Question 770904
Carolyn drove from her home to school at 40 mph and was 10 minutes late. The next day she drove to school at 60 mph and was 10 minutes early. How far is far Carolyn's school from her home?


Let the required time to reach be t hours

Part I
speed = 40 mph
10 minutes = 10/60 hours = 1/6 hours
Actual time taken = with speed of 40 mph = t+1/6
Distance = speed * time 
{{{40*(t+(1/6))}}}

Part II
speed = 60 mph
time = t-(1/6)

{{{distance = 60 * (t-(1/6))}}}


{{{40*(t+(1/6))= 60 *(t-(1/6))}}}

Distance is same. so equate distance


{{{40*(6t+1)/(6)=60*(6t-1)/(6)}}}

multiply by 6
40((6t+1))=60(6t-1))

240t+40=360t-60

360t-240t=60+40
120t=100
t= 100/120
t= 5/6 hours 

Calculate distance

{{{D= 40*(t+(1/6))}}}
plug t
D= 40*(5/6 +1/6)
=40 miles

Distance from home = 40 miles