Aptitude - Time and Work - Discussion

LCM method:

The LCM of A,B,C is 60 so consider total work to be 60 unit.
Since A take 20 days, he does 3 unit per day.
Similarly, B does 2 unit per day.
Similarly, C does 1 unit per day.

So total unit of work done by 3 combined is 6 unit.

So by Addition of each day of Unit of work done until reaching total work. So the addition of A alone and A+B+C can be summarized as.
A doing work for first 2 days will be 2*3, total Work 6 done.
On 3rd day work done is 6, total work done 12.
Again A working alone for 4th and 5th is 2*3 =6,
Total work 18
On 6th day work done (A+B+C) 6, Total work 24
On 7th and 8th A is 6, total work 30.
On 9th day again (A+B+C) work done 6, TOTAL 36.
On 10 and 11th A alone work was done 6, Total work 42.
On 12th (A+B+C) work done 6, Total work 48.
On 13 and 24th A alone is 6, Total work 54.
On 15th A+B+C is 6, Total work was done 60 unit, work completed.

A's 2 day's work = (1/20x2) = 1/10.

(A + B + C)'s 1 day's work = [(1/20)+(1/30)+(1/60)] = 6/60 = 1/10.

Work done in 3 days = [(1/10)+(1/10)] = 1/5.

Now what they ask "How many days can "A" do the work?"

Here revise the FORMULA NO:2.

That is:

Days from Work:

If A's 1 day's work =1/n,then A can finish the work in n days.

Here ===> [1 DAY's work = NO.of WORK / required "n" Day's do the work ].

We want no.of DAY's;

So rearrange the formula; we get,

No.of WORK = (1DAY's work * "n" DAY's) ------> take eqn no.1.

Now, (1/5) work is done in 3 days.

3 = 1/5.

So apply above the data's in eqn no.1.

1(No.of Work) = (Day's of Work )3 * ("n" DAY's)5.

1(No.of Work) = 15 Day's

So "A" is require 15 Day's to do a 1 work.

"I HOPE U WILL B UNDERSTAND"

There's a simple logic friends,

If a person wrks for n days, his 1 day's work is 1/n.
or the other way round,
if 1/n is the 1 day's work then he works for n days.
& it can be very well known if 1 day's work is known it can be multiplied with any numbers of days you wish to.

Similarely,

Here last statement tell the same i.e. 3days work is 1/5,
so to calculate 1 day's work we need to divide it by 3 wiz gives 1/15.
so if 1/15 is 1 day's work then they together work for 15 days.
& the logic behind 2 days & 3 dats work is :
A's 2 days work is (1/20)*2, i.e 1 day's multiplied by nth day.

but on the "3rd day" they all A,B&C work together wiz equals to 1/10.

Therefore 2 days + 3rd day's wrk gives 3 day's work.

"A" can complete work in 20 days.

A's 1 day work is 1/20.

A's 2 days work is 2/20.

According to the question A work for 2 day's and on the third day.

B & C assists (**helps, not replaced).

So it goes like this.

A's first day + A's second day + (A + B + C) on third day, B & C helped.

i.e 2 1/20+1/20/+(1/20+1/30+1/60).

=>2/20+(1/20+1/30+1/60).

=>2/20+1/10.

= 1/5.

So for first 2 days and combining B and C on third day they will work 1/5th of a work.

All together (A+A+ (A+B+C) ) works for 3 days to get 1/5th of a work so.

If 1/n work is done in 1 days total work is done in 1*n = n days.

The same here, if 1/5 work is done in 3 days total work is done in 5*3 = 15 days.

For 1st 2 days A will do work himself.

A's work in 1st 2 days:

(1/20)*2 = 1/10 or 0.1.

0.1 is 10% of total work done which is 1 or 100%.

On 3rd day B and C will assist A to do work.

A's, B's, and C's work in 3rd day:

(1/20+1/30+1/60)*1 = 1/10 or 0.1.

Again in this day three of them could do only another 10% of work out of 100% or 0.1 out of 1.

For 1st three days A, B, and C accomplished:

1/10+1/10 = 2/10, 0.2 or 20 % of total work out of 100% or 1.

To finish they need to accomplish 100-20 = 80 % or 0.8 of work remaining.

Having 20% or 3 days we need 15 (in which every 3rd day is assisted by B and C) more days (100%/20% = 5 => 5*3 = 15) to achieve total work accomplishment.

Here's a simple explanation :

WORK = (Rate)*(Time) ***time will cancel out***

First Day: Person A does all the work.

His working rate is 1/20 so (1/20)*1day = 1/20 work.

Second Day: Person A does all the work.

His working rate is 1/20 so (1/20)*1day = 1/20 work.

All in all in 2 days is 2/20 or 1/10.

Third Day: Person A is assisted by B and C.

(1/20)+(1/30)+(1/60) times 1 day = 1/10.

All in all in 3 days is (1/10)+(1/10)= 1/5 work is done.

So 4/5 remaining..

Question Says B and C assisted A EVERY third day.

Repeat the process until you get 5/5 or 1 full work that would be on the 15th day, A will complete 5/5 or 1 full work.

Sorry Guys. Can somebody please explain me more clearly.
My doubt is
" when we say A+B+C =1/10.
it means we have considered 1 day's work of A,B,C and the answer 1/10 is total of all 3 together working days"


" so when we say A's 2 day's work = 1/10
it means we are considering only A's 2day's work "

So,finally when we add for 3 days= (1/10) + (1/10)
here 1/10 is A's 2day's work, but when we take 2nd time 1/10 then does'n it mean that we are considering all 3 working together's total 1/10 ????

As the given question is In how many days can A do the work if he is assisted by B and C on every third day?

Hope u can udnerstand what I wrote above ...

A's one day work is 1/20 ,

Therefore A's 2 day work will be (1/20)*2=(1/10),

Now (A+B+C)'s one day work will be = (1/20)+(1/30)+(1/60)=(1/10).

Since A is assisted on every third day of his work then total work done in three days will include A's 2 day's of work plus the work done on third day, that is (A+B+C)'s one day work,

Therefore work done in 3 days will be,A's two days work (1/10 )+ A,B,C's one day work (1/10)= (1/5).

Now in three days 1 out of 5 parts of work is done , and yet 4 parts are to be done there fore 4*3= 12 days and since we have calculated the earlier 1 parts requirement of days i,e 3 days total days will to finish 5 parts will be (12+3)= 15 days.

A work in one day=1/20,
B work in one day=1/30,
C work in one day=1/60.

as A work for 2 days and then on third day B and C do work.
total work done in three days = A 3 days work+B 1 day work+ C one day work,
=3/20+1/30+1/60,
=9/60+2/60+1/60
=12/60.

Total work done in three days =1/5.
3=1/5part.

Let x days required to complete work=1 part.

3/x=1/5/1.
x=3*5=15 days.

As per the question, A works all days but on every three days A gets assisted by B&C.


Upon calculation through LCM method the total workdays =60 and also with the help of LCM it is possible to find A's & B+C each day work i.e. 60/20 =3, [60/30+ 60/60].

Thus A's efficiency for one day work is 3 and B+C =2+1 =3. On every 3days A finishes 9 work but as he is assisted by B&C on every 3rd day their work also gets added with A's work as a result of which in every 3rd day 9+3=12 work gets completed.

Hence as on every 3rd day, 12 work gets completed the entire work of 60 would be completed by [60*3] /12 =15 days.