Aptitude - Pipes and Cistern - Discussion

Considering the LCM of the all the four taps ie 6.

Now considering the capacity of tank be 6 units.

For the 1st half,

As only 1 tap is working so it fills 6/6 units/hr = 1 unit /hr (part filled by tap 1/total capacity of the tank).

As only 1 tap is working for the 1st half I.e. For 3 units of tank (total capacity of tank/2).

Calculations:

1 unit filled in 1 hr.

3 units filled in 3 hrs.

In second half 4 taps of same type are used, therefore it fills 4/6*6 = 4 units/hr (part filled by 4 taps/total capacity of the tank).

Now left out 3 units are filled by the combination of 4 taps.

Calculation:

4 units filled in 1 hr.

3 units filled in 3/4 hrs = 45 mins.

Therefore total time taken = time taken by 1st tap + time taken by combination of 4 similar taps.

= 3 hrs + 3/4 hrs = 3 hrs 45 mins.

Here, given problem is that,

First half of the tank is filled in 3hrs by using only 1 tap.
After that 3 more pipes are opened so, now total working pipes 4(they are similar means takes the same time to fill the tank).

The remaining we have 6 - 3 = 3hrs.
3hrs means = 3 * 60 = 180min
So, equally this time shared by 4 pipes 180/4 = 45 min.
So total time is 3hrs 45 min.

Time taken by 1 tap to fill half of the tank is =3hrs.
part filled by 4 taps in 1 hr=4*1/6=2/3.
then remaining part= 1-2/3=1/3 how is it 1/2 can anybody please tell
and what this ::symbol mean