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.
Non-calculator
Answer key: Algebra worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- (a) 3y² — There is one 3 in the product and the letter y appears twice. A number that multiplies a letter is written in front of it as the coefficient, so the 3 goes at the front. The two ys multiply each other, so the letter carries an index of 2. The shortest correct way of writing the product is therefore 3y². Note that (3y)² would mean 3 × y × 3 × y, which squares the coefficient as well, while 3 + y² turns a multiplication into an addition and y³ treats the coefficient as a third factor of y.
- (d) £2000 — This is an arithmetic sequence with first term £500 and common difference £300. The 6th term is 500 + 5 × 300 = 2000. A candidate who uses 6 lots of the increase instead of 5 gets 500 + 6 × 300 = 2300. A candidate who forgets to add the first year's profit at all gets 5 × 300 = 1500. A candidate who miscounts the number of increases as 4 instead of 5 gets 500 + 4 × 300 = 1700.
- (c) x ≤ 4 — Subtract 9 from both sides: −2x ≥ 1 − 9 = −8. Divide both sides by −2, flipping the inequality: x ≤ 4. A candidate who divides by −2 but forgets to flip the inequality gets x ≥ 4. A candidate who divides −8 by −2 but keeps a negative sign gets x ≤ −4. A candidate who miscalculates 1 − 9 as −10 gets x ≤ 5.
- (c) The cost, in pounds, for each extra gigabyte of data used — In C = 15 + 2g, the number multiplying g is the gradient, which gives the extra cost for each extra unit of g — here, £2 for each extra gigabyte. 'The fixed monthly fee, in pounds' describes the constant term 15, not the coefficient of g. 'The total number of gigabytes included in the plan' misreads the coefficient as a quantity of data rather than a cost per gigabyte. 'The cost … for each extra 2 gigabytes' doubles the unit the coefficient actually applies to — it is the cost for each single extra gigabyte.
- (a) x = 3/2 or x = −3/2 — Method: the equation has an x² term and a number but no x term, so make x² the subject and then take the square root of both sides, keeping the negative root as well as the positive one. Working: adding 9 to both sides of 4x² − 9 = 0 gives 4x² = 9, and dividing both sides by 4 gives x² = 9/4. Square-rooting the top and the bottom of 9/4 gives 3/2, so x = 3/2 or x = −3/2, and each value checks out because 4 × 9/4 − 9 = 0. Answer: x = 3/2 or x = −3/2. The distractors: x = 3 or x = −3 comes from square-rooting both sides of 4x² = 9 without first dividing by the 4, so the coefficient of x² is ignored; x = 9/4 or x = −9/4 comes from stopping at x² = 9/4 and writing that value down as x, leaving the square root undone; x = 3/2 only comes from taking the positive square root of 9/4 and losing the negative solution.
- (a) 30 US dollars — £24 is 6 times £4, and the same scale factor applies to US dollars, so multiply 5 by 6: 5 × 6 = 30, giving 30 US dollars. Multiplying £24 by 4/5, the pounds-per-dollar ratio, instead of by 5/4, the dollars-per-pound ratio, gives 24 × 4/5 = 19.2, applying the ratio the wrong way round and giving 19.20 US dollars. Adding the extra £20 straight onto the 5 as if £1 equalled 1 US dollar, 5 + 20 = 25, gives 25 US dollars. Multiplying £24 by 5 without first dividing by the original £4, 24 × 5 = 120, gives 120 US dollars.
- (c) x = −2 or x = 4 — The solutions of x² − 2x − 8 = 0 are the x-values where y = 0. From the table, y = 0 when x = −2 and when x = 4, so those are the two solutions. Distractor origins: x = −1 or x = 3 picks the pair of x-values that give equal (but non-zero) y-values instead of y = 0; x = 0 or x = −8 mixes up an x-value with its corresponding y-value; x = −2 only reads off one of the two roots and misses the other.
