Aptitude - Time and Work - Discussion

Discussion Forum : Time and Work - General Questions (Q.No. 3)
3.
A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
12 days
15 days
16 days
18 days
Answer: Option
Explanation:

A's 2 day's work = 1 x 2 = 1 .
20 10

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

Work done in 3 days = 1 + 1 = 1 .
10 10 5

Now, 1 work is done in 3 days.
5

Whole work will be done in (3 x 5) = 15 days.

Discussion:
362 comments Page 2 of 37.

RAJA said:   1 year ago
Why have to take 3 days of work?
(35)

Abishek Aryal said:   2 years ago
In 1 day A can do 1/20.
In 3 days A can do 3/20.

In 1 day B can do 1/30.
In 1 day c can da 1/60

In 3 days ( 3 day of A+ 1 day of B + 1 day of C) = 1/5.

Now,
1/5 in 3 days.
1 in 15 days.
(72)

Subham said:   2 years ago
A can do 20.
B can do 30.
C can do 60.

Total work (LCM of Total time)= 60
Now one day work
A's One day work 3.
B's One day work 2.
C's One day work 1.
Now A 2 days work 3 × 2 = 6.
Now B and c join after 3 days (2+1)×3 = 9
Now 6 + 9 = 15 (Ans).
(114)

Naren kumar B said:   3 years ago
A = 20,
B = 30,
C = 60.
Find 1/A + ((1/B + 1/C)/3).
= 15 => Answer.
(195)

Aya adel said:   3 years ago
a=1/20
b=1/30
c=1/60.

So a + b + c = 1/10.

Work done in 3 days =1/20 + 1/20 + 1/10 = 1/5.

The second 3 days will be 2/5,
Third 3 days 3/5,
Fourth 3 days 4/5,
Fifth 3 days 5/5,

So, A needs 3*5 = 15 days.
(137)

Mayur S said:   3 years ago
As LCM(20, 30, 60) = 60.

A's rate = 3, B's rate = 2 and C' rate = 1.
As work = 60 units

A will work for 2 days = 2 x 3 = 6.
B and C will join A on third day = (2+3+1) x 1 = 6
Work done in 3 days = 12 units.
Work done each day = 4.

The time required for 60 units of work at a rate of 4 is 15 days.
(389)

Harshal sonwane said:   4 years ago
1/5 work is done in 3 days.

The total work 1 done in x days.
Simply cross multiply;
x*1/5 = 3 * 1.
x/5 = 3,
x = 15.
(104)

Prem Prasidda said:   4 years ago
A = 1/20.
B = 1/30.
C = 1/60.

Now, 1/20+1/30+1/60=5/60 = 1/12.
Here, 1 days efficiency of A is 3, B is 2 and C is 1.
Now,
A can do the work in first day= 3,
" in the second day=3+3=6,
" in third day = 6+6=12,
So, in three days, A completed the work 12.
Now,
12 work = 3 days.
60 work = 3/12*60= 15 days.
(102)

Sawaira ikram said:   5 years ago
Work done by A = 1/20.
Work done by B= 1/30.
Work done by C= 1/60.
Now, work done by A in 2 days is= (1/20) * 2 = 1/10.
And work done by A,B,C in one day= 1/20 + 1/30 + 1/60 = (3+2+1)/60 = 6/60 = 1/10.

And the work done in first 3 days = 1/10 + 1/10 = 1/5.

So we can say that 1/5 part of work is done= in 3 days.
And 1 complete work is done in=. 3/(1/5) = 3*5 = 15 days.
(126)

Suresh said:   5 years ago
@Monoj.

Best explanation.
(24)


Post your comments here:

Your comments will be displayed after verification.