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.Make x the subject of the formula y = 2x − 9.y = 2x − 9
- 2.The simultaneous equations x = 2 and x + y = 9 are given. Work out the value of y.
- 3.Six years ago Harry was three times as old as his brother Leo. Harry is now 24 years old. Work out Leo's age now.
- 4.Write down the expression that means the same as m² × m × 3.
- 5.A sequence begins at 60, and each term after that is found by subtracting 7 from the term before it. Work out the 5th term of the sequence.
- 6.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.
- 7.A geometric sequence has first term 3 and common ratio 2. Work out the 5th term of the sequence.
- 8.A car's fuel tank contains 50 litres when it starts a journey. After travelling for 2 hours, it contains 38 litres, and fuel is used at a constant rate. Work out the rate at which fuel is used, in litres per hour.
- 9.Solve the inequality 3x + 6 ≤ 0.
- 10.A candle is 30 cm tall when lit and burns at a constant rate. After burning for 5 minutes, its height is 20 cm. The height h cm after t minutes is given by h = 30 − mt. Work out the value of m.
- 11.The graph of y = 20/x is drawn for x > 0. As the value of x increases, what happens to the value of y?
- 12.The first four terms of a sequence are 5, 8, 11, 14. Work out an expression, in terms of n, for the nth term.
- 13.x = 5. Work out the value of 3x² − 4.
- 14.A candle is 30 cm tall and burns down at a steady rate of 1 cm per hour. Write down the function for the height of the candle y, in centimetres, after x hours.
- 15.Simplify p + p + p.
Answer key
- (a) x = (y + 9) / 2 — Method: undo the subtraction of 9 first, then undo the multiplication by 2. Working: y = 2x − 9, so adding 9 to both sides gives y + 9 = 2x, then dividing both sides by 2 gives x = (y + 9) / 2. The value x = (y − 9) / 2 comes from a sign error, keeping the −9 instead of moving it to +9. The value x = 2(y + 9) comes from multiplying by 2 instead of dividing. The value x = y / 2 + 9 comes from dividing by 2 before adding 9, the wrong order of operations, instead of dividing the whole bracket.
- (b) y = 7 — Method: one of the equations already gives the value of x, so that value is substituted into the other equation. Working: replacing x with 2 in x + y = 9 gives 2 + y = 9, so y = 9 − 2 = 7. Answer: y = 7, and the pair x = 2 with y = 7 satisfies both equations. The distractors: y = 11 comes from rearranging x + y = 9 as y = 9 + x and adding instead of subtracting; y = 9 comes from ignoring the x term and reading the total as the value of y; y = 4.5 comes from halving 9, as though x and y had to be equal.
- (a) 12 — Method: write both ages as they were six years ago, call Leo's present age x, and form an equation from the comparison at that time. Working: six years ago Harry was 24 − 6 = 18 and Leo was x − 6, so 18 = 3(x − 6); expanding gives 18 = 3x − 18, adding 18 to both sides gives 36 = 3x, and dividing by 3 gives x = 12. Checking: six years ago Harry was 18 and Leo was 6, and 18 = 3 × 6. Answer: 12. The distractors: 6 comes from solving 18 = 3(x − 6) as far as Leo's age six years ago, 18 ÷ 3 = 6, and giving that as his age now; 8 comes from dividing Harry's present age by 3, 24 ÷ 3 = 8, using the multiple at the wrong moment in time; 36 comes from stopping at 3x = 36 and giving 36 as Leo's age.
- (d) 3m³ — Method: multiply the powers of m by adding their indices, then bring the number coefficient to the front. Working: m² × m has indices 2 and 1; add them to get 3, giving m³, then × 3 gives 3m³. Answer: 3m³. 3m² comes from multiplying the indices instead of adding them: 2 × 1 = 2, giving m², then × 3 = 3m². m³ comes from correctly combining the m's but dropping the coefficient 3. m⁶ comes from multiplying the index by the coefficient instead of writing the coefficient in front: taking the 2 in m² and the 3 to give m raised to the power 2 × 3, which is m⁶, with the lone m left out.
- (c) 32 — The terms are 60, 53, 46, 39, 32 — each found by subtracting 7 from the term before, so the 5th term is 32. Subtracting 7 five times from the first term instead of four times, 60 − 7 × 5 = 25, treats the first term as if it came before the sequence starts. Adding 7 four times instead of subtracting, 60 + 7 × 4 = 88, uses the wrong operation. Stopping one term early gives the 4th term, 39.
