St. george's university, grenada Interview Questions, Process, and Tips

share= 5:7
i.e.,
=9600*5/12
= 4000

4:2

427 is the answer
Because
385 means (3+5=8) or (8-3=5) ,
462 means (4+2=6) or (6-2=4) ,
572 means (5+2=7) or (7-5=2) ,
396 means (6+3=9) or (9-3=6) ,
671 means (6+1=7) or (7-6=1) ,
264 means (4+2=6) or (6-2=4) ,
but only 427 is the number in which if we do addition of any of its two digits (e.g 4+2 or 4+7) then its answer doesn’t come 3rd digit
like if we do 4+2 then answer will be 6 so the number should be 426 but it is not
or if we do 7-4 then answer will be 3 (not 2)
so the right answer is 427

Use 3pt. Formula
( (x-x1)/(x2-x1) )=( (y-y1)/(y2-y1) )
We get 150

– 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

((15 X m) + 23)/(m+1)= 16
solving , we get number of matches (m) as 7 (excluding last
match)
so he shud hit 39 runs in last innings to ake average to 18

4 days

B.25

ans is C.
let H be present age of husaband.W be present age of wife.
H+W=91.
now let diff between their ages be x.that is H-W=x.
now when husband is W yaers old, his wife must be W-x years
old.and it is given that H=2(W-x). so 2W-H=2x. and H-
W=x.eliminating x we get 4W=3H. but H+W=91, so solving thse
two H=52 W=39.

13. One group of each = 2+3+5 = 10 and 1 remainder for each = 1+1+1 = 3. 10+3 = 13

500

Let three consecutive even numbers be 2n,2n+2,2n+4
product=2n(2n+2)(2n+4)
If n=1,
product =2(4)(6)=48.
So, all the numbers will be divisible by 48 if change n.

Conclusion 1 follows

Therefore, the number of possible 13 digit numbers using 1, 2, 3, 4, 5 which are divisible by 4

= x 3 + x 2

= (3 + 2)

=

= 244140625 ways

15km/hr