Roland2rule 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)

BTGmoderatorRO 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)

OA is C

We are given that Bob bikes at a rate of x miles per hour and walks at a rate of y miles per hour. If we let the distance between his home and school be d, then the distance he bikes is d/2, and the distance he walks is d/2.

Thus, his biking time is (d/2)/x = d/(2x), and his walking time is (d/2)/y = d/(2y). Since he spent t total hours biking and walking:

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

Multiplying the entire equation by 2xy, we have:

dy + dx = 2xyt

d(y + x) = 2xyt

d = 2xyt/(x + y)

Answer: C


What do I need to solve this, I need the help of an expert here. Thanks