Question 769626
<pre>
Because of bad weather, a bus driver reduced the usual speed along a 150-mile bus route by 10 mph. The bus arrived only 30 minutes later than its usual arrival time. How fast does the bus usually travel?
Ans:
Let the bus' normal speed be x mph.
Normal time taken to cover 150 miles = 150/x hours (since time = distance / speed)
If he reduces the speed by 10 mph, the new speed = x - 10
Time taken at the slower speed = 150/(x - 10) hours
It is given that the new time is 0.5 hours (30 minutes) more than the normal 
time.
So we get the equation
{{{150/(x - 10) = 150/x + 0.5}}}
{{{150*x = 150*(x - 10) + 0.5*x*(x - 10)}}}
{{{150*x = 150*x - 1500 + 0.5*x^2 - 5*x}}}
{{{0.5*x^2 - 5*x - 1500 = 0}}} or
{{{x^2 - 10*x - 3000 = 0}}}
This is a standard quadratic equation which can be solved through factorization
{{{(x - 60)*(x + 50) = 0}}}
The two solutions are:
x - 60 = 0 i.e. x = 60
x + 50 = 0 i.e. x = -50
Since x cannot be negative, x = 60
Normal speed = {{{highlight(60)}}} mph
Check for correctness:
At normal speed, time taken = 150/60 = 2.5 hours
Reduced speed = 60 - 10 = 50
Time taken at 50 mph = 150/50 = 3 hours which is 30 minutes more than 2.5 hrs!
So the answer is correct. Hope you got it :)
</pre>