Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 11)
11.
A can finish a work in 18 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?
Answer: Option
Explanation:
| B's 10 day's work = |  |
1 | x 10 |  |
= | 2 | . |
| 15 | 3 |
| Remaining work = | |
1 - | 2 | |
= | 1 | . |
| 3 | 3 |
| Now, | 1 | work is done by A in 1 day. |
| 18 |
 |
1 | work is done by A in | |
18 x | 1 | |
= 6 days. |
| 3 | 3 |
Discussion:
63 comments Page 2 of 7.
Ajay Singh said:
1 decade ago
See as per my knowledge of solving such question it should be in this way:
A's 1 day work=1/18,
B's 1 day work=1/15,
Both (A+B)'s 1 day work =1/18+1/15=1/10,
So their 10 days of work together will be=1/10*10=1,
Thus,remaining work =1-1=0,
So, in my opinion this question is incorrect.
Please reply if I am wrong and give possible solution.
A's 1 day work=1/18,
B's 1 day work=1/15,
Both (A+B)'s 1 day work =1/18+1/15=1/10,
So their 10 days of work together will be=1/10*10=1,
Thus,remaining work =1-1=0,
So, in my opinion this question is incorrect.
Please reply if I am wrong and give possible solution.
UDAY.H.P said:
1 decade ago
Consider the full work to be done = 1
A can complete his work in 15 days but at the same speed he will work only for 10 days that is = 2/3
then the remaining work is = 1-(2/3)
A's working speed is =1/18 per day
that means aA can compleate this total work in 18 day
but the remaining work is 1/3 so 1/3*18 that is=6days
A can complete his work in 15 days but at the same speed he will work only for 10 days that is = 2/3
then the remaining work is = 1-(2/3)
A's working speed is =1/18 per day
that means aA can compleate this total work in 18 day
but the remaining work is 1/3 so 1/3*18 that is=6days
Anonymous said:
1 decade ago
The question is flawed. Both A and B can do the work together in less than 10 days. Consider revising the wording of the problem.
A[WorkPerDay] = 0.05556.
B[WorkPerDay] = 0.06667.
Cumulatively in 10 days the total is 1.2223, which is more than enough to finish the work.
A and B finish the work in 8 days.
A[WorkPerDay] = 0.05556.
B[WorkPerDay] = 0.06667.
Cumulatively in 10 days the total is 1.2223, which is more than enough to finish the work.
A and B finish the work in 8 days.
Willoy said:
1 decade ago
The question is flawed.
Both A and B finish the job in 8 days.
Compute for work done rather than remaining:
A[work per day] = 0.05556.
B[work per day] = 0.06667.
A+B[8 days work] = 0.97784.
A+B[10 days work] = 1.2223 <--- job over done.
B left the job because he's not doing anything anymore.
Both A and B finish the job in 8 days.
Compute for work done rather than remaining:
A[work per day] = 0.05556.
B[work per day] = 0.06667.
A+B[8 days work] = 0.97784.
A+B[10 days work] = 1.2223 <--- job over done.
B left the job because he's not doing anything anymore.
Gouri said:
7 years ago
A=> 18days.
B=> 15days.
B left after 10 days of work then the remaining work of B have to be done by A,
L.C.M. of 18 & 15 =>90.
The efficiency of A is 5 and of B is 6,
work done by B is 6*10=60,
then 90-60=30 (here 90 is L.C.M.)
then work done by A is 30/5 = 6days.
B=> 15days.
B left after 10 days of work then the remaining work of B have to be done by A,
L.C.M. of 18 & 15 =>90.
The efficiency of A is 5 and of B is 6,
work done by B is 6*10=60,
then 90-60=30 (here 90 is L.C.M.)
then work done by A is 30/5 = 6days.
Amit said:
1 decade ago
See ;
B can work 1/15 part of the work in one day and he has worked 10 days means - 1/15*10 = 2/3 part of work has been completed. Hence remained work is 1-2/3 = 1/3.
Given,
A can complete whole work in 18 days.
Hence,
1/3 part of work will be completed in 18*1/3 = 6 days.
B can work 1/15 part of the work in one day and he has worked 10 days means - 1/15*10 = 2/3 part of work has been completed. Hence remained work is 1-2/3 = 1/3.
Given,
A can complete whole work in 18 days.
Hence,
1/3 part of work will be completed in 18*1/3 = 6 days.
Nav raj bhatta said:
4 years ago
Easy method:
A = 18.
B = 15.
Now total work = LCM of A & B which is 90,
A's per day work=90/18 = 5
B's per day work=90/15 = 6.
Now B's work in 10 days = 10 * 6 = 60,
Remaining work=90-60 = 30,
Now the time taken by A to finish the remaining work = 30/5 = 6 days.
A = 18.
B = 15.
Now total work = LCM of A & B which is 90,
A's per day work=90/18 = 5
B's per day work=90/15 = 6.
Now B's work in 10 days = 10 * 6 = 60,
Remaining work=90-60 = 30,
Now the time taken by A to finish the remaining work = 30/5 = 6 days.
(42)
Larisa said:
2 years ago
@All.
Here is the Easier method:
Efficiency = A : 18 = 5 > 90.
B : 15 = 6 > 90.
Total work = 90:
B work for 10 days so = 6× 10 = 60,
90 - 60 = 30,
A's per day efficiency = 5 so , 30 ,÷ 5 = 6,
So 6 days is the answer.
Here is the Easier method:
Efficiency = A : 18 = 5 > 90.
B : 15 = 6 > 90.
Total work = 90:
B work for 10 days so = 6× 10 = 60,
90 - 60 = 30,
A's per day efficiency = 5 so , 30 ,÷ 5 = 6,
So 6 days is the answer.
(13)
Ashok said:
9 years ago
Verification of answers:
B can do 10 days work in = 10/15.
A can do one day work = 1/18.
So we can verify the options like to became A+B = 1.
Given That,
B left the work after days so.
10/15+ x*1/18 = 1.
By verifying result: x = 6.
B can do 10 days work in = 10/15.
A can do one day work = 1/18.
So we can verify the options like to became A+B = 1.
Given That,
B left the work after days so.
10/15+ x*1/18 = 1.
By verifying result: x = 6.
Jay said:
1 decade ago
See, B complete his work in 15 days so split in 3 so we get 1:1:1.
Then he done the work of 10 days, so only last one remaining.1/3.
Now, A finish the whole work in 1/18. He need to done work in 1/3.
So 18*1/3 = 6 days.
Then he done the work of 10 days, so only last one remaining.1/3.
Now, A finish the whole work in 1/18. He need to done work in 1/3.
So 18*1/3 = 6 days.
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