Vatglobal (vat it group) Interview Questions, Process, and Tips

I am looking for a qualified chartered accountant in my profile who is known for his excellence in the field.

The series is 15, 16, 34, 105, 424, 2124, 12576
1) (15*1) + 1 = 16
2) (16*1) + 2 = 34
3) (34*3) + 3 = 105
4) (105*4) + 4 = 424
5) (424*5) + 5 = 2125
Clearly, they gave 2124 instead of 2125, so it is the wrong number in the series.

.

A

6561 → 6560/2 [-1]
3280
1640
820
410
205 [+1] =206
103 → [-1] = 102
51 [+1] = 52
26
13 [-1] = 12
6
3 [+1] 4
2
1

total sum = 205 Rs.

33

Ok

8+1=9
9+2=11
11+3=14
14+4=18
18+5=23
and finally
23+6=29

sample space=36
prob of one of the dice getting face 6
i.e n={(1,6)(2,6)(3,6)(4,6)(5,6)(6,1)(6,2)(6,3)(6,4)
(6,5)(6,6)}=11
p=n/s
p=11/36
ans:11/36

1:2

– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.

– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m

To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec

In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m

at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m

If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)

Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.

For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.

This will take an additional (9.6 + 2) seconds.

Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds

75/8 days

LCM=48
Then add remainder 3
ans =51