O-LevelMathematicsAlgebra and graphs4024/21 · May/June 2016 · Q812 Marks

(a) p = (8-5q)/q (b) [Figure 8.1] The lengths of the parallel sides of trapezium A are x cm and (x – 2) cm. The lengths of the parallel sides of trapezium B are x cm and (x + 3) cm. The height of trapezium A is H cm and the height of trapezium B is h cm. The area of each trapezium is 15 cm².

(a)(i)Find p when q = 2.6.[1]
(a)(ii)Express q in terms of p.[2]
(b)(i)Show that H = 15/(x-1) and h = 30/(2x+3).[2]
(b)(ii)Find an expression in terms of x for the difference in height, H – h, between trapezium A and trapezium B, and show that it simplifies to 75/((x-1)(2x+3)).[3]
(b)(iii)(a)The difference in height is 1.5 cm. Show that 2x² + x − 53 = 0.[2]
(b)(iii)(b)Find x, giving your answer correct to 2 decimal places.[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2016). The mark or mark type is shown in brackets.

[1] (a)(i) -1.92 (3......
[2] (a)(ii) q = 8/(p+5)
[2] (a)(ii) M1 for qp = 8-5q or p = 8/q - 5
[2] (b)(i) H and h correctly derived
[2] (b)(i) M1 for correct substitution in the formula for the area of a trapezium.
[3] (b)(ii) 75/((x-1)(2x+3)) correctly derived
[3] (b)(ii) M1 for (15(2x+3)-30(x-1))/((x-1)(2x+3)) soi
[3] (b)(ii) B1 for 30x + 45-30x + 30 soi
[2] (b)(iii)(a) Equation correctly derived.
[2] (b)(iii)(a) B1 for 75/((x-1)(2x+3)) = 1.5
[2] (b)(iii)(b) 4.90
[2] (b)(iii)(b) B1 for ((-1 +/- sqrt(1^2 - 4 * 2 * (-53))) / (2 * 2)) soi

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