Higher June 2019 Paper 6 Q14

14 The length of the longest diagonal of a cube is 25 cm.

Calculate the total surface area of the cube. [5]

Mark scheme

Answer	Marks	Part marks and guidance
1250 nfww	5	M4 for [\(6x^2 =\)] 2 × 625 or B4 for final answer 1244 to 1250.05 OR M1 for \(3x^2\) oe or 625 and M1 for \(3x^2 = 625\) oe and A1 for [\(x =\)] \(\sqrt{\dfrac{625}{3}}\) or \(\dfrac{25\sqrt{3}}{3}\) oe or 14.4 to 14.434 soi (14.4 to 14.434 seen implies M1M1A1) and M1 for 6 × their \(x^2\) If 0 scored, SC1 for starting from \(x^2 = 25\) and final answer 150 or starting from \(2x^2 = 25\) and final answer 75	Special cases: Starting from \(3x^2 = 25\) oe soi M1M0 for \(3x^2 = 25\) A1 for [\(x =\)] \(\sqrt{\dfrac{25}{3}}\) or \(\dfrac{5\sqrt{3}}{3}\) oe or 2.88 to 2.89 soi (2.88 to 2.89 seen implies M1M0A1) M1 for 6 × their \(x^2\) soi by 50 Starting from \(2x^2 = 625\) oe soi M1M0 for \(2x^2 = 625\) A1 for [\(x =\)] \(\sqrt{\dfrac{625}{2}}\) or \(\dfrac{25\sqrt{2}}{2}\) or 17.6 to 17.7 soi (17.6 to 17.7 seen implies M1M0A1) M1 for 6 × their \(x^2\) (1875 as final answer implies M1M0A1M1A0) Starting from \(x^2 = 625\) oe soi M1M0 for \(x^2 = 625\) A0 (equation has been simplified and it is a more substantial error) M1 for 6 × their \(x^2\) (3750 as final answer implies M1M0A0M1A0)