Printable · GCSE Foundation · ages 14-16
Algebra worksheet — GCSE Foundation
Fifteen questions across the algebra statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Algebra worksheet — GCSE Foundation
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- 1.The formula for the circumference of a circle is C = 2πr, where r is the radius. Make r the subject of the formula.
- 2.Expand 4(2x − 3).
- 3.Work out the value of 3n − 2 when n = 1
- 4.Which expression is equivalent to 0.5(4x + 6) − x?
- 5.Simplify p + p + p.
- 6.A graph shows the total monthly cost of a mobile phone tariff against the number of minutes of calls. The line starts at £12 and stays level until 200 minutes, then rises by 5p for each further minute of calls. Use the graph to work out the total cost of a month in which 250 minutes of calls are made.
- 7.Tom is asked to classify the statement 5(2x − 3) = 10x − 15. Which statement about it is correct?
- 8.A hot air balloon is at a height of 20 m and rises at a steady rate. After 4 minutes it is at a height of 52 m. Work out the rate at which the balloon rises, in metres per minute.
- 9.Two brothers have ages that add up to 30. The elder brother is 6 years older than the younger brother. Work out the age of the younger brother.
- 10.Work out the value of 3x² − 4x when x = −2.
- 11.Make x the subject of the formula y = 4(x − 6).
- 12.Solve the simultaneous equations 3x + y = 14 and x + y = 6. Work out the value of x.
- 13.A sequence starts at 2. Each term after that is found using the rule "double the previous term, then add 1". Work out the 4th term of the sequence.
- 14.In the expression 5x² − 3x + 7, which of these is a single term of the expression?
- 15.A student says that 5(x + 2) is equivalent to 5x + 2. Which statement explains why the student is wrong?
Answer key
- (a) r = C / (2π) — Method: undo the multiplication by 2π by dividing both sides by 2π. Working: C = 2πr, so dividing both sides by 2π gives r = C / (2π). The value r = C / π comes from dividing by π only and forgetting the factor of 2 in 2π. The value r = 2πC comes from multiplying by 2π instead of dividing. The value r = C − 2π comes from subtracting 2π instead of dividing by it.
- (c) 8x − 12 — Multiply each term inside the bracket by 4: 4 × 2x = 8x and 4 × (−3) = −12, so 4(2x − 3) = 8x − 12. A candidate who forgets to multiply the second term by 4 gets 8x − 3. A candidate who makes a sign error, treating 4 × (−3) as +12, gets 8x + 12. A candidate who adds 4 to the bracket instead of multiplying gets 2x + 1.
- (b) 1 — Method: replace the letter with its value, work out the multiplication first and the subtraction afterwards. Working: 3n = 3 × 1 = 3, so the expression becomes 3 − 2, which is 1. Answer: 1. The distractors: 3 comes from substituting into 3n but stopping before the 2 is taken away; −1 comes from subtracting the wrong way round and working out 2 − 3; 5 comes from adding the 2 instead of subtracting it, giving 3 + 2.
- (c) x + 3 — Expand the bracket: 0.5(4x + 6) = 2x + 3. Then subtract the x: 2x + 3 − x = x + 3. The option 2x + 3 comes from expanding the bracket correctly but then forgetting to subtract the x at all. The option x + 6 comes from forgetting to multiply the 6 inside the bracket by 0.5 (treating it as 2x + 6), then subtracting x. The option 3x + 3 comes from adding the x instead of subtracting it: 2x + 3 + x = 3x + 3.
- (b) 3p — p + p + p means three lots of p added together, and repeated addition of the same term is written as a coefficient: 3p. Writing p³ mistakes the repeated addition for repeated multiplication, as if the expression had been p × p × p. Writing 3 + p adds the number of terms (3) onto p as a separate constant, instead of writing 3 as a coefficient of p. Writing just p forgets to count the terms at all, as though repeating the same letter makes no difference. The simplified expression is 3p.
- (a) £14.50 — Method: the first 200 minutes are already covered by the £12, so only 250 − 200 = 50 minutes are charged extra. Extra charge = 50 × £0.05 = £2.50. Total cost = £12 + £2.50 = £14.50. Distractor origins: £24.50 charges 5p for all 250 minutes instead of only the 50 minutes over 200 (250 × £0.05 = £12.50, plus £12 = £24.50); £12.50 makes that same slip of charging all 250 minutes but then forgets to add the £12 monthly fee; £13.50 works out 250 − 200 wrongly as 30 extra minutes instead of 50 (30 × £0.05 = £1.50, plus £12 = £13.50).
