Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 3)
3.
A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
Answer: Option
Explanation:
A's 2 day's work = | ![]() |
1 | x 2 | ![]() |
= | 1 | . |
20 | 10 |
(A + B + C)'s 1 day's work = | ![]() |
1 | + | 1 | + | 1 | ![]() |
= | 6 | = | 1 | . |
20 | 30 | 60 | 60 | 10 |
Work done in 3 days = | ![]() |
1 | + | 1 | ![]() |
= | 1 | . |
10 | 10 | 5 |
Now, | 1 | work is done in 3 days. |
5 |
Whole work will be done in (3 x 5) = 15 days.
Discussion:
357 comments Page 5 of 36.
Billal Hossain said:
5 years ago
Solution:
A's 3 days work = 3/20.
(B+C)'s 3rd day or 1 day work= 1/20.
(A+B+C)'s 3 days work = (3/20+1/20).
= 1/5.
A's 1/5 part work in 3 days.
1 part ---> 15 days.
A can do the wok in 15 days (Ans).
A's 3 days work = 3/20.
(B+C)'s 3rd day or 1 day work= 1/20.
(A+B+C)'s 3 days work = (3/20+1/20).
= 1/5.
A's 1/5 part work in 3 days.
1 part ---> 15 days.
A can do the wok in 15 days (Ans).
Poonam said:
5 years ago
Here is a solution:
Efficiency A = 3 , B = 2, C = 1 and A+B+C = 6.
Now,
Work done in format ie A can do a work in 2day and next 1day with B&c .
So.
Eff * time = work.
A = 3 * 2 = 6.
A + B + C = 6 * 1 = 6.
Means work done in 3 days is 6 + 6 = 12 work.
Days -------- work
3 -------- 12
? ------- 60 (total work)
Days =(60*3) / 12.
The answer is = 15 days.
Efficiency A = 3 , B = 2, C = 1 and A+B+C = 6.
Now,
Work done in format ie A can do a work in 2day and next 1day with B&c .
So.
Eff * time = work.
A = 3 * 2 = 6.
A + B + C = 6 * 1 = 6.
Means work done in 3 days is 6 + 6 = 12 work.
Days -------- work
3 -------- 12
? ------- 60 (total work)
Days =(60*3) / 12.
The answer is = 15 days.
(1)
RJ Rishi Winston said:
5 years ago
As per the question, A works all days but on every three days A gets assisted by B&C.
Upon calculation through LCM method the total workdays =60 and also with the help of LCM it is possible to find A's & B+C each day work i.e. 60/20 =3, [60/30+ 60/60].
Thus A's efficiency for one day work is 3 and B+C =2+1 =3. On every 3days A finishes 9 work but as he is assisted by B&C on every 3rd day their work also gets added with A's work as a result of which in every 3rd day 9+3=12 work gets completed.
Hence as on every 3rd day, 12 work gets completed the entire work of 60 would be completed by [60*3] /12 =15 days.
Upon calculation through LCM method the total workdays =60 and also with the help of LCM it is possible to find A's & B+C each day work i.e. 60/20 =3, [60/30+ 60/60].
Thus A's efficiency for one day work is 3 and B+C =2+1 =3. On every 3days A finishes 9 work but as he is assisted by B&C on every 3rd day their work also gets added with A's work as a result of which in every 3rd day 9+3=12 work gets completed.
Hence as on every 3rd day, 12 work gets completed the entire work of 60 would be completed by [60*3] /12 =15 days.
Sameer said:
5 years ago
Fraction of work done in one by A, B, and C is,
1/20, 1/30 and 1/60.
In first 2 days, fraction of work completed by A is 2/20.
Third day, B and C are assisting A then fraction of work completed on 3rd day will be
1/20 + 1/30+ 1/60 = 6/60 i.e 1/10.
So total fraction of work done after 3 days.
2/20+1/10=4/20 i.e. 1/5.
So complete work will be completed in 5 days.
1/20, 1/30 and 1/60.
In first 2 days, fraction of work completed by A is 2/20.
Third day, B and C are assisting A then fraction of work completed on 3rd day will be
1/20 + 1/30+ 1/60 = 6/60 i.e 1/10.
So total fraction of work done after 3 days.
2/20+1/10=4/20 i.e. 1/5.
So complete work will be completed in 5 days.
Vivek said:
6 years ago
I can't get 3rd step.
How do 1/10+1/10 came and how it is divided by 5?
How do 1/10+1/10 came and how it is divided by 5?
Samiksha koli said:
6 years ago
@All.
I couldn't get this. Because A is assisted by B & C on the third day only.
So, A B C will together work for first two days. And for last day A will work alone because he is assisted by B & C for third day. So total work of three days Will be A+B+C=1/10 for 2 days it will be 1/5 and A's last day's work will be 1/20.
So, the total work of three days will be1/4. So total work has to be done in 3*4=12 days.
Correct me, if I am wrong.
I couldn't get this. Because A is assisted by B & C on the third day only.
So, A B C will together work for first two days. And for last day A will work alone because he is assisted by B & C for third day. So total work of three days Will be A+B+C=1/10 for 2 days it will be 1/5 and A's last day's work will be 1/20.
So, the total work of three days will be1/4. So total work has to be done in 3*4=12 days.
Correct me, if I am wrong.
16cs1045 said:
6 years ago
10 %for A in first 2 day 10%(5+3.33+1.66)5 for A in third day and 5 for B and C in third similarly for other 12 day work will complete. Therefore the total time required is 15 day.
Common man said:
6 years ago
3days = 1/5.
6days = 2/5.
9days = 3/5.
12days = 4/5.
15days = 5/5 = 1 total work done is complete.
6days = 2/5.
9days = 3/5.
12days = 4/5.
15days = 5/5 = 1 total work done is complete.
Jayshri said:
6 years ago
How got the 3rd step (1/10+1/10)?
G.prathyusha said:
6 years ago
To complete 1/5th of work - 3 days.
Hence to complete 5/5th of work - x days.
By cross multiply, you will get x value as 15 days.
Hence to complete 5/5th of work - x days.
By cross multiply, you will get x value as 15 days.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers