Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 15)
15.
Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:
Answer: Option
Explanation:
| Part filled in 2 hours = | 2 | = | 1 |
| 6 | 3 |
| Remaining part = |  |
1 - | 1 |  |
= | 2 | . |
| 3 | 3 |
 (A + B)'s 7 hour's work = |
2 |
| 3 |
| (A + B)'s 1 hour's work = | 2 |
| 21 |
C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }
| = | |
1 | - | 2 | |
= | 1 |
| 6 | 21 | 14 |
C alone can fill the tank in 14 hours.
Discussion:
44 comments Page 5 of 5.
Dipali Toraskar said:
4 years ago
Can you explain about the Remaining part?
(1)
Abir said:
3 years ago
Three pipes can fill a pipe in 6 hours,
(A+B+C) * 6 -----> (1)
Again, 3 workers work for 2 hours altogether so,
(A+B+C) * 2 , But The rest of the work done by A+B together, (A+B) *7
so. (A+B+C) *2 + (A+B) * 7 -----> (2)
ATQ,
(A+B+C) *6 = (A+B+C)*2 + (A+B) * 7.
Solving the equation we will get = C : A+B :: 3 : 4 (put this value in 1 or 2).
putting the value in 1 we will get,
(4+3)*7 = 42, this is the total work.
C = 42/3 = 14 Hours (ans)
(A+B+C) * 6 -----> (1)
Again, 3 workers work for 2 hours altogether so,
(A+B+C) * 2 , But The rest of the work done by A+B together, (A+B) *7
so. (A+B+C) *2 + (A+B) * 7 -----> (2)
ATQ,
(A+B+C) *6 = (A+B+C)*2 + (A+B) * 7.
Solving the equation we will get = C : A+B :: 3 : 4 (put this value in 1 or 2).
putting the value in 1 we will get,
(4+3)*7 = 42, this is the total work.
C = 42/3 = 14 Hours (ans)
(7)
L BEHERA said:
3 years ago
In this case, take LCM of 6 and 7, that is 42 and assume total manpower day is 42, the efficiency of A+B+C is 42/6 = 7.
A, B, C did for 2 days that is 7 * 2 = 14 MD complete, remaining 42-14=28 MD done by A+B in 7hr, So, A+B efficiency is 28/7 = 4.
So, C's efficiency is 7 - 4 = 3.
C can complete total work by 42/3=14 days.
A, B, C did for 2 days that is 7 * 2 = 14 MD complete, remaining 42-14=28 MD done by A+B in 7hr, So, A+B efficiency is 28/7 = 4.
So, C's efficiency is 7 - 4 = 3.
C can complete total work by 42/3=14 days.
(24)
Gaurav Gowardhan said:
10 months ago
A .+ B + C = 6 hrs
2 hrs = 1/3 work,
remaining = 2/3 work.
A+B, 2/3 work= 7 hrs,
Total work 3/3 i.e 1 = 7 hrs + 7/2 hrs = 21/2 hrs
Now consider LCM of A + B + C & A+B.
i.e 6 & 21/2 is 42
Taking efficiency: A + B + C = 7units
A+B = 4 units.
So C's efficiency= A+ B+C -( A+ B).
= 7-4 = 3.
Total work is LCM i.e 42 & efficiency of C is 3.
So total time= 42/3 = 14.
2 hrs = 1/3 work,
remaining = 2/3 work.
A+B, 2/3 work= 7 hrs,
Total work 3/3 i.e 1 = 7 hrs + 7/2 hrs = 21/2 hrs
Now consider LCM of A + B + C & A+B.
i.e 6 & 21/2 is 42
Taking efficiency: A + B + C = 7units
A+B = 4 units.
So C's efficiency= A+ B+C -( A+ B).
= 7-4 = 3.
Total work is LCM i.e 42 & efficiency of C is 3.
So total time= 42/3 = 14.
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