Wasaya Airways Interview Questions, Process, and Tips

first fill 3quart pail and pour in 5quart pail.fill again 3
quart pail and fill remaining 2 quarts in 5 quart
pail.hence 1 quart remains in 3quart pail.empty 5quart pail
and pour 1 quart from 3 quart pail. also add another 3
quart from 3 quart pail. hence 1 + 3 quarts = 4 quarts.

Actual price is RS.60 and (plus)he got profit of 10%. So 60+ 10%(60) = 66

(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

Matches played: 60.
Matches won: 30% of 60 => (60*(30/100)) = 18 matches.
Iterative approach:
On adding 1 to matches played and matches won, on every iteration until the win percentage gets to 50. So
19 / 61 = 0.3114754098360656
20 / 62 = 0.3225806451612903
21 / 63 = 0.3333333333333333
22 / 64 = 0.34375
…
…
…
…
Similarly,
41 / 83 = 0.4939759036144578
42 / 84 = 0.5

So, after 60th match 24 more matches has to be played and won to get 50% average winning rate.

5*(root2)

A

3×8=24=2xt ===> t=12hours

.

45

3/7

C