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

 

50 men can build a tank in 40 days 

Assume 1 man does 1 unit of work in 1 day

Then the total work is 50×40 = 2000 units

50 men work in the first 10 days and completes 50×10 = 500 units of work

45 men work in the next 10 days and completes 45×10 = 450 units of work

40 men work in the next 10 days and completes 40×10 = 400 units of work

35 men work in the next 10 days and completes 35×10 = 350 units of work

So far 500 + 450 + 400 + 350 = 1700 units of work is completed and

Remaining work is 2000 - 1700 = 300 units

30 men work in the next 10 days. In each day, they does 30 units of work.

Therefore, additional days required = 300/30 =10

Thus, total 10+10+10+10+10 = 50 days required.

After 10 days, the remaining food would be sufficient for the 1000 students for 20 more days

-->If 1000 more students are added, it shall be sufficient for only 10 days (as the no. of students is doubled, the days are halved).

(20 x 18) men can complete the work in in one day.

one man's one day work = 1/360

(18 x 15) women can complete the work in 1 day

1 woman's one day work = 1/270

So, required ratio = 1270:1360= 4:3

L's 10 days work = 115×10=23

Remaining work = 1-23=13

Now,  work is done by K in one day = 1/18

1/3 work is done by K in  18×13 = 6 days

Raghu can complete the work in (12 x 9)hrs = 108 hrs.

 Arun can complete the work in (8 x 11)hrs = 88 hrs.

Raghu's 1 hrs work = 1/108 and Arun's 1 hrs work = 1/88

(Raghu + Arun)'s 1 hrs work = 1108+188=492376

 
So, both Raghu and Arun will finish the work in 237649 hrs

Number of days of 12 hours each=237649×112 = 19849=4349 days

Work done by P and Q in the first two hours, working alternately

= First hour P + Second hour Q

⇒14+112=13

work is completed in 2 hours

Then, the total time required to complete the work by P and Q working alternately=2 x 3= 6hours

Thus, work will be completed at 3pm.

Given K=4L 

-->K+L = 4L+L = 5L

These 5L people can complete the work in 24 days, which means L alone can do the work in (24 x 5)=120 days.

Hence, K alone can do the work in 120/4= 30 days.

One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.

C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.

Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.

Remaining Work = 7/10, which was done by A,B and C in the initial number of days. 

Number of days required for this initial work = 7 days. 

Thus, the total numbers of days required = 4 + 7 = 11 days.