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Question Stats:
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Painters A and B can paint a house working alone in 20 and 30 days respectively. They started painting a house together but then A left after a number of days but then rejoined B before the job was completed. If B worked alone for 5 days and then A and B together completed the work in 4 days, after how many days of working together, did A leave B? A. 4 B. 5 C. 6 D. 7 E. 8
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Re: D0130
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08 Dec 2015, 05:01
I prefer assuming total work before solving such questions. Helps me to avoid fractions. ALONE A finishes work in 20 days B finishes work in 30 days Taking LCM of 20 and 30 (Or any convenient common mulitple) > 120 So if total work is 120, then to finish 120 units of work as per their current rate A finishes work 120/20 = 6 units per day B finishes work 120/30 = 4 units per day Both together finish 6 + 4 = 10 units per day. Coming back to what's asked  B alone worked for 5 days, so 4*5 = 20 units of total 120 units is done Together they worked for 4 hours before finishing the work = 10*4 = 40 units. We've got 40 + 20 = 60 units done. 12060 = 60 units pending. This work is nothing but the work that A & B did together before A left B alone to suffer. Hence 60/10 = 6 days! +Kudos, if this helped!
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15 Sep 2014, 23:13
Official Solution:Painters A and B can paint a house working alone in 20 and 30 days respectively. They started painting a house together but then A left after a number of days but then rejoined B before the job was completed. If B worked alone for 5 days and then A and B together completed the work in 4 days, after how many days of working together, did A leave B? A. 4 B. 5 C. 6 D. 7 E. 8 The rate of A is \(\frac{1}{20}\) job/day; The rate of B is \(\frac{1}{30}\) job/day; The combined rate of A and B is \(\frac{1}{20}+\frac{1}{30}=\frac{1}{12}\) job/day. In 5 days that B worked alone (2nd stage) \(5*\frac{1}{30}=\frac{2}{12}\) of the job was done; In 4 day that A and B worked together (3rd stage) \(4*\frac{1}{12}=\frac{4}{12}\) of the job was done; So, in the 1st stage of the work \(1\frac{2}{12}\frac{4}{12}=\frac{6}{12}\) part of the job was done by A and B together, which at the rate of \(\frac{1}{12}\) job/day took them 6 days. Answer: C
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Re: D0130
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15 Feb 2016, 05:24
nishantdoshi wrote: I think this the explanation isn't clear enough, please elaborate. Hi, Lets divide the work in two parts, one when B worked alone and second when both worked together.. Since B can finish work in 30 days, His 1 day work = 1/30 of total.. so in 5 days, he will finish= 5/30= 1/6 of work..
now second part, when both work together.. we should not confuse in two different time frames, that is one at the start and other in the end.. we just require how many days both worked together.. one day work of both= 1/20 + 1/30 = 5/60= 1/12.. so they have to finish 11/6 work= 5/6.. it will be done in (5/6)/(1/12)= 5*12/6= 10.. so both work for 10 days together.. In the second part they worked for 4 days to finish the work.. so they worked for 104=6 days initially.. ans 6C
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I will go with this approach,
rate of A= 1/20 job per day rate of B = 1/30 job per day
when they work together the no of days required to complete the task = 1/20 + 1/30 = 3+2/60 = 5/60 = 1/12
when A and B were working together let us assume that after X days A left the working while , so the work completed in X days by both of them = x * 1/12= x/12
so the work remaining = 1x/12 = 12x/12
Now B worked for 5 days , so total work done by B in 5 days = 5*1/30 = 1/6
now the total work remaining = total work remaining after x days  work done alone by B in 5 days = 12x/12  1/6 = 10x/12
now A rejoined the work and they both completed the remaining work in 4 days remaining work = 4 days of work for A and B 10x/12 = 4/12 10x = 4 x= 6
so the no of days after which A left working together is 6



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Re: D0130
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29 Jan 2017, 09:18
let for x days A worked with B before he left Now A worked for X+ 4 days B worked for X+4+5 ie. X+9 days.
(X+4)/20 + (x+9)/30 =1 solve this you will get X=6



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Re: D0130
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30 Jan 2017, 08:15
Bunuel wrote: Painters A and B can paint a house working alone in 20 and 30 days respectively. They started painting a house together but then A left after a number of days but then rejoined B before the job was completed. If B worked alone for 5 days and then A and B together completed the work in 4 days, after how many days of working together, did A leave B?
A. 4 B. 5 C. 6 D. 7 E. 8 Given from the prompt:A & B worked together + B worked alone + A& B joined again to complete work Combined rate = 1/20+ 1/30 = (3+2)/60=5/60= 1/12 job/day Let T= time for both A & B together before A leaving B To solve the question:Use the standard formula Work = rate * time Following the sequence stated by prompt T *1/12 + 5 *1/30 + 4 *1/12 = 1 .............( the work here is one house) T /12 + 1/6 + 1/3 = 1 T /12 + 3/6= 1 T /12 = 1/2 .......T=6 Answer: C



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Re D0130
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19 Apr 2016, 22:32
I think this is a highquality question and I agree with explanation.



