For two years having same calendar must have two common conditions: both having same length in terms of number of days and first day of the week.
The year 1991 has 365 days,,that is 1 odd day , year 1992 has 366 days , i.e. 2 odd days , year 1993 has 365 days i.e. 1 odd day. the years 1994,1995,1996 have 1 odd day each.
the sum of odd days so calculated from year 1990 to 1996 .
(1+2+1+1+1+1)=7 odd days. 5
hence, the year 1997 will have same calendar as that of year 1990.
Given that, 30 % of A = 20 % of B
? A/B = 20/30 = 2/3
? A : B = 2 : 3
Let initial quantity be Q, and final quantity be F
F = Q(1 - 8/Q)
=> Q = 20
? log5[(x2 + x ) / x] = 2
? log10(x + 1) = 2
? x + 1 = 25
? x = 24
∴ Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).
We know that speed is inversely proportional to time.
Given that, (Speed of A ) : (speed of B ) = 2 : 7
?(Time taken by A ) : (Time taken by B ) = 1/2 : 1/7 = 7 : 2
Let th distance = D
and usual speed = V
According to the question,
D/(3V/4) - D/V = 2
? D/V = 3 x 2 = 6
Time taken to cover the distance with usual speed = 6 h
?2n = 64
? 2n/2 = 26
? n/2 = 6
? n = 12
Distance = 160 km
Relative Speed = 8 + 2 = 10
Time = Distance/Relative speed = 160/10 = 16 h
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.