Let the distance between Delhi and Kanpur be y km .
Suppose the train leaving from Delhi is A and the train leaving from Kanpur B
A's speed = y/ (10 a.m - 5 a.m) = y/5 km/hr.
B's Speed = y/ (2 p.m - 7 a.m) = y/7 km/hr.
Since B starts two hours later than A, the distance already covered by A at the start of B = 2y/5 km.
Remaining distance = y- 2y/5 = 3y/5 km.
Relative speed of approach of two trains = (y/5 + y/7)
= 12y/35 km/hr.
Time taken to cover the remaining distance by both trains = (3y / 5) / (12y / 35)
= (3/5) x (35/12) = 7/4 hrs.
= 1 hr. 45 min.
? The two train will meet at (7 a.m + 1hr.45 min)
= 8.45 a.m
(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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