A and B are two laborers who have been given the task to build a house

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

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
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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

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

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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

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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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

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!!

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

+ 1 kudos for you.

Matt

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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.

"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.

Let's say it took 't' days to build by A, then B must have worked 't-1' days as B is a destroyer and it has to be completed by A
So,
t/20 + t-1/30=1
t=58 days
therefore, total days to build= t + t-1= 58+57=115 days
Answer is B.

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.

Is it necesaary that A does 3 units on last day, isnt 3 units the maximum work he can do in a day? What if the amount of work left was less than 3 units?

Easiest way to get through this problem is to set up a Value for the total work and then follow the pattern of building.

We can not just simply take (Rate of Build - Rate of Breaking Down) because that assumes they are working together, independently and simultaneously, at their own respective rates.

Here, the builder spends a whole day building (1/20)th of the house.

The next day the breaker breaks down (1/30)th of the house.


Let total work = 600

A = (1/20) (600) = 30 units built each day

B = (1/30) (600) = -20 units broken down each day

Track the cumulative sum each day to see how far along the house is:

Need to get to 600 = fully built house:
Day 1: A builds 30———-30
Day 2: B breaks down 20 ——- 10
Day 3: A builds 30 ——— 40
Day 4; B breaks down 20 ——— 20
Day 5: A builds 30 ——— 50
Day 6: B breaks down 20 ——— 30


We will reach a consecutive Odd Integer day in which A has built up the house (then the next day B will take down 20 units). We want to find that day.

Notice that for each consecutive ODD Day, the cumulative house is built up by 10 units a day, starting with Day 1 = 30

Find the Number of consecutive Odd Integers needed to get to the 600 fully built house

30 ——- 40———50 .......... —— 600

(Subtract 30 from each term)

0 ——— 10 ——-20 ........ ——— 570

(Now divide each number by 10)

0 ——— 1 ——— 2 ....... ———- 57

Including 0, the house will be built on:

57 + 1 = 58th Consecutive Odd Integer Day


Day 1—- 2K + 1 = 1 ——- when k = 0

Day 2 ——2K + 1 = 2 —— when k = 1

To find the 58th consecutive odd integer, plug in k = 57

2 (57) + 1 = 115

On the 115th day, A will have finished the house.

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