Aptitude - Time and Work - Discussion

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

Given 180/1800.

For 20, 30 & 60 the common multiple is 1800 (LCM).

Then you'll get the sum of 180/1800.

This is how I got 1/10 for the work done by A, B & C in 1 day.

Hi friends.

I just want to know why we are taking 3 days work for 1/10 + 1/10.

Why didn't we take first B + C one day work is 1/30 + 1/60 = 1/20?

Please explain me.

At the last step 1/5 * 3 we must do reciprocal right?

Or just multiple and get 15 for the sake of the answer in the options. Please tell me.

How A's 2 day's work = 1/10? Please explain the step.

A = 20, B = 30 &C = 60.

Total work (LCM of 20, 30, 60) = 60.

A's 1day work= 60/20 = 3units.
B's 1day work= 60/30 = 2.
C's 1day work= 60/60 = 1.

1st day A will do the work =3 unit.
2nd day again A will do the work = 3.
In 3rd-day, A + B + C will do the work = 3 + 2 + 1 = 6.

After 3 days total work = 12 (3 + 3 + 6).

After 6 days total work = 24.

Similarly;

After 12 days total work = 48.

15 days total work = 60.

So total no days to complete the work (60) is 15 days.

3 day's work is 1/5.

So, 1 day's work is = 1/15.

Total no of day's to complete the work is (reciprocal of 1day's work) = 15 day's.

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?

@Nikitha.

Superb explanation.

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.