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let,first num be x and second num be y
hcf =264
lcm=44
given equation
x/2=44 i.e x=88
product of 2 numbers is equal to lcm*hcf
88*y=44*264
y=132
Example 1:
Assign, A=20, B=10, C=5, D=5(Because C is equal to D as
given), E=1.
A/B = 20/10 = 2. So A/B = 2
A/C = 20/5 = 4. So A/C = 4
A/E = 20/1 = 20. So A/E = 20
Therefore “A/E is Greatest”
Example 2:
Assign, A=100, B=50, C=20, D=20(Because C is equal to D as
given), E=10.
A/B = 100/50 = 2. So A/B = 2
A/C = 100/20 = 5. So A/C = 5
A/E = 100/10 = 10. So A/E = 10
Therefore “A/E is Greatest”
B. The Secretary, Ministry of Finance is the right answer
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
6
let the third no be x ,
then the first no is 3x ,
second no is 2*3x ie 6x
average is (3x+6x+x )/3 = 20
10x/3 = 20
x= 60/10 = 6
third no is 6 , second is 6x = 36, first no is 3x = 18
largest no is 36 i.e second no
12 %
After the mixing, 1lt of the mixture has 0.8lt of milk and 0.2lt of water. 1lt of milk costs Rs.12 so 0.8lt of milk costs 0.8 x 12 = 9.6
He sells the mixture at Rs.15,
so profit%=(15-9.6)/9.6 x 100 = 56.25%
Let n be the number of days it takes A and B, working together, to finish. And we know B=A+10 and B=3A, so:
3A=A+10
2A=10
A=5
Then B=15
So:
1/A + 1/B=1/n where n is the total amount of days. So:
1/5 + 1/15=1/n
3n+n=15
n=15/4 days
100
the answer is 34
Assume Q can do the work alone in Q days:
work of a day by both P and Q =>
(1/15) + (1/Q ) = 1/6
solving this Q = 10.
30
Smart Question. What was the total gain by both of them, not Just Krishan
Speed = (45×518)m/sec=252m/sec
Total distance covered = (360+140) m = 500 m.
∴Required time= 500×225sec = 40 sec.
brother
3
D. 17 : 3