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by AndreaV424 » Tue Sep 08, 2009 9:39 am
Q: A policeman chases after a thief who has a 500m head start. They run at constant speeds, with the policeman running 1 km every 6 minutes and the thief 1 km every 10 minutes. How long will policeman take to catch up with the thief?

A. 3 1/2 min
B. 7 1/2 min
C. 10 min
D. 15 min
E. 17 1/2 min

1. I know I'll need to convert miles to km. (1.96 km = 1 mile).
2. I know policeman's rate is 1/6 and thief's rate is 1/10.

What is the best way to set up an equation for problems similar to this?

Thank you!
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by Nermal » Tue Sep 08, 2009 9:50 am
Doesn't the thief have a 500 meter head start?
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by antondesh » Tue Sep 08, 2009 10:00 am
Yes, I think "m" stands for "meters" in this problem so no conversion to miles necessary :P

Here's how I would work through it:

The unknown is time = x
The rate of the cop is 1/6th of a km per minute
The rate of the thief is 1/10th of a km per minute
But the thief has 1/2 of a km head start (500m)

So:

1/6*x - (1/10*x + 1/2) = 0
x/6 - x/10 - 1/2 = 0
5x/30 - 3x/30 = 1/2
2x/30 = 1/2
4x = 30
x = 7.5 minutes = 7 and 1/2 minutes

Is that the answer?
Last edited by antondesh on Tue Sep 08, 2009 11:54 am, edited 2 times in total.
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by bharathh » Tue Sep 08, 2009 10:01 am
Shucks! You don't have to beat yourself up for something like this.

One thing you may want to do is to convert everything to a base where things are easy to calculate.

You have distance and speed ... you need time.

First i convert the speeds to a common unit ... kmph...

If the policeman runs 1 km in 6 mins. He will run 10 km in 1 hr. So his speed is 10kmph.

Similarly the thief will run 6 km in that same hr.

The difference is speed is 4kmph

The extra distance the policeman has to cover is 500m = 0.5 km

So to catch the thief, the time he will take is 0.5/4 = 1/8 of an hr or 1/8 * 60 min = 7.5 mins
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