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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)
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• 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.
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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
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.
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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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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.
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: (1/20-1/30)= 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).
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
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:
A-B+A-B+......A-B+A =1
=> 1/20-1/30+1/20-1/30...+1/20-1/30=1
Had the work been done for 3 days .. the condition would have been - A-B+A => 2A-B Had the work been done for 5 days .. the condition would have been A-B+A-B+A=> 3A-2B Had the work been done for 7 days .. the condition would have been A-B+A-B+A-B+A=> 4A-3B
this means - (n+1)A-nB=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
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 (t-1)] = 1
Solve for T (A's days) -> 58 Calculate T-1 (B's days) -> 57 Total days for house to be built = 58 + 57 = 115
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
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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