(a) Given that P = (\(\frac{rk}{Q} - ms\))\(^{\frac{2}{3}}\) (i) Make Q the subject of the relation; (ii) find, correct to two decimal places, the value of ...

Answer Details

(a)(i) Make Q the subject of \(P = \left(\dfrac{rk}{Q} - ms\right)^{\frac{2}{3}}\).

Raise both sides to the power \(\tfrac{3}{2}\):

\[P^{\frac{3}{2}} = \frac{rk}{Q} - ms.\]

\[\frac{rk}{Q} = P^{\frac{3}{2}} + ms \Rightarrow Q = \frac{rk}{P^{\frac{3}{2}} + ms}.\]

(a)(ii) With \(P = 3, m = 15, s = 0.2, k = 4, r = 10\):

• \(ms = 15 \times 0.2 = 3\)
• \(rk = 10 \times 4 = 40\)
• \(P^{\frac{3}{2}} = 3^{1.5} = \sqrt{27} = 5.196\)

\[Q = \frac{40}{5.196 + 3} = \frac{40}{8.196} = 4.88\ (\text{2 d.p.}).\]

(b) Given \(\dfrac{x + 2y}{5} = x - 2y\). Cross-multiplying:

\[x + 2y = 5(x - 2y) = 5x - 10y.\]

\[12y = 4x \Rightarrow \frac{x}{y} = \frac{12}{4} = 3.\]

Therefore \(x : y = 3 : 1\).