Answer:
a) 9000 L
[tex]\textsf{b) (i)} \quad \textsf{Tap }P: \quad \dfrac{9000}{x}\:\sf minutes[/tex]
[tex]\textsf{b) (ii)} \quad \textsf{Tap }Q: \quad \dfrac{9000}{x+5}\:\sf minutes[/tex]
c) Proof below.
[tex]\textsf{d)} \quad x=\dfrac{-5+5\sqrt{31} }{2}=11.41941...[/tex]
e) 5 hours 23 minutes
Step-by-step explanation:
Part (a)
The aquarium can be modeled as a rectangular prism.
Note: 1 m³ = 1000 L
[tex]\begin{aligned}\textsf{Volume of rectangular prism} & = \sf length \times width \times height\\\\\implies \textsf{Volume of aquarium} & = \sf 3\:m \times 2.5\:m \times 1.2\:m\\& = \sf 9 \:\:m^3\\& = \sf 9000\:L\end{aligned}[/tex]
Part (b)
[tex]\boxed{ \sf Time = \dfrac{Volume}{rate}}[/tex]
[tex]\textsf{(i)} \quad \textsf{Tap }P: \quad \dfrac{9000}{x}\:\sf minutes[/tex]
[tex]\textsf{(ii)} \quad \textsf{Tap }Q: \quad \dfrac{9000}{x+5}\:\sf minutes[/tex]
Part (c)
4 hours = 4 × 60 minutes = 240 minutes
If it takes 4 hours longer to fill the aquarium with water using tap P as compared to tap Q, then:
[tex]\begin{aligned}\textsf{Tap }P - 240 & = \textsf{Tap }Q\\\\\implies \dfrac{9000}{x} -240 & = \dfrac{9000}{x+5}\\\\\dfrac{9000 - 240x}{x} & = \dfrac{9000}{x+5}\\\\(9000-240x)(x+5) & =9000x\\\\9000x + 45000 -240x^2-1200x & = 9000x\\\\-240x^2-1200x+45000 & =0\\\\-120(2x^2+10x-375) & =0\\\\2x^2+10x-375 & =0\end{aligned}[/tex]
Part (d)
Quadratic Formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
Therefore:
[tex]a=2, \quad b=10, \quad c=-375[/tex]
Substitute these values into the formula and solve for x:
[tex]\implies x=\dfrac{-10 \pm \sqrt{10^2-4(2)(-375)} }{2(2)}[/tex]
[tex]\implies x=\dfrac{-10 \pm \sqrt{100+3000} }{4}[/tex]
[tex]\implies x=\dfrac{-10 \pm \sqrt{3100} }{4}[/tex]
[tex]\implies x=\dfrac{-10 \pm \sqrt{100 \cdot 31} }{4}[/tex]
[tex]\implies x=\dfrac{-10 \pm \sqrt{100}\sqrt{31} }{4}[/tex]
[tex]\implies x=\dfrac{-10 \pm 10\sqrt{31} }{4}[/tex]
[tex]\implies x=\dfrac{-5\pm 5\sqrt{31} }{2}[/tex]
As the rate is positive only,
[tex]\implies x=\dfrac{-5+5\sqrt{31} }{2}=11.41941...\sf \:litres\:per\:minute[/tex]
Part (e)
If the rate that tap P fills the aquarium with water is x litres per min, and the rate the tap Q fills the aquarium with water is (x+5) litres per min, the rate that both taps fill the aquarium is:
[tex]\begin{aligned}\implies \sf Rate & = x+(x+5)\\& = 2x + 5\\& = 2\left(\dfrac{-5+5\sqrt{31} }{2}\right)+5\\& = -5+5\sqrt{31}+5\\& = 5\sqrt{31}\:\: \sf litres\:per\:min\end{aligned}[/tex]
Therefore, if both taps are turned on together, the time it takes to fill the 9000L aquarium is:
[tex]\implies \sf Time & = \dfrac{9000}{5\sqrt{31} }\:\: \sf minutes \end{aligned}[/tex]
[tex]\implies \sf Time & =323.2895436... \:\: \sf minutes \end{aligned}[/tex]
[tex]\implies \sf Time & =5\:hours\:23.2895436... \:\: \sf minutes \end{aligned}[/tex]
[tex]\implies \sf Time = 5\: h\: 23\: min[/tex]
Note we really should round up to the nearest minute, even though 23.289... rounded is 23, as if we round down, the aquarium will not quite be filled with water.
