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16, 25, 36, 72, 144, 196, 225
C. 21000
11
120
The numbers that lie between 100 and 1000 which are divisible by 14 are 112, 126,140 …,994
a = 112; l = 994, d = 14
n= (l−a)/d+1
= (994-112)/14+1
= 64
Sn=n/2(l+a)
= 64/2(994+112)
= 32*1106
= 35392
5:6::7:10::6:5
=((5/6)/(7/10)/(6/5))
=((5/6)*(6/5)/(7/10))
=(1/(7/10))
=10/7
answer is 28
>2+8=10
>reverse of 28 is 82
>subtract 28 from 82
>82-28=54
the number is 28
Son in law
Mohammad Gauri
Ravi do a work in 30 days
Prakash do a work in 40days
then both work together 1/30+1/40=70/1200
then solve values 120/7 is 17(1/7) days
Out of 10 persons, 4 are graduates; so, (10 – 4) = 6 are under-graduates.
If there is no restriction, any three can be chosen from the ten in (10C3) = 120 ways.
Now, if all three chosen are under-graduates; it can take place in (6C3) = 20 ways.
Therefore, the probability that there will be no graduate among the three chosen = (20 / 120) = (1 / 6).
Therefore, the probability that there will be at least one graduate among the three chosen = {1 – (1 / 6)} = (5 / 6) = 0.8333.
45 km/hr
625
2, 6, 12, 20, 30, 42, 56, (…..)
Difference between 2 and 6 is = 4
Difference between 6 and 12 is = 6
Difference between 12 and 20 is = 8
Difference between 20 and 30 is = 10
Difference between 30 and 42 is = 12
Difference between 42 and 56 is = 14
So ne number will be with 16 Difference i .e 72
Therefore Answer will be 72
but how?
35 is not the factor of 120 so…35 is the answer
One day work = 1 / 20
One mans one day work = 1 / ( 20 * 75)
Now:
No. Of workers = 50
One day work = 50 * 1 / ( 20 * 75)
The total no. of days required to complete the work = (75 * 20) / 50 =
30
Let’s say,
I have x coins of 50 paise and (80-x) coins of 100 paise,so the equation is like this ,
50x + (80-x)*100 = 64*100
x = 32
So ,I have 32 coins of 50 paise
72
Because it is not a square number