- (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) 48 — Each term is found by multiplying the previous term by the common ratio, 2: 3, 6, 12, 24, 48 — the 5th term is 48. A candidate who mistakes the common ratio for a common difference, and adds 2 four times, would reach 3+4×2=11. A candidate who works out the multiplier 2⁴=16 but forgets to multiply it by the first term would give 16. A candidate who multiplies one time too many (finding the 6th term instead of the 5th) would reach 3×2⁵=96.
- (c) 6 — Method: rate = amount of fuel used ÷ time taken. Working: fuel used = 50 − 38 = 12 litres, time taken = 2 hours, so rate = 12 ÷ 2 = 6 litres per hour. Answer: the rate is 6 litres per hour. 44 comes from finding the average of the two fuel amounts, (50 + 38) ÷ 2, instead of the fuel used. 12 comes from working out the fuel used but forgetting to divide by the time taken. 19 comes from dividing the remaining fuel, 38, by the time taken instead of the fuel used.
- (b) x ≤ −2 — Method: take the number term off both sides, then divide by the coefficient of x; the sign turns round only when you divide BY a negative number, and here you divide by 3. Working: subtracting 6 from both sides of 3x + 6 ≤ 0 gives 3x ≤ −6; dividing both sides by 3, which is positive, gives x ≤ −2. Answer: x ≤ −2. The distractors: x ≥ −2 comes from turning the sign round because the right-hand side has become negative, which is not the rule; it is the sign of the divisor that matters; x ≤ 2 comes from moving the 6 across without changing its sign, giving 3x ≤ 6; x ≤ −18 comes from multiplying both sides by 3 instead of dividing by it.
- (a) 2 — The candle's height falls from 30 cm to 20 cm, a drop of 10 cm, over 5 minutes, so m = 10 ÷ 5 = 2. 10 comes from using the drop in height but forgetting to divide by the time. 0.5 comes from dividing the time by the drop instead of the drop by the time (5 ÷ 10). 4 comes from dividing the final height by the time, 20 ÷ 5, instead of using the drop in height.
- (b) y decreases towards zero but never reaches it — Method: as x gets larger, dividing 20 by a bigger number gives a smaller result, so y decreases; because 20/x can never be exactly zero for any positive x, the curve gets closer to zero without ever reaching it, so y decreases towards zero but never reaches it. Distractor origins: 'y increases towards a limit but never reaches it' has the relationship backwards, treating y as increasing when it is actually decreasing; 'y decreases at a steady rate and reaches zero' wrongly assumes the graph behaves like a straight line that eventually hits zero; 'y stays the same however large x becomes' wrongly assumes there is no change in y at all.
- (d) 3n + 2 — The common difference is 3 (8−5=3), so the expression starts 3n. To match the first term when n=1, 3×1+c=5, so c=2: the nth term is 3n+2. A candidate who uses the first term itself as the constant, instead of first term minus the common difference, would write 3n+5 (giving 8, 11, 14, 17 — one term too high throughout). A candidate who omits the constant term altogether would write just 3n (giving 3, 6, 9, 12, not matching the sequence at all). A candidate who adds the common difference to n instead of multiplying would write n+3 (giving 4, 5, 6, 7, far too small).
- (c) 71 — Method: square x first, then multiply by 3, then subtract 4, following the order of operations. Working: x² = 5² = 25; 3 × 25 = 75; 75 − 4 = 71. Answer: 71. 221 comes from squaring (3x) as a whole first: (3 × 5)² = 225, then − 4 = 221, squaring the coefficient along with x. 75 comes from correctly working out 3x² but forgetting to subtract the 4. 3 comes from subtracting the 4 from x before squaring: (5 − 4)² × 3 = 3, doing the operations in the wrong order.
- (c) y = −x + 30 — Method: in a linear model the value at the start is the constant term and the steady rate of change is the gradient, which is negative when the quantity is falling. Working: at x = 0 the candle is 30 cm tall, so the constant term is 30; it loses 1 cm every hour, so the gradient is −1 and the height after x hours is y = −x + 30. Answer: y = −x + 30. The distractors: y = x + 30 comes from taking the rate as +1 and making the candle grow rather than shrink; y = 30x − 1 comes from building the model correctly as 30 − 1x and then copying it into the form y = mx + c with the two numbers left where they stood, so the starting height 30 ends up multiplying x and the hourly 1 is left behind as the constant being taken away; y = −30x + 1 comes from reading the two numbers the other way round, taking 30 cm per hour as the rate and 1 cm as the starting height, which burns 30 cm an hour from a candle only 1 cm tall.
- (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.
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