- Epsom and st helier university hospitals nhs trust General Aptitude Interview Questions
- Epsom and st helier university hospitals nhs trust Trainee Interview Questions
- Epsom and st helier university hospitals nhs trust Personal Questions round Interview Questions
- Epsom and st helier university hospitals nhs trust HR Interview Questions
- Epsom and st helier university hospitals nhs trust Lead Interview Questions
No of men employed = N
No of days to finish the work = 9 days
No of men after increase = (N + 10)
No of days to finish the work = 6 days
Equating mandays
9N = (N+10)*6
9N — 6N = 60
3N = 60
N = 20
No of men employed = 20
let the no be x
then given that the difference between the number and 3/5th
of the number is 50
therefor , x-3/5x=50
2x=250
x=125 is the answer
1×2×…100=100!
Number of zeros in product of n numbers =[5n]+[52n]+[53n]+…
Number of zeros in product of 100 numbers =[5100]+[52100]+[53100]
where [.] is greatest integer function
=[20]+[4]+[0.8]=20+4=24
Total game played= 60
%won =30%
Total won= 60*30/100 i.e. 18
now team plays x games and win all of those to increase the
average to 50%.
So,
(60+x)*50/100=18+x
(60+x)/2=18+x
60+x=36+2x
24=x
So the final answer is 24.
Solution # 15-18
Sitting solution be- D T M B H R
Ans 15 :Mr. D
ans 16 : 1 2 3
Ans 17 : 3
Ans 18 : Mr B Suffering from anemia.
answer is one
x+8y=20, x=-3y
(-3y)+8y=20
5y=20
y=4
The answer for y is 4
350m
Kerala
avg of 10nos.=23==>23*10=230
if each no increased by 4 ==> 4*10=40
then new avg is giveen by : 230+40=270
270/10=27
hence the new avg =27
740
40
x – 30 = 1/4 x || *4
4x – 120 = x || -x + 120
3x = 120 || /3
x = 40
25
Example 1:
Assign, A=20, B=10, C=5, D=5(Because C is equal to D as
given), E=1.
A/B = 20/10 = 2. So A/B = 2
A/C = 20/5 = 4. So A/C = 4
A/E = 20/1 = 20. So A/E = 20
Therefore “A/E is Greatest”
Example 2:
Assign, A=100, B=50, C=20, D=20(Because C is equal to D as
given), E=10.
A/B = 100/50 = 2. So A/B = 2
A/C = 100/20 = 5. So A/C = 5
A/E = 100/10 = 10. So A/E = 10
Therefore “A/E is Greatest”