Question

Two tapes A and B fill a water tank in 6 and 9 hours respectively another Tapsee at and sympathy the whole tank in 12 hours how long they take to fill the tank if all the three taps are opened simultaneously ?

Answer 1

Answer:

Pipe A can fill a tank in 6 hrs.

It fills in 1 hr = (V/6) [V = total volume of tank]

Similarly, Pipe B fills in 1 hr = (V/9)

And, Pipe C empties (V/2) in 12 hrs.

So, Pipe C empties in 1 hr = (V/24)

Now if all the taps are open simultaneously,

The tank gets filled in 1 hr = (V/6) + (V/9) - (V/24) = (17V/72)

(17V/72) volume of tank gets filled in 1hr.

Therefore, (3V/4) volume of tank gets filled in = (72/17)*(3/4) = 54/17 = 3 hrs. 10 min. 35 sec.

Step-by-step explanation:

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Answer 2

Correct Question:-

Two tapes A and B fill a water tank in 6 and 9 hours respectively whereas another tape C empties the whole tank in 12 hours. How long they take to fill the tank if all the three taps are opened simultaneously ?

Required Answer:-

Time taken by A to fill the tank = 6 hours.

• Portion of tank filled in 1 hour = 1/6

Time taken by B to fill the tank = 9 hours

• Portion of tank filled in 1 hour = 1/9

Time taken by C to empty the tank = 12 hours

• Portion of tank emptied in 1 hour = 1/12

If all the three taps are opened simultaneously. The amount of water filled/emptied in 1 hour:

$= \dfrac{1}{6} + \dfrac{1}{9} - \dfrac{1}{12}$

~Negative sign for the the third pipe because it is emptying the tank.

$= \dfrac{6 + 4 - 3}{36}$

$= \dfrac{7}{36}$

In 1 hour, 7/36 portion of tank is being filled. Hence, time taken by the three taps to fill the tank = 36/7 hours.