Aptitude - Time and Work - Discussion

@Tabbu.

Here 1 means the whole part.

So subtract the work done from the whole work to get remaining work.

The LCM of 15 and 20 is 60 which means total work done=60. Then the efficiency of a & b are 4 & 3 respectively(one-day efficiency). Now if both work together their efficiencies must be added which is =7 and therefore 7*4=24.

Is the work was done by them for 4 days which means the leftover is 60-24=36.

Am I right?

A can do work in 15 days=1/15.
B can do work in 20 days=1/20.
both (A+B)=1/15+1/20=20+15/300=35/300=7/60.
(A+B)Work together in 4 days=7/60*4 in this 60 will be cancelled 4 table in 15.

Then; times=>7/15.
The remaining work will be(1-7/15).
=>15-7/15.
=>8/15.

Please explain this step 1-7/15.

Very simple,
15+20=35,
15*20=300,
35/300=7/60,

work is 1 so=1-7/15.
=15-7/15,
Ans is = 8/15.

The total work of A& B is = 60.
we can get 60 by taking LCM of it.

A's one day work = 4.
B's one day work = 3.

Both A & B 's one day work=7.
A's 4days work=4*4=16.
B's 4 days work=4*3=12.
So, A&B work=28.

The remaining work=60-28=32.
= 32/60,
= 8/15.

A's 1 day's work = 1/15.
B's 1 day's work = 1/20.

(A+B)'s 1 day's work= (1/15+1/20),
L.C.M of 15,20 is 60 then,
15*4=60, 20*3=60,
4 + 3/60=7/60.

Can you explain the step 7÷60?

A's 1 day's work = 1/15.
B's 1 day's work = 1/20.

(A+B)'s 1 day's work= (1/15+1/20),
= 20+15/300(cross multiplication),
= 35/300,
= 7/60.

(A+B)'s 4 day's work= 7/60 * 4,
= 7/15(60/4 is division it is 15).

Therefore remaining work=1 - 7/15,
= 15-7/15,
= 8/15.

@Fazeena.

The work completed in 4 days is 7/15. Therefore remaining work is 1-(7/15). So, total work is 1.