Unit 5
6 marks · NormalA rectangle has a length of cm (to the nearest cm) and a width of cm (to the nearest cm).
Write down the upper and lower bounds of the length and width.
Calculate the upper and lower bounds for the area of the rectangle. Give your answers with units.
A quadratic function represents the relationship between two variables. Solve .
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) Length: 7.95 cm ≤ length < 8.05 cm; Width: 2.5 cm ≤ width < 3.5 cm · must show both lower and upper bounds for length and width, using correct rules
(b) Upper bound: 8.05 × 3.5 = 28.175 cm²; Lower bound: 7.95 × 2.5 = 19.875 cm² cm^2 · must show upper bound using largest values, lower bound using smallest values
(c) x = 3 or x = -1 · must show both roots found by factorisation
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Applies correct bounds rule for either length or width.
A1 (a) · 1 mark · Correct bounds given for both the length and width.
M2 (b) · 1 mark · Uses correct pairs of bounds for area calculation. (follow-through)
A2 (b) · 1 mark · Both upper and lower bound areas correct with units. (follow-through)
M3 (c) · 1 mark · Sets up factorisation for quadratic equation.
A3 (c) · 1 mark · Both solutions to quadratic found correctly.
Worked solution
a) Length: to nearest 0.1 cm, so lower bound = 8.0 - 0.05 = 7.95 cm; upper bound = 8.0 + 0.05 = 8.05 cm. Width: to nearest cm, so lower bound = 3 - 0.5 = 2.5 cm; upper bound = 3 + 0.5 = 3.5 cm.
b) Area upper bound = 8.05 cm × 3.5 cm = 28.175 cm²; Area lower bound = 7.95 cm × 2.5 cm = 19.875 cm².
c) x^2 - 2x - 3 = 0. This factors as (x - 3)(x + 1) = 0, so x = 3 or x = -1.