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.n is an integer. Which of these expressions is always an even number?
- 2.Harry is 14 years old and Mia is 8 years old. Will Harry ever be exactly twice as old as Mia? Give a reason for your answer.
- 3.A taxi fare, in pounds, for a journey of m miles is given by the formula F = 3 + 2.5m. A journey costs £15.50. Work out the distance travelled, m, by first rearranging the formula to make m the subject, then substituting F = 15.50.
- 4.The graph of y = f(x) has a maximum turning point at (5, 8). Which of these correctly gives the corresponding turning point on the graph of y = −f(x + 1), and its type?
- 5.A rectangle has width x cm. Its length is 3 cm more than its width. The perimeter of the rectangle is 38 cm. Form an equation and solve it to find x.
- 6.Solve the inequality x² − 5x + 6 < 0.
- 7.A point has coordinates (x, y). In which two quadrants is the product x × y positive?
- 8.The curve y = 2x² − 12x + 7 has a minimum point. By completing the square, find the value of x at which the minimum occurs.y = 2x² − 12x + 7
- 9.Describe a sequence of two transformations that maps the graph of y = x² onto the graph of y = −(x − 5)².y = x²
- 10.Solve x² − 2x − 24 = 0.
- 11.A proof sets out to show that the sum of the squares of two consecutive odd numbers, written as 2n + 1 and 2n + 3, is always 2 more than a multiple of 8. Four attempts to expand (2n + 1)² + (2n + 3)² and reach a conclusion are shown below. Which attempt correctly proves this claim?
- 12.A rectangle has length (2x + 5) cm and width (x − 2) cm. Work out an expression, in terms of x, for the perimeter of the rectangle. Give your answer in its simplest form.
- 13.The first five terms of a quadratic sequence are 4, 7, 12, 19, 28. Work out an expression, in terms of n, for the nth term.
- 14.Work out the coordinates of the reflection of (6, −3) in the y-axis.
- 15.Three consecutive integers add up to 72. Using n for the smallest integer, form an equation and solve it to find the smallest of the three integers.
Answer key
- (a) n² + n — n² + n = n(n + 1), the product of two consecutive integers. One of any two consecutive integers is always even, so their product is always even, whatever whole number n is. n² − n + 1 = n(n − 1) + 1 is always ODD, since n(n − 1) is even and adding 1 makes it odd — the opposite of what's asked. 2n + 1 is always odd by definition, not even. n² + 1 is not always even at all: it depends on whether n is odd or even, and testing n = 2 gives 5, which is odd.
- (c) No — that moment has already passed — Method: call the number of years from now x, add x to both ages, form the equation from the comparison and then interpret the value of x that comes out. Working: in x years Harry will be 14 + x and Mia will be 8 + x, so 14 + x = 2(8 + x); expanding gives 14 + x = 16 + 2x, and subtracting x and 16 from both sides gives x = −2. A negative value of x places the moment two years in the past, when Harry was 12 and Mia was 6 and 12 = 2 × 6, so it is not something still to come. Answer: no — that moment has already passed. The distractors: the claim that it has never happened and never will comes from reaching x = −2 and reading a negative number of years as no solution at all, when x = −2 does not say that no such moment exists but says where it is — two years before now; the claim that it happens when Harry is 16 comes from doubling Mia's present age, 2 × 8 = 16, and reading that as the age Harry has to reach; the claim that it happens when Harry is 20 comes from expanding 2(8 + x) as 8 + 2x, doubling only the x, which gives x = 6.
- (d) 5 — Method: rearrange the formula to make m the subject, then substitute F = 15.50. Working: F = 3 + 2.5m, so subtracting 3 from both sides gives F − 3 = 2.5m, then dividing by 2.5 gives m = (F − 3) / 2.5. Substituting F = 15.50: m = (15.50 − 3) / 2.5 = 12.50 / 2.5 = 5. The value 6.2 comes from dividing 15.50 by 2.5 without subtracting the fixed £3 first. The value 3.2 comes from dividing first and subtracting afterwards, in the wrong order: (15.50 / 2.5) − 3 = 3.2. The value 7.4 comes from adding £3 instead of subtracting it before dividing: (15.50 + 3) / 2.5 = 7.4.
