So if u travel 5 km uphill and then 5 km on plain and then 5 km on
downhill then time taken
by you will be equal to 15 km on the plain road
This is not the truth.
5/72 + 5/64 + 5/56 - 15/64 = 5/72+5/56-10/64 = 10/63-10/64 > 0
(that is solely due avg of speed of downhill and uphill is = speed on
plain road)
This only leads to:
if u travel 5 hrs uphill and then 5 hrs on plain and then 5 hrs on
downhill then distance traveled
by you will be equal to travel 15 hrs on the plain road.
On 2010-9-15 15:07, rahul patil wrote:
the solution seems to be simple.
Just try to imagine what is happening
You have a road with downhill and uphill.
So if u travel 5 km uphill and then 5 km on plain and then 5 km on
downhill then time taken
by you will be equal to 15 km on the plain road(that is solely due avg
of speed of downhill and uphill is = speed on plain road)
so the from A to B we reach 40 min earlier due to there more downhill
road.
while from A to B it is uphill.
So let us take x km as the road distance which is not plain.
t1 = time to travel x on downhill = x/72
t2 = time to travel x on uphill = x/56
but as given 40min = 2/3 hr = x/56 - x/72
so, x= 168.
so it will take 3 hrs to climb while travelling from B to A and plain
road distance = 5/3 * 64 = 106.67 km
dist = 168 + 106.67
On Wed, Sep 15, 2010 at 8:21 AM, Terence <[email protected]
<mailto:[email protected]>> wrote:
You could also get a unique solution if the car has speed of 72 63 56
in downhill, plain and uphill respectively.
I think the speed Vd, Vp, Vu was chosen so that 2Vp = Vd + Vu.
But for unique solution, it ought to be 2/Vp = 1/Vd + 1/Vu.
Under this condition, we can get the unique S=x+y+z:
From
x/Vd + y/Vp + z/Vu = T1
x/Vu + y/Vp + z/Vd = T2
We get (1/Vu+1/Vd)(x+z)+2/Vp*y = T1+T2
Apply 2/Vp = 1/Vd + 1/Vu, then 2/Vp(x+y+z)=T1+T2
S=x+y+z = Vp(T1+T2)/2
On 2010-9-15 9:31, Gene wrote:
This isn't right. Dropping both y terms is the same as
setting y to
zero. The answer you get is correct, but there are many
others as has
been said.
You could get a unique solution if the route were constrained
to be
monotonic (level and up or else level and down).
On Sep 14, 4:28 pm, Minotauraus<[email protected]
<mailto:[email protected]>> wrote:
Actually the solution is unique. The middle part with the
Ys is the
same and therefore can be omitted out. Now you are left with
2 equations and 2 unknowns.
I used time in minutes and I have x = 1.28, z = 0.30476
units (y can
be found out).
I guess the trick was 1. to write the equations that Yan did
and 2. to recognize that the plain part is the same and
hence can be
cancelled.
On Sep 14, 3:31 am, Yan Wang<[email protected]
<mailto:[email protected]>> wrote:
actually, there are many solutions, just pick up one
from them...
On Tue, Sep 14, 2010 at 3:23 AM, Abhilasha jain
<[email protected]
<mailto:[email protected]>> wrote:
how can u solve 3 variables using 2 equations?
On Tue, Sep 14, 2010 at 3:44 PM, Yan
Wang<[email protected]
<mailto:[email protected]>> wrote:
x/72 + y/64 + z/56 = 4
&
x/56 + y/64 + z/72 = 4+2/3
find a solution to this ...
On Tue, Sep 14, 2010 at 2:31 AM,
bittu<[email protected]
<mailto:[email protected]>> wrote:
Amazon Interview Question for Software
Engineer / Developers
A car has speed of 72 64 56 in downhill,
plain and uphill
respectively . A guy travels in the car
from Pt. A to pt. B in 4 Hrs
and pt. B to pt. A in 4 Hrs and 40 min.
what is the distance between A
and B?
Regards
Shashank
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