A can complete a project in 20 days and B can complete the

Assume to total work to be 60 Units ( LCM of 20 & 30)
A does 3 units per day ( =60/20)
B does 2 units per day ( =60/30)
B works for 10 days alone work done = 2*10 = 20 units
A and B work together for the rest of the time
work left = 40 units
A and b together to 5 units oer day ( 2+3 = 5)
days required to do 40 units = 40/5 = 8

Total = 10+8 = 18

I am getting 27.. please tell me where i am going wrong.

The rate of A 1/20 job per day;
The rate of B 1/30 job per day:

If both are working together: 1/20 + 1/30 = 1/12, which means A and B working together will complete the project in 12 days.

if A left 10 days before the completion means they worked for 2 days together and B worked for 10 days more, so,
2*1/12 = 1/6 work done together => 5/6th project is remaining which wud be done by B,

B completes the project in.....30 days
B completes 5/6th project in......30*5/6=25

So, total of 25+2 = 27 days.

Please correct me if i am going in wrong direction.

A = w/20
B = w/30

A+B = w/20+w/30

Rate*Time = Work
(A+B)* (t-10) = W (A left the job 10 days before the total day)

solving we get t=28 days

putting total days = y-10= 28-10 = 18 days

I think I am wrong somewhere...

d is the total days to complete the project.
A worked for d-10 days while b worked the whole d days.

Setup the Rate Equation like this \((Rate of A) (d -10) + (Rate of B) (d) = 1\)

\((\frac{1}{20}) (d-10) + (\frac{1}{30}) d = 1\), 1 represents 100% of the project
\(3d -30 + 2d = 60 ==> 5d = 90 ==> d = 18 days\)

Answer: A

I used logic and back solved to get this answer. We know the combined rate 12 days (1/30) + (1/20). So eliminate choice E as we know that A is stopping ten days earlier. Next look at 28 days and the 26.67 days. If we were to subtract 10 from each we would get something greater than 12 days, so we can eliminate those because if they worked together for more than 12 days the project would be complete. Next I looked at the 16 days subtracted out 10 days do this means they worked together for six days, completing half of the work (6 * (1/12)) and b would have 10 days (1/30) to complete the other half which is obviously not possible. Hence 18 is the answer

A -> 1 day does 1/20 jobs
B -> 1 day does 1/30 jobs
=> (1/20+1/30)*t + 10*1/30=1 -> t=8 -> Time: 8+10=18 -> A

1. (no.of days worked by A)/ (time normally taken by A working alone) + (no.of days worked by B)/ (time normally taken by B working alone) =1
2. Let B work on the project for x days which is also the time for completion of the project
3. (x-10)/20 +x/30 = 1, So x=18 days

We are given that A can complete a project in 20 days and B can complete the same project in 30 days. Thus, the rate of A is 1/20 and the rate of B is 1/30. Since A quits 10 days before the project is completed, we can let t = the time in days worked by B, and thus, (t - 10) = the time in days worked by A. To determine t, we calculate the work done by A and the work done by B, using the work formula: work = rate x time.

work of A + work of B = 1

(1/20)(t - 10) + (1/30)(t) = 1

(t - 10)/20 + t/30 = 1

Multiplying the entire equation by 60, we have:

3t - 30 + 2t = 60

5t = 90

t = 18

Answer: A

I did a very silly calculation mistake in this question-
(3+2)(x-10) + 2*10 = 60 should give us the number of days...

Hi Bunuel,

I understand the method but when I applied this on my own, I placed 1/T on the RHS of the equation instead of 1. Can you please help me understand why do we have 1 instead of 1/T on RHS?

1/t is 1/(time), so rate.

The left hand side is (rate)(time) + (rate)(time) = (job done) + (job done) so it cannot equal to (rate), it should be the total (job done), which is 1.

[quote="154238"]A can complete a project in 20 days and B can complete the same project in 30 days. If A and B start working on the project together and A quits 10 days before the project is completed, in how many days will the project be completed?

