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 5 of 7.
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.
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.
SAGAR said:
1 decade ago
10*1/15+x*1/18 = 1.
By solving you get x = 6 days.
By solving you get x = 6 days.
Sayali said:
1 decade ago
@Rucha.
How you calculate that rate?
How you calculate that rate?
Deepak Patgar said:
1 decade ago
It will be easy if we use chocolate method.
A takes 18 days and B takes 15 days. LCM of 18 and 15 is 90.
Let 90 be the total number of chocolates.
A takes 18 days to eat 90 chocolates. so A eats 5 chocolates per day.
B takes 15 days to eat 90 chocolates. So B eats 6 chocolates per day.
in 10 days B finishes 6*10 = 60 chocolates.
Remaining chocolates are 30.
So to eat 30 chocolates A need 30/5 = 6 days.
Ans = 6 days.
A takes 18 days and B takes 15 days. LCM of 18 and 15 is 90.
Let 90 be the total number of chocolates.
A takes 18 days to eat 90 chocolates. so A eats 5 chocolates per day.
B takes 15 days to eat 90 chocolates. So B eats 6 chocolates per day.
in 10 days B finishes 6*10 = 60 chocolates.
Remaining chocolates are 30.
So to eat 30 chocolates A need 30/5 = 6 days.
Ans = 6 days.
Raja said:
1 decade ago
LCM 90 brief explain?
Left the job B what mean?
Left the job B what mean?
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.
Ruchi Jain said:
1 decade ago
Well the way I solve this question,
REMAINING IS 30 which is to be done by A and A's rate is 5 30/5 = 6.
A B B
T 18 15 10
R 5 6 6
W 90 90 60
REMAINING IS 30 which is to be done by A and A's rate is 5 30/5 = 6.
Swapnika said:
1 decade ago
A done in 18 says so we can do in this way also:
18*5 //therefore (15-10=5).
18*5 = 90.
90/15 = 6.
18*5 //therefore (15-10=5).
18*5 = 90.
90/15 = 6.
Meher said:
1 decade ago
I don't understand the logic of multiplying 18*1/3.
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