Expanding brackets and substitution – Year 7 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Replace x by 5 and y by \(-3\): \(4\times5-2\times(-3)=20-(-6)=20+6=\)26. Subtracting a negative number is the same as adding. (1 mark)
   2. \(x^{2}=25\) and \(y^{2}=(-3)^{2}=9\), because a negative number squared is positive. So \(25-9=\)16. (1 mark)
   3. Work out the bracket first: \(5+(-3)=2\). Then multiply by 3: \(3\times2=\)6. (1 mark)
   4. The top is \(xy=5\times(-3)=-15\). Dividing by 3 gives \(-15\div3=\)\(-5\), negative because the signs are different. (1 mark)

   Keep negative values in brackets when substituting; writing \(-3^{2}\) in part (b) would wrongly give \(-9\).

2. Answer to Question 2 (4 marks)

   1. Multiply both terms inside the bracket by 4: \(4\times3x=12x\) and \(4\times(-2)=-8\). The answer is \(12x-8\). (1 mark)
   2. Expand each bracket: \(2(x+5)=2x+10\) and \(3(x-1)=3x-3\), giving \(2x+10+3x-3\) (1 mark). Collect like terms: \(2x+3x=5x\) and \(10-3=7\), so the answer is \(5x+7\) (1 mark).
   3. Factorising is the reverse of expanding. The highest common factor of 12 and 18 is 6, so take 6 outside the bracket: \(12y\div6=2y\) and \(18\div6=3\). The answer is \(6(2y-3)\). (1 mark)

   Check (c) by expanding: \(6\times2y-6\times3=12y-18\). Writing \(2(6y-9)\) is not fully factorised and scores no mark.

3. Answer to Question 3 (4 marks)

   1. Substitute \(h=4\) into the formula: \(C=5+3\times4\). Multiplication comes before addition, so \(C=5+12=\)£17. (1 mark)
   2. Put \(C=29\) into the formula to form the equation \(5+3h=29\) (1 mark). Subtract 5 from both sides: \(3h=24\). Divide both sides by 3: \(h=\)8 hours (1 mark).
   3. The 5 is not multiplied by h, so it does not change with time. It is a fixed charge of £5 that is paid however long the bicycle is hired for, while the 3 is the cost of £3 for each hour. (1 mark)

   Check (b): \(5+3\times8=5+24=29\), which matches the cost given.

4. Answer to Question 4 (4 marks)

   1. The area of a rectangle is length × width, which gives \(x(x+4)\) (1 mark). Multiply x by each term in the bracket: \(x\times x=x^{2}\) and \(x\times4=4x\). The area is \(x^{2}+4x\) m² (1 mark).
   2. The perimeter is two widths plus two lengths: \(2x+2(x+4)=2x+2x+8=\)\(4x+8\) metres. (1 mark)
   3. Substitute \(x=5\) into the area: \(5^{2}+4\times5=25+20=45\). This equals the given area of 45 m², so \(x=5\) works. The flowerbed is then 9 m long and 5 m wide. (1 mark)

   With \(x=5\) the perimeter is \(4\times5+8=28\) m, and \(2(9+5)=28\) confirms it.

5. Answer to Question 5 (4 marks)

   1. One family pack costs \(2a+3c\) pounds, so 40 packs bring in \(40(2a+3c)\) (1 mark). Multiply each term by 40: \(40\times2a=80a\) and \(40\times3c=120c\). The expanded expression is \(80a+120c\) (1 mark).
   2. Substitute \(a=8\) and \(c=5\): \(80\times8+120\times5=640+600=\)£1 240. (1 mark)
   3. The target is £1 500, so the shortfall is \(1500-1240=\)£260. (1 mark)

   Check (b) with the unexpanded form: one pack costs \(2\times8+3\times5=31\) pounds, and \(40\times31=1240\). Both forms give the same total, showing the expansion was correct.

Total : 20 marks