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5/55*100=9.09% defective eggs
75%
c
17,19,23,29
4400+(97 leap year )=4497
33
500 ways
Let Suvarna, Tara, Uma and Vibha be S,T,U,V respectively
initially in the beginning each persons share be
V = x U = y T = z
S = w = (x+y+z+32) Reason: She has to double others share, so she should have each and everyone’s share and still should be left out with 32
after 1st Round of game
S loses and is out with 32 and doubles the others share
V = 2x U = 2y T = 2z
After 2nd Round of game
T loses and is out with 32 and doubles the others share
V = 4x U = 4y
This means T had 2z = 2x + 2y + 32
After 3rd round of game
U looses and is out with 32 and doubles others share
V = 8x
This means U initially has 4y = 4x + 32
In the end V = 8x = 32
Solving this we get x = 4, y = 12, z = 32 and w = 80
There fore Suvarna had highest share in the beginning
First write equations from info:
(A) (Mon + Tue + Wed)/3 = 111 Rearrange as ——–> Tue + Wed = 111 – Mon
(B) (Tue + Wed + Thu)/3 =102 Rearrange as ——–> Tue + Wed = 102 – Thu
(C) Thu = 0.8(Mon)
Substitute equation C into B:
(B) Tue + Wed = 102 – 0.8(Mon)
At this point I changed the values for clearer algebra:
Mon = x
Tue + Wed = y
Re-write equations A & B with new values:
(A) y = 111 – x
(B) y = 102 – 0.8x
Solve simultaneous equations:
111 – x = 102 – 0.8x
111 – 102 = x – 0.8x (Re-arraged)
9 = 0.2x
x = 45
Thus, Mon = 45C
Thu = 0.8(45)
Thu = 36C
So the answer is it was 36C on Thursday
445, 221, 109, 46, 25, 11, 4
C.46
4+7
11+(7*2)
25+(7*2*2)
53+(7*2*2*2)
109+(7*2*2*2*2)…..
40
A certain sum amounts to Rs. 1725 in 3 years
and amounts to Rs.1875 in 5 years
so interest of 2 years = 1875 -1725
= 150
so interest of 1 year = 75
so interest of 3 years = 75 × 3 =225 rs
so , Principal = Amount – SI
= 1725 – 225
= 1500 rs
now ,
S.I. = P × N × R /100
75 = 1500 × 1 × R /100
R = 75 / 15
R = 5%
24 days
q
4