Printable · GCSE Higher · ages 14-16
Algebra worksheet — GCSE Higher
Fifteen questions across the algebra statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Non-calculator
Algebra worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.A plumber charges a £40 call-out fee plus £25 for each hour worked. The total charge, £C, for a job lasting h hours is given by C = 40 + 25h. Work out the charge for a job lasting 3 hours, and name what kind of statement C = 40 + 25h is.
- 2.Describe a sequence of two transformations that maps the graph of y = x² onto the graph of y = −(x − 5)².y = x²
- 3.A proof that (n + 3)² − (n − 3)² is always a multiple of a certain number begins: Line 1: (n + 3)² − (n − 3)² = (n² + 6n + 9) − (n² − 6n + 9). Which expression correctly completes Line 2?
- 4.Make x the subject of the formula y = 4(x − 6).
- 5.Work out the period of the graph of y = cos x, in degrees.y = cos(x)
- 6.Solve the simultaneous equations 5x − 2y = 16 and 3x + 2y = 16. Work out the value of x.
- 7.The graph of y = sin x is transformed onto the graph of y = sin x + 1. Which statement correctly describes the transformation and the new range of the graph?y = sin(x)
- 8.The graph of y = sin x is reflected in the x-axis to give the graph of y = g(x). Work out the value of g(90), where x is measured in degrees.y = sin(x)
- 9.Solve the inequality 2(x − 1) ≤ 8.
- 10.A point has coordinates (x, y), where x + y = 0 and x is not 0. In which two quadrants could this point lie?
- 11.A geometric sequence has first term 4 and common ratio √2. Work out the sum of the first three terms of the sequence, giving your answer in the form a + b√2.
- 12.For 0 ≤ x ≤ 360, the equation sin x = 0.5 has two solutions. What are they?
- 13.A company charges for hiring chairs for a day. Hiring 1 chair costs £12, hiring 2 chairs costs £19, hiring 3 chairs costs £26, and hiring 4 chairs costs £33. Work out an expression, in terms of n, for the cost in pounds of hiring n chairs for a day.
- 14.A straight line passes through the points (2, 5) and (4, 11). Work out the gradient of the line.
- 15.A zip-line runs from a platform 25 m high to a landing point 5 m high, over a horizontal distance of 40 m. Work out the gradient of the zip-line.
Answer key
- (b) £115, and C = 40 + 25h is a formula — Substitute h = 3 into the rule, multiplying before adding. The hours cost 25 × 3 = 75, and adding the call-out fee gives 40 + 75 = 115, so the charge is £115. The rule itself links two different quantities, C and h, and lets one be worked out from the other, so it is a formula; an expression would have no equals sign in it. Adding the fee before multiplying gives 65 × 3 = 195, which charges the call-out fee three times over, and stopping at 40 + 25 leaves £65, the charge for a single hour.
- (b) Translate +5 in x, then reflect in the x-axis. — Translating y = x² by 5 units in the positive x-direction gives y = (x − 5)². Reflecting this in the x-axis, which replaces y with −y, gives y = −(x − 5)², matching the target. Using a translation of −5 in x instead gives y = (x + 5)², and reflecting that in the x-axis gives y = −(x + 5)² — the sign inside the bracket is wrong. Reflecting in the y-axis first does nothing to y = x², since (−x)² = x², so translating afterwards only reaches y = (x − 5)² with no negative sign at all. Translating by 5 units in y instead of x gives y = x² + 5, and reflecting that in the x-axis gives y = −x² − 5, a different curve altogether — a vertical shift does not create the (x − 5)² term the target equation needs.
- (b) 12n — Distributing the minus sign across the second bracket gives n² + 6n + 9 − n² + 6n − 9, and the n² terms and the +9/−9 cancel, leaving 6n + 6n = 12n. Writing 18 comes from only negating the first term of the second bracket, n², and treating the −6n and +9 as unchanged, which gives n² + 6n + 9 − n² − 6n + 9 = 18. Writing 2n² + 18 comes from adding the two brackets instead of subtracting them, (n² + 6n + 9) + (n² − 6n + 9) = 2n² + 18. Writing 6n comes from correctly negating the bracket but then only counting one of the two 6n terms, missing that they add rather than cancel.
- (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.
- (c) 360° — The graph of y = cos x repeats itself every 360°, so its period is 360°. 180° is the period of y = tan x, not y = cos x — confusing the two graphs gives this. 90° is the horizontal distance from a maximum of y = cos x to the next point where the graph crosses the x-axis, a quarter of the period, mistaken for the whole period. 720° comes from counting the interval containing two full repeats of the graph, from one maximum to the second repeat of that maximum, instead of the first.
