Solving one-step and two-step equations – Year 7 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Undo "add 8" by subtracting 8 from both sides: \(x=21-8=\)13. Check: \(13+8=21\). (1 mark)
   2. Undo "divide by 5" by multiplying both sides by 5: \(y=7\times5=\)35. Check: \(35\div5=7\). (1 mark)
   3. Two steps, undone in reverse order. Add 4 to both sides: \(3a=18\). Then divide by 3: \(a=6\). Check: \(3\times6-4=14\). (1 mark)
   4. Multiply both sides by 3 to clear the fraction: \(n+2=18\). Subtract 2: \(n=16\). Check: \((16+2)\div3=6\). (1 mark)

   The balance idea is the key: whatever is done to one side must be done to the other so the equation stays true.

2. Answer to Question 2 (4 marks)

   1. Expand the bracket: \(4x-8=20\) (1 mark). Add 8 to both sides: \(4x=28\). Divide by 4: \(x=7\) (1 mark). A quicker route is to divide both sides by 4 first, \(x-2=5\), which gives the same answer.
   2. The unknown is on both sides, so subtract \(2x\) from both sides to keep the x terms on the left: \(3x+3=18\) (1 mark). Subtract 3: \(3x=15\). Divide by 3: \(x=5\) (1 mark).

   Check (a): \(4(7-2)=4\times5=20\). Check (b): left side \(5\times5+3=28\) and right side \(2\times5+18=28\), so both sides balance.

3. Answer to Question 3 (4 marks)

   1. Call the number n. Multiplying by 5 gives \(5n\), subtracting 8 gives \(5n-8\), and the result is 27: \(5n-8=27\). (1 mark)
   2. Undo the subtraction by adding 8 to both sides: \(5n=35\) (1 mark). Undo the multiplication by dividing by 5: \(n=7\) (1 mark).
   3. Follow the instructions with 7: \(7\times5=35\), then \(35-8=27\). This is the answer stated in the puzzle, so my number is correct. (1 mark)

   Working backwards gives the same result: start from 27, add 8 to get 35, then divide by 5 to get 7. Writing \(5(n-8)\) in part (a) would be wrong, because the subtraction happens after the multiplication.

4. Answer to Question 4 (4 marks)

   1. The total cost is the joining fee plus £6 for each of the n lessons: \(15+6n=63\). (1 mark)
   2. Subtract the fee from both sides: \(6n=48\) (1 mark). Divide by 6: \(n=8\) lessons (1 mark). Check: \(15+6\times8=15+48=63\).
   3. For 8 lessons Ravi's club costs \(5+7\times8=5+56=£61\), and Hamza's club costs £63. Ravi's club is cheaper by \(63-61=\)£2. (1 mark)

   The answer depends on the number of lessons. Ravi's club has the cheaper fee but dearer lessons, so it is cheaper only up to 9 lessons; at 10 lessons both clubs cost £75.

5. Answer to Question 5 (4 marks)

   1. The perimeter is two lengths plus two widths: \(2(3x+2)+2(x+4)=6x+4+2x+8=8x+12\). The perimeter is 52 cm, so \(8x+12=52\). (1 mark)
   2. Subtract 12 from both sides: \(8x=40\). Divide by 8: \(x=5\). (1 mark)
   3. Substitute \(x=5\): the length is \(3\times5+2=17\) cm and the width is \(5+4=9\) cm (1 mark). The area is \(17\times9=\)153 cm² (1 mark).

   Check: \(2(17+9)=2\times26=52\) cm, matching the perimeter in the question. Note that area is in cm² while perimeter is in cm; mixing the units loses the final mark.

Total : 20 marks