SOLUTION: John walks the 15 meters from the front door of his house to his moped at a constant rate of speed. Having forgotten to lock the front door, he runs the 15 meters back to the fro

Question 100914: John walks the 15 meters from the front door of his house to his moped at
a constant rate of speed. Having forgotten to lock the front door, he runs the 15 meters
back to the front door along the same path. His speed running is 2 m/sec faster than
his speed walking. If the total time to walk to the car and run back to the door was 8
seconds, what was John�s walking speed?

Answer by TP(29) (Show Source): You can put this solution on YOUR website!
Let w be the walking speed, then w+2 must be the running speed (since we're told that his running speed is 2m/s more than his walking speed).
Let the walking time be t seconds and the running time be T seconds

Now using these facts: distance= average speed*time,
the distance is 15 metres in each case( walking and running),
the factors of 15 are 5 and 3 and
t+T= 8 (given)

we can find w BY INSPECTION as follows:
Since 15=3*5=w*t
and
15=5*3=(w+2)*T then it follows that w=3,t=5 and T=3.
So the walking speed,w,is 3m/sec.ANS

You can of course solve the question using algebra but why do that with such a trivial question.
In an exam situation time is of the essence and a solution BY INSPECTION is totally valid provided you show what it is based upon.
Using algebra and the definitions given above a possible solution might be as follows:
15=w*t so t=15/w
Similarly,15=(w+2)*T so T=15/(w+2).
But t+T=8 and so (15/w)+(15/(w+2))=8
This leads to (1/w)+(1/(w+2))=8/15
Adding the two algebraic fractions together we get:
(w+2+w)/w(w+2) = 8/15
Hence 2w+2 = 8 (i) and w(w+2) = 15 (ii)
Considering 2w+2 = 8 it follows that w+1 = 4 (dividing both sides by 2)
And so w = 3
That is, the walking speed is 3m/sec.ANS
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