Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 13)
13.
A and B can together finish a work 30 days. They worked together for 20 days and then B left. After another 20 days, A finished the remaining work. In how many days A alone can finish the work?
Answer: Option
Explanation:
(A + B)'s 20 day's work = | ![]() |
1 | x 20 | ![]() |
= | 2 | . |
30 | 3 |
Remaining work = | ![]() |
1 - | 2 | ![]() |
= | 1 | . |
3 | 3 |
Now, | 1 | work is done by A in 20 days. |
3 |
Therefore, the whole work will be done by A in (20 x 3) = 60 days.
Discussion:
63 comments Page 1 of 7.
Abinesh said:
7 months ago
A + B = 30 days
Then A + B works for 20 days.
This shows that they have completed 2/3 works.
Now 1/3 of the work is done by A in 20 days.
So, 1/3 * 1/20,
= 60 days.
Then A + B works for 20 days.
This shows that they have completed 2/3 works.
Now 1/3 of the work is done by A in 20 days.
So, 1/3 * 1/20,
= 60 days.
(4)
SamDon Sk said:
2 years ago
Assume painting a room A + B can complete in 30 days.
But they work for only 20 days and b left till now they have painted most of the room.
The remaining days left they can do work together is 10 days.
But A worked for 20 days to complete that 10 days of work MEANS.
A had worked DOUBLE days as they can work together.
So, A can alone can complete the work is DOUBLE of together work that is 30 * 2 = 60.
But they work for only 20 days and b left till now they have painted most of the room.
The remaining days left they can do work together is 10 days.
But A worked for 20 days to complete that 10 days of work MEANS.
A had worked DOUBLE days as they can work together.
So, A can alone can complete the work is DOUBLE of together work that is 30 * 2 = 60.
(28)
Anitha said:
2 years ago
A + B = 1/30
B only work 20 days after leaving;
B=1/20,
A+1/20 = 1/30
A = 1/20-1/30
A = 30-20/600
A = 10/600
A = 1/60
So, A Worked alone 60 days.
B only work 20 days after leaving;
B=1/20,
A+1/20 = 1/30
A = 1/20-1/30
A = 30-20/600
A = 10/600
A = 1/60
So, A Worked alone 60 days.
(44)
Ancy said:
2 years ago
A+B' s can finish work = 30 days
if,
A+B work together = 20 days
Therefore, 30:20 =3:2 (3 is the total work & 2 is together work)
= 3-2 = 1 (1 is remaining A's work)
A's remaining work is 20 days
= 20 * 3 = 60 days.
if,
A+B work together = 20 days
Therefore, 30:20 =3:2 (3 is the total work & 2 is together work)
= 3-2 = 1 (1 is remaining A's work)
A's remaining work is 20 days
= 20 * 3 = 60 days.
(18)
HB Rajputh said:
2 years ago
I'm not getting it, please help me by explaining in detail.
(5)
Piyush Srivastava said:
3 years ago
@All.
Here is the solution.
Together A + B takes 30 days to finish the work. So let us assume the total work to be 300 units(it can be any multiple of 30 to make the calculation easy) .
Therefore per day work efficiency of A+B together = 300/30 (work done = per day efficiency*time , so efficiency = work done/time taken) .
Thus A + B one day efficiency= 10 units.
Now ATQ, B left after 20 days, which means A + B worked together for 20 days. Therefore work done by A + B together in 20 days = 20 *10 = 200 (work done = one day efficiency * no of days worked )
Now remaining work = 300 - 200 = 100 units (total work - work done by A + B together)
Now A takes 20 days to finish the remaining work. Thus day efficiency of A = 100/20 = 5 units per day. ( Again using work done / time taken = per day work efficiency. Unitary method)
Thus per day efficiency of A = 5 units per day.
Now time is taken by A to finish the whole work alone = 300/5 = 60 days.. (total work / per day efficiency) . Hence A takes 60 days to finish the whole work alone.
