SOLUTION: ROSE DROVE HER CAR FROM HER HOME TO THE PROVINCE AND BACK,A TOTAL DISTANCE OF 120KM.HER AVERAGE SPEED RETURNING WAS 3KPH SLOWER THAN HER AVERAGE SPEED GOING TO THE PROVINCE.IF HER

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Question 162019: ROSE DROVE HER CAR FROM HER HOME TO THE PROVINCE AND BACK,A TOTAL DISTANCE OF 120KM.HER AVERAGE SPEED RETURNING WAS 3KPH SLOWER THAN HER AVERAGE SPEED GOING TO THE PROVINCE.IF HER TOTAL DRIVING TIME WAS 9 HOURS ,WHAT WAS HER AVERAGE SPEED IN GOING TO THE PROVINCE?
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ROSE DROVE HER CAR FROM HER HOME TO THE PROVINCE AND BACK, A TOTAL DISTANCE OF
120KM. HER AVERAGE SPEED RETURNING, WAS 3KPH SLOWER THAN HER AVERAGE SPEED
GOING, TO THE PROVINCE. IF HER TOTAL DRIVING TIME WAS 9 HOURS,
WHAT WAS HER AVERAGE SPEED IN GOING TO THE PROVINCE?
:
From the given information we can assume that one way; 120/2 = 60 km
:
Let s = her speed going
and
(s-3) = her speed returning
:
Write a time equation: Time = dist%2Fspeed
Go time + return time = 9 hrs
60%2Fs + 60%2F%28%28s-3%29%29 = 9
:
Eliminate the denominators, multiply equation by s(s-3), results:
60(s-3) + 60s = 9(s(s-3))
;
60s - 180 + 60s = 9(s^2 - 3s)
:
120s - 180 = 9s^2 - 27s
:
0 = 9s^2 - 27s - 120s + 180
A quadratic equation:
9s^2 - 147s + 180 = 0
This can be factored to
(3s - 45)(3s - 4) = 0
Two solutions
3s = 4
s = 4%2F3
and
3s = 45
s = 15 km/h this seems to be the most reasonable
:
Speed going: 15 km/h; and speed returning 12 km/h
:
Check in original equation
60/15 + 60/12 =
4 + 5 = 9 hrs