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Junior | Next Rank: 30 Posts
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by harry_x1 » Fri Jun 29, 2007 11:09 pm
John walking at a speed of 20 km/h reaches his college 10 mins late. Next time he increases his speed by 5 km/h, but finds that he is still late by 4 mins. What is the distance of his college from his house.

(a) 20 km
(b) 6 km
(c) 12 km
(d) none of these
Last edited by harry_x1 on Sat Jun 30, 2007 6:30 am, edited 1 time in total.
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by f2001290 » Fri Jun 29, 2007 11:56 pm
Harry - Please post the options also.
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by blah45 » Sat Jun 30, 2007 9:02 am
I'm getting 10km.

What is option (E)? if (E) is not 10km then my answer would be (D).

Answer please
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by jrbrown2 » Sat Jun 30, 2007 12:21 pm
We know that the distance to college remains constant.
D = Distance
T = Time to reach college on time

We know that @ 20 km/h John was 10 minutes late
D = 20km/h(T+10)

We know that @ 25 km/h John was 4 minutes late
D = 25km/h(T+4)

Equating the 2 equations

20(T+10) = 25(T+4)
T = 20 Minutes

Use one of those equations to determine distance

20km/h(20+10 min)(1h/60 min) = 10 km
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