Three pipes A, B and C can fill a tank in 6 hours. After working at i

Question

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. How many hours will take C alone to fill the tank ?

A. 10
B. 12
C. 14
D. 16
E. 18

A. 10
B. 12
C. 14
D. 16
E. 18

chetan2u · Accepted Answer

Hi,

INFO 
1) ONE hour work of all three is 1/6
2) after two hours, C is closed

INFERENCE 
 1) In 2 hours all three would do 2/6 of work =1/3 of work..
2) remaining 2/3 rd of work is done by all three in 4 hours and by ONLY a and B in 7 hrs..

SOLUTION.. 
 therefore lets find how much time C will take to complete this--
one hour work- 1/4-1/7=3/28..
so C takes 28/3 hour to complete 2/3 work..
so C will complete the entire work in 28/3 * 3/2=14 days..
C

Abhishek009 · Answer

Let the total capacity of the tank be 42 units (LCM of 2,6 & 7)

Efficiency of ABC is 42/7 = 6units/hr

In 2 hours A B and C fill up 14 units of the tank , thus 28 units are left.

A and B fill up 28 units in 7 hours thus efficiency of A & B is 4 units

We know , A+B+C = 7 and A + B = 4

So, Efficiency of C is 3 units/hr

Thus the time required to fill up the entire tank is 42/3 = 14 hours.

gps5441 · Answer

Combined rate of work
1/a+1/b+1/c=1/6
Work done by 3 in 2 hrs=2x1/6=1/3
Work left to be done
1-1/3=2/3.
This work is done by b&c in 7 hrs. 
Thus Rate of work is
Work/time=2/3/7=2/21
Therefore 
1/a-2/21=1/6. Solve the equation a=14 hrs

Posted from my mobile device

mvictor · Answer

i solved this way...
total work to be done = 6 parts. 
all together can do 1 part in 1 hour.
2 hours passed - 2 parts were done. so left 4 parts to finish.
4 parts out of total 6 parts is 2/3.
now..A and B finish filling 2/3 in 7 hours. 2/3 divide by 7 = 2/21. this is the rate for A and B.
to find the rate for C, subtract from 1/6 (rate of all three pipes) the rate of A and B (2/21)
1/6 - 2/21 (multiply first fraction by 7, second by 2) = 7/42 - 4/42 = 3/42
now 3/42 is the rate of C. it can fill 3 pools in 42 hours, or one in 14 hours.
answer is C.

generis · Answer

1) Combined rate: A, B, and C fill one tank in six hours:

\frac{1}{A} + 
\frac{1}{B} + 
\frac{1}{C} = \frac{1}{6}

2) Amt of work finished, work remaining. They work at that rate for 2 hours.   r*t = W

They finish: (\frac{1}{6}* 2) = \frac{2}{6}= \frac{1}{3}  tank

Remaining work:  (1 - \frac{1}{3})= \frac{2}{3}  tank

3) Rate of A and B? C stops. A and B do remaining  \frac{2}{3}  in 7 hrs
 
A and B's combined rate?   W/t = r

\frac{(\frac{2}{3})}{7}  =  \frac{2}{21}

That is,  \frac{1}{A} + 
\frac{1}{B} = 
\frac{2}{21}

4) Find C's rate. Numbers aren't as bad as they look.

\frac{1}{A} + 
\frac{1}{B} + 
\frac{1}{C} = \frac{1}{6}

\frac{2}{21} + 
\frac{1}{C} = \frac{1}{6}

Multiply each term by (6 * 21) = 126

12 + \frac{126}{C} = 21

\frac{126}{C} = 9

\frac{1}{C} = \frac{9}{126}=  C's rate

5) C's time? 
When work is 1, flip rate to get time. C's time:  \frac{126}{9}  = 14 hours

OR   (\frac{W}{r} = t)  :

\frac{1}{(\frac{9}{126})}

\frac{126}{9}
= 14  hours

Answer C

KarishmaB · Answer

A, B and C fill the tank in 6 hrs so their combined rate is 1/6 tank/hr
In 2 hrs, they will fill 1/3rd of the tank.

Now leftover 2/3rd of the tank is filled by A and B in 7 hrs. So combined rate of A and B = (2/3)/7 = 2/21 tank/hr

Rates are additive so,
Rate of C = Rate of A,B and C - Rate of A and B = 1/6 - 2/21 = 3/42 = 1/14 
Hence, C alone will fill 1 tank in 14 hrs.

Answer (C)

SchruteDwight · Answer

I) Combined, equal time: 6(A+B+R)
II) Combined, unequal time: 9A+9B+2R

Sine both I) & II) do the same work: 6(A+B+R)=9A+9B+2R –> R=3/4(A+B) = the rate we are looking for is 3/4 of the rate of A & B combined

since total rate is 1/6=x+3/4*x –> 3/4*x = 1/14 –> t=14

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