Ucl centre for blockchain technologies (cbt) Interview Questions, Process, and Tips

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)

sister

area is given by Area of Rectangle = Length of Rectangle*(sqrt((Diagonal of Rectangle^2)-(Length of Rectangle^2)))
so substituting we get area=4*(sqrt((25-16)))
area=4*(sqrt(9))
since area cannot be negative hence we take sqrt of 9 as 3 (as sqrt of 9 is 3 and -3 as well)
area=4*3=12m^2

56.25%

a)0
b)206
c)0
d)250
e)39
f)92

North-east

let x be lenghth of candles.

thicker one—- in 1 hr 1/6th will go.
thinner one— in 1 hr 1/4th will go.

let he lit for n hrs.

so in n hrs thicker one goes n/6.
and thinner one goes n/4.

so remainini length are x-n/6 and x-n/4.so x-n/6=2(x-n/4)
implie n=3x. if x=1 n=3hrs.

D

40

3630