Delta dental insurance company Interview Questions, Process, and Tips

18th

cut a sphere in all 3 axis with center point as common point
of all axis.
u would be getting 2 lines and an arc
in other words
sphere is cut into 3 cuts in x,y,z directions
it is only similar

Assume there are 52 weeks in one year.
Since he supposed to have a new order for every two weeks, he
needs 52/2 = 26 orders to break the office record.

Now after 28 weeks,he has got only 28/2 -6 = 8 orders.

Hence,he needs 26-8 = 18 new orders in the remaining 24= 52-28 weeks to break the office record.

Compute 24 orders/18 weeks = 4/3 orders/week ,
we see that averagely he has a new order for
every 4/3 weeks in the remaining weeks to break the office record.

Friday

= (1 * 5! * 2!) * 3 + ( 2! * 5 ! * 1) * 3 + ( 3! * 5! ) + ( 5! * 3! )

Thanq saddha

d

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

1/x

7:8

12

Let the number of males be given the name M.
Let the number of females be given the name F.
If 15 females are absent, then M will be twice that of
present females.
This means that M = 2 * (F – 15)
M = 2 * F – 30.
or 2 * F – M = 30.
Now if in addition to the 15 females being absent, we also
have 45 males being absent,
then this gives the equation,
(F – 15) = 5 * (M – 45)
which simplifies to
F – 15 = 5 * M – 225
5 * M – F = 210
Pulling the equations together, we get
5 * M – F = 210
-M + 2 * F = 30
Multiply the first equation by 2, and keep the second
equation as is.
10 * M – 2 * F = 420
– M + 2 * F = 30
Add the equations.
9 * M = 450
M = 50
Verify answer.
Calculate F
from – M + 2 * F = 30
-50 + 2 * F = 30
2 * F = 30 + 50
F = 40.
If 15 females are absent, then number of males will be twice
that of females.
40 – 15 = 25.
50 = 2 * 25. Confirmed.
If also 45 males were absent, then female strength would be
5 times that of males.
Female strength is 25 due to the 15 females being absent.
50 – 45 = 5.
25 = 5 * 5. Confirmed.

105/29 mins

1 : 2

Assume Q can do the work alone in Q days:

work of a day by both P and Q =>
(1/15) + (1/Q ) = 1/6
solving this Q = 10.