31st Dec, 2011 = (2010 years + Period from 1.1.2011 to 31.12.2011)
Odd days in 2000 years = 0
10 years has 2 leap years + 8 ordinary years.
Number of odd days in 10 years ( 2 x 2 + 8) = 5 odd days.
31st,Dec => complete year of 2011 (non-leap year) = 1 odd day.
Total number of odd days = (0 + 5 + 1) = 6.
Given day is saturday.
4th observation i.e., 8 is the median.
47/10000 = .0047
.02 =(2/100) x 100% =2%
Let N / 11 = 233
Then, N = 233 x 11 = 2563
? Missing digit is 5.
121012 = 12 x 10084 + 4
? remainder = 4
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
? = 750.0003 ÷ 19.999
? ? ? 750 ÷ 20
? ? ? 375 ? 38
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