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 22 of 36.
Mareeswari said:
10 years ago
Sir I can't understand the A is working two days please explain.
Manju verma said:
10 years ago
Thank you for the solution its really easy way to understand.
Jibinlal said:
10 years ago
1/5 work is done in 3 days. Remaining work is 4/5.
1/5+1/5+1/5+1/5 = 3+3+3+3 = 12 days.
Total = 12 days (4/5th work)+3 days (1/5th work). So 15 days.
1/5+1/5+1/5+1/5 = 3+3+3+3 = 12 days.
Total = 12 days (4/5th work)+3 days (1/5th work). So 15 days.
Tarun said:
10 years ago
Anyone explain this question with LCM method.
Manju said:
10 years ago
LCM OF 20, 30, 60 is 60. Work out LCM method that means 10*2*3=60.
Abhishek said:
10 years ago
Let A does work in x days.
Let B does work in y days.
y - x = 60.
3x = y.
3x-x = 60.
x = 30.
y = 90.
Hence, both will do work in (1/30+1/90) = 6/135.
No of days = 135/6.
22.5 = 45/2.
Answer = 22.1/2.
Let B does work in y days.
y - x = 60.
3x = y.
3x-x = 60.
x = 30.
y = 90.
Hence, both will do work in (1/30+1/90) = 6/135.
No of days = 135/6.
22.5 = 45/2.
Answer = 22.1/2.
Chittu said:
10 years ago
B and C can complete this work in 20 days.
B and C one day work will be = (1/20).
Instead of assuming every third day B and C will join A in work, let us reduce the efficiency of B and C by 3 so this makes B and C one day as 1/60.
So now doing the calculations.
(1/20)+(1/60) = 1/15. Hence they can complete the work in 15 days.
Please let me know in case of any clarifications needed.
B and C one day work will be = (1/20).
Instead of assuming every third day B and C will join A in work, let us reduce the efficiency of B and C by 3 so this makes B and C one day as 1/60.
So now doing the calculations.
(1/20)+(1/60) = 1/15. Hence they can complete the work in 15 days.
Please let me know in case of any clarifications needed.
Chirag gupta said:
10 years ago
I read this but understood nothing please explain in easiest way.
Shivam said:
10 years ago
Let n be the no. of total days required;
A's one day work = 1/10.
B's = 1/30, C's = 1/60.
So n(1/10)+n/3(1/30+1/60) = 1.
From here, we get n = 15.
A's one day work = 1/10.
B's = 1/30, C's = 1/60.
So n(1/10)+n/3(1/30+1/60) = 1.
From here, we get n = 15.
Akash said:
10 years ago
Hey why in last we multiplied 1/5 with 3?
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