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 3 of 7.
Tani said:
9 years ago
Two men can do a piece of work in 6 hours and 4 hours. After the first has worked for 2 hours he is joined by another man. By when should the work be completed?
Can anyone please solve this question. I am facing a lot of problem in this question. This is from composite mathematics. The answer at the back is 1 hour 36 minutes and I am getting 1 hour 33 Minutes.
Can anyone please solve this question. I am facing a lot of problem in this question. This is from composite mathematics. The answer at the back is 1 hour 36 minutes and I am getting 1 hour 33 Minutes.
Sowndarya said:
9 years ago
Simple & super @Arsh.
Akshay silgari said:
9 years ago
I don't understand the logic of multiplying 18*1/3. Please help me in this.
Abhilasha said:
7 years ago
I have a question that while B is working for 10 days is A remains idle? because nothing is mentioned about that in the question.
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.
MIDHUN said:
6 years ago
1/18 WORK DONE IN 1 DAY.
1/3 WORK DONE IN X DAY.
X*1/18 = 1/3,
X = 18/3 = 6.
1/3 WORK DONE IN X DAY.
X*1/18 = 1/3,
X = 18/3 = 6.
Veereshachar said:
6 years ago
A can do it in 18 days.
B can do it in 15 days.
Take LCM of those 18 and 15 I.e., 90.
Consider total work as 90,
ThenA's efficiency = total work/time taken by A.
=>90/18=5.
Similarly for B=>90/15=6.
So B's 10days work is 10*6=60.
Remaining work is 90-60=30.
Time taken by A to complete a work is =remaining work/efficiency of A.
=>30/5 = 6days.
B can do it in 15 days.
Take LCM of those 18 and 15 I.e., 90.
Consider total work as 90,
ThenA's efficiency = total work/time taken by A.
=>90/18=5.
Similarly for B=>90/15=6.
So B's 10days work is 10*6=60.
Remaining work is 90-60=30.
Time taken by A to complete a work is =remaining work/efficiency of A.
=>30/5 = 6days.
Mahadev said:
6 years ago
How to calculate the remaining work that is 1/3?
Please, anyone, clear it.
Please, anyone, clear it.
Iqbal said:
6 years ago
TAKE LCM of 18 & 90 and divide by the days
5-A-18.
6-B-15.
First B do work for 10 days, means 10*6 = 60 already done left work(30) is done by A alone so divide 30/5=6.
And 6 is the answer
5-A-18.
6-B-15.
First B do work for 10 days, means 10*6 = 60 already done left work(30) is done by A alone so divide 30/5=6.
And 6 is the answer
Moni said:
9 years ago
A can do a work in P days and B can do the same work in 2P days, A stop working after some days B alone complete the work in R days. How many days it takes to complete the work?
Can anyone solve this problem?
Can anyone solve this problem?
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