<PRE><B>Question 493396</B>
a boy on his bicycle intends to arrive at a certain time to a town that is 30 km away from his home .
after riding 10 km, he rested for half an hour and as a result he was obliged to ride the rest of the trip 2km/hr faster .
what was his original speed?
:
Let s = his original speed
then
(s+2) = his faster speed
:
Write a time equation; Time = dist/speed
:
Required time = faster time + half an hr 
{{{30/s}}} = {{{30/((s+2))}}} + .5
multiply by s(s+2)
s(s+2)*{{{30/s}}} = s(s+2)*{{{30/((s+2))}}} + .5s(s+2)
:
Cancel the denominators
30(s+2) = 30s + .5s^2 + 1s
30s + 60 = 30s + .5s^2 + s
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Arrange as quadratic equation on the right
0 =.5s^2 + 30s - 30s + s - 60 
.5s^2 + s - 60 = 0
:
multiply by 2 (make the coefficient of s^2 = 1, easier to factor)
s^2 + 2s - 120 = 0
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Factors to
(s+12)(s-10) = 0
:
the positive solution
s = 10 km/hr is his original speed
;
:
:
Check this by finding the actual time for each scenario
30/10 = 3hrs
30/12 = 2.5 hrs + half hr rest = 3hrs</PRE>