- (b) 4 — Adding the two equations: the y-terms, −2y and +2y, cancel, and the x-terms combine to 5x + 3x = 8x; the right-hand sides add to 16 + 16 = 32. This gives 8x = 32, so x = 4. A candidate who adds only one of the right-hand sides, instead of both, would get 8x = 16, so x = 2. A candidate who divides 32 by 4 instead of 8 would get x = 8. A candidate who subtracts the equations instead of adding them, getting 2x − 4y = 0, and then wrongly assumes y = 0, would get x = 0.
- (c) Translate +1 in y; new range 0 ≤ y ≤ 2 — Adding 1 to sin x translates the graph 1 unit in the positive y-direction, and every y-value on the range increases by 1: −1 + 1 = 0 and 1 + 1 = 2, giving a new range of 0 ≤ y ≤ 2. A translation in the x-direction would not add anything to the y-values, so 'Translate +1 in x; range unchanged' correctly leaves the range at −1 ≤ y ≤ 1 for that (wrong) transformation, but the transformation itself is not what y = sin x + 1 shows. 'Translate +1 in y; new range −2 ≤ y ≤ 0' correctly spots the translation but subtracts 1 from each bound instead of adding it. 'Reflect in the x-axis' mistakes the transformation for a reflection rather than a translation; reflecting sin x in the x-axis does leave the range at −1 ≤ y ≤ 1 unchanged in size, but that is not the transformation y = sin x + 1 actually applies.
- (a) −1 — Reflecting y = sin x in the x-axis gives y = −sin x, so g(x) = −sin x. Since sin 90° = 1, g(90) = −1. Reading sin 90° = 1 and forgetting to apply the reflection gives 1. Misreading the angle as 0° instead of 90° gives sin 0° = 0, so 0. Confusing sin 90° with sin 30° = 0.5, then reflecting it, gives −0.5.
- (b) x ≤ 5 — Method: divide out the bracket first, then undo the number term; the inequality sign turns round only if both sides are multiplied or divided by a negative number. Working: dividing both sides of 2(x − 1) ≤ 8 by 2 gives x − 1 ≤ 4, and 2 is positive so the ≤ is unchanged; adding 1 to both sides gives x ≤ 5. Answer: x ≤ 5. The distractors: x ≤ 3 comes from subtracting 1 from 4 instead of adding 1 to both sides; x ≤ 4 comes from stopping at 8 ÷ 2 = 4 and never undoing the −1 inside the bracket; x < 5 comes from reading ≤ as a strict inequality, which wrongly leaves the boundary value out of the solution set.
- (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.
- (a) 12 + 4√2 — a₁ = 4, a₂ = 4√2, and a₃ = a₁ × r² = 4 × (√2)² = 4 × 2 = 8. The sum of the first three terms is 4 + 4√2 + 8 = 12 + 4√2. 8 + 4√2 comes from leaving out a₁ and adding only a₂ + a₃ = 4√2 + 8. 36 + 4√2 comes from squaring the whole second term instead of applying the ratio to the first term: (4√2)² = 32 used as a₃, giving 4 + 4√2 + 32 = 36 + 4√2. 4 + 12√2 comes from using r³ instead of r² for the third term: 4 × (√2)³ = 4 × 2√2 = 8√2, giving 4 + 4√2 + 8√2 = 4 + 12√2.
- (d) 30° and 150° — The first solution is x = 30°, since sin 30° = 0.5. The graph of y = sin x is symmetrical about x = 90° between 0° and 180°, so the second solution is 180° − 30° = 150°. Adding 180° to the first solution instead of subtracting it from 180° gives 30° and 210°, but sin 210° = −0.5, not 0.5. Reflecting the first solution about x = 90° by adding 30° to 90° instead of subtracting from 180° gives 30° and 120°, but sin 120° = √3/2, not 0.5. Misremembering the standard value and using sin 45° = 0.5 instead of sin 30° = 0.5 gives 45° and 135°, but sin 45° = √2/2, not 0.5.
- (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.
- (c) 3 — Gradient = (change in y) ÷ (change in x) = (11 − 5) ÷ (4 − 2) = 6 ÷ 2 = 3. A candidate who puts the change in x over the change in y instead would get 2 ÷ 6 = 1/3. A candidate who subtracts the y-coordinates in the reverse order, but not the x-coordinates, would get (5 − 11) ÷ (4 − 2) = −3. A candidate who adds the coordinates instead of subtracting them would get (11 + 5) ÷ (4 + 2) = 16/6 = 8/3.
- (b) −0.5 — Method: gradient = change in height ÷ horizontal distance, and the height decreases so the change is negative. Working: change in height = 5 − 25 = −20, horizontal distance = 40, so gradient = −20 ÷ 40 = −0.5. Answer: the gradient is −0.5. 0.5 comes from dropping the negative sign, ignoring that the zip-line descends. 2 comes from inverting the gradient, dividing the horizontal distance by the drop in height instead of the other way round. −0.8 comes from dividing the drop by the platform height, 25, instead of by the horizontal distance, 40.
Build your own mix at the worksheet builder.