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:
357 comments Page 31 of 36.
Roshan said:
7 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:
7 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:
7 years ago
Thank you @Vikas, very good explanation.
Dhara jawale said:
7 years ago
Please explain the second step in detail.
Shashank Tripathi said:
7 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:
7 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:
7 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.
Riya said:
7 years ago
If we use the LCM method.
We get the 1 day's work of A= 3; B= 2; c= 1. & total work done by a,b,c together on 3rd day will be 6 , therefore; 60/6 = 10 &
2 days work by A= 3*2=6 , 60/6= 10.
Therefore 3 days work = 10 + 10 =20.
Please explain the solution after this.
We get the 1 day's work of A= 3; B= 2; c= 1. & total work done by a,b,c together on 3rd day will be 6 , therefore; 60/6 = 10 &
2 days work by A= 3*2=6 , 60/6= 10.
Therefore 3 days work = 10 + 10 =20.
Please explain the solution after this.
Shashikumar g said:
7 years ago
Nice explanation, Thanks @Shashank Tripathi.
T baba said:
7 years ago
@ALL.
As per my calculation.
If it takes A 1/20 a day then two days= 1/20+1/20=2/20=1/10
If B and C Joins A it will be 1/20+1/30+1/60=6/60=1/10
So for 3 days, it will require A two days work plus A+B+C work= 1/10+1/10 as shown above=1/5.
If 1/5 of the total work is 3 days, then 1 divided by 1/5( bearing in mind that 1/5 is 3 days)=5
then 5 *3=15 days.
As per my calculation.
If it takes A 1/20 a day then two days= 1/20+1/20=2/20=1/10
If B and C Joins A it will be 1/20+1/30+1/60=6/60=1/10
So for 3 days, it will require A two days work plus A+B+C work= 1/10+1/10 as shown above=1/5.
If 1/5 of the total work is 3 days, then 1 divided by 1/5( bearing in mind that 1/5 is 3 days)=5
then 5 *3=15 days.
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