Forming and solving equations - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 28 exam-style questions on Forming and solving equations for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

28 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

1 mark

A car travels at an average speed of \(s\) km/h for 45 minutes.

Write down an expression, in terms of \(s\), for the distance, in kilometres, that the car travels.

Model answer

45 minutes \(= 0.75\) hours
Distance \(= 0.75s\) km

Explanation

Distance = speed × time, but the time must be in hours because the speed is in km/h. 45 minutes is \(\frac{45}{60} = 0.75\) hours. Writing \(45s\) is the common mistake.

Mark scheme

1 mark
B1 for \(0.75s\)
oe, e.g. \(\dfrac{3s}{4}\) or \(\dfrac{45s}{60}\); do not accept \(45s\)

Question 2

3 marks

Kai, Lina and Max each have a mobile phone bill.
Lina's bill is £8.40 more than Kai's bill.
Max's bill is 1.5 times Lina's bill.
The total of the three bills is £130.20

Work out the amount of Max's bill.

Model answer

Kai: \(x\), Lina: \(x + 8.40\), Max: \(1.5(x + 8.40) = 1.5x + 12.60\)
\(x + x + 8.40 + 1.5x + 12.60 = 130.20\)
\(3.5x + 21 = 130.20\)
\(3.5x = 109.20\)
\(x = 31.20\)
Lina: £39.60
Max: \(1.5 \times 39.60 = 59.40\)
Max's bill is £59.40

Explanation

Choose one unknown (here Kai's bill) and write the other bills in terms of it, keeping \(x + 8.40\) in a bracket when you multiply by 1.5. Add the three expressions, set the total equal to £130.20 and solve. Remember to answer the question asked: the value of \(x\) is Kai's bill, not Max's.

Mark scheme

1 mark
M1 for expressions for the bills, e.g. Kai \(x\), Lina \(x + 8.40\), Max \(1.5(x + 8.40)\)
oe, e.g. starting from Lina's bill \(y\): Kai \(y - 8.40\), Max \(1.5y\)
1 mark
M1 for forming and simplifying the equation, e.g. \(x + x + 8.40 + 1.5(x + 8.40) = 130.20\) leading to \(3.5x + 21 = 130.20\), or for \(x = 31.20\)
oe; using Lina's bill: \(3.5y - 8.40 = 130.20\) or \(y = 39.60\)
1 mark
A1 for £59.40
cao; accept £59.4; correct answer with no working scores full marks

Question 3

3 marks

A gym charges a joining fee and then a fixed amount each month.
Ali pays a total of £427.40 for his first 12 months.
Bea pays a total of £676.20 for her first 20 months.

Work out the joining fee.

Model answer

\(j + 12m = 427.40\)
\(j + 20m = 676.20\)
Subtract: \(8m = 248.80\), so \(m = 31.10\)
\(j = 427.40 - 12 \times 31.10 = 427.40 - 373.20 = 54.20\)
The joining fee is £54.20

Explanation

Both totals include the same joining fee, so subtracting the equations removes it: the extra 8 months cost £248.80, so one month costs £31.10. Substitute back to find the joining fee. Check with Bea: \(54.20 + 20 \times 31.10 = 676.20\).

Mark scheme

1 mark
M1 for forming two equations, e.g. \(j + 12m = 427.40\) and \(j + 20m = 676.20\)
oe
1 mark
M1 for subtracting to get \(8m = 248.80\), or for \(m = 31.10\)
oe
1 mark
A1 for £54.20
cao; accept £54.2; correct answer with no working scores full marks

Question 4

5 marks

Chloe cycles 36 km from her home to a lake at an average speed of \(v\) km/h.
She cycles home along the same route at an average speed that is 3 km/h slower.
The journey home takes 25 minutes longer than the journey to the lake.

Work out the value of \(v\).
Give your answer correct to 3 significant figures.

Model answer

Time to the lake \(= \dfrac{36}{v}\) hours, time home \(= \dfrac{36}{v - 3}\) hours, 25 minutes \(= \dfrac{5}{12}\) hour
\(\dfrac{36}{v - 3} - \dfrac{36}{v} = \dfrac{5}{12}\)
Multiply by \(12v(v - 3)\):
\(432v - 432(v - 3) = 5v(v - 3)\)
\(1296 = 5v^2 - 15v\)
\(5v^2 - 15v - 1296 = 0\)
\(v = \dfrac{15 \pm \sqrt{225 + 25920}}{10} = \dfrac{15 \pm \sqrt{26145}}{10}\)
\(v = 17.669...\) or \(v = -14.669...\)
Speed is positive, so \(v = 17.7\) (3 s.f.)

Explanation

Use time = distance ÷ speed for each journey; the slower journey takes longer, so subtract the faster time from the slower time. Change 25 minutes into hours (\(\frac{25}{60}\)) because the speeds are in km/h. Multiply through by the common denominator to get a quadratic, and reject the negative solution because a speed cannot be negative.

Mark scheme

1 mark
P1 for an expression for a time, e.g. \(\dfrac{36}{v}\) or \(\dfrac{36}{v - 3}\)
oe; correct answer with no working scores full marks
1 mark
P1 for a correct equation, e.g. \(\dfrac{36}{v - 3} - \dfrac{36}{v} = \dfrac{25}{60}\)
oe, e.g. \(\dfrac{5}{12}\) or 0.4166... for 25 minutes
1 mark
P1 for rearranging to a 3-term quadratic, e.g. \(5v^2 - 15v - 1296 = 0\)
oe, e.g. \(v^2 - 3v - 259.2 = 0\)
1 mark
M1 for a correct substitution into the quadratic formula
e.g. \(\dfrac{15 \pm \sqrt{15^2 - 4 \times 5 \times (-1296)}}{10}\); ft their 3-term quadratic
1 mark
A1 for 17.7
awrt 17.7 (17.669...); the negative solution must be rejected or ignored

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