3 pipes A,B,C together can fill tank in 4hrs. After 1hr, C is turned off and tank is filled in 6 more hrs..?

Question

Find time in which C alone can fill the empty tank.

My Try:
upto 1hr, 1/4th of total work is done by 3pipes together. Then 3/4th of remaining work is done by A,B. So, 1/4(efficiency of ABC)+3/4(Eff. of AB)=1
So, 1/4(1/4)+3/4(1/8)=1 => 5/32=1...eh?

Where am i wrong? Tell me the related concept..

Please also explain where am i wrong in My Try ??

I DONT need solution. Just tell me
1)Where am i wrong?
2)What are various concepts for this type of ploblem?(a site/link of concepts)

Answer 1

You can think of it like this:

It normally takes three hours to fill the remaining 3/4. Now it takes six hours. So:

rate A + rate B +rate C = 2(rate A + rate B)

rate C = rate A + rate B

2 rate C = rate A + rate B + rate C

That means all of them together are twice as efficient as C alone.

Alternatively, C is half as efficient as the three of them. So if it takes 4hrs with all three, it takes 8 with C alone.


### Added, due to request for further explanation:

Your mistake is that you don't sum 1/4 the efficiency of one and 3/4 the other. That will not give you 1.

There is no "related concept" except that you've got to keep things straight. Figure out how quick AB is, by using the partial tank and the time that takes:

3/4 tank in 6 hours with A, B

Now look at this:
4/4 tank in 4 hours with A, B, C
3/4 tank in 3 hours with A, B, C

And notice that it takes twice as long if you remove C, so eff. C = (1/2) eff A,B,C.

That means it takes twice as long with just C. 8 hours.

Answer 2

4 (A+B+C)= 1 4hrs with ABC to get 1 tank
6 (A+B) =.75 6hrs for A&B to fill 3/4 tank
6 (A+B) =3/4 or 4(A+B)=1/2 so 1/2 + 4C=1 C=1/8
C fills 1/8 of the tank in an hour
C alone fills the tank in 8 hrs

Answer 3

14

Answer 4

Let f = full tank, and A, B, C = rate at which respective pipes flow.
Then f = 4A + 4B + 4C, or C = f/4 - A - B
and f = C + 7A + 7B, or C = f - 7A - 7B.
multiply first eqn by 7 ==> 7C = 7/4f - 7A - 7B
second eqn = C = f - 7A - 7B
so 6C = 3/4f. We need f = 8C, so C can fill tank in 8 hours working alone.

Answer 5

A+B+C X60 = 25%
A+B X360 = 75%
therefore find out rate which A+B works per hour

= 75% / no of hours
1/6 of 75% = 12.5 % per hour
therefore C's rate is also 12.5% per hour as A+B (12.5%) + C = 25%

therefore C would fill tank on its own in 8 hours

Answer 6

Let V=volume of tank
3/4V=6A+6B
3V=24A+24C
V=8A+8B
Also: V=4A+4B+4C
Set V=V : 8A+8B=4A+4B+4C
4A+4B=4C
A+B=C
Therefore V=4A+4B+4C= 4(A+B) +4C

V=4(C)+4C =8C

C can filkl the tank in 8 hours