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?
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:
361 comments Page 31 of 37.
Kapil dev said:
9 years ago
day-1, day-2, day-3.
A. A. A+B+C.
Every 3rd-day b & c joining.
So 3(3A+B+C)=3(1/20)+1/30+1/60,
=9+2+1/60,
=1/5,
3A+B+C=1/3*5.
=1/15.
15 days work is complited.
A. A. A+B+C.
Every 3rd-day b & c joining.
So 3(3A+B+C)=3(1/20)+1/30+1/60,
=9+2+1/60,
=1/5,
3A+B+C=1/3*5.
=1/15.
15 days work is complited.
Vaibhav More said:
8 years ago
As A does work in 20 days.
B does in 30 days.
C does in 60 days.
Take the lcm and you will get total work = 60 parts.
No of parts A does work = 60/20 = 3 part / day.
Similarly B = 2 parts /day and c= 1 parts/day.
If A does work for 2 days he covers 3+3 parts = 6 parts.
And every third day B & C joins and then A + B +C = 6 parts on 3rd day.
Total work from first day to third day = 6 + 6 parts= 12 parts.
Total work = 60 parts.
Therefore no of days require = 5 * 3 = 15 days.
B does in 30 days.
C does in 60 days.
Take the lcm and you will get total work = 60 parts.
No of parts A does work = 60/20 = 3 part / day.
Similarly B = 2 parts /day and c= 1 parts/day.
If A does work for 2 days he covers 3+3 parts = 6 parts.
And every third day B & C joins and then A + B +C = 6 parts on 3rd day.
Total work from first day to third day = 6 + 6 parts= 12 parts.
Total work = 60 parts.
Therefore no of days require = 5 * 3 = 15 days.
Sangi said:
8 years ago
Why 3 is multiplied with 5 at last?
Please explain.
Please explain.
Anu said:
8 years ago
I really didn't get the last step where they said whole work will be done in 15 days. Can anyone explain how it comes?
Roshan said:
8 years ago
@ALL. In simple method.
1/5 work is done in 3 days,
2/5 work is done in 6 days,
3/5 work is done in 9 days,
4/5 work is done in 12 days,
5/5 work is done in 15 days.
1/5 work is done in 3 days,
2/5 work is done in 6 days,
3/5 work is done in 9 days,
4/5 work is done in 12 days,
5/5 work is done in 15 days.
Kiran said:
8 years ago
My answer is 17.
.
LCM = 60,
Efficiency of A = 3/day,
Efficiency of B = 2/day,
Efficiency of C = 1/day.
.
3+3+6+3+3+6+3+3+6+3+3+6+3+36+3+3= 60,
Total 17 days.
.
LCM = 60,
Efficiency of A = 3/day,
Efficiency of B = 2/day,
Efficiency of C = 1/day.
.
3+3+6+3+3+6+3+3+6+3+3+6+3+36+3+3= 60,
Total 17 days.
Tejeswar Sai said:
8 years ago
Thank you @Vikas, very good explanation.
Dhara jawale said:
8 years ago
Please explain the second step in detail.
Shashank Tripathi said:
8 years ago
A-> 20 days,
B->30 days,
C->60 days,
L.C.M of (20,30,60)=60.
Work = 60 unit.
so the capacity of A = 60/20 = 3 unit /day.
so the capacity of B = 60/30 = 2 unit /day.
so the capacity of C = 60/60 = 1 unit /day.
day 1 : A -> 3 unit/day.
day 2 : A -> 3 unit/day.
day 3 : A+B+C -> 3+2+1 unit/day-> 6 unit/day (since on 3rd day helped by B and C)
Sum all 3 days capacity = 3+3+6 = 12 unit in 3 days.
12 unit in 3 days -> equation (1)
so for 60 unit work multiply by 5 in equation (1)
60 unit work in 15 days.
So A can do in 15 days.
B->30 days,
C->60 days,
L.C.M of (20,30,60)=60.
Work = 60 unit.
so the capacity of A = 60/20 = 3 unit /day.
so the capacity of B = 60/30 = 2 unit /day.
so the capacity of C = 60/60 = 1 unit /day.
day 1 : A -> 3 unit/day.
day 2 : A -> 3 unit/day.
day 3 : A+B+C -> 3+2+1 unit/day-> 6 unit/day (since on 3rd day helped by B and C)
Sum all 3 days capacity = 3+3+6 = 12 unit in 3 days.
12 unit in 3 days -> equation (1)
so for 60 unit work multiply by 5 in equation (1)
60 unit work in 15 days.
So A can do in 15 days.
Shashank Tripathi said:
8 years ago
A-> 20 days,
B->30 days,
C->60 days,
L.C.M of (20,30,60)=60.
Work = 60 unit.
so the capacity of A = 60/20 = 3 unit /day.
so the capacity of B = 60/30 = 2 unit /day.
so the capacity of C = 60/60 = 1 unit /day.
day 1 : A -> 3 unit/day.
day 2 : A -> 3 unit/day.
day 3 : A+B+C -> 3+2+1 unit/day-> 6 unit/day (since on 3rd day helped by B and C)
Sum all 3 days capacity = 3+3+6 = 12 unit in 3 days.
12 unit in 3 days -> equation (1)
so for 60 unit work multiply by 5 in equation (1)
60 unit work in 15 days.
So A can do in 15 days.
B->30 days,
C->60 days,
L.C.M of (20,30,60)=60.
Work = 60 unit.
so the capacity of A = 60/20 = 3 unit /day.
so the capacity of B = 60/30 = 2 unit /day.
so the capacity of C = 60/60 = 1 unit /day.
day 1 : A -> 3 unit/day.
day 2 : A -> 3 unit/day.
day 3 : A+B+C -> 3+2+1 unit/day-> 6 unit/day (since on 3rd day helped by B and C)
Sum all 3 days capacity = 3+3+6 = 12 unit in 3 days.
12 unit in 3 days -> equation (1)
so for 60 unit work multiply by 5 in equation (1)
60 unit work in 15 days.
So A can do in 15 days.
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