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A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re
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18 May 2016, 05:43
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A, B, C working alone finish constructing a house in 4, 6 and 12 hrs. resp. A starts alone and works for 1 hr. He is then joined by B and they work together for 1 hr. Finally C joins them and they all work together to finish the work. If they start working at 9 am, at what time will they finish? A. 11.30 am B. 11.40 am. C. 12.00 pm D.11.10 am E. 11.20am Let's take the LCM of 4,6, and 12. LCM = 36 We know, R* T = W (For A) R * 4 = 36 R =9 (For B) R * 6 = 36 R =6 (For C) R * 12 = 36 R =3 A+B+C combined rate = 18 Now, A completes 9 hrs of work in 1 hour. A and B combined completes 9+6 hrs of work in 1 hour. Total work completed is 24. Remaining work is 3624 = 12. Rate of A+B+C = 18. Therefore, R * T = W ==> 18 * t = 12 ==> t = 12/18 = 2/3hrs [2/3 * 60 = 40 min] 40 min + 1+1 hour = 2 hour 40 min.
therefore, answer is b. Is this a sub600 level question?
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A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re
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18 May 2016, 08:13
bimalr9 wrote: A, B, C working alone finish constructing a house in 4, 6 and 12 hrs. resp. A starts alone and works for 1 hr. He is then joined by B and they work together for 1 hr. Finally C joins them and they all work together to finish the work. If they start working at 9 am, at what time will they finish?
A. 11.30 am B. 11.40 am. C. 12.00 pm D.11.10 am E. 11.20am
Let the total work be 12 units Efficiency of A = 3 units/ hour Efficiency of B = 2 units/ hour Efficiency of C = 1 units/ hour Quote: A starts alone and works for 1 hr.............If they start working at 9 am So, A completes 3 units of work by 10 am Quote: He is then joined by B and they work together for 1 hr. So, A and B completes 5 units of work by 11 am Quote: Finally C joins them and they all work together to finish the work. Work left is 4 units ( 12  3  5 ) and combined efficiency of A, B and C is 6 units So, time required to complete 4 units will be 4/6 => 4/6*60 = 40 minutes... So they will require 40 minutes after 11 am to complete the remaining work... Hence answer will be (B) 11:40 am !

This problem doesn't require the use of the formula R* T = W So do not involve in the concept of RTW unnecesarily 
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Re: A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re
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18 May 2016, 08:38
A's 1 hr work= 1/4 B's 1 hr work= 1/6 C's 1hr work= 1/12 A+B+C's 1 hr work = 1/4 +1/6+1/12= 1/2 Now A worked alone for 1 hr. That means he did 1/4th of the work. A and B worked together for 1 hr. So, they both did 1/4+1/6= 10/24 of the total work. After 2 hrs the total work done was= 1/4 +10/24= 16/24 And, the remaining work was= 8/24= 1/3 All three of them did 1/3 work together. If they do 1/2 work in 1 hr, they will do 1/3 work in 2/3*60= 40 mins Total time taken to finish the job 1hr +1hr+40 mins= 2hr 40 mins They will finish the job at 11:40 am. IMHO, this is not a sub 600 level question.
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Re: A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re
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18 May 2016, 10:10
bimalr9 wrote: A, B, C working alone finish constructing a house in 4, 6 and 12 hrs. resp. A starts alone and works for 1 hr. He is then joined by B and they work together for 1 hr. Finally C joins them and they all work together to finish the work. If they start working at 9 am, at what time will they finish? A. 11.30 am B. 11.40 am. C. 12.00 pm D.11.10 am E. 11.20am Let's take the LCM of 4,6, and 12. LCM = 36 We know, R* T = W (For A) R * 4 = 36 R =9 (For B) R * 6 = 36 R =6 (For C) R * 12 = 36 R =3 A+B+C combined rate = 18 Now, A completes 9 hrs of work in 1 hour. A and B combined completes 9+6 hrs of work in 1 hour. Total work completed is 24. Remaining work is 3624 = 12. Rate of A+B+C = 18. Therefore, R * T = W ==> 18 * t = 12 ==> t = 12/18 = 2/3hrs [2/3 * 60 = 40 min] 40 min + 1+1 hour = 2 hour 40 min.
therefore, answer is b. Is this a sub600 level question? Say they work for t fraction of an hour after 11. (Rate*Time) for each = (2 + t)/4 + (1 + t)/6 + t/12 = 1 = Total Work 6t/12 = 1/3 t = 2/3 hrs = 40 mins Time at which the work gets done = 11:40
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Re: A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re
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18 May 2016, 10:27
From the question we know :
Ra x 4 = 1 => Ra = 1/4 Rb x 6 = 1 => Rb = 1/6 Rc x 12 = 1 => Rc = 1/12
stage 1 : Ra x 1hour = 1/4 x 1hour = 1/4 = 6/24
stage 2 : (Ra + Rb) x 1hour = (1/4 + 1/6) x 1hour = 10/24
stage 3 : (Ra + Rb + Rc ) x Time = remaining work
remaining work = 1  (6/24 + 10/24) = 8/24
So, (1/4 + 1/6 + 1/12) x Time = 8/24 12/24 x Time = 8/24 Time = 2/3 = 40 min
Finally, Stage 1 + stage 2 + stage 3 = 1h + 1h + 40min = 2h40min 9h + 2h40 = 11h40, answer choice B



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Re: A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re
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24 May 2016, 11:15
I like putting things in a table , that helps to proceed without making a mistake.
A , B and C can do 1/4, 1/6 and 1/12 each individually and respectively.
so Time___Person involved_____Word done_________Remaining 9:00_______A________________0________________1 10:00____A (B just joins)_______1/4_______________3/4 11:00__A,B,(C just joins)__1/4 + (1/4 +1/6) = 2/3__1  (1/4 + (1/4 +1/6)) = 1/3
So 1/3 is done by A+B+C, who can do 1/4+1/6+1/12 = 1/2 in an hour, so would need, 2/3 hour for remaining, which is 2/3* 60min = 40 mins. Therefore 11:40. (Excuse me for my formatting.)



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A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re
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24 May 2016, 14:26
let t=total time 1/4+5/12+(t2)(1/2)=1 t=2 2/3 hrs=2 hrs 40 min 9 am+2 hrs 40 min=11:40 am



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Re: A,B,C working alone finish constructing a house in 4, 6 and 12 hrs. re
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06 Jun 2017, 16:44
bimalr9 wrote: A, B, C working alone finish constructing a house in 4, 6 and 12 hrs. resp. A starts alone and works for 1 hr. He is then joined by B and they work together for 1 hr. Finally C joins them and they all work together to finish the work. If they start working at 9 am, at what time will they finish?
A. 11.30 am B. 11.40 am. C. 12.00 pm D.11.10 am E. 11.20am We are given that A, B, and C, working alone, finish constructing a house in 4, 6, and 12 hours respectively. Thus, we can let the rate of A = 1/4, the rate of B = 1/6, and the rate of C = 1/12. We can let x = the number of hours the three people work together when C joins in. Thus, the number of hours A works is 2 + x, the number of hours B works is 1 + x, and the number of hours C works is x. We can now determine x with the following equation: work of A + work of B + work of C = 1 (2 + x)/4 + (1 + x)/6 + x/12 = 1 Multiplying the entire equation by 12, we have: 3(2 + x) + 2(1 + x) + x = 12 6 + 3x + 2 + 2x + x = 12 6x = 4 x = 4/6 = 2/3 hour = 40 minutes Since A started at 9 a.m., they finished constructing the house 2 hours and 40 minutes later, or 11:40 a.m. Answer: B
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