QUT Interview Questions, Process, and Tips

ans: .1432 inch
1-3/4=1/4
1-1/12=11/12
1/8,1/4,11/12 are broken off
so
5/8
5/8/4
5*11/8/4/12
remaining .1432inch

It is assumed to be constant. Now, to cross past the pole, the train should cover a distance of x meters. Now, the time taken by the train to cross a platform of length 100 m is 25 seconds. Hence, the length of the train is 150 m.

2hr 24min

5 trains

GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC

I can join your company with in one month that is my notice period time

34

Ans: 60 kph

Suppose Person meets the train everyday at 3 PM at Station A.

His speed is 12kph.

So normally he reaches 5 km before the meeting point (pt B) at (5/12 hr = 25 min before) 2:35PM.

But if he is late by 30 min, then he will reach that point (pt B) by 3:05 PM.

Train is traveling at its normal speed so it covers the distance of 5 Km in 5 min starting from Station A and reaches the meeting point (pt B) at 3:05 PM.

So speed of the train is 5KM/5min = 60 kph.

find out what is the problem and taking the bad out always works.

From the first statement
Speed=distance/time
=>240/24=10m/s
so speed=10m/s
From the second statement
distance=length of the platform+length of the train=650+240=890m
Time=distance/speed
here, the speed of the train is already calculated
=>890/10=89s
So the time taken by the train to cross the platform is 89 seconds

390

Either 4 or (-)4

let no of total crockery =x
2x/3 crockery broken
x/2 handle broken
x/4 both brocken
2x/3+x/2-x/4 = no of total broken(crockery or handle)
=11x/12
unbroken =x-11x/12=x/12
x/12=2(given in question)
x=24

995