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A and B are two laborers who have been given the task to build a house [#permalink]
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Updated on: 31 Mar 2017, 13:01
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A and B are two laborers who have been given the task to build a house. A can build a house in 20 days while B can build it in 30 days. But B went rogue and instead of building the house, he starts breaking it at the same rate. If A starts the work on the 1st day and they work alternately, in how many days will the house get built? (Assume once the house gets build, they will stop working) A. \(60\) days B. \(115\) days C. \(120\) days D. \(140\) days E. \(121\frac{1}{3}\) days
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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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Updated on: 31 Mar 2017, 13:01
The official solution has been posted. Looking forward to a healthy discussion..
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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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28 Mar 2017, 21:35
Answer is B, right? after 118 days, A and B build 59/60, it takes A 1/3 days to build 1/60 of the house. Totally it takes 118+1/3 days



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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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trangphamthuy91 wrote: Answer is B, right? after 118 days, A and B build 59/60, it takes A 1/3 days to build 1/60 of the house. Totally it takes 118+1/3 days Hey, The answer is choice B but the options have been modified a bit. Please check. Thanks, Saquib Quant Expert eGMATAiming to score Q50 or higher in GMAT Quant? Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes. Register
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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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Updated on: 31 Mar 2017, 13:02
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Solution • Let us assume that building the house is equivalent to 60 units of works (LCM of 20 and 30 days).
o Thus, efficiency of A = \(\frac{(Work)}{(Time)}\) \(= \frac{60}{20} = 3\) units/day. o And Efficiency of B \(= \frac{60}{30} = 2\) units/day. • As per the question:
o A starting working on the first day and on the second day B comes and breaks the house at his own rate.
o Thus, the pattern is like this ABABABABAB…..ABABA... • From the above we can infer that:
o On day 1 A build 3 units. (As its efficiency is 3 units per day)
o On day 2 B breaks 2 units. (As its efficiency is 2 units per day)
o Thus, in 2 days only 1 unit of house is built.
o But we need to build 60 units.
AND (important!), we need to ensure that the last day A comes and build the wall.
The reason for this is that on the last day, you cannot break the house and still make it! The house will be built on the last day when A comes to work. o Thus, we know that we can build 1 units in 2 days.
o To build 57 units(the last three unit will be done by A), we will take (57*2) = 114 days.
o On the 114th day, B had come to break it.
Thus, the next day A will come and complete the remaining portion of 3 units in \(3\) days. o Hence the total time taken would be \(115\) days. • And the correct answer is Option B. Thanks, Saquib Quant Expert eGMATAiming to score Q50 or higher in GMAT Quant? Attend this webinar on 2ndApril to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes. Register
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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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31 Mar 2017, 04:16
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But the premise says that once the work gets completed they no longer work and after accomplishing 57 units of work in 114 days,'A' will come and complete those remaining 3 units in a day's time.Thus, altogether they would take 115 days instead



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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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31 Mar 2017, 13:04
attari92 wrote: But the premise says that once the work gets completed they no longer work and after accomplishing 57 units of work in 114 days,'A' will come and complete those remaining 3 units in a day's time.Thus, altogether they would take 115 days instead Hey Attari92, That is an excellent point that you have put forward! And you are correct. I somehow missed that while making the question and posting the solution. My bad! I have modified the solution and the options. Once again, awesome observation! Thanks, Saquib Quant Expert eGMATAiming to score Q50 or higher in GMAT Quant? Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes. Register
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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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15 May 2017, 19:20
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EgmatQuantExpert wrote: Q. A and B are two laborers who have been given the task to build a house. A can build a house in 20 days while B can build it in 30 days. But B went rogue and instead of building the house, he starts breaking it at the same rate. If A starts the work on the 1st day and they work alternately, in how many days will the house get built? (Assume once the house gets build, they will stop working)
A. \(60\) days B. \(115\) days C. \(120\) days D. \(140\) days E. \(121\frac{1}{3}\) days Options are quite straightforward. I was looking for an odd number of days, as it is given that they will stop as soon as the house is built. Only option (B) fits this criterion. In all other options there is some destruction from laborer B



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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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03 Jun 2017, 00:32
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B cannot work in the last day because he is a demolisher and the house won't be fully built. Hence A must work in the last day to sort the wreckages out by B in the previous day and finish building the house. As A is starting the work and he is finishing it, the number of days must be odd. Only option B fits.
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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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11 Jun 2017, 13:06
Probably I don't understand the question, as it is stated my solution is: (1/201/30)= 1/60 in 2 days > in order to build 59/60 it takes 118 days, than A comes and finish the work in 1/3 days. Therefore the solution should be 118(1/3).
Expert comment really appreciated.
Thanks, Matt



