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:
359 comments Page 1 of 36.
Harikrishnan said:
4 weeks ago
A--> 20 days
B--> 30 days LCM = 60 (Total Units)
C--> 60 days.
A's one day work 3 units.
B's one day work 2 units.
C's one day work 1 unit.
2 days A alone will work, every third day B and C will join with A.
(2×3) = 6 units + (1 × 6) = 6 units.
For one cycle, 12 units for 3 days,
So, total units 60/12 = 5 days.
5 × 3 = 15 days.
B--> 30 days LCM = 60 (Total Units)
C--> 60 days.
A's one day work 3 units.
B's one day work 2 units.
C's one day work 1 unit.
2 days A alone will work, every third day B and C will join with A.
(2×3) = 6 units + (1 × 6) = 6 units.
For one cycle, 12 units for 3 days,
So, total units 60/12 = 5 days.
5 × 3 = 15 days.
(10)
Muthu Kumar said:
1 month ago
Thanks all for explaining the answer.
(2)
Madhu said:
5 months ago
Yes, I think 20 is the correct answer.
(11)
Suram Ramana Reddy said:
6 months ago
A = 20/60 -> 3 units
B = 30/60 -> 2 units
C = 60/60 -> 1 unit
= 6 units.
L.C.M=60
B and C help on every third day, which means A can do the work alone for 2 days.
2 days A work = 3x3 = 9 units + 3rd day(A+B+C) 6 = 15.
B = 30/60 -> 2 units
C = 60/60 -> 1 unit
= 6 units.
L.C.M=60
B and C help on every third day, which means A can do the work alone for 2 days.
2 days A work = 3x3 = 9 units + 3rd day(A+B+C) 6 = 15.
(52)
Krishna said:
8 months ago
Here, A alone complete the whole work then answer is 20. Because total work is 60 and A's efficiency is 3.
Then 60/3 = 20 days.
Then 60/3 = 20 days.
(12)
Deepa said:
8 months ago
LCM of A, B, C's works = 60,
And the work done by A in 1 day = 3 (60/20),
so in 2 days work done by A is = 3*2= 6,
And work done by A+B+C in 1 day = 6 [(60/20) + (60/30) + (60/60)],
So total work done for 3 days = 12,
So the total number of days required to complete 60 units of work = 60 * 3/12 = 15.
And the work done by A in 1 day = 3 (60/20),
so in 2 days work done by A is = 3*2= 6,
And work done by A+B+C in 1 day = 6 [(60/20) + (60/30) + (60/60)],
So total work done for 3 days = 12,
So the total number of days required to complete 60 units of work = 60 * 3/12 = 15.
(17)
Shreemanjesh said:
10 months ago
A = 20, B = 30, c = 60,
LCM = 60,
A = 3,
b = 2,
C = 1.
A = 3 × 3 = 9 + (abc)6 = 15.
LCM = 60,
A = 3,
b = 2,
C = 1.
A = 3 × 3 = 9 + (abc)6 = 15.
(53)
RAJA said:
10 months ago
Why have to take 3 days of work?
(25)
Abishek Aryal said:
12 months 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.
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.
(61)
Subham said:
1 year 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).
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).
(101)
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