- (c) Second and fourth — If x + y = 0 then y = −x, so x and y always have opposite signs, one positive and one negative. A point with a negative x and a positive y lies in the second quadrant, and a point with a positive x and a negative y lies in the fourth quadrant, so the point lies in the second or the fourth. A candidate who reads x + y = 0 as meaning x and y have the same sign picks First and third, which is where x × y is positive, not where x + y = 0. A candidate who decides that y must be the positive coordinate picks the two quadrants above the x-axis, First and second. A candidate who decides that y must be the negative coordinate picks the two quadrants below the x-axis, Third and fourth.
- (d) 110 — Method: form the equation 15 + 0.08m = 23.80, where m is the number of minutes, then solve for m. Working: subtract the fixed fee: 0.08m = 23.80 − 15 = 8.80. Divide by the cost per minute: m = 8.80 ÷ 0.08 = 110. Answer: 110 minutes. 1.1 comes from using 8 instead of 0.08 as the cost per minute, forgetting to convert pence to pounds. 297.5 comes from dividing the whole bill by the cost per minute without subtracting the fixed fee first, 23.80 ÷ 0.08. 485 comes from adding the fixed fee to the bill instead of subtracting it, before dividing by the cost per minute, (23.80 + 15) ÷ 0.08.
- (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) 12.5 — Method: apply the term-to-term rule to the term just written, and keep the exact value even when halving does not give a whole number. Working: the term before the one wanted is 25, and halving it means working out 25 ÷ 2, which is 12 with 1 left over to share, giving a half. Answer: 12.5. The distractors: 12 comes from halving 25 and then cutting the result down to a whole number; 0 comes from treating the sequence as one with a constant difference and taking 25 away from 25; 6.25 comes from halving twice and giving the term after the next one.
- (d) 7n + 5 — Method: find the rate charged per extra chair, then find the fixed part of the cost that fits hiring 1 chair. Working: the cost rises by £7 for each extra chair (19 − 12 = 7, 26 − 19 = 7, 33 − 26 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 12, so c = 5. Answer: the cost in pounds is 7n + 5. The value 7n comes from ignoring the fixed part of the charge entirely. The value 7n + 12 comes from using the cost of 1 chair as the fixed part directly, without subtracting the per-chair rate first. The value 12n + 7 comes from swapping the roles of the cost of hiring 1 chair, £12, and the rate per extra chair, £7 — using the total for one chair as the coefficient of n and the rate as the fixed part.
- (a) 5k — Method: adding a letter to itself repeatedly means counting how many of that letter you have, written as a coefficient in front of the letter. Working: k + k + k + k + k is five lots of k, written 5k. Answer: 5k. k⁵ comes from treating repeated addition as repeated multiplication (raising to a power) instead of counting copies. 5 + k comes from adding the count of terms (5) to a single k instead of multiplying. k/5 comes from dividing instead of counting how many k's there are.
- (b) −5 — Method: a point whose x-coordinate is 0 lies on the y-axis, so its y-coordinate is the constant c; once m and c are both known the equation can be written down and x substituted into it. Working: the line passes through (0, 1), so c = 1 and the equation is y = −3x + 1; substituting x = 2 gives y = −3 × 2 + 1 = −6 + 1 = −5. Answer: −5. The distractors: 7 comes from ignoring the minus sign on the gradient and working out 3 × 2 + 1; 5 comes from working out 3 × 2 = 6 and then using the minus sign to take the constant off the product, 6 − 1 = 5; −7 comes from subtracting the constant instead of adding it, −6 − 1 = −7.
- (c) y = 3x + 7 — Method: the gradient is the change in y divided by the change in x with both taken in the same order, and a point whose x-coordinate is 0 gives the constant straight away because it lies on the y-axis. Working: m = (7 − 1) ÷ (0 − (−2)) = 6 ÷ 2 = 3; the point (0, 7) lies on the y-axis, so c = 7 and the line is y = 3x + 7. Answer: y = 3x + 7. The distractors: y = −3x + 7 comes from taking the y-difference as 1 − 7 while taking the x-difference as 0 − (−2), so the two subtractions run in opposite orders; y = 3x + 1 comes from using the y-coordinate of (−2, 1) as the constant instead of the point that actually lies on the y-axis; y = (1/3)x + 7 comes from writing the gradient upside down as the change in x over the change in y, 2 ÷ 6.
Build your own mix at the worksheet builder.