Rohit crosses a 700 metres wide road in 35 seconds. What is his speed (in km/hr)?

 

Suppose A, B and C take x, x/2 and x/3 respectively to finish the work. 

Then, (1/x + 2/x + 3/x) = 1/2 

=> 6/x = 1/2      => x = 12 

So, B takes 6 hours to finish the work.

We have M1 D1 H1 / W1 = M2 D2 H2 / W2 (Variation rule)
(60 x 30 x 8)/ 150 = (35 x D2 x 6) / 200
D2 = (60 x 30 x 8 x 200) / (150 x 35 x 6) => D2 = 91.42 days =~ 91.5 days.

Ratio of times taken by A and B = 160 : 100
A can do the work in 12 days
Let B can do the work in "D" days
=> 160:100 = 12 : D
=> D = 12 x 100/160 = 7.5 hrs

A can do the work = 18 days
B can do the work = 18/2 = 9 days
(A + B)'s 1 day work = 1/18 + 1/9 = 1/6
=> In 3 days = 3x1/6 = 1/2 work is completed.

1 boy's 1 day work = 1/70

1 girl's 1 day work = 1/140

(5 boys + 10 girls)'s 1 day work = (5/70 + 10/140) = (1/14 + 1/14) = 1/7

5 boys and 10 girls will complete the work in 7 days.

Ratio of times taken by A and B = 100 : 130 = 10 : 13.
Suppose B takes x days to do the work.

Then, 10 : 13 :: 23 : x     =>     x = ( 23 x 13/10 )     =>     x = 299 /10.

A's 1 day's work = 1/23 ;
B's 1 day's work = 10/299 .

(A + B)'s 1 day's work = ( 1/23 + 10/299 ) = 23/299 = 113 .

Therefore, A and B together can complete the work in 13 days. 

For 30 days he receives

30 x 25 = 750
               

Actually he gets 425

Therefore, Fine is 750 - 425 =  325

=> No of days absent = 325/(25+7.50)= 10

Let 1 man's 1 day work = x and 1 woman's 1 day work = y.
Then, 4x + 6y = 1/8 and 3x + 7y = 1/10
Solving these two equations, we get:
x = 11/400 and y = 1/400
10 woman's 1 day work = (1/400 x 10) = 1/40.

Hence, 10 women will complete the work in 40 days.