Alternative Method 4 3 + = 4 × x 5(3 + x) = 4 × 9 M1 3 + = x 15 + 5x = 36 15(b) M1dep Cont. = 3 x – 5x = 36 15 M1dep – or 4.2 A1 oe Q Additional Guidance Mark 15(b) M1 M1dep 3+ =9 x 4 4 5 M1dep =9 3− x 4 5 4 A1 x= If students use a different equation than that chosen in part (a) they can only score a maximum of 3 out of 4, unless they choose the correct equation
15(b) Cont. Alternative Method 4 5(3 + x) = 4 × 9 M1 3 + x = 4 × 15 + 5x = 36 M1dep 3 + x = 5x = 36 – 15 M1dep x = – 3 or 4.2 A1 oe Q Additional Guidance Mark 15(b) 3+ x =9 4 4 5 x =9 3− 4 5 4 x= If students use a different equation than that chosen in part (a) they can only score a maximum of 3 out of 4, unless they choose the correct equation M1 M1dep M1dep A1
November-2014-Paper-2-Mark-Scheme1.docx