Total distance = d

d/2x + d/2y = t (time = dist/speed -- d/2 because he travels halfway at x and the other half at y)

t* 4xy = d(2x+2y)

d = 2xyt/x+y

sparkle6 wrote: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)

[spoiler]Answer: C[/spoiler]

Let d be the distance from home to school
time = distance / speed

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

We can solve for d to get the ans

Edit - Posters above did it way way simpler than I did so follow them.

Let: x = speed of Bob biking
y = speed of Bob walking
t = time for Bob to get to school
t1 = time for Bob to bike to halfway point (where his bike breaks down)
t2 = time for Bob to walk the rest of the way from the halfway point
d = total distance Bob has to travel

We know that x*t1 + y*t2 = d --> this says that the time Bob is biking times his biking speed plus the time Bob is walking times his walking speed is equal to the total distance.

We also know that t1 + t2 = t --> it's intuitive that both times will equal the total time.

Finally we know that d/2 = x*t1 = y*t2 --> Bob went 1/2 the distance on his bike and half the distance walking.

From equation 3, t1 = d/(2*x). Plug this into equation 2 to get:
d/(2*x) + t2 = t
t2 = t - d/(2*x)

Plug this equation and our equation 3 result into equation 1 to get:
x*d/(2*x) + y*(t - d/(2*x)) = d

It's going to take some simplification to get the answer here and solve for d:

d/2 + y*t - d*y/(2*x) = d
y*t = d*(1 - 1/2 + y/(2*x))
d = y*t/(1/2 + y/(2*x))

This is not an answer choice so we can simplify further by finding a common denominator for the bottom part of the fraction:

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

The answer is C