Centenary bank Interview Questions, Process, and Tips

Given:

In a group of 15 students,

7 have studied Latin,

8 have studied Greek,

3 have not studied either.

To find:

The number of students who studied both Latin and Greek.

Solution:

In a group of 15 students, have studied Latin, 8 have studied Greek, 3 have not studied either.

Therefore,

n(A∪B) = 15 – 3

n(A∪B) = 12

7 have studied Latin,

n(A) = 7

8 have studied Greek,

n(B) = 8

n(A∩B) is the number of students who studied both Latin and Greek.

n(A∩B) = n(A) + n(B) – n(A∪B)

n(A∩B) = 7 + 8 – 12

n(A∩B) = 15 – 12

n(A∩B) = 3

The number of students who studied both Latin and Greek is 3

Final answer:

3 of them studied both Latin and Greek.

Thus, the correct answer .3

Tree

8400 ? 120 x 15 + 150 = ?
ans:
8400 % 120 x 15 + 150 = ?
1200

let the average of 25 person be x

total age of a person=average×no.of person

total age of 25 person=X×25

=25X

when a new person of 46kg come average decrease by 5kg(X-5)

Total age of 26 person =25x+46

Average age of 26 person=X-5

A/Q

.25X+46=26(X-5)

25X+46=26X-130

26X-25X=130+46

X =176

hence the av. of 25 person be 176kg

1min = 60 seconds
6 min=6*60=360seconds
4 sec=10
360sec=360*10 /4=900

Ans = 8
Use simple box method
[1][2][2][2]= 1x2x2x2 =8
Logic >
once place have 2 number (2,6)
Tens place have also 2 number (7,3)
So that number is divisible by 4
Now, 2 numbers fixed at once and tens place left 2 numbers which is choice at hundredth place
Now 3 number fix left 1 number which is placed at thousands place
Total number form is 4×4 = 16
But divisible by 4 is 1x2x2x2 =8

There should be 5(2+2+1) decimal places in the answer and the decimal place should end with 2(3*2*2=12) so the correct answer is a, no need for calculator. You can rule out all the other options because it does not satisfy this condition.

2.01

9 sec=4 times
x sec=12 times
x=(9*12)/4=27 sec

1.5 km/hour

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