St. joseph's indian school Interview Questions, Process, and Tips

A 1×1 cube in the middle of an edge of the 3×3 cube will
have two faces painted. A cube has 12 edges, so the answer
is 12.

24.
exp: 0 is formed by multiplying 5 with 2. so first we find
how many 5’s and 2’s are there in 100!.
No of 5’s : 100/5=20/5=4/5.
20 +4 =24.
No of 2’s : 100/2=50/2=25/2=12/2=6/2=3/2=1/2.
50 +25 +12 +6 +3 +1.
It has 24 5’s and 2’s.
so the no. of zeros=24.

128 is incorrect. There is a difference of multiples of 3 in the number pattern, which starts at 24 and reduces by 3 each time. Therefore the correct pattern would read 190, 166, 145, 127, 112, 100, 91.
this would leave the pattern with differences in the pattern 24, 21, 18, 15, 12, 9. The next number in the pattern would be 6 lower than 91. 91 – 6 = 85. Then 85 -3 – 82, followed by 82 – 0 = 82. After this, if -3 were taken the pattern would start to rise so 82 – – 3 = 85.

1 ball – 4 runs (1st batter – 94+4=98)
2 ball – 3 runs and one run is short run during running between the wicket (1st batter- 98+3-1 =100)and strike changes for 2nd batter
3 ball – 6 runs ( 2nd batter 94+6=100)

A. Rs 140
7+8+11=26
B: 7/26*520= 140

2

8 days

first 10 odd num are 1,3,5,7,9,11,13,15,17,19
sum=100
avg=100/10
avg of 10 odd num =10

1(100-1)+ 2(100-2)+…..99(100-99)
N=100
solution: N(N-1)
100(99)= 9900

47

A will finish work in 8 days (4days with B and 4 days with C).

The probability will be
white ball = 5/9
red ball = 4/9
The answer can be 5/9 and 4/9 or 1

145.2

12

Mm