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 3 of 7.
Rahul said:
3 months ago
@All.
Here is the explanation for the answer.
As per the question, A=2B, A=2C.
equating, we get B = 3C/2,
A+B+C = 1/2.
Substitute the derived relations in the above equation, and the answer will be somewhere around 6.333333.
So, the correct answer is 6 days.
Here is the explanation for the answer.
As per the question, A=2B, A=2C.
equating, we get B = 3C/2,
A+B+C = 1/2.
Substitute the derived relations in the above equation, and the answer will be somewhere around 6.333333.
So, the correct answer is 6 days.
Mahendra said:
3 years ago
Assume total work iso no of chairs from the give no of chair made per one day as per efficiency
A : B : C = 1 : 2 : 3.
So total chairs is 6 in one day.
Total days they together : 2 days.
So, total works = 6 * 2 = 12.
B alone can take = 12/2 = 6 days.
A : B : C = 1 : 2 : 3.
So total chairs is 6 in one day.
Total days they together : 2 days.
So, total works = 6 * 2 = 12.
B alone can take = 12/2 = 6 days.
(36)
Punitsw said:
10 years ago
In simple words:
A:B:C = 1:2:3 (Total 6 equivalent to A).
Now A, B and C together complete work in 2 days (i.e in 12 man days of A). So if A, B or C do the work alone, the required days would be:
A = 12 days.
B = 6 days.
C = 4 days.
A:B:C = 1:2:3 (Total 6 equivalent to A).
Now A, B and C together complete work in 2 days (i.e in 12 man days of A). So if A, B or C do the work alone, the required days would be:
A = 12 days.
B = 6 days.
C = 4 days.
KOTHA SRAVANI said:
1 decade ago
Hai friends,
Taking the values of A,B,C as x, x/2, x/3.
1/x + 2/x + 3/x = 1/2.
On calculating we get,
6/x = 1/2.
Therefore, x = 12;
On substituting X = 12 in main equation.
1/12 + 2/12+ 3/12 = 1/2.
Hence, the value of B = 1/6.
Taking the values of A,B,C as x, x/2, x/3.
1/x + 2/x + 3/x = 1/2.
On calculating we get,
6/x = 1/2.
Therefore, x = 12;
On substituting X = 12 in main equation.
1/12 + 2/12+ 3/12 = 1/2.
Hence, the value of B = 1/6.
Dipika said:
10 years ago
If C takes x days to complete a work A will take 3x days.
If B takes y days to complete a work A will take 2y days.
But 3x = 2y, so x = 2/3y.
As per ques, 1/2y+y+2/3y = 1/2.
Therefore y = 6.
B takes 6 days to complete the work.
If B takes y days to complete a work A will take 2y days.
But 3x = 2y, so x = 2/3y.
As per ques, 1/2y+y+2/3y = 1/2.
Therefore y = 6.
B takes 6 days to complete the work.
Ajay said:
7 years ago
Time ratio of A,B,C=6:3:2.
Efficiency ratio=1:2:3.
It takes 2 days for all of them to complete the work. So, Total work=(1+2+3)*2=12.
Time taken by B=12/2 = 6days.
Time taken by A=12/1 = 12days.
Time taken by C=12/3 = 4days.
Efficiency ratio=1:2:3.
It takes 2 days for all of them to complete the work. So, Total work=(1+2+3)*2=12.
Time taken by B=12/2 = 6days.
Time taken by A=12/1 = 12days.
Time taken by C=12/3 = 4days.
(3)
Subhas said:
1 decade ago
A can complete 1 unit/day.
B can complete 2 unit/day.
C can complete 3 unit/day.
Total works complete together by all in 1 days = 6 unit/day.
So, they complete in two days = 6X2 = 12 units.
So B takes 12/2 = 6 days to complete.
B can complete 2 unit/day.
C can complete 3 unit/day.
Total works complete together by all in 1 days = 6 unit/day.
So, they complete in two days = 6X2 = 12 units.
So B takes 12/2 = 6 days to complete.
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?
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.
Krishman Varges said:
1 decade ago
This can be solved using physics. Force=Power/time. A, B and C does the same amount of work, hence, force is a constant. The power of three differs. Hence 1/t=f/p where f is a constant and t=2. 1/2=1/x+2/x+3/x.
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