A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in 55% of the time. How many workers were there in the group?

(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

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.

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