Question

Bob bikes to school every day at a steady rate of x miles per hour. On a particular day, Bob had a flat tire exactly halfway to school. He immediately started walking to school at a steady pace of y miles per hour. He arrived at school exactly t hours after leaving his home. How many miles is it from the school to Bob's home?

A) (x + y) / t

B) 2(x + t) / xy

C) 2xyt / (x + y)

D) 2(x + y + t) / xy

E) x(y + t) + y(x + t)

Answer 1

Answer:
`C) (2xyt)/((x+y))`

Step-by-step explanation:We start with the total time t and we define it like this

`t=t_(1) +t_(2)`

being
`t_(1)` the time he was on x speed and
`t_(2)` the time he was on y speed

Now for the distance we have the velocity equation
`Velocity=(distance)/(time)` and in this excercise we would have the two equations

`x=(d/2)/(t_(1))`

`y=(d/2)/(t_(2))`

then

`t_(1)=(d/2)/(x)`

`t_(2)=(d/2)/(y)`

Next

`t=(d/2)/(x)+(d/2)/(y)`

this we simplify to get

`d= (2xyt)/((x+y))`