easy or hard part 3?

Question

rose drove her car from her home to the province and back, a total distance of 120 km. her average speed returning was 3 kph 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?
joel drives his car for 20 km at a certain speed. at a distance, he increases his speed by 5 km per hour and drives for an additional 20 km. if the total trip taken was one hour , what is his original speed?

Answer 1

Alright,

1) THE ROSE DRIVE
We have the distance to be travelled as 120/2 = 60km.
Let s be the average speed while going to the province
And the return trip is s-3 average speed.
time taken to go to the province is 60/s = t1
time taken to return from the province is 60/(s-3) = t2

The sum of them is given as 9 hrs.
ie., t1 + t2 = 9
=> 60/s + 60/(s-3) = 9
=> 60(1/s + 1/(s-3) ) = 9
=> (s-3 + s)/s(s-3) = 9/60
=> 120s - 180 = 9s^2 - 27s
=> 40s - 60 = 3s^2 - 9s
=> 3s^2 - 49s + 60 = 0
=> 3s^2 - 45s - 4s + 60 = 0
=> 3s(s - 15) - 4(s - 15) = 0

hence s = 15 or s 4/3 both in units km/hr

2) JOEL ON WHEELS
The the original speed as s then the time taken for first is 20/s
The next 20km is speed (s-5) and the time taken is 20/(s-5)

both add up to 1 hr
thats 20/s + 20/(s-5) = 1
=> 20 ( s-5 + s) = s(s-5)
=> 40s - 100 = s^2 - 5s
=> s^2 - 45s + 100 = 0
here a = 1 b = -45 c =100
we use the formula x ={ -b +/- sqrt( b^2 - 4ac)}/2ac
2ac = 200
sqrt(b^2 - 4ac) = sqrt(2025 - 400) = sqrt(1624) = 40.3

the roots are x = (45 + 40.3 )/200 => x1 = 0.4765
and x = (45 - 40.3)/200 => x2 = 0.0235

I think I messed up the second one .

Peace out.

Answer 2

Pradnumya is right for the first. for the second equation i get S^2-35s-100=0 then solve for s.

Answer 3

These are even easier then number 2, but again, without a good math background most people are not going to understand the questions enough to try to answer them.