Two water taps together can fill a tank in 3 1/3 hours. The tap of

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Q) Two water taps together can fill a tank in 3 $\frac{1}{3}$ 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.5

Ans:

It is given that,
Time taken by both water taps together = 9 $\frac{3}{8} = \frac{75}{8}$ hrs

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

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

In the question, it is given that time taken by larger diameter pipe is 5 hrs less than smaller diameter pipe
Therefore, hence, time taken by larger diameter pipe is = X – 5 hrs

Since smaller diameter pipe fills the tank of volume V in X hrs
Hence, the volume filled by smaller diameter pipe in 1 hr = $\frac{V}{X}$

Similarly, larger diameter pipe fills the tank of volume V in Y hrs
Hence, the volume filled by larger diameter pipe in 1 hr = $\frac{V}{X - 5}$

Therefore, the volume filled by both pipes together in 1 hr = $\frac{V}{X} + \frac{V}{X - 5}$

= $\frac{V(X + X - 5)}{X (X - 5)} = \frac{V(2 X - 5)}{X (X - 5)}$

Now, $\frac{V(2X - 5)}{X(X - 5)}$ volume of tank is filled by both pipes in 1 hr

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

= $\frac{V}{\frac{V(2X - 5)}{X(X - 5)}}$ = $\frac{X (X - 5)}{2X - 5}$

Its given that the both pipe fill the tank in $\frac{10}{3}$ hrs

$\therefore \frac{X(X - 5)}{2X - 5} = \frac{10}{3}$

∴ 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 = $\frac{5}{3}$ hrs

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

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

and (ii) $\frac{5}{3}$ – 5 = $\frac{- 10}{3}$ hrs

Here, we reject X = $\frac{5}{3}$ 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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