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6*6=36 medals
6 days
1st day=1+35/7=6 remaining 30 medals
2nd day=2+28/7=6 remaining 24 medals
3rd day=3+21/7=6 remaining 18 medals
…
6th day 6 medals
let,first num be x and second num be y
hcf =264
lcm=44
given equation
x/2=44 i.e x=88
product of 2 numbers is equal to lcm*hcf
88*y=44*264
y=132
pass English =80%
fail=100-80=20%
pass Math = 70%
fail in Math=100-70=30%
fail in both=10%
total fail students= fail in Eng+fail in Math-common
= 20+30-10=40%
if 40% fail then 60% will pass
let total students=x
hence
60% 0f (total students)=144
60/100 of (x)= 144
x=(144×100)/60
X=240
total students=240
let first digit be ‘X’
then 5th digit is ‘3X’
let 2nd digit be ‘Y’
then 3rd digit is ‘Y-3’
and 4th digit is ‘Y+4’
then the no is ‘(X)(Y)(Y-3)(Y+4)(3X)’
from the above we can say 3X<=9
so X<=3 and any of the digit in the number is <=9
and also given that 3 pairs sum is 11...
so make trial and error..
if X=1...any of the no is 10 which is wrong trial....
if X=2...then let Y+4=9 ==> Y=5
then no is 25296
first pair 2+9=11
second pair 2+9=11
third pair 5+6=11
now the answer is 25296
friendly environment, supportive colleagues and place where i’ll have to work.
1.5
Triangular Prism
let d distance and s be speed.
d/7 = x and d/5 = x+12
solving we get d=210.
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”
(N * 1.1) * 0.9 = 7920
N=8000
Answer is 33%
{35 % of all the employees are men.
so 65 % (100-35) of all the employees are women.
35*20%=7% of the men attended the annual company picnic and
65*40%=26% of the women attended the annual company picnic.
so 33%(7+26) of all the employeeswent to the picnic.}
data inadequate