The star gold coast Interview Questions, Process, and Tips

Answer is 274 .

Let the required term be x.

Pattern :- 1) Find the difference between consecutive terms as :

2 – 1 = 1

4 – 2 = 2

13 – 4 = 9

31 – 13 = 18

112 – 31 = 81

2) Now write the above numbers obtained by differences in series as :

1, 2, 9, 18, 81, y

(Suppose y be next term after 81.)

3) Look the difference series carefully and it follows the below pattern as :

(9 * 2)/1 = 18

(18 * 9)/2 = 81

(81 * 18)/9 = 162 = y (the next term after 81)

Thus the required answer is :

x = 112 + y

x = 112 +162 =274

7770-7077=693

B

I am Srilaxmi. I born and bought-up in WARANGAL. My father Agriculture cum politician and my Mom homemaker. I blessed with three brothers and one sister. My elder brother was married and working as a TA in MPDO office at Warangal. My first younger brother was also married and working with Sushee Infra Pvt Ltd as a Asset Manager at Hyderabad. Younger brother and sister are into their studies. Coming to me I completed my MBA from ICFAI university in 2008 and worked as a Audit assistant in accounting firm for two years. I had actively participated in bank audits i.e. is in CBI (Concurrent audit) and SBI(Statutory audit) during my services. Later I pursued CS Executive and attempted for five times then I give up to clear my M.Com. At present I was studying postal studies of CS Executive programme. My hobbies are reading books and listening to music.

Balls- B1, B2, B3, B4, B5, B6, B7, B8, B9.
Group1 – (B1, B2, B3), Group2 – (B4, B5, B6), Group3 – (B7, B8, B9)
Now weigh any two groups. Group1 on left side of the scale and Group2 on the right side.
When weighing scale tilts left – Group1 has a heavy ball or right – Group2 has a heavy ball or balanced – Group3 has a heavy ball.
Lets assume Group 1 has a heavy ball.
Now weigh any two balls from Group1. B1 on left side of the scale and B2 on right side.
When weighing scale tilts left – B1 is the heavy or tilts right – B2 is the heavy or balanced – B3 is the heavy.

90 degree

300%

99/1020

Petroleum

One day’s work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.

C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.

Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.

Remaining Work = 7/10, which was done by A,B and C in the initial number of days.

Number of days required for this initial work = 7 days.

Thus, the total numbers of days required = 4 + 7 = 11 days.

.

1 1 2+4
2 2 1+4
3 1+2 4
4 4 1+2
5 4+1 2
6 4+2 1
7 4+2+1 0

13km