- (a) (4, −8), a minimum point — The transformation x → x + 1 inside f translates the graph 1 unit to the LEFT, so the x-coordinate becomes 5 − 1 = 4. The minus sign in front of f reflects the graph in the x-axis, so the y-coordinate becomes −8, and a reflection in the x-axis turns every maximum into a minimum, so (4, −8) is a minimum point. Writing '(4, −8), a maximum point' gets the coordinates right but forgets that a reflection in the x-axis swaps maximum and minimum points. Writing '(6, −8), a minimum point' comes from translating 1 unit to the RIGHT instead of the left — f(x + 1) always moves the graph in the negative x-direction. Writing '(4, 8), a minimum point' gets the x-coordinate and the type right, but forgets to actually negate the y-coordinate, even though it does correctly reclassify the point as a minimum.
- (b) 8 — The perimeter is 2(x + (x + 3)) = 4x + 6, so 4x + 6 = 38, which gives 4x = 32 and x = 8. A candidate who forgets the '+3' and treats the rectangle as a square, solving 2(2x) = 38, gets x = 9.5. A candidate who forgets to double the sum of the sides, solving 2x + 3 = 38, gets x = 17.5. A candidate who uses 3x instead of x + 3 for the length, solving 2(x + 3x) = 38, gets x = 4.75.
- (c) 2 < x < 3 — Factorise x² − 5x + 6 = (x − 2)(x − 3), giving roots x = 2 and x = 3. Since the coefficient of x² is positive, the graph is a U-shape that dips below the x-axis between its roots. So x² − 5x + 6 < 0 for 2 < x < 3. Distractor routes: x < 2 or x > 3 takes the region OUTSIDE the roots, where the graph is above the x-axis, the opposite of what is wanted. 2 ≤ x ≤ 3 uses ≤ instead of the strict < the question asks for, wrongly including the roots themselves, where the expression equals zero, not less than zero. −3 < x < −2 comes from factorising as (x + 2)(x + 3), reversing the sign of both roots.
- (b) the first and the third — Method: the sign of a product depends only on the signs of the two numbers multiplied: like signs give a positive product and unlike signs a negative one, so look for the quadrants in which both coordinates carry the same sign. Working: in the first quadrant x and y are both positive, and positive times positive is positive; in the third quadrant both are negative, and negative times negative is positive as well; in the second quadrant x is negative while y is positive, and in the fourth x is positive while y is negative, so each of those gives a negative product. Answer: the first and the third. The distractors: 'the second and the fourth' comes from applying the rule for a NEGATIVE product, unlike signs, to a positive one; 'the first and the second' comes from testing only the y-coordinate and keeping the quadrants where the height is positive; 'the first and the fourth' comes from testing only the x-coordinate in the same way.
- (d) 3 — 2x² − 12x + 7 rewrites as 2(x² − 6x) + 7, then as 2[(x − 3)² − 9] + 7, which simplifies to 2(x − 3)² − 11, since −2 × 9 + 7 = −11. The bracket (x − 3)² is zero when x = 3, so the minimum occurs at x = 3. Treating the shift as b/a instead of b/(2a) — using 6 instead of 3 — gives x = 6, which is wrong. Reading the bracket's sign directly without negating it gives x = −3, wrong, because (x − 3)² is zero at x = 3, not x = −3. Reading off the coefficient of x itself, −12, and calling that the answer skips the completing-the-square process entirely and gives x = −12, which is wrong because b is not the turning point's x-coordinate under any circumstance. Always check: substituting your value of x should make the bracketed term equal to zero, and nothing else.
- (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.
- (a) x = 6 or x = −4 — We need two numbers that multiply to −24 and add to −2: these are −6 and 4, since −6 × 4 = −24 and −6 + 4 = −2. So x² − 2x − 24 = (x − 6)(x + 4) = 0, giving x = 6 or x = −4. A candidate who swaps the signs, using 6 and −4 the wrong way round in the brackets, gets x = −6 or x = 4. A candidate who picks the wrong factor pair, 8 and −3 (which multiply to −24 but add to +5, not −2), gets (x + 8)(x − 3) = 0 and answers x = −8 or x = 3. A candidate who makes both factors negative gets x = −6 or x = −4, which would require the constant term to be +24, not −24.
