Problem Solving

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)

imo:C

d=rt

1st-
d/2 = xt1 ----(a)

2nd-
d/2 = yt2 ----(b)

from (a) and (b)

t1+t2=t = 2xyt/x+y

thus C

I agree with raunekk , answer should be C.

OA - C. Could someone show me how to solve this problem by plugging in values
THanks!

Let d be the total distance

d/2 covered wiht speed x miles/hr
d/2 covered with speed y miles per hour

distance/speed = time

So t = d/2 /x + d/2 / y
t = d/2x + d/2y
t = dy+dx / 2xy
t = d(x+y) / 2xy

d = 2xyt/(x+y)

C)

Yes, the answer should be C

beater wrote:OA - C. Could someone show me how to solve this problem by plugging in values
THanks!

Let x=20, y=10, total distance = 40 miles (long way to school, right?)

therefore, t=3 (20/20 + 20/10) and the answer we get when we plug in values should be 40miles.

A. (x + y) / t
(20+10) / 3 = 10 Incorrect

B. 2(x + t) / xy
2(20+3) / 20*10 = 46/300 Incorrect

C. 2xyt / (x + y)
(2*20*10*3) / (20+10) = 1200/30 = 40 CORRECT

I probably could have chosen easier to apply numbers but it works.

When a person/object travels the same distance at two different speeds say a and b then the average speed is

2ab/(a+b)

here bob travels the same distance ... first half at x mph and second half at y mph then avg speed = 2xy(x+y)

time = t

distance = speed* time = 2xyt/(x+y)