Demica Interview Questions, Process, and Tips

4) QEOKTWG

4/5

Let the journey distance be : x miles
Now x/40+x/30=8 (time=distance/speed)
Solving this for x we get. x=960/7
therefore, x=137(approx)
(137.14 exact)

to find the root of f(x) = 0;

petticoat

d

c

(x**2 – 6* x + 5) = (x-1)*(x-5)
(x**2 + 2 * x + 1) = (x + 1) * (x+1) = (x+1)**2
For what x is (x-1)*(x-5)/( (x+1)**2) a minimum?
One way to answer this question is by using calculus.
Take the derivative, and set to zero.
Since this is a fraction of polynomials, and a fraction is
zero only if it’s numerator is zero, we need calculate only
the numerator of the derivative and set it to zero.
The numerator of the
Derivative of (x-1)*(x-5)/( (x+1)**2) is
( (x-1) + (x-5) ) ( x+1)**2 – (x-1)(x-5)( 2 (x+1) )
= (2 x – 6) (x+1)**2 – (2) (x-1)(x-5) (x+1)
= 0
Divide through by 2 (x+1)
(x-3)(x+1) – (x-1)(x-5) = 0
(x**2 – 2 x – 3 ) – (x**2 – 6 x + 5) = 0
x**2 – x**2 – 2 x + 6 x – 3 – 5 = 0
4 x – 8 = 0
x = 2
Plugging in x = 2 into the original
(x**2-6*x+5)/(x**2+2*x+1)
gives us (2**2 – 6 * 2 + 5)/(2**2 + 2*2 + 1)
= (4 – 12 + 5) / (4 + 4 + 1) = -3/9 = -1/3
Least value is -1/3

A’s speed=6 mph ,B’s speed=8 mph
Let, after x hrs, they will meet.
so, the distance traveled by A in x hrs should be the same as the distance traveled by B in (x-1/2)hrs [as B started the journey after 30 min of A]
Thus, 6x=8(x-1/2)[as distance=speed*time]
=>8x-6x=4
=>2x=4
=>x=2
after 2 hrs they will meet so time=(9+2)=11.00 a.m

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