Question 1208878
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x = the man's current speed in kph
y = time it takes to drive 100 km going at a speed of x kph


Let's consider the scenario where the man goes at his present speed.
distance = rate*time
d = r*t
100 km = (x kph)*(y hr)
100 = xy
y = 100/x
This will be used to substitute in later.



Now consider when he goes 10 kph faster, so his travel duration is 30 minutes = 30/60 = 0.5 hours less. 
His old speed x is now x+10 kph.
The old time duration y is now y-0.5 hours.
So,
d = r*t
100 = (x+10)*(y-0.5)
100 = (x+10)*(100/x - 0.5) ........... replace y with 100/x


I'll let the student handle the scratch work from here.
Solving for x should give x = -50 and x = 40. 
Ignore the negative solution because a negative speed makes no sense.


If the man travels at 40 kph, then he needs 100/40 = 2.5 hours to travel the 100 km.
If he travels instead at 40+10 = 50 kph, then he needs 100/50 = 2 hours, which is 0.5 hrs = 30 min early compared to the 2.5 hour figure. This confirms we have the correct present speed.



Answer: <font color=red>40 kph</font>
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