- (c) (2n + 1)² + (2n + 3)² = (4n² + 4n + 1) + (4n² + 12n + 9) = 8n² + 16n + 10 = 8(n² + 2n + 1) + 2, and n² + 2n + 1 is an integer, so the sum is always 2 more than a multiple of 8. — Expand each square carefully: (2n + 1)² = 4n² + 4n + 1 and (2n + 3)² = 4n² + 12n + 9, since the cross term is 2 × 2n × 3 = 12n. Adding gives 8n² + 16n + 10, and factorising out 8 from every term that can hold one gives 8(n² + 2n + 1) + 2; since n² + 2n + 1 is always an integer, the sum is always 2 more than a multiple of 8. The attempt reaching 8(n² + 2n) + 10 has the correct expansion but stops the factorisation one step early — it never pulls a further 8 out of the 10 (10 = 8 + 2), so 'always 10 more than a multiple of 8' should be reduced to 'always 2 more than a multiple of 8'. The attempt reaching 2(4n² + 8n + 5) also has the correct expansion, and the factorisation is true, but 'always even' only shows the sum is a multiple of 2 — being even is necessary but nowhere near sufficient to be a multiple of 8, and the argument never finds the extra factor of 4. The fourth attempt makes an expansion slip, using (2n + 3)² = 4n² + 9 instead of 4n² + 12n + 9 — dropping the 12n cross term entirely — so it works from the wrong expression 8n² + 4n + 10 throughout, and no amount of correct working afterwards can recover the right conclusion.
- (b) 6x + 6 — Perimeter = 2[(2x + 5) + (x − 2)] = 2(3x + 3) = 6x + 6. A candidate who adds the length and width but forgets to double for the perimeter gets 3x + 3. A candidate who makes a sign error and adds 2 instead of subtracting it before doubling gets 2[(2x + 5) + (x + 2)] = 6x + 14. A candidate who multiplies the length and width instead of adding them, confusing the perimeter formula with the area formula, and then doubles that product, gets 2(2x + 5)(x − 2) = 4x² + 2x − 20.
- (d) n² + 3 — First differences: 3, 5, 7, 9. Second differences: 2, 2, 2, so the sequence is quadratic and the coefficient of n² is half the second difference: a = 2 ÷ 2 = 1. Subtracting n² (1, 4, 9, 16, 25) from the terms (4, 7, 12, 19, 28) leaves 3, 3, 3, 3, 3, a constant, so the nth term is n² + 3. Using the second difference itself as a, without halving it, gives 2n² + 3. Finding a = 1 correctly but then dropping the constant remainder gives n². Treating the first first difference (3) as a common difference and building a linear formula a + (n − 1)d = 4 + 3(n − 1) gives 3n + 1, which fits only the first term.
- (a) (−6, −3) — Reflecting in the y-axis changes the sign of the x-coordinate and keeps the y-coordinate the same: (−6, −3). (6, 3) comes from reflecting in the x-axis instead, which changes the sign of the y-coordinate. (−6, 3) comes from reflecting in both axes. (6, −3) comes from not applying the reflection at all.
- (d) 23 — Method: let the smallest integer be n, so the three consecutive integers are n, n + 1 and n + 2. Form the equation n + (n + 1) + (n + 2) = 72. Working: simplify the left side: 3n + 3 = 72, so 3n = 69, giving n = 23. Answer: 23. 24 comes from dividing 72 by 3 directly, 72 ÷ 3 = 24, which finds the middle integer rather than realising the three numbers differ. 25 comes from correctly finding n = 23 but reading off the largest integer, n + 2, instead of the smallest as asked. 21 comes from dividing first and subtracting after, (72 ÷ 3) − 3, instead of subtracting 3 before dividing by 3.
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