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 16 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.
Shakti said:
1 decade ago
Last step. 1/5th of work completed in 3 days.
We have to find 1 or whole work can completed in how many days.
1/5th work= 3 days.
1 work= 3/ (1/5) = 3*5= 15 days.
We have to find 1 or whole work can completed in how many days.
1/5th work= 3 days.
1 work= 3/ (1/5) = 3*5= 15 days.
T.S said:
5 years ago
@Vijay
We know that the whole work done is 1. here 3 days work is 1/5.
So,
if, 1/5 work=3days.
then, 1 work= ?
Cross multiply we get ?= 3 x 5= 15days.
We know that the whole work done is 1. here 3 days work is 1/5.
So,
if, 1/5 work=3days.
then, 1 work= ?
Cross multiply we get ?= 3 x 5= 15days.
(1)
Emiley said:
9 years ago
A can do apiece of work in 30 days. B in 50 days and C in 40 days. If A is assisted by B on one day and by C on the next day alternately work will be completed in?
Mahesh dubey said:
1 decade ago
Its very easy.
If we cancel 6/60 = 1/10.
And 1/10 + 1/10 = 2/10 (because upper term is add and lower term is common, so when we cancel 2/10 then you got = 1/5.
If we cancel 6/60 = 1/10.
And 1/10 + 1/10 = 2/10 (because upper term is add and lower term is common, so when we cancel 2/10 then you got = 1/5.
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.
Chitra said:
1 decade ago
Work done in 3 days = 1/5.
Work done in next 3 days = 1/5+1/5.
Work done in 5 days = 1/5+1/5+1/5+1/5+1/5 = 5/5 = 1 (task completed).
Hence 3*5 is used.
Work done in next 3 days = 1/5+1/5.
Work done in 5 days = 1/5+1/5+1/5+1/5+1/5 = 5/5 = 1 (task completed).
Hence 3*5 is used.
Santosh said:
5 years ago
A =20 Days
B=30 Days
C=60 Days
LCM = A,B,C
LCM = 60
A = 60/20 =3.
B = 60/30 = 2.
C = 60/60 =1.
AFTER 3 DAYS B, C JOINED.
Total =3 + 3(3 + 1) = 15 DAYS.
B=30 Days
C=60 Days
LCM = A,B,C
LCM = 60
A = 60/20 =3.
B = 60/30 = 2.
C = 60/60 =1.
AFTER 3 DAYS B, C JOINED.
Total =3 + 3(3 + 1) = 15 DAYS.
Balaram said:
1 decade ago
Thank you everybody for participating in the discussion and making clarification for all my unanticipated doubts.... Nikita Prasad gave good explanation.
Jibinlal said:
10 years ago
1/5 work is done in 3 days. Remaining work is 4/5.
1/5+1/5+1/5+1/5 = 3+3+3+3 = 12 days.
Total = 12 days (4/5th work)+3 days (1/5th work). So 15 days.
1/5+1/5+1/5+1/5 = 3+3+3+3 = 12 days.
Total = 12 days (4/5th work)+3 days (1/5th work). So 15 days.
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