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 33 of 37.
Rishi said:
8 years ago
How can you say the total work is 1?
Sha said:
8 years ago
How does the last step is (3*5=15)?
How come 3 here?
How come 3 here?
Dipankar said:
8 years ago
12-60.
1-5,
So 3=5*3=15.
1-5,
So 3=5*3=15.
Prem K said:
8 years ago
A's 1 day's work is 1/20.
A's 3 day's work is 3/20.
(B+C)'s 1 day's work is 1/30+1/60 = 1/20.
B and C will work on every third day means A will work three days and B and C will work for 1 day.
(A+B+C)'s 3 day's work = A's 3 day's work + (B+C)'s 1 day's work,
= 3/20 + 1/20,
= 1/5.
3 day's work is 1/5 then, whole work will be done in (3*5) = 15 days.
A's 3 day's work is 3/20.
(B+C)'s 1 day's work is 1/30+1/60 = 1/20.
B and C will work on every third day means A will work three days and B and C will work for 1 day.
(A+B+C)'s 3 day's work = A's 3 day's work + (B+C)'s 1 day's work,
= 3/20 + 1/20,
= 1/5.
3 day's work is 1/5 then, whole work will be done in (3*5) = 15 days.
Gayatri Dhokne said:
8 years ago
Take the LCM of A, B, C which is 60.
For LCM we have to multiply A, B, C by 3,2,1 respectively, this is work done per day by each of them.
For the first 2 days, only A works so, 2*3=6.
On the 3rd day A is assisted by B&C so,total work done in one day is 6units,
Therefore, the work they have done in 3days is 6+6=12,
Total work is done =60.
Days require to complete the work by A,B, C =60 ÷ 12=5days.
Days required by A to complete work = 5*3 = 15days.
For LCM we have to multiply A, B, C by 3,2,1 respectively, this is work done per day by each of them.
For the first 2 days, only A works so, 2*3=6.
On the 3rd day A is assisted by B&C so,total work done in one day is 6units,
Therefore, the work they have done in 3days is 6+6=12,
Total work is done =60.
Days require to complete the work by A,B, C =60 ÷ 12=5days.
Days required by A to complete work = 5*3 = 15days.
Sravya Suryadevara said:
8 years ago
Here in question, it is mentioned that A alone worked for 2 days, on the third day B and C assisted him.
So,
A's 2 days work= 2(1/20) = 1/10.
On third day three together worked.
i.e.,
3rd day work = (1/20)+(1/30)+(1/60) = 1/10.
Work done in 3 days= (1/10)+(1/10) = 1/5.
Total work is 3*5 = 15 days.
So,
A's 2 days work= 2(1/20) = 1/10.
On third day three together worked.
i.e.,
3rd day work = (1/20)+(1/30)+(1/60) = 1/10.
Work done in 3 days= (1/10)+(1/10) = 1/5.
Total work is 3*5 = 15 days.
Ajaykumarchevuri said:
8 years ago
A-20days
B-30days.
C-60days
LCM is 60
So,
A one day work 3 units (20*3)
B one day work 2 units(30*2)
C one day work 1 unit (1*60)
Then; 6 units ( both A+B+C one day work on every third day)
So A 2 days work = 3units *2 = 6 units
A+B+C on 3 Rd day = 6 units*1= 6 units
So, weget 12 units.
So 12/60 = 1/5 (A 2 days work +(A+B+C) work on a third day/total work)
So 3days * 1/5 = 15 days.
Here the answer.
B-30days.
C-60days
LCM is 60
So,
A one day work 3 units (20*3)
B one day work 2 units(30*2)
C one day work 1 unit (1*60)
Then; 6 units ( both A+B+C one day work on every third day)
So A 2 days work = 3units *2 = 6 units
A+B+C on 3 Rd day = 6 units*1= 6 units
So, weget 12 units.
So 12/60 = 1/5 (A 2 days work +(A+B+C) work on a third day/total work)
So 3days * 1/5 = 15 days.
Here the answer.
Sruthi said:
8 years ago
LCM of A,B & C is 60.
so the total work to be complete is 60.
A->3units/day{60/20}.
B->2units/day{60/30}.
c->1unit/day(60/60}.
A's 2days work-> 3*2 = 6units.
3rd day A is assisted by B & C -> A + B + C.
A + B + C 1 day work->3 + 2 + 1 = 6.
2days A alone->6units.
3rd day->6 {A + B + C work}.
60/12 * 3 = 15.
so the total work to be complete is 60.
A->3units/day{60/20}.
B->2units/day{60/30}.
c->1unit/day(60/60}.
A's 2days work-> 3*2 = 6units.
3rd day A is assisted by B & C -> A + B + C.
A + B + C 1 day work->3 + 2 + 1 = 6.
2days A alone->6units.
3rd day->6 {A + B + C work}.
60/12 * 3 = 15.
Sai Kumar Reddy Nossam said:
8 years ago
Common multiple of A,B & C is 120.
A's one day work =120/20 = 6 units,
B's one day work =120/30 = 4 units,
C' one day work = 120/60 = 2 units,
work done in three days by A, B, C= 24 units.
So, the total work = (120/24) * 3 = 15 days.
A's one day work =120/20 = 6 units,
B's one day work =120/30 = 4 units,
C' one day work = 120/60 = 2 units,
work done in three days by A, B, C= 24 units.
So, the total work = (120/24) * 3 = 15 days.
Nazam said:
8 years ago
How did (1/10+1/10) = 1/5 came?
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