Question

Three taps P,Q,R fill up a tank independently in 5 hr,10 hr and 15 hr respectively. Initially the tank is empty and exactly one pair of tap is open during each hour and every pair of taps is open at least for one hour. What is the minimum number of hours required to fill the tank.

Answer 1

Answer:

$\frac{35}{9}$hrs

Step-by-step explanation:

Independently tap P can fill up a tank in 5hr,

So in 1 hr, tap P can fill 1/5th of the tank.

Independently tap Q can fill up a tank in 10hr

So in 1 hr, tap Q can fill 1/10th of the tank.

Independently tap R can fill up a tank in 15hr

So in 1 hr, tap R can fill 1/15th of the tank.

In 1 hr, tap P and tap Q together can fill 3/10th of a tank.

In 1 hr, tap P and tap R together can fill 4/15th of a tank.

In 1 hr, tap Q and tap R together can fill 1/6th of a tank.

Given that exactly one pair of tap is open during each hour and every pair of taps is open for atleast for one hour, hence atleast 3 hours with taps PQ being open, then PR being open and QR being open to fill a tank,since tank needs to be filled in minimum hours to fill rest of tank taps PQ pair would be open since they fill the tank fast.

Indivdually if each pair is open for 1 hr, in 3 hrs (3/10 + 4/15 + 1/6)th of tank would be filled i.e., 11/15th of tank is filled. Now, the remaining 4/15th of tank can be filled in minimum hours by opening pair PQ,

Since pair PQ takes 10/3hr to fill a tank,

to fill 4/15th of a tank, it takes 4/15 * 10/3 = 8/9hr

Hence, minimum number of hours required are 3$\frac{8}{9}$ hrs or $\frac{35}{9}$hrs.