Question

There are two taps opening into a tank. If both are opened, the tank would be full in 18 minutes. The time taken for it to fill with only the large tap open is 15 minutes less than the time to fill with only the small tap open. What is the time taken to fill the tank only with the small tap open?

Answer 1

Hint: Assume the time taken by the small tap to alone fill the tank to be ‘x’ minutes. Then with the given data write the time taken to fill the tank by the larger tap alone. After this, form a quadratic equation in x and solve it.

Complete step-by-step answer:
Let the time needed to fill the whole tank with only the smaller tap open be ‘x’ minutes. It is given that the time taken for the tank to fill with only the larger tap open is 15 minutes less than the time to fill with only the small tap open.
Therefore, we can write the time needed to fill the whole tank with only the larger tap open as ( $ x-15 $ ) minutes.
Let the volume of the tank be V.
If the time needed for filling the tank of volume V with only a smaller tap is ‘x’ minute, then the volume that is filled in one minute is $ \dfrac{V}{x} $ .
Similarly, the volume that is filled in one minute with only the larger tap open is $ \dfrac{V}{x-15} $ .
It is said that when both the taps are open, the tank is filled in 18 minutes. This means that opening the taps fill the tank of volume V in 18 minutes. $ $
If the smaller can fill a volume of $ \dfrac{V}{x} $ units in one minute then in 18 minutes it will fill a volume of $ \dfrac{V}{x}\times 18 $ units.
Similarly, in 18 minutes the larger tap will fill a volume of $ \dfrac{V}{x-15}\times 18 $ units.
Therefore,
 $ \Rightarrow \left( \dfrac{V}{x}\times 18 \right)+\left( \dfrac{V}{x-15}\times 18 \right)=V $
 $ \Rightarrow 18\left( \dfrac{1}{x}+\dfrac{1}{x-15} \right)=1 $
 $ \Rightarrow 18\left( \dfrac{x-15+x}{x(x-15)} \right)=1 $
 $ \Rightarrow 18\left( \dfrac{2x-15}{{{x}^{2}}-15x} \right)=1 $
 $ \Rightarrow 36x-270={{x}^{2}}-15x $
 $ \Rightarrow {{x}^{2}}-51x+270=0 $
We can write the above equation as $ (x-6)(x-45)=0 $ .
This means that either $ (x-6)=0 $ or $ (x-45)=0 $ .
 $ \Rightarrow x=6 $ or $ x=45 $ .
However, if $ x=6 $ , then the time taken by the larger to fill the tank alone is $ x-15=6-15=-9 $ .
And we know that time cannot be negative. Hence, $ x=6 $ is discarded.
Therefore, the time taken to fill the tank only with the small tap open is 45 minutes.
So, the correct answer is “45 minutes.”.

Note: To solve the quadratic equation that we got in the above solution, we found the factors of the expression $ {{x}^{2}}-51x+270 $ . However, it is difficult to find the factors when the coefficients are larger. That time you can use other methods like completing the square method or quadratic formula for solving the quadratic equation.