1MA0 Higher November 2015 Paper 2 Q20
20 Here are two watch faces, A and B.
Diagram NOT accurately drawn
Both watch faces are circular with radius 2 cm.
The materials used to make both watch faces have the same thickness.
A is made entirely of plastic.
B has a \(20^\circ\) sector of metal and a \(340^\circ\) sector of plastic.
The ratio of the cost per cm\(^2\) of the metal to the cost per cm\(^2\) of the plastic is 3 : 2
Work out the ratio of the cost of the materials for A to the cost of the materials for B.
Give your answer in its simplest form.
You must show all your working. (4)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(\pi 2^2 \times 2 = 8\pi\) \(\pi 2^2 \times \tfrac{20}{360} \times 3 + \pi 2^2 \times \tfrac{340}{360} \times 2\) \(8\pi : \tfrac{74}{9}\pi\) 72 : 74 Or \(\tfrac{20}{360} \times 3 + \tfrac{340}{360} \times 2 = \tfrac{37}{18}\) \(2 : \tfrac{37}{18}\) 36 : 37 | 36 : 37 | 4 | M1 method to find relative cost of design A e.g. \(\pi 2^2\), \(k \times \pi 2^2\) M1 for a complete method to find the relative cost of the watch face for design B e.g. \(\pi 2^2 \times \tfrac{20}{360} \times \tfrac{3}{5} + \pi 2^2 \times \tfrac{340}{360} \times \tfrac{2}{5}\) M1 (dep on M1, M1) for the cost of design A : cost of design B A1 cao Or M1 for method to find fraction of the sectors in design B \(\tfrac{20}{360}, \tfrac{340}{360}\) M1 for complete method to find the relative cost of the watch face for design B M1 (dep on M1,M1) for the cost of design A : cost of design B A1 cao |