A local market trader prices bags of a speciality spice using \(P = 3x^{\frac23}\), where \(x\) is the mass of a bag in grams and \(P\) is the price in penc...

The pricing rule \(P=3x^{\frac23}\) is not proportional (\(P\) does not simply double when \(x\) doubles), because raising to the power \(\frac23\) scales the mass down before it is multiplied by 3; each part below either applies this rule directly or reverses it to find a mass from a price.

(a) At \(x=27\): \(P = 3\times27^{\frac23} = 3\times\left(27^{\frac13}\right)^{2} = 3\times3^{2} = 3\times9 = 27\) pence. [2 marks]

(b) Starting from \(x^{\frac23}=16\), raise both sides to the power \(\frac32\) (the reciprocal of \(\frac23\)) to isolate \(x\): \(x = 16^{\frac32} = (\sqrt{16})^{3} = 4^{3} = 64\) g. [3 marks]

(c) The 8 g bag costs \(3\times8^{\frac23} = 3\times4 = 12\) pence, so a bag priced at double this costs \(24\) pence. Setting up the equation: \(24 = 3m^{\frac23}\), so \(m^{\frac23}=8\), as required. Solving, \(m = 8^{\frac32} = (\sqrt8)^{3} \approx 22.6\) g (3 s.f.). [3 marks]

(d) No: doubling \(x\) changes the price by a factor of \(2^{\frac23}\approx1.59\), not by a factor of 2, because \(P\) is proportional to \(x^{\frac23}\) rather than to \(x\) itself, so equal percentage increases in mass do not give equal percentage increases in price. [2 marks]

Exam tip: whenever a quantity is proportional to a power of \(x\) other than \(x^{1}\), doubling \(x\) never doubles that quantity - work out the actual scale factor, \(2^{\text{power}}\), instead of assuming it is 2.