Clark, schaefer, hackett & co. Interview Questions, Process, and Tips

A=P+SI——(1)
sum of money after 30 years = Double the money
A=2P———-(2)

Equate 1 and 2
P+SI=2P
SI=2P-P
SI=P————-(3)

WKT, SI= PTR/100

Equate 3 and SI

P=PTR/100
P=P*30*R/100
R=P*100/P*30
R=100/30 or 10/3 or 3(1/3) %

I think experience because if experience brings pay. If someone having experience in any work then he/she can work easily. Although pay is important but not more then experience

1600 years contain 0 odd day.
300 years contain 1 odd day.
94 years = (23 leap years + 71 ordinary years)
= (46 + 71) odd days
= 117 odd days, i.e., 5 odd days
Days from 1st January 1995 to 28th February 1995
= (31 + 28) days = 59 days
= (8 weeks + 3 days) = 3 odd days
∴ Total number of odd days
= (0 + 1 + 5 + 3) = 9 odd days i.e., 2 odd days.
So, the required day is Tuesday.

Simple interest Formula:
A=P(1+rt)
Therefore,
815=P(1+3r)
P+3r=815

854=P(1+4r)
P+4r=854
Solve the sums using Elimination method.
P+4r=854
P+3r=815
r=39.
Toget Principal
P+4r=854
P+4*39=854
P=854-117
P=698

B

32

13

42m

Capacity of the tank =(12 x 13.5) liters =162 liters.

Capacity of each bucket =9 liters

Number of buckets needed = 162/9 =18.

if rs 12 and rs 15 per meter then 750*12+750*15

The answer cannot be determined as there is a particular formula where the consecutive numbers start.

Ans :15
X be a present age of mudit
After 18 year age of mudit is X+18
4 year ago age was X-4
As per question,
X+18=3(X-4)
X+18=3X-12
3X-X=18+12
2X=30
X=15

Explanation:

The number of ways of selecting three men, two women and three children is:

= ⁴C₃ * ⁶C₂ * ⁵C₃
= (4 * 3 * 2)/(3 * 2 * 1) * (6 * 5)/(2 * 1) * (5 * 4 * 3)/(3 * 2 * 1)
= 4 * 15 * 10
= 600 ways.

A.