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16 Jul 2016, 09:56
I think this is a highquality question and I agree with explanation.



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22 Aug 2016, 11:00
I think this is a highquality question and I agree with explanation. Great question. I freaked out when I first faced it but was able to solve it within 2 mins! Thanks



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Re D0130
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29 Jan 2017, 06:35
I think this is a poorquality question and I don't agree with the explanation. 1/20 + 1/30 is not 1/12



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Re: D0130
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29 Jan 2017, 09:20
Abosse0 wrote: I think this is a poorquality question and I don't agree with the explanation. 1/20 + 1/30 is not 1/12 Actually it is exactly 1/12. 1/20 + 1/30 = 3/60 + 2/60 = 5/60 = 1/12.
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18 May 2017, 01:58
I think this is a highquality question.



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07 Jul 2017, 11:38
I think this is a highquality question and I agree with explanation.



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Re: D0130
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28 May 2018, 10:49
I solved by setting up a table and following the story in the problem.
W R T Painter A 60 60 / 20 = 3 20 Painter B 60 60 / 30 = 2 30 combined 60 3 + 2 = 5 12
Painter B worked alone for 5 days at a rate of 2 (things) per day for a total of 10 (things) Painters A and B worked together for 4 days at a rate of 5 (things) per day for a total of 20 (things)
1 whole house = 60 (things)
60  10  20 = 30 (things) that A and B worked on together at their combined rate
30 / 5 = 6 days



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Re: D0130
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05 Oct 2018, 00:21
Let's pick up a Number which is the LCM of both of A and B's rates, for eg. 60. Here, let 60 be the total units of work to be done.
So, if A takes 20 days to finish 60 units, it means he accomplishes 60/20= 3 units per day. Similarly, if B takes 30 days to finish 60 units, he accomplishes 60/30= 2 units per day
Now, the total work done by them in 1 day= 3+2= 5 units.
For the 5 days B worked by himself, he gets 5*2 units= 10 units done. The 4 days A and B work together in the end, they get 4* 5 units= 20 units done. Total work done= 30 units. Leftover work= 6030 units= 30 units. This is the work that A and B must have done together, before A left.
So, given the combined rate, A and B together must have done 30 units in 30/5= 6 days!



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Re: D0130
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21 Aug 2019, 14:55
Understood the solutions after some effort. Nice solutions above. thanks guys.



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Re: D0130
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01 Dec 2019, 08:48
Bunuel wrote: Painters A and B can paint a house working alone in 20 and 30 days respectively. They started painting a house together but then A left after a number of days but then rejoined B before the job was completed. If B worked alone for 5 days and then A and B together completed the work in 4 days, after how many days of working together, did A leave B?
A. 4 B. 5 C. 6 D. 7 E. 8 Units of Work = 600 A can do 30 per day and B can do 20 per day. We know B worked for 5 days alone so that 100 units. Work left = 500 We know both worked for 4 days together to finish the job. So 4*50=200. Work left= 300 Now this work was completed by both before A left. So 300/50=6 days. Hence C
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Here is my approach:
Assuming a is the amount of work painter A can do in 1 day, b is the amount of work painter B can do in 1 day.
Painters A and B can paint a house working alone in 20 and 30 days respectively => 20a = 30b => a=1.5b
Assuming X is the day A and B worked together before A left. The total amount of work no matter whether they did together is the same, so: X(a+b) + 5b + 4(a+b) = 30b <=> X x 2.5b + 5b + 4 x 2.5b = 30b <=> X = 6
Hence C
I don’t like working with fraction so this approach suits my taste.
I’m a GMAT beginner, I’m sorry if my explanation is not very good.
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Re: D0130
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17 Jan 2020, 20:38
Great question, disagree that it is a 500 level question, upper 600 or 700 I'd say Q48.










