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?
5

5 1
2

6
8
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

Therefore 1 work is done by A in ( 18 x 1 ) = 6 days.
3 3

Discussion:
63 comments Page 2 of 7.

Harsha said:   5 years ago
Nice, Thanks @Siva.

Priyanka Sengar said:   5 years ago
@All.

A=18
B=15
B=10
A=?

So 18:15::x:10
X = 12
12 ÷ 2 = 6.

Bantu Venkatesh said:   6 years ago
15days-----------18days.
5days------------ ?
So, 5/15 * 18 = 6days.
(3)

Kamu said:   6 years ago
Can't get it. Please explain me clearly.

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

Mahadev said:   6 years ago
How to calculate the remaining work that is 1/3?

Please, anyone, clear it.

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.

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.

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.

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.


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