Griffin hospital Interview Questions, Process, and Tips

Each PAIR of stations means a PAIR of tickets (A to B and B to A)

2(old stations)(new stations) + 2(new stations)(new stations – 1) = 46

(N * X) + (X * (X – 1)) = 23

factoring ___ X (N + X – 1) = 23

23 is a prime number with only two factors, 1 and 23

so N = 23 and X = 1

1/5

55967

( a ) BQDCJCMF
“TERMINAL” split it into the half we get “TERM” “INAL”

Now the first half is decreased by one and the next half is
increased by one, so we get:

“SDQLJOBM” (pls note in the ques we have “SDQIJOBM” where
‘L’ shud have come instead of ‘I’)

so “CREDIBLE” is to “BQDCJCMF”

10000*8:10000*12
2:3

5x=25000
x=5000

p:q
10000:15000

108

12%

5*2=10
10*5=50
50*10=500
500*50=25000
So odd one is 5000

2:3

=> 3s = d
=> 5(s-9) = d

therefore
5s-45=3s
2s = 45
s=22.5

hence
d= 67.5

3*2-1=5
5*2-2=8
8*2-3=13
13*2-4=22
22*2-5=39
39*2-6=72
so series is 3,5,8,13,22,39,72
24 should be replaced by 22
24 is wrong…….

the answer is 45 km/h
Define x:

Let the total distance be x.

Find the time needed in terms of x for speed at 30 km/h:
Find the time needed in term of x for 40 km./h:
Find the difference in time:
Solve x:
Find the total time needed to for the whole journey:
Given that he was 40 mins late:
Find the minimum speed needed:
Answer: The minimum speed is 45 km/h

Monday =y cheques
Tuesday =3y cheques
wednesday=4000 cheques
totaling to 16000 cheques
therefore y+3y+4000=16000 cheques
4y+4000=16000 cheques therefore 4y=16000-4000 cheques= 4y=12000cheques
therefore y=12000/4=3000 cheques
monday=3000 cheques
tuesday=3×3000=9000cheques

Assuming the youngest has X years, so we have
X+(X+3)+(X+6)+(X+9)+(X+12)=50
5X+30=50
5X=20
X=4
The youngest has 4 years