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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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11 Jun 2017, 13:27
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We will not need to do the same for 118 days. Doing this for 114 days will complete 57/60 of the work as per your login and on the 115th day when A does 1/20th of the work 57/60 + 1/20 = 60/60(Work finished) Hope it helps!! Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app
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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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11 Jun 2017, 13:46
pushpitkc wrote: We will not need to do the same for 118 days. Doing this for 114 days will complete 57/60 of the work as per your login and on the 115th day when A does 1/20th of the work 57/60 + 1/20 = 60/60(Work finished) Hope it helps!! Sent from my ONEPLUS A3003 using GMAT Club Forum mobile appThansks Pushpitkc. You are right. Now I get IT. + 1 kudos for you. Matt Posted from my mobile device



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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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12 Aug 2017, 05:20
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Let A be the work done per day by A ( 1/20) and B be the work done by B per day (1/30) As work done must be equal to 1, we will start the work with A and end the work with A, as once the house is constructed work will stop:
AB+AB+......AB+A =1
=> 1/201/30+1/201/30...+1/201/30=1
Had the work been done for 3 days .. the condition would have been  AB+A => 2AB Had the work been done for 5 days .. the condition would have been AB+AB+A=> 3A2B Had the work been done for 7 days .. the condition would have been AB+AB+AB+A=> 4A3B
this means  (n+1)AnB=1 (n+1)/20  n/30=1 => n = 57 ( no of days B will work)A now, A works one day more than B( from above)=> n+1=58
Total no of days  58+57=115 days.



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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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02 Oct 2017, 10:35
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"Assume once the house gets build, they will stop working" A will keep building the house while B breaks it the next day. Therefore, when A finally completes the house, "they will stop working". Thus, B will be working one day less than A eventually as shown below:
Make,Break,Make,Break...Make,Break,MAKE
Equation:
Work of A  Work of B = 1 [1/20 x T]  [1/30 x (t1)] = 1
Solve for T (A's days) > 58 Calculate T1 (B's days) > 57 Total days for house to be built = 58 + 57 = 115



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A and B are two laborers who have been given the task to build a house [#permalink]
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16 Oct 2017, 11:05
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EgmatQuantExpert wrote: A and B are two laborers who have been given the task to build a house. A can build a house in 20 days while B can build it in 30 days. But B went rogue and instead of building the house, he starts breaking it at the same rate. If A starts the work on the 1st day and they work alternately, in how many days will the house get built? (Assume once the house gets build, they will stop working)
A. \(60\) days
B. \(115\) days
C. \(120\) days
D. \(140\) days
E. \(121\frac{1}{3}\) days This is a beautiful question. A builds the house at a rate of \(\frac{1}{10}\) and b breaks the house at a rate of \(\frac{1}{30}\). So, in 2 days, the work completed will be \(\frac{1}{20}\)\(\frac{1}{30}\)=\(\frac{1}{60}\). But remember, A started first, so the house will be completed a figure that is less than 120. So, B



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A and B are two laborers who have been given the task to build a house [#permalink]
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21 Oct 2017, 18:01
I can think about two approach below. Please check it out. It may be helpful.
A can build a house in 20 days. And B can break it in 30 days.
METHOD 1 Lets take LCM of (20,30) = 60 . 60 be the unit of work to do. A works on 1st day and B on 2nd day and so on.
Day 1 In 1 day A can finish  60/20 = 3 unit of work. Day 2 In 1 day B can break 60/30 = 2 unit of work.
So after day 2. Total work that was done  1 unit of work. Implies, it will take 120 days to complete  60 unit of work.
=> Here is the most important part of the question. => (Assume once the house gets build, they will stop working)
Lets think about it. Try to go in reverse gear. B completed  120 days > 60 unit of work. A completed  119 days > 62 unit of work = 60+2 (work done by B on 120th day). B completed  118 days > 59 unit of work = 62 3 (work done by B on 119th day). A completed  117 days > 61 unit of work = 59 + 2 (work done by B on 118th day). B completed  116 days > 58 unit of work. = 61  3 (work done by A on 117th day). A completed  115 days > 60 unit of work. = 58 + 2 (work done by B on 116th day).
Answer  B => 115 days.
METHOD 2 We can do it directly in a go if we use our visualization skill.
Lets take LCM of (20,30) = 60 . 60 be the unit of work to do. A works on 1st day and B on 2nd day and so on.
Day 1 In 1 day A can finish  60/20 = 3 unit of work. Day 2 In 1 day B can break 60/30 = 2 unit of work.
So after day 2. Total work that was done  1 unit of work. Implies, it will take 120 days to complete  60 unit of work.
But we need to realize that there is possibility of A completing the work before 120 DAYS as A is doing positive work. Lets consider after 110 days. After 110 days. Total work done is  55. After 112 days. Total work done is  56. After 114 days. Total work done is  57.
Now hold on. A can finish 3 unit of work in a day. => After 115 days. Total work done is  60.
Answer B => 115.



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Re: A and B are two laborers who have been given the task to build a house [#permalink]
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07 Jan 2018, 13:09
Let's say it took 't' days to build by A, then B must have worked 't1' days as B is a destroyer and it has to be completed by A So, t/20 + t1/30=1 t=58 days therefore, total days to build= t + t1= 58+57=115 days Answer is B.




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