Northern health and social care trust Interview Questions, Process, and Tips

Initially (2*33 + 1*24)/(2+1) = 30p a gm
So, cost for 100g = 100*30 = 3000p

Later, (1*33 + 2*24)/(1+2) = 71/3 p a gm.
So, cost for 100g = 100*(71/3) = 2366.67p

He saves 3000 – 2366.67 = 633.33p

Concept of weighted average.

20 minutes

@ Aprameya Shyam
length=x does not imply breadth=36-x
2x+4y=36.
3 possible answers are
(2,8) -> 16sqcm.
(4,7) -> 28sqcm.
(6,6) -> 36sqcm

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

d / (x+1/2) = 4/5 * t
and d / (x-1/2) = 5/2 + t
solving we get d = 15.

Tuesday

50 paise coins= 400 = Rs 200
20 paise coins = 600 = Rs 120
10 paise coins= 800 = Rs 80

x

30 days of earning =300 rs
No of days he was absent = x

Equating the abv
300 – 10 x – 2 x = 216
After solving for x you will get 7days

at least 2, at most 4….
2 if you get lucky the first 2 times you draw a jelly bean, 4 would be if in the first 3 tries you get one of each color and the next draw will get you 2 of the same color

Might be using dual E1 in that case will be much more helpful.

11

Answer: 3121 gold coins
Let total no of coins be M
Let the disbursement D to each son:

D1 = 1 + (M – 1)/5 = (M + 4)/5
D2 = 1 + ( M – D1 -1)/5 = (D1) * 4/5
D3= (D2) * 4/5
D4= (D3) * 4/5
D5= (D4) * 4/5
Total disbursements to sons=
= ∑D= (M+4)*1/5[ 1+4/5+(4/5)(4/5)+ (4/5)(4/5)(4/5)+(4/5)(4/5)(4/5)(4/5) ]
= (2101/3125)*(M+4)
Thus balance left for daughters =M-{(2101/3125)*(M+4)}
=(1024M-8404)/3125
This balance should be a positive integer ( assuming M and all disbursements are full coins )
Thus 1024M-8404 should be a multiple of 3125….so….
1024M – 8404 = N*3125 where N is an integer
Using Python code:
n=int(input(“Enter num n: “))
X=int()
a=int()
a=0
X=’ ‘
for a in range(0,n+1):
a=a+1
X= (3125*a + 8404)/1024
if (3125*a + 8404)% 1024== 0:
print(X,a)
Enter num n: 10000
3121.0 1020
6246.0 2044
9371.0 3068
12496.0 4092
15621.0 5116
18746.0 6140
21871.0 7164
24996.0 8188
28121.0 9212
We get minimum value of N = 1021 and M = 3121 gold coins

south east

c

C)1440
2*2=4
4*3=12
12*4=48
48*5=240
240*6=1440

TEACHER is written as VGCEJGT, how would DULLARD ?
T value is 19
V value is 21 ,
here we get to know that there is difference of 1 letter between them.
Answer: FWNNCTF.