Aptitude - Time and Work - Discussion

10 %for A in first 2 day 10%(5+3.33+1.66)5 for A in third day and 5 for B and C in third similarly for other 12 day work will complete. Therefore the total time required is 15 day.

@All.

I couldn't get this. Because A is assisted by B & C on the third day only.

So, A B C will together work for first two days. And for last day A will work alone because he is assisted by B & C for third day. So total work of three days Will be A+B+C=1/10 for 2 days it will be 1/5 and A's last day's work will be 1/20.

So, the total work of three days will be1/4. So total work has to be done in 3*4=12 days.

Correct me, if I am wrong.

I can't get 3rd step.

How do 1/10+1/10 came and how it is divided by 5?

Fraction of work done in one by A, B, and C is,

1/20, 1/30 and 1/60.

In first 2 days, fraction of work completed by A is 2/20.

Third day, B and C are assisting A then fraction of work completed on 3rd day will be
1/20 + 1/30+ 1/60 = 6/60 i.e 1/10.

So total fraction of work done after 3 days.
2/20+1/10=4/20 i.e. 1/5.

So complete work will be completed in 5 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.

Here is a solution:

Efficiency A = 3 , B = 2, C = 1 and A+B+C = 6.

Now,
Work done in format ie A can do a work in 2day and next 1day with B&c .
So.
Eff * time = work.

A = 3 * 2 = 6.
A + B + C = 6 * 1 = 6.
Means work done in 3 days is 6 + 6 = 12 work.
Days -------- work
3 -------- 12
? ------- 60 (total work)

Days =(60*3) / 12.
The answer is = 15 days.

Solution:

A's 3 days work = 3/20.
(B+C)'s 3rd day or 1 day work= 1/20.
(A+B+C)'s 3 days work = (3/20+1/20).
= 1/5.

A's 1/5 part work in 3 days.
1 part ---> 15 days.

A can do the wok in 15 days (Ans).

3A----20
2B----30
1C----60 this come from L. C. M------60.

Add:all efficiency A+B+C=3+2+1=6.
Per day works of A+B+C=60/6=10.
3days work of A+B+C.

3 = (3+3per day work of A +6 per day work of a+b+c).
3 = 12(Tw=60 if we want 60in the table of 12 then we need to * ...12*5 and also *right side)
5*3 = 12 * 5.

15day = 60. A can do the work in 15 days.

A =20 Days
B=30 Days
C=60 Days

LCM = A,B,C
LCM = 60
A = 60/20 =3.
B = 60/30 = 2.
C = 60/60 =1.

AFTER 3 DAYS B, C JOINED.

Total =3 + 3(3 + 1) = 15 DAYS.

Step 1: Work is done by A+B+C in one day can be calculated as 1/20 + 1/30 + 1/60 = 1/10.
Step 2: For the first two days only A is doing work alone so the work is done by A in two days will be 2*1/20= 1/10.
Step 3: Now total work is done in 3 days by A B & C will be = 1/10+1/10= 1/5.
Step 4: But we need a complete work to be done so we have to convert 1/5 to 1 therefore if we do 1/5 work 5 times then work will be completed (1/5 x 5=1).

Therefore 5 x 3 = 15.

Hope you will understand this solution.