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 3 of 7.
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...
Ims said:
7 years ago
X=20,Y=12.
SO LCM=60
X=3 unit /day.
Y=5 unit /day.
According to the question;
x work for 4 days alone so 3 *4 = 12.
So 60-12=48.
X+y=8 units.
Therefore, 48÷8=6.
So, 6+4(days of A alone works) = 10.
SO LCM=60
X=3 unit /day.
Y=5 unit /day.
According to the question;
x work for 4 days alone so 3 *4 = 12.
So 60-12=48.
X+y=8 units.
Therefore, 48÷8=6.
So, 6+4(days of A alone works) = 10.
(4)
ANURAG said:
1 decade ago
(x and y)'s one day work= 2/15
Therefore total number of days for the total work= 15/2
And,total number of days for 4/5 of the work=( 15/2 )*(4/5)
Hence total time = 6+4 = 10 days
Therefore total number of days for the total work= 15/2
And,total number of days for 4/5 of the work=( 15/2 )*(4/5)
Hence total time = 6+4 = 10 days
Prienka said:
7 years ago
Why we are calculating total time taken i.e 6+4 days. The question was asked how long did the work last?
Should it be 6 days only right? Anybody can clear my confusion?
Should it be 6 days only right? Anybody can clear my confusion?
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.
Padma said:
1 decade ago
A+B can work
1/20+1/12=8/60.
A can work =4*1/20=1/5.
Remaining work=1-1/5=4/5.
A+B work to days 1/n formula=60/8.
60/8*4/5=6 days.
Total day taken will be=6+4=10.
1/20+1/12=8/60.
A can work =4*1/20=1/5.
Remaining work=1-1/5=4/5.
A+B work to days 1/n formula=60/8.
60/8*4/5=6 days.
Total day taken will be=6+4=10.
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.
Amarnath said:
1 decade ago
Sorry @Zia Don't apply formulas it loses confidence for others. Whatever they know will apply if try tell in simple answer don't tell with formulas.
Zia said:
1 decade ago
Hi, I have a shortcut.
(X-a)*Y/X+Y [X=20,Y=12,a=4]
= (20-4)*12/20+12.
= 16*12/32.
= 12/2.
= 6.
So, total time taken (6+4) = 10 days.
(X-a)*Y/X+Y [X=20,Y=12,a=4]
= (20-4)*12/20+12.
= 16*12/32.
= 12/2.
= 6.
So, total time taken (6+4) = 10 days.
Kavya said:
1 decade ago
Short cut: A=20, B=12.
4 days left: 20-4 = 16.
B joined : (1/20) + (1/12) = 15/2.
Therefore: (16/20)(15/2) = 6.
Total days = 6+4 = 10.
4 days left: 20-4 = 16.
B joined : (1/20) + (1/12) = 15/2.
Therefore: (16/20)(15/2) = 6.
Total days = 6+4 = 10.
(1)
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