14 An oil tanker can hold \( 1200 \mathrm{~cm}^{3} \) of oil. Pump \( A \) alone, can empty the tanker in \( x \) hours. Pump \( B \) alone, can empty the tanker in \( (x+3) \) hours. (a) Find in terms of \( x \), the amount of oil, (i) that Pump \( A \) can empty per minute. (ii) that Pump \( B \) can empty per minute. (b) When both pumps are used at the same time, they can empty the oil tanker in 7 hours. Form an equation in \( x \) and show that it simplifies to (c) Solve the equation \( x^{2}-11 x-21=0 \), leaving your answers correct to 2 (decimal places. (d) Find, the rate of flow of Pump \( B \) in cubic centimetres per second.

To tackle this fun-filled oil pumping puzzle, let’s dive in! First, we’ll kick off with (a): (i) Since Pump \( A \) can empty the tanker in \( x \) hours, we convert hours to minutes (where 1 hour = 60 minutes). Hence, Pump \( A \) empties \( \frac{1200 \, \text{cm}^3}{x \cdot 60} = \frac{20}{x} \, \text{cm}^3 \) per minute. (ii) For Pump \( B \), which takes \( x + 3 \) hours, it will empty \( \frac{1200}{(x + 3) \cdot 60} = \frac{20}{x + 3} \, \text{cm}^3 \) per minute. Now onto (b): When both pumps are working together, their combined emptying rate equals the total capacity divided by time (7 hours), which gives us: \[ \frac{20}{x} + \frac{20}{x + 3} = \frac{1200}{7 \cdot 60} \] This simplifies to: \[ \frac{20}{x} + \frac{20}{x + 3} = \frac{200}{7} \] Combining terms leads to a swift dance of fractions to arrive at an equation in \( x \). Now for (c): To solve the quadratic equation \( x^2 - 11x - 21 = 0 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -11 \), and \( c = -21 \). This will yield two possible values of \( x \) (approximately): \( x \approx 13.87 \) and \( x \approx -1.87 \) (we'll discard the negative as time can’t go backwards!). Finally in (d), the rate of flow for Pump \( B \) in cubic centimeters per second is calculated by first finding the oil it pumps per minute as derived previously, then converting to seconds: \[ \text{Flow rate of Pump } B = \frac{20}{x + 3} \, \text{cm}^3 \text{ per minute} = \frac{20}{x + 3 \, \text{cm}^3 \text{ per minute} \cdot \frac{1}{60} \text{ min/sec} = \frac{20 \cdot 60}{x + 3} \text{ cm}^3 \text{ per second} \] Substituting \( x \) gives the precise flow rate. Isn’t math just marvelous?