Also, the people who want to know how much time will B take to do the whole work alone, the answer is again 60 days. Why?
Because per day efficiency of A + B together was 10 units, also after calculating A's per day efficiency was found out to be 5 units, thus B's per day efficiency= 10 - 5 = 5 units per day.
So it's the same as A, therefore B will also take 60 days to finish the whole work alone.
Here is the solution.
Together A + B takes 30 days to finish the work. So let us assume the total work to be 300 units(it can be any multiple of 30 to make the calculation easy) .
Therefore per day work efficiency of A+B together = 300/30 (work done = per day efficiency*time , so efficiency = work done/time taken) .
Thus A + B one day efficiency= 10 units.
Now ATQ, B left after 20 days, which means A + B worked together for 20 days. Therefore work done by A + B together in 20 days = 20 *10 = 200 (work done = one day efficiency * no of days worked )
Now remaining work = 300 - 200 = 100 units (total work - work done by A + B together)
Now A takes 20 days to finish the remaining work. Thus day efficiency of A = 100/20 = 5 units per day. ( Again using work done / time taken = per day work efficiency. Unitary method)
Thus per day efficiency of A = 5 units per day.
Now time is taken by A to finish the whole work alone = 300/5 = 60 days.. (total work / per day efficiency) . Hence A takes 60 days to finish the whole work alone.
Also, the people who want to know how much time will B take to do the whole work alone, the answer is again 60 days. Why?
Because per day efficiency of A + B together was 10 units, also after calculating A's per day efficiency was found out to be 5 units, thus B's per day efficiency= 10 - 5 = 5 units per day.
So it's the same as A, therefore B will also take 60 days to finish the whole work alone.
(10)
Sandip said:
3 years ago
Still, I can't understand this. Can anyone help me to get it?
(4)
Arun said:
3 years ago
Why the reaming work is -1? Explain it please.
(1)
Arun said:
3 years ago
a+b=30 days ==>1/30 amount of work in a single day.
a and b worked for 20 days, (1/30)*20 = 2/3.
after b left, a completed the remaining 1-(2/3)=>1/3 amount of work in 20 days.
------for 20 days-------
a will done 1/3==>0.33 amount of work.
------if its 40 days----
a will done 2/3==>0.66 amount of work.
------if its 60 days------
a will done 3/3==>1 amount of work.
So, the answer is, A will take 60 days to complete the total(ie:1) amount of work.
Hope this helps.
a and b worked for 20 days, (1/30)*20 = 2/3.
after b left, a completed the remaining 1-(2/3)=>1/3 amount of work in 20 days.
------for 20 days-------
a will done 1/3==>0.33 amount of work.
------if its 40 days----
a will done 2/3==>0.66 amount of work.
------if its 60 days------
a will done 3/3==>1 amount of work.
So, the answer is, A will take 60 days to complete the total(ie:1) amount of work.
Hope this helps.
(7)
ATHI said:
4 years ago
In efficiency method;
A+B = 1/30.
(B only work for 20 days that mean A & B work together only 20 days. The remaining 20 days A alone do it).
Now,
A+B = 20 days
A (alone)= 20 days.
Total work:
20(A+B)+ 20(A) = 1, (1 IS TOTAL WORK )
20(1/30)+20A = 1,
(2/3)+20A = 1,
2+60A = 3,
60A = 3-2,
60A = 1.
A = 1/60.
i. e, A = 60days.
A+B = 1/30.
(B only work for 20 days that mean A & B work together only 20 days. The remaining 20 days A alone do it).
Now,
A+B = 20 days
A (alone)= 20 days.
Total work:
20(A+B)+ 20(A) = 1, (1 IS TOTAL WORK )
20(1/30)+20A = 1,
(2/3)+20A = 1,
2+60A = 3,
60A = 3-2,
60A = 1.
A = 1/60.
i. e, A = 60days.
(5)
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