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Time Speed Distance method

Expert replies
by axat » Mon Jun 15, 2009 10:04 am
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?
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Source: — Problem Solving |

by ghacker » Mon Jun 15, 2009 10:14 am
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
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by axat » Mon Jun 15, 2009 10:29 am
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?
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Re: Time Speed Distance method

by ssmiles08 » Mon Jun 15, 2009 11:27 am
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)
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by Stuart@KaplanGMAT » Mon Jun 15, 2009 11:30 am
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.
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by cramya » Mon Jun 15, 2009 2:24 pm
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
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by tohellandback » Mon Jun 15, 2009 7:38 pm
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
The powers of two are bloody impolite!!
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by rah_pandey » Tue Jun 16, 2009 12:40 am
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
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