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Jeff drove to work from this home, averaging 40 miles per ho [#permalink]
04 Sep 2013, 23:16

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Difficulty:

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Question Stats:

62% (02:55) correct
38% (02:36) wrong based on 173 sessions

Jeff drove to work from this home, averaging 40 miles per hour and was 12 minutes late. The next day he left home for work at the same time, took the same route, averaging 48 miles per hour, and was 7 minutes late. How far in miles is it from Jeff's home to his work?

Re: Jeff drove to work from this home, averaging 40 miles per ho [#permalink]
04 Sep 2013, 23:22

1

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fozzzy wrote:

Jeff drove to work from this home, averaging 40 miles per hour and was 12 minutes late. The next day he left home for work at the same time, took the same route, averaging 48 miles per hour, and was 7 minutes late. How far in miles is it from Jeff's home to his work?

a) 20.0 b) 24.5 c) 30.0 d) 37.5 e) 40.0

LET DISTANCE = D

if T IS REQUIRED TIME TO REACH .

T = \(\frac{D}{40} - \frac{12}{60} = \frac{D}{48} - \frac{7}{60}\) SOLVING D = 20

hence A _________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

Re: Jeff drove to work from this home, averaging 40 miles per ho [#permalink]
04 Sep 2013, 23:48

2

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fozzzy wrote:

I didn't understand the equations but great method! Please elaborate..

\(\frac{12}{60}\) the reason you are subtracting is because its \(\frac{12 minutes}{60 minutes}\) late?

Let say jeff need to reach in time T. LET distance to office = d now first day time taken by jeff = distance/speed = d/40 this time is 12 minutes more than T therefore T = d/40 -12/60

SIMILARLY second day time taken by jeff = d/48 This is 7 minute more than T Therefore T = d/40-7/60

NOW EQUATING both equation of T. \(\frac{d}{40} -\frac{12}{60} = \frac{d}{40}-\frac{7}{60}\)

solving D = 20

hope it helps _________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

Re: Jeff drove to work from this home, averaging 40 miles per ho [#permalink]
04 Sep 2013, 23:55

fozzzy wrote:

Jeff drove to work from this home, averaging 40 miles per hour and was 12 minutes late. The next day he left home for work at the same time, took the same route, averaging 48 miles per hour, and was 7 minutes late. How far in miles is it from Jeff's home to his work?

Re: Jeff drove to work from this home, averaging 40 miles per ho [#permalink]
05 Sep 2013, 07:36

1

This post received KUDOS

Expert's post

praneeth4u wrote:

Jeff drove to work from this home, averaging 40 miles per hour and was 12 minutes late. The next day he left home for work at the same time, took the same route, averaging 48 miles per hour, and was 7 minutes late. How far in miles is it from Jeff's home to his work?

a) 20.0 b) 24.5 c) 30.0 d) 37.5 e) 40.0

still i didn't understood why 12/60 and 7/60.....no where they've mentioned 60min right... please answer

The point is that the rates are given in miles per hour. So, we are converting 12 and 7 minutes into hours:

Re: Jeff drove to work from this home, averaging 40 miles per ho [#permalink]
09 Sep 2013, 05:46

I tried a different method here: I used the following thought process...if the two day trips differed in time by 5 min (12 min late vs. 7 min late), then we can set up the two time equations and set them equal to 5 min or 1/12 hrs.

d/40 - d/48 = 1/12 (5 min); common denominator is 240, so we have (6d-5d) / 240 = 1/12 hr.

if we solve for d, we get: d / 240 = 1/12; 12d = 240 therefore, d = 20 miles.

Re: Jeff drove to work from this home, averaging 40 miles per ho [#permalink]
09 Sep 2013, 08:05

1

This post received KUDOS

blueseas wrote:

fozzzy wrote:

I didn't understand the equations but great method! Please elaborate..

\(\frac{12}{60}\) the reason you are subtracting is because its \(\frac{12 minutes}{60 minutes}\) late?

Let say jeff need to reach in time T. LET distance to office = d now first day time taken by jeff = distance/speed = d/40 this time is 12 minutes more than T therefore T = d/40 -12/60

SIMILARLY second day time taken by jeff = d/48 This is 7 minute more than T Therefore T = d/40-7/60

NOW EQUATING both equation of T. \(\frac{d}{40} -\frac{12}{60} = \frac{d}{40}-\frac{7}{60}\)

solving D = 20

hope it helps

I have a quick question.

For T = d/40 -12/60, why wouldn't we add 12/60 (i.e. 12 minutes) to the d/40? After all, doesn't his time increase with his slower speed of d/40?

Thanks!

Edit: I think I got it!

Let's pretend he needs to get home in 60 minutes. The time he takes on the first day works out to be 72 minutes. From this we subtract 12 minutes to even out the equation. "T" represents his time if he is on time. d/40 represents his time on the day in question (which will be greater than his regular time) and 12/60 or 7/60 represents the extra time he took, if subtracted from his slower time would represent his normal time.

Does that make sense?

First day: t = d/40 - 12/60 Second day: t = d/48 - 7/60

d/40 - 12/60 = d/48 - 7/60 d/40 = d/48 + 5/60

(LCM of 40 and 48 is 240)

d = 40*(d/48) + 40*(5/60) d = 40d/48 + 200/60 d = 200d/240 + 800/240 d = (200d+800)/240 240d = 200d+800 40d = 800 d = 20

Re: Jeff drove to work from this home, averaging 40 miles per ho [#permalink]
12 Oct 2013, 21:19

1

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Let 't' is the usual time taken(without getting late)... Since the ration of speed in both case is 40:48 ie 5:6 , the ratio of time taken will be reverse ie 6/5. SO t+12 / t+ 7 = 6/5 giving t = 18 mins. substitute in any - D= 40*(18+12)/60 = 20 miles divide by 60 to change into hour...

Re: Jeff drove to work from this home, averaging 40 miles per ho [#permalink]
14 Oct 2013, 21:54

Let T be the time when Jeff is not late for work

When Jeff is 12 minutes or \(\frac{1}{5} hours\) late, \(Time= T+\frac{1}{5}\) \(Distance= 40(T+\frac{1}{5})\)-------------------------------------------------------------------------------------(1)

When jeff is 7 minutes or \(\frac{7}{60} hours\) late, \(Time= T+\frac{7}{60}\) \(Distance= 48(T+\frac{7}{60})\)------------------------------------------------------------------------------------(2)

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