Problem solving and GCSE exam practice – Year 11 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \(360\div12=30\); £150 and £210 (1 for the part, 1 for both) Add the parts: \(5+7=12\). One part is \(360\div12=30\), so the shares are \(5\times30\) and \(7\times30\). Check: \(150+210=360\).
   2. \(0.12\times210=\) £25.20 (2) 12% as a decimal is 0.12, so multiply the larger share, £210, by 0.12.

   In (b), a non-calculator method is to find 10% (£21) and 2% (£4.20) and add them, giving £25.20 again. A common error in (a) is to divide £360 by 5 and by 7.

2. Answer to Question 2 (4 marks)

   1. \(5x-10=3x+6\), so \(2x=16\) and \(x=8\) (1 each) Expand the bracket, then subtract \(3x\) and add 10 to both sides. Check: \(5\times6=30\) and \(24+6=30\).
   2. \(y=x-2\), so \(3x+2x-4=16\), \(x=4\), \(y=2\) (1 each) Make \(y\) the subject of the second equation and substitute into the first. Check: \(12+4=16\) and \(4-2=2\).

   In (a), the check substitutes \(x=8\): \(5(8-2)=30\) and \(3\times8+6=30\). In (b), elimination also works: double \(x-y=2\) and add.

3. Answer to Question 3 (4 marks)

   1. \(\tan x=\dfrac{30}{50}\) (1), \(x=\tan^{-1}(0.6)\) (1), \(31.0^\circ\) (1) The height of the tower is opposite the angle and the 50 m distance is adjacent, so we use tangent.
   2. \(\sqrt{900+2500}=\sqrt{3400}=\) 58.3 m (1) Use Pythagoras with the two shorter sides 30 m and 50 m.

   Check (b) with trigonometry: \(30\div\sin31.0^\circ\approx58.3\) m. A common error in (a) is to use \(\tan^{-1}\left(\tfrac{50}{30}\right)\), the angle at the top.

4. Answer to Question 4 (4 marks)

   1. Let the numbers be \(2n+1\) and \(2n+3\) (1). Sum \(=4n+4\) (1) \(=4(n+1)\) (1), which is a multiple of 4 (1). Consecutive odd numbers differ by 2. Since \(n+1\) is a whole number, \(4(n+1)\) is a multiple of 4 for every whole number \(n\). For example, \(5+7=12=4\times3\).

   A proof must work for every value, so examples alone score no marks. Using \(n\) and \(n+2\) is not enough either, because \(n\) might be even.

5. Answer to Question 5 (4 marks)

   1. Perimeter \(=2\times(40+25)=130\) (1); \(130-3=\) 127 m (1) The fence goes round the perimeter, but the 3 m gap for the gate is not fenced, so subtract 3.
   2. \(127\times14.50=\) £1 841.50 (2) Multiply the length of fence by the price per metre.

   In (a), a common error is to add only \(40+25\), giving half the perimeter. In (b), using a calculator, \(127\times14.5=1841.5\), written as £1 841.50 in money.

Total : 20 marks