Aptitude - Time and Work - Discussion

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.

Thank you @Inna Reddy Chilakala.

@Shweta.

Can you please elaborate last line?

How come 4/5?

Please explain.

x/20 + (x-4)/12 = 1,
12x + 20x - 80 = 240,
32x = 320,
x = 10 days.

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?

Clear and detailed explanation. Thank you.

4/20+x/20+x/12=1.

X=6 so total days=6+4=10days.

X"s one day work is 1/20.
so he worked 4 days alone which is,
4* 1/20 = 1/5 remember the Confusion avoider formula below,
(Days * single day"s work( together or alone, this case alone by X) = completed work) remember this
when we know days RHS is Completed work, if Days not known RHS is Pending work
1/5 work is completed out entire work 1
entire work - completed work = 1-1/5 = 4/5 or 80% which is pending work
pending work is 4/5 which is done by X and Y
one day's work of X and Y is 1/20 + 1/12 = 2/15 this is 1 day"s work and Inverse is the Days. If you ask WHY see X 1/20 is one day's work of X, inverse which is 20 is the number of days. Is it clear?

15/2 * 4/5= 6 days
Use this formula.

Days * single day's work ( this case by X and Y) = remaining or completed work.
Days we have to find * 2/15 = 4/5 which gives 4/5* 15/2 =6.

Now 6 days for 80% or 4/5 of the job by x and Y and 4 days for initial 1/5 or 20% of the job by x.

So, total 10 days.

Can we do like this?

(4/20)+(x/12)=1.
It give 9.6 and dats means 10days.