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:
359 comments Page 16 of 36.
Aariz said:
10 years ago
Hey, guys. I have some confusion in this question. Is there anyone from Delhi, please guide me to solve this question.
Preethu said:
10 years ago
Why do we multiply it by 5?
How we get the answer as 15? Please help me.
How we get the answer as 15? Please help me.
Jenni said:
10 years ago
How to multiply and solve this equation?
1/20 + 1/30 + 1/60
1/20 + 1/30 + 1/60
Jemish said:
10 years ago
My new approach to explaining the answer.
=> A is working continuously.
=> B and C join at every 3rd day.
A's efficiency is used as it is,
But B and C's efficiency is used 1/3rd part considering a task.
thus,
A--> 1/20
B--> 1/3 * (1/30)
c--> 1/3 * (1/60)
Now simple.
To find out days required = 1/20 + 1/90 + 1/180 = 1/15.
Answer: 15 days required.
=> A is working continuously.
=> B and C join at every 3rd day.
A's efficiency is used as it is,
But B and C's efficiency is used 1/3rd part considering a task.
thus,
A--> 1/20
B--> 1/3 * (1/30)
c--> 1/3 * (1/60)
Now simple.
To find out days required = 1/20 + 1/90 + 1/180 = 1/15.
Answer: 15 days required.
Shruthi said:
10 years ago
Please explain the 2nd step => (A + B + C) 's 1 day's work.
How it becomes 6/60.
How it becomes 6/60.
Ompal. said:
10 years ago
A, B, and C person can a do work 20, 40 and 60 days respectivily. When they works alternative. They compeleted work how many days.
Shridhar said:
10 years ago
Got it on 3 day B & C is helping A.
Shridhar said:
10 years ago
A as done the work for 2 days, he is asserted to B and C in 3rd, he as not done the work at 3rd day. Then why 1/20 + 1/30 + 1/60?
Sudhansu sekhar said:
10 years ago
Why applied 1/10 + 1/10?
Samrat said:
10 years ago
Note: A alone can do its work in 20 days.
So A can do the work in 1 day is 1/20.
2nd day A can do work = (1/20)2 = 1/10.
3rd day A, B and C work together.
Therefore A's 1 day work = 1/20, B's 1 day work = 1/30, C's 1 day work = 1/60.
Together A, B and C do work in 1 day = 1/20 + 1/30 + 1/60.
= (3+2+1)/60.
= 6/60 = 1/10.
So after 3 days A's done = 1/10 + 1/10.
= 2/10 = 1/5.
= (1/5)X days = 1.
X = 5.
1/5 work = 3 days.
(1/5)X days = (3)5 days = 15 days.
So A can do the work in 1 day is 1/20.
2nd day A can do work = (1/20)2 = 1/10.
3rd day A, B and C work together.
Therefore A's 1 day work = 1/20, B's 1 day work = 1/30, C's 1 day work = 1/60.
Together A, B and C do work in 1 day = 1/20 + 1/30 + 1/60.
= (3+2+1)/60.
= 6/60 = 1/10.
So after 3 days A's done = 1/10 + 1/10.
= 2/10 = 1/5.
= (1/5)X days = 1.
X = 5.
1/5 work = 3 days.
(1/5)X days = (3)5 days = 15 days.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers