Question

A cistern can be filled by two taps A and B in 12 hours and 16 hours respectively.
The full cistern can be emptied by a third tap C in 8 hours. If all the taps are
turned on at the same time, in how much time will the empty cistern be filled up
completely?​

Answer 1

Answer:

Total time taken is 48 hours.

Step-by-step explanation:

A = 12 h

B = 16 h

C = -8 h

Let the total time taken be T.

Time taken by each tap = T/12 ; T/16 ; T/8

T/12 + T/16 - T/8 = 0

LCM = 48

$\frac{4T+3T-6T}{48}$ = 0

T/48  = 0

T = 48

Answer 2

Answer:

Therefore, the Cistern will be filled up completely in 48 hours, if all the three taps are opened at the same time.

Step-by-step explanation:

Given that:- A cistern can be filled by two taps A and B in 12 hours and 16 hours respectively The full cistern can be emptied by a third tap C in 8 hours.

To find out:- If all the taps areturned on at the same time, in how much time will the empty cistern be filled up completely?

Solution:-

The time taken by A to fill the cistern = 12 hours

The time taken by B to fill the Cistern = 16 hours

The time taken by C to fill the Cistern = 8 hours.

Now, we have to find their first hour work,

∴ A's 1 hour's work = 1/12

B's 1 hour's work = 1/16

C's 1 hour's work = -1/8

∴ (ABC)'s 1 hour's work = (1/12)+(1/16)-(1/8)

LCM of 12,16 and 8 is 48.

= (3+4-6)/84

= (7-6)/48

= 1/48

Therefore, the Cistern will be filled up completely in 48 hours, if all the three taps are opened at the same time.

:)