First, we find out the day on 1.03.2005. Period upto 1.3.2005.
= (2004 yr + Period from 1.1.2005 to 1.3.2005)
? Odd day in 1600 yr = 0
Odd day in 400 yr = 0
4 yr = (1 leap year + 3 ordinary years)
= (1 x 2 + 3 x 1) = 5 odd days
Number of days between 1.1.2005 to 1.3.2005,
= January + February + March
= 31 + 28 + 1
= 60 days = (8 weeks + 4 days)
= 4 odd days
Total number of odd days
= (0 + 0 + 5 + 4) = 9
= 1 weak + 2 odd days
? 1.3. 2005 was Tuesday.
So, Saturday will fall on 5.3.2005.
Hence, Saturday lies on 5th, 12th, 19th, and 26th of March 2005.
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
Subtract 20, 25, 30, 35, 40, 45 from successive numbers. So 0 is wrong.
NA
Each previous number is multiplied by 2.
? 8 m shadow means original height = 12 m
? 1 m shadow means original height = 12/8 m
? 100 m shadow means original height = (12/8) x 100 m
= (6/4) x 100 = 6 x 25 = 150 m
Let 8% of 96 = y of 1/25
? (8 x 96)/100 = y/25
? y = (8 x 96 x 25)/100 = 192
Since the principal is not given, so data is inadequate.
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