It is true that there are infinitely many solutions (or zero
solutions) (x, y, z) in any case.
But what we are interested here is S=x+y+z.
Apply y = S-x-z, you get:
S/Vp+x(1/Vd-1/Vp)+z/(1/Vu-1/Vp) = T1
S/Vp+x(1/Vu-1/Vp)+z/(1/Vd-1/Vp) = T2
Adding the two, you get
2S/Vp + (x+z)(1/Vd+1/Vu-2/Vp) = T1+T2
When 1/Vd+1/Vu-2/Vp = 0,
S = Vp(T1+T2)/2, no matter what x and z are.
In this condition, All the solution (S,x,z) forms a line: (the
intersection line of above 2 planes)
x-z = (T2-T1) / (2(1/Vu-1/Vp))
S = Vp(T1+T2)/2.
which is parallel to x-z plane.
On 2010-9-16 6:32, Gene wrote:
No. Two linear equations in three unknowns will always yield many
solutions (or zero solutions). These are essentially plane
equations. Two planes intersect in a line (unless they are
parallel). You might get a de facto unique solution for some values
of Vu, Vd, Vu, T1, T2 from the constraints x,y,z>= 0. It would have
to lie on an axis.
For example, you can aways pick z and find x and y. Using your
notation,
x/Vd + y/Vp + z/Vu = T1
x/Vu + y/Vp + z/Vd = T2
Subtract to get:
x(1/Vd - 1/Vu) + z(1/Vu - 1/Vd) = T1 - T2
then
x = [ (T1 - T2) - z(1/Vu - 1/Vd) ] / (1/Vd - 1/Vu)
So now you can pick any z and get x. Once you have both of these,
plug them in here:
y = Vp (T1 - x/Vd - z/Vu)
As long as x,y,z>= 0, you are in business.
On Sep 14, 10:51 pm, Terence<technic....@gmail.com> 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<anike...@gmail.com> 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<wangyanadam1...@gmail.com> wrote:
actually, there are many solutions, just pick up one from them...
On Tue, Sep 14, 2010 at 3:23 AM, Abhilasha jain
<mail2abhila...@gmail.com> wrote:
how can u solve 3 variables using 2 equations?
On Tue, Sep 14, 2010 at 3:44 PM, Yan Wang<wangyanadam1...@gmail.com> 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<shashank7andr...@gmail.com> 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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