It takes 6 beavers 10 hours to build a certain dam, working at a unifo

[color=#0000ff]Each beaver can build 1/60 dam per hour
=> From 12 to 6PM: 6 beavers can build 1/60 * 6 * 6 = 6/10 dam

From the next 3 hours: 7/60+8/60+9/60 = 4/10 dam is built more

=> Total hours: 9 => Ans: E

Number of man hours required = 6*10 = 60.
6 workers work for 6 hrs --> Number of man hours completed = 6*6 = 36
Man hours remaining = 60 - 36 = 24
Effort for the next 3 hours = 7 + 8 + 9 = 24. Work gets completed at 9.00 PM

Answer: E

it takes (6)(10)=60 beaver hours to build dam
(6)(6)=36 beaver hours completed from noon to 6:00pm
60-36=24 beaver hours remaining at 6:00pm
t=time required
ratio of cumulative beaver hours to time=(t+13)/2
24/t=(t+13)/2
t=3 hours
9:00pm

6 beavers can complete in 6 hours 36/60 work
at the end of 7 p.m 36/60 + 7/60
at the end of 8 36/60 + 7/60 + 8/60
at the end of 9 60/60; the work will be completed.

Got the same answer with a different approach -

Rate of 1 beaver = 1/60
Amount of work done by 6 beavers till 6 pm (in 6 hours) = 6/60 *6 = 6/10
Amount of work left = 4/10
No. of beavers working between 6 - 7pm = 7
Hence, the amount of work done between 6 - 7 = 7/60
and between 7-8 = 8/60
and between 8-9 = 9/ 60
Adding these 3 we get 4/10 (which was the amount of work left after 6 pm)
Hence the work gets done at 9.

Answer - E

This is a more comprehensive approach using the formula for 'sum of an arithmetic series'. This can be applied to solve similar problems where a greater number of hours are needed for completing the work from the point where one worker is added to the workforce every hour. For instance if, say, 19 additional hours are needed to complete the work, the process of step-by-step compilation of the amounts of work done each hour would be time consuming.

By 6:00 pm, the 6 beavers will have completed 3/5 of the work of the dam so 2/5 of work remains to be done. From this point on, 1 beaver is added to the workforce (WF) every hour until the dam is finished.
1 beaver can do 1/60 of the work in 1 hour. Let 'n' be the number of hours needed after 6:00 pm to finish the dam.

The strength of the WF in each of the 1st, 2nd and 'n'th hours are 7, 8 and {7+(n-1)1}** and the corresponding portions of the work done are (1/60)*7, (1/60)*8 and (1/60)*{7+(n-1)1} respectively.
(** Refer to the formula for the 'n'th term of an Arithmetic Series)
The total portion of work done in 'n' hours = (1/60){7+8+...(n-1)1} = (1/60){n*[7+(7+n-1)]/2} = (1/60){n(13+n)/2}
Since the total portion of work completed in 'n' hrs is equal to the remaining (2/5)th:

(1/60){n(13+n)/2} = (2/5)...> n^2+13n-48=0...> n^2 + 16n - 3n - 48 = 0...> n(n+16)-3(n+16)...> (n-3)(n+16)=0...> n=3.

Therefore, 3 hours are needed (after 6:00 pm) to complete - the dam will be finished at 9:00 pm. ANS: E

Use units of work - 6 beavers for 10 hrs means 60 units of work are required.

From 10 noon to 6 pm, 6 beavers complete 6*6 = 36 units of work. Leftover are 60 - 36 = 24 units of work. 

In the next few hours, the units of work completed each hour will keep increasing by 1 i.e. 7 units + 8 units + 9 units. These add up to 24 units ( 7 + 8 + 9 = 24) so work will be complete by 9 PM.

Answer (E)

Here are some work-rate discussions:
https://youtu.be/88NFTttkJmA

https://anaprep.com/arithmetic-applying ... -concepts/

https://anaprep.com/arithmetic-work-rat ... s-and-all/

https://anaprep.com/arithmetic-avoiding ... work-rate/

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