Dupaco community credit union Interview Questions, Process, and Tips

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.

GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC

Ok

1×2×…100=100!
Number of zeros in product of n numbers =[5n​]+[52n​]+[53n​]+…
Number of zeros in product of 100 numbers =[5100​]+[52100​]+[53100​]
where [.] is greatest integer function
=[20]+[4]+[0.8]=20+4=24

Calculation:
⇒ If 1000 divided by 112, the remainder is 104. ⇒ 112 – 104 = 8 ⇒ If 8 is added to 1000 it will become the smallest four-digit number and a multiple of 112. ⇒ 1000 + 8 = 1008 ∴ The required result will be 1008.

I believe it will be 30 km/hr

24 times

2nd

copy cat

64

2100

17