Foundation June 2025 Paper 3 Q23
23 Ben is trying to make \(m\) the subject of \(p = \dfrac{m}{3} + 5\)
Here is his working.
\[\begin{aligned} p - 5 &= \frac{m}{3} \\ 3 \times p - 5 &= m \\ m &= 3p - 5 \end{aligned}\]Ben’s answer is wrong.
(a) What mistake has Ben made? (1)
(b) Factorise fully \(2x^3y + 4xy^2\) (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| Mistake identified | C1 | for identifying the mistake Acceptable examples \(p - 5\) should be multiplied by 3 (\(-\))5 should be multiplied by 3 All of left side / everything should be multiplied by 3 He failed to multiply the 5 as well He should have / didn’t put brackets around the \(p - 5\) (The \(3p - 5\)) should be \(3p - 15\) (The \(-5\)) should be \(-15\) / (the 5) should be 15 (The answer should be) \(m = 3p - 15\) / \(m = 3(p - 5)\) He only times the \(p\) by 3 Not acceptable examples The first line should be \(3p = m + 5\) He should have multiplied everything Ben didn’t divide \(p - 5\) by 3 He failed to multiply the \(p - 5\) He failed to multiply the (\(-\))5 He only times the \(p\) He should have multiplied by 3 first He only multiplied one side by 3 He needs to get rid of the fraction Should have used brackets Just circling the \(3 \times p - 5\) and / or the \(m = 3p - 5\) Needs to multiply the 5 by \(-3\) He should have done \(p - 5 \times 3\) |
| Answer | Mark | Mark scheme |
|---|---|---|
| \(2xy(x^2 + 2y)\) | B2 | for \(2xy(x^2 + 2y)\) oe eg \(2(x^2 + 2y)xy\) |
| (B1 | for \(2x(x^2y + 2y^2)\) or \(2y(x^3 + 2xy)\) or \(xy(2x^2 + 4y)\) or for correctly identifying the HCF in the factorisation of the form \(2xy(ax^2 \pm \ldots)\) or \(2xy(\ldots \pm by)\) where \(a\) and \(b\) are integers or \((x^2 + 2y)\) as a factor eg \(2x(x^2 + 2y)\)) |
Additional guidance
\(\ldots\) can be numerical or algebraic but not equal to 0 or absent