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
e-GMAT

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The official solution has been posted. Looking forward to a healthy discussion..
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Hey,

The answer is choice B but the options have been modified a bit. Please check.

Thanks,
Saquib
Quant Expert
e-GMAT

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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
e-GMAT

Aiming 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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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.

Posted from my mobile device

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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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

Total no of days - 58+57=115 days.

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.

The conventional way of solving it gives us 60 days (if they work every day). (1/(1/20-1/30) = 60)
But they work alternately, so the work wil take something like 120 days. Really it will be fewer days, because A laborer starts firstm the only option is B - 115 days.