4MA0 Higher January 2015 Paper 3 Q17
17 \(P\) is directly proportional to \(q^3\)
\(P = 270\) when \(q = 7.5\)
(a) Find a formula for \(P\) in terms of \(q\) (3)
(b) Work out the positive value of \(q\) when \(P = q\) (2)
| Scheme | Marks |
|---|---|
| \(P = kq^3\) | M1 |
| \(270 = k(7.5)^3\) oe or \(k = \dfrac{270}{7.5^3}\) oe | M1 |
| \(P = \dfrac{16}{25}q^3\) | A1 |
| (3) |
Notes
M1: Allow \(kP = q^3\) oe
Do not allow \(P = q^3\)
M1: for correct substitution in a correct equation.
Implies first M1
A1: \(P = 0.64q^3\) oe with \(P\) the subject
Award M2A1 if \(P = kq^3\) on answer line and \(k\) evaluated as \(\frac{16}{25}\) in part (a) or part (b)
| Scheme | Marks |
|---|---|
Eg \(q^2 = \dfrac{25}{16}\) or \(\dfrac{1}{q^2} = \dfrac{16}{25}\) or \(1 = \dfrac{16}{25}q^2\) or \(q^2 = \dfrac{1}{0.64}\) Or \(P^2 = \dfrac{25}{16}\) or \(\dfrac{1}{P^2} = \dfrac{16}{25}\) or \(1 = \dfrac{16}{25}P^2\) or \(P^2 = \dfrac{1}{0.64}\) | M1 |
| \(1\frac{1}{4}\) | A1 |
| (2) | |
| (5 marks) |
Notes
M1: Correct equation involving \(q^2\) or \(p^2\)
ft \(k\) from an equation of the form \(P = kq^3\) if \(k \neq 1\)
A1: \(\dfrac{5}{4}\), 1.25 [ignore 0 or negative value.]