(A) 18 days
(B) 27 days
(C) 26.67 days
(D) 16 days
(E) 12 days

In 10 days B can do 1/3 of work

So A and B together have to do 2/3 of work

X(1/12) = 2/3
X= 8

So total 18 days.

Good question



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A completes a project in 20 days which means A completes 5% of the project in a day.

B completes the same project in 30 days which means A completes 3.34% of the project in a day.

In a day, A&B together completes 8.34% of the project.

A quits 10 days before the project is completed, which means B completed 33.4% of the project on his own (10 days * 3.34% each day)

Therefore A&B together completed 66.6% of the project, so they must have worked for 8 days together. (66.6% / 8.34%)

So total time to finish the project = 10 + 8 = 18 days.

Ah I'm not convinced by this Bunuel since this is not how real life is. The answer 27 makes more sense to me because of the way the question is phrased. In real life, how can you know how much time it takes for a project to be completed? You base on actual A's and B's productivity. Given the fact that A's productivity is x/20 work per day, and B is x/30 work per day, we can estimate the days it'll take A and B to finish the work. (12 days)

Then you decide to work together and set a date, "this is the day that it'll be finished." Then after 2 days, A decides to be an asshole and takes a vacation in the Maldives. There are 10 days left before the project is completed - but that's no longer the case since it'll take longer for B to finish the work by himself. The amount of work left is 5/6. Now with the productivity of x/30, B has to slave longer to finish up that 5/6 work left. We take 5x/6 divided by x/30, and we have 25 days. It could have been done in 12 days if A did not leave after 2 days, but now it's done in 27 days.

The way that you did it requires that A notifies everyone that A leaves after 2 days BEFORE they even start working together. However, the way the question is phrased is that they've already started working together and 12 is the amount of time it takes to complete the project.
"If A and B start working on the project together and A quits 10 days before the project is completed"

We are asked how many days project needs to be completed, so we mark that with D;
We also know that if we combine A's and B's done works we'll get one whole so that gives us: Rate*Time=Work
(1/20 * (T-10)) + 1/30*T= 1; => (T-10)/20 + T/30= 1;
(3(T-10) + 2T)/60; => (5T - 30)/60 => (T - 6)/12 = 1;

T = 12+6 = 18 Days.

Instead of taking the work as 1 unit, we take it as the LCM of 20 and 30 = 60 units.

Since A can complete the work in 20 days, per day he does 60/20 = 3 units of work
Since B can complete the work in 30 days, per day he does 60/30 = 2 units of work

From here, we can proceed in 2 ways:

Method 1:

Let the number of days in which the project is completed be \(n\)
Thus, B worked for the entire \(n\) days
Work done by B = \(2n\) units

A worked for \((n - 10)\) days
Work done by A = \(3(n - 10)\) units

Since total work in 60 units:

\(2n + 3(n - 10) = 60\)
\(n = 18\) days

Answer A


Method 2:
Since A quit 10 days before completion, B must have worked alone for those 10 days
Work done by B in those 10 days = 10 * 2 = 20 units

Since total work is 60 units, A and B must have completed the remaining 60 - 20 = 40 units together
Work done by A and B in 1 day = 3 + 2 = 5 units
Thus, time for which A and B worked to compete 40 units = 40/5 = 8 days

Thus, total project duration = 8 + 10 = 18 days

Answer A

Self-made Formula

For the given scenario:
Time needed = b(a+q)/(a+b) = 30(20+10)/(20+30) =18 days.

Opposite Scenario:
If B quits instead of A, the change would be only in the numerator. 'a' and 'b' will be reversed.

Time needed = a(b+q)/(a+b) = 20(30+10)/(20+30) = 16 days.

Here, a = # of days A needs to complete the job alone
b = # of days B needs to complete the job alone
q = # of days one person/machine quit before completing the job after starting together.