To determine the day of the week for 8th February 2004, we need to consider the number of days between 8th February 2004 and 8th February 2005. Here's how we can solve this:
The day of the week on 8th February 2004 was Sunday.
(A + B)'s 2 day's work = 2 x (1/3) = 2/3
Remaining work = 1 - (2/3) = 1/3
A will complete 1/3 work in 2
A will complete 1 work in 6
A's 1 days work = 1/6
B's 1 day's work = (1/3) - (1/6) = 1/6
? B will take 6 days to complete the work alone.
Distance covered in one revolution = total distance travelled / total number of revolution.
= ( 88 x 1000) / 1000 m
= 88 m
We know that the distance covered in one revolution = circumference of the wheel.
? ?d = 88
? 22d / 7 = 88
? d = 28 m
So, 36 is wrong.
Given expression = (.896 x .752 +.896 x .248) / (.7 x .034 + .7 x.966)
=.[896 x (.752+.248)] / [.7 x (.034+.966)]
= .(896 x 1) / ( .700 x 1)
=896/700
= 1.28
Average speed =2 x 40 x 60 / ( 40+ 60)
= 4800 / 100
= 48 km/hr
Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
After 61 days, it will be Saturday.
9548 16862 = 8362 + x + 7314 x = 16862 - 8362 ----- = 8500 16862 -----
We know that
BG = SI on TD = (240 x 30 x 1 x 1)/100
= ? 72
BG = BD - TD
? BD = BG + TD
= 72 + 240
= ? 312
If previous year is leap year then calendar of May is similar to July
In tossing a coin 2 times the sample space is 4 i,e (H, H), (H, T), (T, H), (T, T)
(1) If A1 denotes exactly one head
then A1 = {(H, T) (T, H) }
So, P(A1 ) = 2/4 = 1/2
(2) If A denotes at least one head
then A = {(H, T) (T, H) (H, H)}
? A = {(H, T) (T, H) (H, H )}
? P(A) = 3/4
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