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.
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Algebra worksheet — GCSE Higher
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- 1.For the equation 2x + 3 = 11, and the inequality 2x + 3 > 11, which statement correctly compares their solutions?
- 2.To rearrange the formula y = 3x − 8 to make x the subject, Priya writes: 'Add 8 to both sides to get y + 8 = 3x, then divide both sides by 3 to get x = y + 8 ÷ 3.' Work out the correct expression for x.y = 3x − 8
- 3.ABCD is a parallelogram. A has coordinates (−3, 1), B has coordinates (2, 1) and C has coordinates (4, 4). Work out the coordinates of D.
- 4.Which of these is a factor of 15x + 20?
- 5.The formula for converting a temperature from degrees Celsius, C, to degrees Fahrenheit, F, is F = 1.8C + 32. Work out F when C = 20, and identify what kind of mathematical statement F = 1.8C + 32 is.
- 6.A cinema sells adult tickets for £10 and child tickets for £6. A family group buys 9 tickets in total, spending £74. Work out how many adult tickets were bought.
- 7.A company's weekly profit, P thousand pounds, when it makes x thousand items is modelled by P = x² − 8x + 12, for x ≥ 0. Work out the values of x for which the company makes a loss.
- 8.A point has coordinates (x, y), where x + y = 0 and x is not 0. In which two quadrants could this point lie?
- 9.The first five terms of a quadratic sequence are 2, 5, 10, 17, 26. Work out the next term in the sequence.
- 10.Factorise fully 56x − 24
- 11.The curve y = −x² + 6x − 5 has a maximum point. Use completing the square to find its coordinates.y = −x² + 6x − 5
- 12.A rectangular field has length (3a + 2) metres and width (a − 1) metres. Write a simplified expression for the perimeter of the field.
- 13.A lift has a weight limit: it must carry no more than 630 kg. Let w be the total weight, in kg, carried by the lift. Which inequality describes this?
- 14.Solve the inequality 5 − x ≤ 2.
- 15.The formula connecting distance (d), speed (s) and time (t) is d = st. Make s the subject of the formula.
Answer key
- (a) The equation has one solution; the inequality has many. — Method: solve each statement. From 2x + 3 = 11, 2x = 8, so x = 4 — a single value. From 2x + 3 > 11, 2x > 8, so x > 4 — every number greater than 4 makes the inequality true, so there are many solutions. So the equation has one solution and the inequality has many. Distractor origins: swapping the two round gives the range to the equation and the single value to the inequality; saying both have exactly one solution treats the > sign as if it were an = sign; saying both have many solutions treats the equation as if it were an inequality.
- (d) x = (y + 8) / 3 — Method: add 8 to both sides, then divide the whole side by 3. Working: y = 3x − 8, so adding 8 to both sides gives y + 8 = 3x, then dividing both sides by 3 gives x = (y + 8) / 3. The value x = y + 8 / 3 is Priya's version, which comes from dividing only the 8 by 3 instead of dividing the whole expression y + 8 by 3. The value x = (y − 8) / 3 comes from a sign error, keeping −8 instead of moving it to +8. The value x = 3(y + 8) comes from multiplying by 3 instead of dividing.
- (d) (−1, 4) — In parallelogram ABCD the side DC is parallel and equal to the side AB, so D = C − AB. The vector from A to B is (2 − (−3), 1 − 1) = (5, 0), so D = (4 − 5, 4 − 0) = (−1, 4). A candidate who adds this vector to C instead of subtracting it gets (4 + 5, 4 + 0) = (9, 4). A candidate who subtracts A's coordinates from C's rather than the vector AB, and drops the minus sign on −3 while doing so, works out (4 − 3, 4 − 1) and gets (1, 3). A candidate who makes only the y-part of that slip, working out 4 − 1 instead of 4 − 0, gets (−1, 3).
- (a) 5 — A factor of the whole expression must divide every term exactly. 15x + 20 = 5(3x + 4), so 5 is a factor. Distractor origins: 3 divides 15x exactly but does not divide 20 exactly; 4 divides 20 exactly but does not divide 15x exactly; 10 also divides 20 exactly but does not divide 15x exactly, so it is a factor of only one term, not of the whole expression.
- (d) 68, and it is a formula — Substituting C = 20 gives F = 1.8 × 20 + 32 = 36 + 32 = 68. The statement F = 1.8C + 32 is a formula because it expresses a general relationship between F and C that holds for any value of C, not just C = 20. A candidate who gets the correct value but calls it an equation is confusing "has an equals sign" with "is an equation" — an equation is only true for particular value(s) of the unknown, whereas this relationship is used to calculate rather than to be solved. A candidate who forgets to multiply C by 1.8 and just adds 32 to 20 gets 52. A candidate who multiplies correctly but forgets to add 32 gets 36.
- (c) 5 — Method: let a be the number of adult tickets, so the number of child tickets is 9 − a. Form the equation 10a + 6(9 − a) = 74. Working: expand the bracket: 10a + 54 − 6a = 74, so 4a = 20, giving a = 5. Answer: 5 adult tickets. 4 comes from correctly finding the number of child tickets but reporting it instead of the number of adult tickets asked for. 1.25 comes from a sign error expanding the bracket, writing 10a + 54 + 6a = 74 instead of subtracting, which gives 16a = 20. 7.4 comes from dividing the total takings by the adult ticket price only, ignoring the children entirely, 74 ÷ 10.
