Aptitude - Time and Work - Discussion

One days work of X=1/20.
One days work of Y=1/12.
Four days work is. 4[1/X = 1/20],
=> 4/X = 4/20,
=>4/X. = 1/5
1 is the whole work.
1-1/5 = 4/5.
Then 4/5 work is to be done.

X's and Y's one day work is = 1/20+1/12=2/15.

Let "K" be the number of days taken by X and Y to complete the work.

1 day --->2/15.
K days --->4/5.
4/5 ×1 = K× 2/15.
4/5 ×15/2 = K.
K = 2×3 = 6 days.
4 days work of X.
6 days work of X and Y.

4 + 6 =10 days.

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.

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.

@All.

4/5 is the remaining work which is divided by 2/15 which is 1-day work by x and y to find the no. Of days i.e.4/5/2/15.

Can you please explain 4/5 * 15/2?

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.

@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.

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.

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?

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.