The answer
(a) \(10.5\) g
(b)(i) base \((20-2x)\) by \((30-2x)\), height \(x\)
(ii) \(0 < x < 10\)
(c) \(\approx 9.8\) g
O-Level E-Math 2018 Paper 2 Question 10 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 10 of the O-Level E-Math 2018 Paper 2. It tests mass from area and density, in the Functions & graphs (volume model, graphical solution) area. It is worth 11 marks: (a) 2, (b) 1 + 1, (c) 7. It is a worded / diagram-based question, so open your Ten-Year Series (TYS) or the official paper at this question, then follow our full worked solution below.
(a) Cutting a 5 cm square from each corner removes \(4 \times 5^2 = 100 \text{ cm}^2\), leaving an area of \(20 \times 30 - 100 = 600 - 100 = 500 \text{ cm}^2\) of cardboard. The density is \(\dfrac{210 \text{ g}}{1 \text{ m}^2} = \dfrac{210}{10\,000} = 0.021 \text{ g/cm}^2\), so the mass \(= 500 \times 0.021 = 10.5\) g.
(b)(i) After the four corner squares of side \(x\) are removed and the flaps are folded up, the base of the box has length \(30 - 2x\) (two squares removed from the 30 cm side) and width \(20 - 2x\) (two removed from the 20 cm side), and the height of the box is \(x\). So volume \(= x(20 - 2x)(30 - 2x)\).
(ii) The cut length must be positive, so \(x > 0\); and the shorter side \(20 - 2x\) must stay positive for a box to form, so \(x < 10\). Hence \(0 < x < 10\).
(c) The amount of cardboard used is the area \(A = 600 - 4x^2\), which decreases as \(x\) increases. So to use the least cardboard while keeping \(V \geqslant 900\), Huan should take the largest \(x\) for which the volume is still at least \(900 \text{ cm}^3\). Plotting \(V = x(20 - 2x)(30 - 2x)\) against \(x\) for \(0 < x < 10\) and drawing the line \(V = 900\), the curve meets \(V = 900\) at \(x \approx 2.3\) and \(x \approx 5.8\) (the maximum volume, about \(1056 \text{ cm}^3\), is near \(x = 3.9\)), so \(V \geqslant 900\) for \(2.3 \leqslant x \leqslant 5.8\). The largest value is \(x \approx 5.8\). Then the area is \(A = 600 - 4(5.8)^2 \approx 600 - 134.6 = 465.4 \text{ cm}^2\), so the mass \(\approx 465 \times 0.021 \approx 9.8\) g. *(Read from the drawn graph; accept roughly \(9.7\) to \(9.9\) g.)*
Answer: (a) \(10.5\) g
(b)(i) base \((20-2x)\) by \((30-2x)\), height \(x\)
(ii) \(0 < x < 10\)
(c) \(\approx 9.8\) g
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
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Genius Plus Academy · O-Level & IP Mathematics
Our O-Level E-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.
It is a mass from area and density question from Functions & graphs (volume model, graphical solution), worth 11 marks: (a) 2, (b) 1 + 1, (c) 7.
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