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 Post subject: Maths HelpPosted: Tue Sep 18, 2007 3:14 pm
A father left home at 2-00am and travels @ 60miles per hour. Ten minutes later his son leave home and was travelling @80 miles per hour.

How far have the father gone at the moment his son begins his chase?

At what time will the son catch up with his father?

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 Post subject: Posted: Tue Sep 18, 2007 3:25 pm

Joined: Mon Jun 18, 2007 2:32 pm
Posts: 6966
Location: East Kent
1st part, if he is travelling at 60miles per hour,he travels a mile a minute..

therefore after 10 minutes, he has done 10 miles...

If his son travels at 80 miles an hour.

80 = 60 + 20 minutes

60mph = 1 mile per minute
20 mph = 1/3 mile per minute

80 mph = 1 1/3 miles per minute, so he travels 4 miles in 3 minutes.

I;'ll get back to you with teh answer once I've drawn a diagram!

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 Post subject: Posted: Tue Sep 18, 2007 3:37 pm

Joined: Fri Mar 17, 2006 5:12 pm
Posts: 1301
Location: Birmingham
This is always an interesting one.

I always tend to use algebra and S=D/T

1st part is as described by yoyo123

For 2nd part:-

At the point they cross they will have travelled the same distance.

Hence, D = T x S

Therefore Tf x Sf = Ts x Ss

But because the son sets out 10 mins (or 10/60 hours) later, hence

Tf = (Ts + 10/60)

Therefore Sf x (Ts + 10/60) = Ss x Ts

And Sf = 60 and Ss = 80

Therefore 60Ts + 10 = 80Ts

or 20Ts = 10

Therefore Ts = 10/20 or 1/2 hour

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 Post subject: Posted: Tue Sep 18, 2007 4:23 pm

Joined: Mon Feb 12, 2007 1:21 pm
Posts: 11954
An easier way is to say -

In the ten minutes before the other starts the father goes 10 miles

Effectively the son is catching up the father at [80-60] = 20 mph

He has 10 miles to do - so thats half an hour - no need for algebra

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