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 5 of 7.
Zeeshan said:
1 decade ago
Well done harsha.
Srikanth Chowdary said:
1 decade ago
Mitra please explain clearly.
Lydia said:
1 decade ago
Please explain the remaining work.
Mitra said:
1 decade ago
(x+4)/20 + x/12 = 1
solving x = 6 days
total days 4+6=10days
solving x = 6 days
total days 4+6=10days
Swetha said:
1 decade 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
Therefore Total work =60
x's 1 day capacity =60/20 =3
Y's 1 day capacity =60/12 =5
Since x alone did work for 4 days,
4*3=12 ,
12 work done in 4 days
there fore remaining work = 60-12 =48
Remaining work was completed by both
remaining work / (x's per day capacity +y's per day capacity)=48/6= 6
therefore remaining work was completed in 6 days
Total work was completed in 4 + 6 = 10 days.
I hope this is useful...
y can do a work in 12 days
To find total work ,take LCM of 20 and 12 i.e 60
Therefore Total work =60
x's 1 day capacity =60/20 =3
Y's 1 day capacity =60/12 =5
Since x alone did work for 4 days,
4*3=12 ,
12 work done in 4 days
there fore remaining work = 60-12 =48
Remaining work was completed by both
remaining work / (x's per day capacity +y's per day capacity)=48/6= 6
therefore remaining work was completed in 6 days
Total work was completed in 4 + 6 = 10 days.
I hope this is useful...
Harsha said:
1 decade ago
In 1 day 1/20 th work.
Let us assume a takes x days to complte it
As B joins 4 days later he has only x-4 days remaining.
They both work respectively with their capacities and finish the one complete work
Hence
(1/20th work) * x days + (1/12th work) * (x-4) days = 1 full work
=> x/20 + (x-4)/12 = 1
=> x = 10days
Let us assume a takes x days to complte it
As B joins 4 days later he has only x-4 days remaining.
They both work respectively with their capacities and finish the one complete work
Hence
(1/20th work) * x days + (1/12th work) * (x-4) days = 1 full work
=> x/20 + (x-4)/12 = 1
=> x = 10days
Vikas said:
2 decades ago
Ramchandra please explain it clearly. It is not understandable.
Ramchandra said:
2 decades ago
Hai frnds this is very long process....
I can find one short cut ..i.e
Suppose x & Y do the work = k days (assume)
y do work = k-1 days
we now that
(x/20)+((x-4)/12))=1 ( By solving )
x=10 days...
I can find one short cut ..i.e
Suppose x & Y do the work = k days (assume)
y do work = k-1 days
we now that
(x/20)+((x-4)/12))=1 ( By solving )
x=10 days...
Neeru said:
2 decades ago
Always total work will be considered as 1, in dat 1/5 th work is already done and the remaining work is 1-1/5 =4/5 dats it.
I hope you understand this.
I hope you understand this.
Nagaraising said:
2 decades ago
Please explain about remaining work?.
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