Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 16)
16.
X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last?
Answer: Option
Explanation:
| Work done by X in 4 days = |  |
1 | x 4 |  |
= | 1 | . |
| 20 | 5 |
| Remaining work = | |
1 - | 1 | |
= | 4 | . |
| 5 | 5 |
| (X + Y)'s 1 day's work = | |
1 | + | 1 | |
= | 8 | = | 2 | . |
| 20 | 12 | 60 | 15 |
| Now, | 2 | work is done by X and Y in 1 day. |
| 15 |
| So, | 4 | work will be done by X and Y in | |
15 | x | 4 | |
= 6 days. |
| 5 | 2 | 5 |
Hence, total time taken = (6 + 4) days = 10 days.
Discussion:
68 comments Page 4 of 7.
Chandra said:
1 decade ago
Because 4 days work done by x. In question they asked total number of days.
Ajith said:
1 decade ago
I this way also it can be thought about:
Assume total days for completion = N.
X 's 1 day's effort = 1/20.
Y's 1 day's effort = 1/12.
(1/20)*N = (1/12)*(N-4)----[i.e, X's N days work = Y's (N-4) days work].
By solving, N = 10 days.
Assume total days for completion = N.
X 's 1 day's effort = 1/20.
Y's 1 day's effort = 1/12.
(1/20)*N = (1/12)*(N-4)----[i.e, X's N days work = Y's (N-4) days work].
By solving, N = 10 days.
Kumutha said:
1 decade ago
Please discuss clearly.
Jyoti said:
1 decade ago
How y=1 /12 please explain?
Akhil said:
1 decade ago
Can anyone explain this problem clearly.
Anusha said:
1 decade ago
a = b+b (0.3).
(1/23) = 1.3 b.
b = (10/{23*13}).
b = (10/299).
a+b = 1/13.
So 13 days.
(1/23) = 1.3 b.
b = (10/{23*13}).
b = (10/299).
a+b = 1/13.
So 13 days.
Sonu jangra said:
10 years ago
x can do a work in 20 days.
y can do a work in 12 days.
To find total work, take LCM of 20 and 12 i.e. 60.
So x+y total 1 day's work = (1/20+1/1/12) = 2/15.
But x work alone = 1/4 day's. Remain work = (1-1/4) = 3/4.
Now, work will be done by x and y = 2/15*3/4 = 1/10.
So 10 day's.
y can do a work in 12 days.
To find total work, take LCM of 20 and 12 i.e. 60.
So x+y total 1 day's work = (1/20+1/1/12) = 2/15.
But x work alone = 1/4 day's. Remain work = (1-1/4) = 3/4.
Now, work will be done by x and y = 2/15*3/4 = 1/10.
So 10 day's.
Chandu said:
10 years ago
X 1 days work = 1/20.
Y 1 days work = 1/12.
X work for 4 days ---> 4/20 = 1/5.
Remaining work to be done--->1-1/5 = 4/5.
After 4 days Y is join with X to work.
So X+Y 1 day's work = (1/20+1/12) = 2/15 ---> X+Y finish work in 15/2 days.
So X+Y finish 4/5 part of work in 15/2*4/5 ----> 6 days.
Total days ---> 4+6 = 10 days.
Y 1 days work = 1/12.
X work for 4 days ---> 4/20 = 1/5.
Remaining work to be done--->1-1/5 = 4/5.
After 4 days Y is join with X to work.
So X+Y 1 day's work = (1/20+1/12) = 2/15 ---> X+Y finish work in 15/2 days.
So X+Y finish 4/5 part of work in 15/2*4/5 ----> 6 days.
Total days ---> 4+6 = 10 days.
(1)
Suraj said:
10 years ago
They said in question in a piece of so how it's possible whole work done by them.
Inna Reddy Chilakala said:
10 years ago
Total work = X.
A = X (begin the work and he is in End of the work).
B = X-4 (After 4 days he joined).
= X/20 + X - 4/12 = 1.
= 3X + 5X - 20/60.
8X = 80.
X = 10.
A = X (begin the work and he is in End of the work).
B = X-4 (After 4 days he joined).
= X/20 + X - 4/12 = 1.
= 3X + 5X - 20/60.
8X = 80.
X = 10.
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