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 1 of 7.
Bikram Mondal said:
5 years ago
X = 20
Y = 12
L.C.M = 60 (Total work)
X efficiency 60÷20 = 3
Y efficiency 60÷12 = 5
Work done by X in 4days = 3 x 4 = 12
X&Y total work efficiency = 3 + 5 = 8
Remaining work completed = 60 - 12/8 = 6
Total time taken = (6+4)days = 10 days.
Y = 12
L.C.M = 60 (Total work)
X efficiency 60÷20 = 3
Y efficiency 60÷12 = 5
Work done by X in 4days = 3 x 4 = 12
X&Y total work efficiency = 3 + 5 = 8
Remaining work completed = 60 - 12/8 = 6
Total time taken = (6+4)days = 10 days.
(10)
Larisa said:
2 years ago
EFFICIENCY:
X = 20 = 3 > 60 (total work)
Y = 12 = 5 >60 (total work).
X started the work & work for 4 days with 3 ( per day work done).
4×3 = 12 ( work done by X in 4 days).
60 - 12 = 48 work remains.
48 ÷ 8 ( 5+3 * & y per day efficiency),
= 6 days + 4 days = 10 days.
X = 20 = 3 > 60 (total work)
Y = 12 = 5 >60 (total work).
X started the work & work for 4 days with 3 ( per day work done).
4×3 = 12 ( work done by X in 4 days).
60 - 12 = 48 work remains.
48 ÷ 8 ( 5+3 * & y per day efficiency),
= 6 days + 4 days = 10 days.
(8)
Jay said:
2 years ago
Let 'p' be the one we should find which is ,
The number of days of total work which is basically X.
Y came after 4 days which is (total days-4) or (p-4).
X and Y can do a piece of work in 20 days and 12 days.
(p/20) + (p-4)/12 = 1
(12*p)+(20*p)-80 = 240.
32 *p =320.
p=10.
The number of days of total work which is basically X.
Y came after 4 days which is (total days-4) or (p-4).
X and Y can do a piece of work in 20 days and 12 days.
(p/20) + (p-4)/12 = 1
(12*p)+(20*p)-80 = 240.
32 *p =320.
p=10.
(5)
Shubham said:
5 years ago
@All.
Simply, the solution is
X takes 20days.
Y takes 12 days.
20 - 60 - 3
12 - 60- 5
So consider X makes 3 chairs and y makes 5 in one day. X worked for 4 days so (3*4) = 12,
So remained chairs 60 - 12 = 48.
Now x+y =3+5=8 so this will complete in (48/8) = 6,
So total 6 + 4 = 10 days.
Simply, the solution is
X takes 20days.
Y takes 12 days.
20 - 60 - 3
12 - 60- 5
So consider X makes 3 chairs and y makes 5 in one day. X worked for 4 days so (3*4) = 12,
So remained chairs 60 - 12 = 48.
Now x+y =3+5=8 so this will complete in (48/8) = 6,
So total 6 + 4 = 10 days.
(5)
Daf said:
6 years ago
The work done by A in 4 days = 4* work done by A in one day.
The work done by A in one day= 1/20.
The work done by a in 4 days= 4/20.
So, the left over work is 1-(4/20) = 16/20.
After 4 days A and B work together.
Work is done by A and B together in one day=(1/20)+(1/12) = 8/60.
No of days needed by A and B to complete the job=(left overwork)/(Work done by A and B together in one day).
= (16/20)/(8/60).
= 6 days.
So total days taken is time taken by A alone + time taken by A and B together.
=6+4.
=10 days.
The work done by A in one day= 1/20.
The work done by a in 4 days= 4/20.
So, the left over work is 1-(4/20) = 16/20.
After 4 days A and B work together.
Work is done by A and B together in one day=(1/20)+(1/12) = 8/60.
No of days needed by A and B to complete the job=(left overwork)/(Work done by A and B together in one day).
= (16/20)/(8/60).
= 6 days.
So total days taken is time taken by A alone + time taken by A and B together.
=6+4.
=10 days.
(4)
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)
Sudharsan said:
6 years ago
Why we are adding +4 to the sum the 6 is the work completed until the last. X+4 is just a time taken, why anyone explain?
(3)
Shanaya said:
4 years ago
Can you please explain 4/5 * 15/2?
(2)
Gouri said:
7 years ago
X=>20 and Y=>12.
L.C.M. of 20&12=>60.
The efficiency of X is 3 and of Y is 5,
3 * 4 = 12 (here 3 is the efficiency of X and 4 is the no: of days he worked alone)
remaining work=60-12=48,
8*x=48 (here 3+5=8),
x=48/8; x=6.
According to the question we want to know how many days work went on
so, 4+6 =10 days.
L.C.M. of 20&12=>60.
The efficiency of X is 3 and of Y is 5,
3 * 4 = 12 (here 3 is the efficiency of X and 4 is the no: of days he worked alone)
remaining work=60-12=48,
8*x=48 (here 3+5=8),
x=48/8; x=6.
According to the question we want to know how many days work went on
so, 4+6 =10 days.
(2)
Tom Cruz said:
7 years ago
x/20 + (x-4)/12 = 1,
12x + 20x - 80 = 240,
32x = 320,
x = 10 days.
12x + 20x - 80 = 240,
32x = 320,
x = 10 days.
(2)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers