Problem Solving

Source: MGMAT

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?

(x + y) / t

2(x + t) / xy

2xyt / (x + y)

2(x + y + t) / xy

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


I want to know what is wrong with my method below:

[spoiler]Let d be the total distance from home to school.

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

Which gives d = (2xyt)/x-y[/spoiler]
That is nowhere in the question, OA however is the very similiar C.
Can anyone please point out my error?

Imagine the distance from home to school is 2d miles

Then (I) cycled half the distance and it took d/x hours

(II) walked half the distance and it took d/y hours

he reached school after " t" hours then t = d/x+d/y

then d = txy/(x+y)

but the total distance is 2d ,

Hence the distance from home to school = 2xyt/(x+y) miles

ghacker wrote:Imagine the distance from home to school is 2d miles

Then (I) cycled half the distance and it took d/x hours

(II) walked half the distance and it took d/y hours

he reached school after " t" hours then t = d/x+d/y

then d = txy/(x+y)

but the total distance is 2d ,

Hence the distance from home to school = 2xyt/(x+y) miles

Thankyou, I see how you arrived at that, and I find it very convenient. But my question remains unanswered, what's wrong with the method that I adopted?

axat wrote:Let d be the total distance from home to school.

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

Which gives d = (2xyt)/x-y

Can anyone please point out my error?

Why are you adding (d/x) to t?



dist./ rate---> time

x / (d/2)----> time (d/2x) for the first half

y / (d/2)----> time (d/2y) for the second half


So in entirety it takes t time to complete the whole distance d.

so you can add the times: (d/2x) +(d/2y) = t and solve for d.

(2yd+2xd)/ 4xy = t

4xyt = 2d(x+y)

2xyt = d(x+y)

2xyt/(x+y)

axat wrote: Thankyou, I see how you arrived at that, and I find it very convenient. But my question remains unanswered, what's wrong with the method that I adopted?

If you explain how you arrived at the initial equation it would be easier to assess where you went wrong.

One other solution:

When equal distances are travelled at different speeds then shortcut for average speed is 2xy/x+y where x,y are the different speeds

t = Distance / speed

t = d / 2xy/x+y

t = d (x+y) / 2xy

d = 2xyt / x+y

plugging in numbers
x=20 miles per hour
distance from home to school=20 miles
y=1 miles per hour
put these values in the answer options and C satisfies

Axat,
I think you have not read the question properly. It does not say on that day Bob took t hrs more than he would have taken had he gone by bike but simply t hours

I hope this clarifies

Rahul