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A Time- distance question..

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by \'manpreet singh » Thu Jun 21, 2012 8:11 am
Hi guys this is a simple problem......however i am bit confused as my answer doesn't match some how..quite strange to me..

A person travels from his home to office at 4km/hr and reaches there 20 min late.If he goes at 6 km/hr he reaches 10 min early.What is the distance b/w his home and office and what is time taken by him if he was to reach on time at his office??
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by Anurag@Gurome » Thu Jun 21, 2012 8:32 am
'manpreet singh wrote:A person travels from his home to office at 4km/hr and reaches there 20 min late.If he goes at 6 km/hr he reaches 10 min early. What is the distance b/w his home and office and what is time taken by him if he was to reach on time at his office?
Say, the distance between his home and his office is D kilometers and he should reach his office in H hours.

Hence, D/4 = (H + 20/60) ---> D = 4(H + 1/3)
And, D/6 = (H - 10/60) ---> D = 6(H - 1/6)

So, 4(H + 1/3) = 6(H - 1/6)
--> 4H + 4/3 = 6H - 1
--> 2H = (1 + 4/3) = 7/3
--> H = 7/6

Hence, D = 6(7/6 - 1/6) = 6

Hence, the distance between his home and his office is 6 kilometers and he should take 1 hours and 10 minutes to reach on time at his office.
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by GMATGuruNY » Thu Jun 21, 2012 11:33 am
'manpreet singh wrote:Hi guys this is a simple problem......however i am bit confused as my answer doesn't match some how..quite strange to me..

A person travels from his home to office at 4km/hr and reaches there 20 min late.If he goes at 6 km/hr he reaches 10 min early.What is the distance b/w his home and office and what is time taken by him if he was to reach on time at his office??

An alternate approach is to GUESS AND CHECK.

Let the distance = 12 kilometers.
Time to travel 12 kilometers at 4kph = 12/4 = 3 hours.
Time to travel 12 kilometers at 6kph = 12/6 = 2 hours.
Time difference = 3-2 = 1 hour.
The time difference given in the problem -- from 20 minutes late to 10 minutes early -- is 1/2 this amount.
Thus, the actual distance must be (1/2)12 = 6 kilometers.

Time to travel 6 kilometers at 6kph = 1 hour.
Since at this rate he arrives 10 minutes early, on time = 1 hour, 10 minutes.
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by dhonu121 » Thu Jun 21, 2012 10:24 pm
Although Mike's approach takes less time, but guessing under time pressure is so counter-intuitive to me. Under time pressure, I would like to start with a concrete approach, which yields the answer in one go!..
In that sense, I like Anurag's approach more. :)
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by \'manpreet singh » Fri Jun 22, 2012 10:56 am
Thanks guys!!!
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by GMATGuruNY » Fri Jun 22, 2012 2:03 pm
'manpreet singh wrote:Hi guys this is a simple problem......however i am bit confused as my answer doesn't match some how..quite strange to me..

A person travels from his home to office at 4km/hr and reaches there 20 min late.If he goes at 6 km/hr he reaches 10 min early.What is the distance b/w his home and office and what is time taken by him if he was to reach on time at his office??
Another approach:

Traveling at 6kph instead of 4kph -- 3/2 the slower speed-- saves 1/2 hour.
Since rate and time are reciprocals, traveling at 3/2 the rate takes 2/3 the time.
Thus, the 1/2 hour time difference = 1/3 of the time traveled at 4kph:
1/2 = (1/3)t
t = 3/2.
Since the time traveled at 4kph is 3/2 hours, the distance = r*t = 4(3/2) = 6 kilometers.
Since 3/2 hours means arriving 20 minutes late, on time = 90-20 = 70 minutes.
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