Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 21)
21.
A takes twice as much time as B or thrice as much time as C to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in:
Answer: Option
Explanation:
| Suppose A, B and C take x, | x | and | x | days respectively to finish the work. |
| 2 | 3 |
| Then, |  |
1 | + | 2 | + | 3 |  |
= | 1 |
| x | x | x | 2 |
 |
6 | = | 1 |
| x | 2 |
x = 12.
So, B takes (12/2) = 6 days to finish the work.
Discussion:
68 comments Page 5 of 7.
Vandana said:
9 years ago
One query, please answer this question:
A, B and C together can finish a piece of work in 12 days, A and C together work twice as much as B, A and B together work thrice as much as C. In what time could each do it separately?
A, B and C together can finish a piece of work in 12 days, A and C together work twice as much as B, A and B together work thrice as much as C. In what time could each do it separately?
Makvana Disha said:
9 years ago
We are also doing this,
A's 1 day work = 1/x, B's 1 day work = 2/x, C's 1 day work = 3/x.
So, (A + B + C) 's one-day work:
= (1/x + 2/x + 3/x).
= (1/x + 5x/x ^ 2).
= 1/x + 5.
= 6x/x = 6 Each of three.
A's 1 day work = 1/x, B's 1 day work = 2/x, C's 1 day work = 3/x.
So, (A + B + C) 's one-day work:
= (1/x + 2/x + 3/x).
= (1/x + 5x/x ^ 2).
= 1/x + 5.
= 6x/x = 6 Each of three.
Rakib said:
9 years ago
Let A = 12 days(twice than B& thrice than C),
B = 6 days,
C = 4 days.
Working together: (12 * 1/2) = 6 days.
Am I Right?
B = 6 days,
C = 4 days.
Working together: (12 * 1/2) = 6 days.
Am I Right?
Raj said:
9 years ago
2A = B.......eqn1.
3A = C.......eqn2.
A + B + C = 2days
A + 2A + C = 2days (from the eqn1),
A + 2A + 3A = 2days(from the run2),
6A = 2days ,
A = 2/6,
A = 1/3 days.
i.e A takes 3days to work.
We have B=2A.
B=2(3),
B=6.
That's B takes 6 days to complete.
3A = C.......eqn2.
A + B + C = 2days
A + 2A + C = 2days (from the eqn1),
A + 2A + 3A = 2days(from the run2),
6A = 2days ,
A = 2/6,
A = 1/3 days.
i.e A takes 3days to work.
We have B=2A.
B=2(3),
B=6.
That's B takes 6 days to complete.
(1)
Ravi U S said:
9 years ago
There are many methods to solve a problem. As I feel Subhas 's method was good, follow it. A can do 1 unit of work in 1 day.
B can do 2 unit of work in 1 day.
C can do 3 unit of work in 1 day.
Totally A+B+C's 1 day's work = 12 units of work.
To do 12 units of work B take 6 days.
B can do 2 unit of work in 1 day.
C can do 3 unit of work in 1 day.
Totally A+B+C's 1 day's work = 12 units of work.
To do 12 units of work B take 6 days.
Ashok Kumar said:
8 years ago
Here is the simple shortcut
Given 2A= B
Given 3A = C
A+B+C=1/2
A+B+3A =1/2 since C=3A
4A+B=1/2
4(B/2) + B =1/2 since 2A=B => A=B/2
6B/2 =1/2 on solve
So final B=1/6 i.e B can do the work alone in 6 days.
Given 2A= B
Given 3A = C
A+B+C=1/2
A+B+3A =1/2 since C=3A
4A+B=1/2
4(B/2) + B =1/2 since 2A=B => A=B/2
6B/2 =1/2 on solve
So final B=1/6 i.e B can do the work alone in 6 days.
Nwbeen said:
8 years ago
Where did 1/2 come from?
Pushpendra said:
8 years ago
Yes, right @Ravi.
Your method of explanation is clear, Thanks @Subhas.
Your method of explanation is clear, Thanks @Subhas.
Raj said:
8 years ago
@Ravi.
Where you get 12 unit work in 1 days and how please explain it?
Where you get 12 unit work in 1 days and how please explain it?
Loki said:
8 years ago
a+2a+3a = 2days.
6a = 2days,
a = 1/3(this is total),
3*2 = 6.
6a = 2days,
a = 1/3(this is total),
3*2 = 6.
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