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
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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

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

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

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).

Expert comment really appreciated.

Thanks,
Matt

Thansks Pushpitkc. You are right. Now I get IT.

+ 1 kudos for you.

Matt

Posted from my mobile device

"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

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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Thank You
Sudhanshu

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.