster distance = 3(x +3)

Then you can add them together to get the total distance.

To Solve for slow speed(x).

3x + 3(x + 3) = 68

3x + 3x + 3×3 = 68

6x + 9 = 68 add like terms

6x = 68 - 9

6x = 59

x = 59/6 = 9.8 mph (slow speed)

Therefore, faster speed,

x + 3 = 12.8 mph (faster speed)

Therefore, the rate of each boys are 9.8 mph (slow speed) and 12.8 mph (faster speed).

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