Let the uphill distance be D1 with speed S1=3miles/hr and downhill distance be D2 with speed 8 miles/hr

D1/3+D2/8=40/60 hrs
=>8D1+3D2=16------Eq(1)

also given she travelled uphill and downhill at an avg of 4 miles/hr
=>total distance/Avg speed=40/60hrs
=>(D1+D2)/4=2/3
=>3D1+3D2=8-------Eq(2)
solving equations 1 and 2
D1=8/5 mile and D2=16/15 miles
=>D1=1 3/5 miles

the formula for average speed s1*s2/s1+s2........can be used only when uphill distance and downhill distances are same which is not in this case..

Hope it is helpful

But aren't the uphill and downhill distances actually the same?
Are we assuming is the same hill no?

Cheers
J

Easy question

3x + 8y / x+y = 4
x + y = 40

So solving y = 8
x = 32

So D is out correct answer choice

Hope it helps
Provide kudos!

Cheers
J

I used weighted avg to get uphill to downhill ratio. U/D = 4/1. i.e. U = 4D.
Uphill to total will be 4/5.

Does 4/5 ratio only apply to the total time OR does it also apply to the total distance.
I mean, can the total distance also be split in this ratio?

I got total dist as 8/3 miles. (total time 40/60 * avg speed 4mph)
will the uphill distance be 4/5 of 8/3...It does not seem so...

Could somebody pls clarify whether 4/5 ratio applies only to the time and why should it not be applied to the total distance?

Thanks

PS - In the above post we have "if total time is 40 minutes then the time spent uphill must be 4 times that spend downhill".
I would like to know whether the same ratio applies to total distance also or just the time.

let D=uphill distance
D/3 hours=uphill time
(4 mph avge speed)(2/3 hour total time)=8/3 miles total distance
8/3-D=downhill miles
(8/3-D)/8=downhill time
D/3+(8/3-D)/8=2/3
D=8/5➡1+3/5 miles

let t=uphill time
3t+8(2/3-t)=8/3 total miles
t=8/15 hours
(3 mph)(8/15 hours)=1 3/5 uphill miles

D=Distance
R=Rate
T=Time
T1= Time 1 = Uphill
T2 = Time 2 = Downtown