Corbion Interview Questions, Process, and Tips

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tan 30=h/x
tan 15=h/100+x
so h/tan 30=(h/tan 15)-100
=33.33app

According to the question:
⇒(x+5)(y−10)=300
⇒xy+5y−10x−50=xy
⇒5×300​−10x−50=0
⇒−150+x2+5x=0
⇒x2+15x−10x−150=0
⇒x(x+15)−10(x+15)=0
⇒x=10 or −15
Price cannot be negative
So, As x>0,x=10.

400

2*3*5*7*11*13*17*19=9699690

127.179(app)
given distance of the train along the wind is 695
and againt the wind is 498
and time = distance/speed
as we know that time is equal in both the cases hence equate
695/s1=498/s2———–(1);
where s1=speed of the plane + speed of the wind
and s2=speed of the plane -speed of the wind
given that speed of the wind is 21k/h
s1=sp+21
s2=sp-21
substitu in eq 1
we get the answer as 27.17(app)

The answer is A)

y1 = 62 Rs/kg
y2 = 72 Rs/kg
y = 64.5 Rs/kg

y2 – y1 = 10 Rs/kg

The distance between the y and y1 is

y – y1 = 64.5 – 62 = 2.5

x1 = (y – y1)/(y2 – y1) = 2.5/10 = 0.25
x2 = 1 – x1 = 1 – 0.25 = 0.75

The target price is calculated by the lever method.

x1 * y2 + x2 * y1 = 0.25 * 72 + 0.75 * 62.5 = 64.5

The ratio is of y1 to y2 is

0.75 : 0.25

Divide by both by 0.25

3 : 1

60*5/18*9=150

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