- (b) An identity, since both sides are equal for every x. — Expanding the left-hand side: 5(2x−3)=10x−15, which is exactly the same as the right-hand side for every value of x, so it is an identity, not an equation that is only true for one particular x. A candidate who treats every equals-sign statement as an equation, without checking whether it holds for all values of x, would choose the equation option. A candidate who mistakes it for a formula is assuming it relates two different letters or quantities, but only x appears — there is no second variable such as area or cost — so it is not a formula. A candidate who mistakes it for an inequality is assuming the two sides are only equal for particular values of x, but expanding shows they are identical for every value, not just some.
- (a) 8 — Method: rate = height gained ÷ time taken. Working: height gained = 52 − 20 = 32 m, time taken = 4 minutes, so rate = 32 ÷ 4 = 8 metres per minute. Answer: the rate is 8 metres per minute. 32 comes from working out the height gained but forgetting to divide by the time taken. 13 comes from dividing the final height, 52, by the time taken instead of the height gained. 5 comes from dividing the starting height, 20, by the time taken instead of the height gained.
- (d) 12 — Method: call the younger brother's age x, write the elder brother's age in terms of x, and form an equation from the total. Working: the elder brother is x + 6, so x + (x + 6) = 30; simplifying gives 2x + 6 = 30, subtracting 6 from both sides gives 2x = 24, and dividing by 2 gives x = 12. Checking: 12 and 18 add up to 30 and differ by 6. Answer: 12. The distractors: 18 comes from solving correctly and then giving the elder brother's age, which is not the age asked for; 15 comes from halving 30 and ignoring the 6-year difference altogether; 24 comes from taking 6 off the total, 30 − 6 = 24, and giving that as an age.
- (b) 20 — 3x² − 4x = 3(−2)² − 4(−2) = 3(4) − (−8) = 12 + 8 = 20. A candidate who makes a sign error on −4x, treating −4 × −2 as −8 instead of +8, gets 12 − 8 = 4. A candidate who squares −2 but keeps the result negative, using 3 × (−4) = −12 for the first term, but correctly works out −4x = −4 × (−2) = 8, gets −12 + 8 = −4. A candidate who squares the whole term 3x, working out (3 × −2)² − 4 × (−2), gets (−6)² + 8 = 36 + 8 = 44.
- (c) x = (y + 24)/4 — Method: expand the bracket first, then undo the operations done to x in reverse order. Working: y = 4(x − 6) = 4x − 24, so y + 24 = 4x, so x = (y + 24)/4. Answer: x = (y + 24)/4. x = (y + 6)/4 comes from expanding the bracket incorrectly, treating 4(x − 6) as 4x − 6 instead of 4x − 24. x = 4y + 96 comes from multiplying by 4 instead of dividing by 4 to undo the multiplication, giving 4(y + 24) = 4y + 96. x = y/4 − 6 comes from dividing by 4 first without expanding the bracket, then subtracting 6 as if the bracket's operation still applied afterwards.
- (d) 4 — Subtracting the second equation from the first: the y-terms cancel, and the x-terms combine as 3x − x = 2x; the right-hand sides give 14 − 6 = 8. This gives 2x = 8, so x = 4. A candidate who subtracts in the wrong order would get 2x = 6 − 14 = −8, so x = −4. A candidate who uses only the first equation, dividing 14 by 3 as if y were 0, would get x ≈ 4.67. A candidate who reports the value of y instead of x would get y = 6 − 4 = 2.
- (b) 23 — Term 1 is 2. Term 2 = 2 × 2 + 1 = 5. Term 3 = 2 × 5 + 1 = 11. Term 4 = 2 × 11 + 1 = 23. A candidate who doubles each term but forgets to add 1 gets 2, 4, 8, 16. A candidate who adds 1 before doubling at each step (the wrong order) gets 2, 6, 14, 30. A candidate who forgets to add 1 only on the final step gets 2 × 11 = 22.
- (c) 5x² — A term is one part of an expression, separated from the others by a + or − sign. The expression 5x² − 3x + 7 has three terms: 5x², −3x and 7. Distractor origins: 5x² − 3x combines two terms together, not a single term; −3x + 7 also combines two terms together; 5x² − 3x + 7 gives the whole expression, not a single term.
- (c) The 5 must multiply both terms inside the bracket, so 5(x + 2) expands to 5x + 10, which is never equal to 5x + 2 for any value of x. — Expanding the bracket correctly, 5(x + 2) = 5x + 10, since the 5 multiplies both the x and the 2. This is never equal to 5x + 2, since that would require 10 = 2. The option claiming 5(x + 2) means 5 × x + 2 ignores that the 5 must multiply the whole bracket, not just the x-term. The option about working out the bracket first with a value of x misunderstands algebraic expansion, which holds for every x, not just specific ones. The option about addition before multiplication misapplies the order of operations to bracket expansion, which always distributes the outer factor over every term inside, whatever x is.
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