Aptitude - Time and Work - Discussion

@All.

Please explain the last step.

I am not understanding A's 2 day's work. Please explain.

A's total work = 1.
A's 1 day work = 1/20 = 0.05.
B's 1 day work = 1/30 = 0.03.
c's 1 day work = 1/60 = 0.01.

B + C combined work together = 0.03 + 0.01 = 0.04.
A's 3 day work = 0.05 + 0.05 + 0.05 + 0.04 = 0.19.
If answer is 15 means = 5 * (0.19) = 0.95 remaining 0.05.

So, the answer is wrong.
In 16th day = (15days work + A's one day work).
= (0.95 + 0.05),
= 1.

So, 16 days is the correct answer.

According to the question the A has been assisted by B and C after the third day.

It means the A has to work for three days and the third day is followed by B and C.

According to me, the answer is [1/10 + 1/10 + 1/20] = 1/4 for 3 days the whole days is 4 * 3 = 12.

3 values there 3persons working to gather assisted A's B + C we calculate A + B + C/2 we get,

A+ B + C = 1/20 + 1/30 + 1/60.

LCM denomineor = 1/10 = 10 * 3 total 30/2 = 15-> 3persons work.

In 3 days they work is 1/5;
Then remaining work is 1 - 1/5 = 4/5,

So using proportions 1/5 : 45 :: 3 : x.

Get 12days.

+ 3days = 15days.

Hi everyone here is my solution.

According to problem (a + b + c) 1 day work = 1/10.
As per the problem A is assisted by B and C every third day so first let us find,

Work done by A in 2 days = 1/20 * 2 = 1/10.
As mentioned in step 3 B and C are assisted on the third day.

(So A's 2-day work, And one important point A also work along with B and C on third day = 1/10 + (1/20 + 1/30 + 1/60) = 1/5.

Its 3 days work done by A B and C = 1/5.

1-day work = x (assume).

By cross multiplication.

We get 3x = 1/5.

x = 1/15.

So its 15 days thank you.

@Nikitha.

Superb explanation.

A can do apiece of work in 30 days. B in 50 days and C in 40 days. If A is assisted by B on one day and by C on the next day alternately work will be completed in?