Q) Two water taps together can fill a tank in 3 hours. The tap of larger diameter takes 5 hours less than the smaller one to fill the tank separately. Find the time (in hrs) which each tap can separately fill the tank.

Ans: 

It is given that,
Time taken by both water taps together = 3 hrs

We need to find out: Time to fill the tank by each pipe individually

Step 1: Let’s consider smaller diameter pipe fills the tank in X hrs and Volume of the tank is V.

∵ it is given that time taken by larger diameter pipe is 5 hrs less than smaller diameter pipe

∴ time taken by larger diameter pipe is = X – 5 hrs

∵ Smaller diameter pipe fills the tank of volume V in X hrs

∴ the volume filled by smaller diameter pipe in 1 hr =

Similarly, ∵ larger diameter pipe fills the tank of volume V in Y hrs

∴ the volume filled by larger diameter pipe in 1 hr =

Therefore, the volume filled by both pipes together in 1 hr =

=

Step 3: Now, volume of tank is filled by both pipes in 1 hr

Therefore, Volume V of the tank will be filled by both pipes in:

= =

Step 4:  ∵ Its given that the both pipe fill the tank in hrs

∴   

∴ 3 X (X – 5) = 10 (2X – 5)

∴ 3 X² – 15 X = 20 X – 50

∴ 3 X² – 35 X + 50 = 0

∴ 3 X² – 30 X – 5 X + 50 = 0

∴ 3X(X – 10) – 5 (X – 10) = 0

∴ (X – 10) (3X – 5)= 0

∴ X = 10 hrs and X = hrs

These are the 2 values of the time taken by smaller diameter pipe

Step 5: Since, time taken by larger diameter pipe is 5 hrs less than smaller diameter pipe,

hence values for time taken by larger pipe will be:

(i) 10 – 5 = 5 hrs for X = 10 hrs

and (ii) – 5 = hrs for X = hrs

Here, we reject X = because value of time taken can not be negative.

Therefore, time taken by smaller diameter pipe is 10 hrs and time taken by larger diameter pipe is 5 hrs.

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