It'd be best to plug in numbers for this problem but in case you're wondering how to solve this problem algebraically:

We're told that Bob bikes to school at a steady rate of x miles per hour. On a particular day, his bike has a flat tire exactly half way to school. Let's assume that the distance from Bob's home to the school is d. On the day that his bike got a flat tire, Bob had already rode has bike for d/2x hours.
When he decided to walk to school in order to cover the other half distance, it took him d/2y hours. The problem tells us that it took Bob t hours to get school after he left his house. So:

d/2x+d/2y=t

solve for d:

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

2xyt/x+y=d

Thanks. I understand yuor explanation. but why it is wrong to assign d to each half, since both halves are equal ? I bet you have solved problems in which a person goes to town a in 20 miles/hour and goes back home in 40 miles/hour and to calculate the total average speed, we say 2d/(d/20 + d/40) gives average speed.

The given explanation is to use variables, is this the best way? Or is it quicker to plug in answers? Please let me know your thoughts.

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:C