Formulae and rearranging – Year 8 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Replace each letter with its value, keeping the negative in brackets: \(v=6+(-2)\times4\) (1 mark). Multiply before adding: \((-2)\times4=-8\), so \(v=6-8=\)\(-2\) (1 mark).
   2. To make x the subject, undo the operations on x in reverse order. First add 8 to both sides: \(y+8=5x\) (1 mark). Then divide both sides by 5: \(x=\dfrac{y+8}{5}\) (1 mark).

   Check (b) with a number: if \(x=2\), then \(y=5\times2-8=2\), and the new formula gives \(x=\dfrac{2+8}{5}=2\). The value comes back unchanged, so the rearrangement is right. In (a) a negative v means the object is moving in the opposite direction.

2. Answer to Question 2 (4 marks)

   1. \(2x+3=11\) is an equation: it is true for just one value of x (\(x=4\)). An expression has no equals sign, and a formula links two or more different quantities. (1 mark)
   2. Divide both sides by 2 to remove the bracket: \(\dfrac{P}{2}=a+b\) (1 mark). Subtract b from both sides: \(a=\dfrac{P}{2}-b\) (1 mark).
   3. Substitute into the rearranged formula: \(a=\dfrac{30}{2}-4=15-4=\)11. (1 mark)

   Check (c): \(2(11+4)=2\times15=30\). The formula \(P=2(a+b)\) is the perimeter of a rectangle of length a and width b.

3. Answer to Question 3 (4 marks)

   1. Substitute \(C=20\): \(F=1.8\times20+32=36+32=\)68 °F. (1 mark)
   2. Undo the operations on C in reverse order. Subtract 32 from both sides: \(F-32=1.8C\) (1 mark). Divide both sides by 1.8: \(C=\dfrac{F-32}{1.8}\) (1 mark).
   3. Use the new formula with \(F=59\): \(C=\dfrac{59-32}{1.8}=\dfrac{27}{1.8}=\)15 °C. (1 mark)

   Check (c) with the original formula: \(1.8\times15+32=27+32=59\). A quick mental estimate also helps: doubling and adding 30 gives \(2\times15+30=60\), close to 59.

4. Answer to Question 4 (4 marks)

   1. Substitute \(d=5\): \(C=25+8\times5=25+40=\)£65. (1 mark)
   2. Subtract 25 from both sides, \(C-25=8d\), then divide by 8: \(d=\dfrac{C-25}{8}\). (1 mark)
   3. Put \(C=70\) into the rearranged formula: \(d=\dfrac{70-25}{8}=\dfrac{45}{8}=5.625\) (1 mark). Josh can only hire for whole days, and 6 days would cost \(25+48=£73\), more than he has. The longest hire is 5 days (1 mark).

   This is a case where the answer must be rounded down even though 5.625 is nearer to 6. Josh would have £5 left over after paying £65 for 5 days.

5. Answer to Question 5 (4 marks)

   1. Set the two costs equal: \(25+8d=15+10d\) (1 mark). Subtract \(8d\) from both sides: \(25=15+2d\). Subtract 15: \(10=2d\), so \(d=\)5 days (1 mark).
   2. For 10 days, Speedy Hire costs \(25+8\times10=£105\) and Handy Hire costs \(15+10\times10=£115\) (1 mark for each cost). Speedy Hire is cheaper, by £10 (the remaining mark for the conclusion).

   Check (a): at 5 days both firms charge £65. Handy Hire has the lower fixed charge, so it is cheaper for hires shorter than 5 days, but its higher daily rate makes it more expensive for longer hires.

Total : 20 marks