Two water taps together can fill a tank in 9 3/8 hours. The tap of

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Q) Two water taps together can fill a tank in 9 $\frac{3}{8}$ hours. The tap of larger diameter takes 10 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:

VIDEO SOLUTION

STEP BY STEP SOLUTION

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

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 10 hrs less than smaller diameter pipe

∴ time taken by larger diameter pipe is = X – 10 hrs

Step 2: ∵ Smaller diameter pipe fills the tank of volume V in X hrs

∴ 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

∴ the volume filled by larger diameter pipe in 1 hr = $\frac{V}{X -10}$

Therefore, the volume filled by both pipes together in 1 hr = $\frac{V}{X} + \frac{V}{X - 10}$

= $\frac{V(X + X - 10)}{X (X-10)} = \frac{V(2 X - 10)}{X (X-10)}$

Step 3: Now, $\frac{V(2X - 10)}{X(X-10)}$ volume of tank is filled by both pipes in 1 hr

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

= $\frac{V}{\frac{V(2X - 10)}{X(X-10)}}$ = $\frac{X (X-10)}{2X -10}$

Step 4: ∵ Its given that the both pipes fill the tank in $\frac{75}{8}$ hrs

∴ $\frac{X(X - 10)}{2X - 10} = \frac{75}{8}$

∴ 8 X (X – 10) = 75 (2X – 10)

∴ 8 X² – 80 X = 150 X – 750

∴ 8 X² -230 X + 750 = 0

∴ 8 X² -200 X – 30 X + 750 = 0

∴ 8X (X – 25) -30 (X – 25) = 0

∴ X = 25 hrs and X = $\frac{30}{8}$ = $\frac{15}{4}$ hrs

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

Step 5: Since, time taken by larger diameter pipe is 10 hrs less than smaller diameter pipe,

hence values for time taken by this pipe will be:

(i) 25 – 10 = 15 hrs for X = 10 hrs

and (ii) $\frac{15}{4}$ – 10 = $\frac{-25}{4}$ hrs for X = $\frac{15}{4}$ hrs

Since value of time taken can not be negative, $\therefore$ X $\neq \frac{15}{4}$

Hence, time taken by smaller diameter pipe is 25 hrs and time taken by larger diameter pipe is 15 hrs.

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