- (b) 2 < x < 6 — Factorise: x² − 8x + 12 = (x − 2)(x − 6), giving roots at x = 2 and x = 6. Since the coefficient of x² is positive, the graph is a U-shape that is negative, below the axis, between its roots. So P < 0 for 2 < x < 6. Distractor routes: x < 2 or x > 6 takes the region outside the roots, which is where P is positive, a profit, the opposite of a loss. 2 ≤ x ≤ 6 includes the endpoints, where P = 0 exactly, break-even rather than a loss, since the inequality is strict. −6 < x < −2 comes from factorising as (x + 2)(x + 6), reversing the sign of both roots.
- (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.
- (c) 37 — The first differences are 3, 5, 7, 9 — they increase by 2 each time (the second difference), so the next first difference is 11, giving 26+11=37. A candidate who repeats the last first difference (9) instead of increasing it would reach 26+9=35. A candidate who increases the difference by 4 instead of 2 would reach 26+13=39. A candidate who adds only the second difference (2) to the last term, instead of the next first difference, would reach 26+2=28.
- (d) 8(7x − 3) — Method: find the highest common factor of the two terms, write it in front of a bracket and divide each term by it. Working: 56 = 8 × 7 and 24 = 8 × 3, so the highest common factor is 8; dividing gives 56x ÷ 8 = 7x and 24 ÷ 8 = 3, and the subtraction sign stays between them. Answer: 8(7x − 3), which multiplies back out to 56x − 24. The distractors: 8(7x + 3) comes from dropping the minus sign of −24 while dividing; 8(56x − 3) comes from dividing only the number term by 8 and leaving 56x untouched inside the bracket; 8(7x − 16) comes from subtracting 8 from 24 instead of dividing 24 by 8.
- (b) x = 3, y = 4 — −x² + 6x − 5 = −(x² − 6x) − 5 = −[(x − 3)² − 9] − 5 = −(x − 3)² + 9 − 5 = −(x − 3)² + 4. Because the coefficient of x² is negative, −(x − 3)² is at most zero, so this turning point is a maximum. Substituting x = 3 gives (x − 3)² = 0, so y = 4, confirming the maximum value 4 at x = 3: turning point (3, 4). Using 6 instead of half of 6 inside the bracket — forgetting to halve before completing the square — lands on turning point (6, 31), wrong, because only half the coefficient belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (−3, 4) — wrong, because (x − 3)² is zero at x = 3, not x = −3. Computing 5 − 9 = −4 instead of 9 − 5 = 4 flips the sign of the constant, giving (3, −4) — wrong, since the constant must be evaluated as 9 minus 5, not 5 minus 9. Whenever the leading coefficient is negative, the turning point is a maximum, not a minimum — check by substituting back into the original equation.
- (c) 8a + 2 — The perimeter of a rectangle is twice the length plus twice the width: P = 2(3a + 2) + 2(a − 1) = (6a + 4) + (2a − 2) = 8a + 2. Answering 5a adds the length and width once each but doubles only one of them, missing that a rectangle has two of each side. Answering 8a + 6 distributes the 2 into (a − 1) correctly as far as 2a, but then adds 2 instead of subtracting it, as though the bracket had been (a + 1). Answering 12a + 8 uses the length for all four sides instead of using the length twice and the width twice, as if the field were a square with side (3a + 2). The perimeter of the field is (8a + 2) metres.
- (d) w ≤ 630 — 'No more than 630 kg' means the weight can be exactly 630 kg or anything less, so the correct inequality is w ≤ 630, using 'less than or equal to' to include the limit itself. Writing w < 630 excludes 630 kg itself, as though the limit could not be reached exactly. Writing w ≥ 630 reverses the direction, describing a minimum weight rather than a maximum. Writing w > 630 both reverses the direction and excludes the boundary value. The inequality describing the lift's weight limit is w ≤ 630.
- (c) x ≥ 3 — Method: collect the number terms first; the x term is negative, so the final step multiplies both sides by −1, and that is the one step that turns the inequality sign round. Working: subtracting 5 from both sides of 5 − x ≤ 2 gives −x ≤ −3; multiplying both sides by −1 turns −x into x and −3 into 3, and because the multiplier is negative the ≤ becomes ≥, so x ≥ 3. Answer: x ≥ 3. The distractors: x ≤ 3 comes from multiplying by −1 without turning the sign round, the commonest slip on this type; x ≤ −3 comes from reading −x ≤ −3 as though the minus sign could simply be rubbed off the left-hand side; x ≥ −3 comes from turning the sign round correctly but leaving the right-hand side at −3 instead of multiplying it by −1 as well.
- (b) s = d / t — Method: undo the multiplication by t by dividing both sides by t. Working: d = st, so dividing both sides by t gives s = d / t. The value s = dt comes from multiplying by t instead of dividing. The value s = t / d comes from inverting the fraction, dividing t by d instead of d by t. The value s = d − t comes from subtracting t instead of dividing by it.
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