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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Answer key: Algebra worksheet — GCSE Higher
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- (a) 18 — The common ratio is 2√3 ÷ 2 = √3. Checking: 6 ÷ 2√3 = √3 and 6√3 ÷ 6 = √3, so the ratio is consistent throughout. The next term is 6√3 × √3 = 6 × 3 = 18, since √3 × √3 = 3. Looking only at the coefficients 2, 2, 6, 6 and continuing them by doubling the last one gives 6 × 2 = 12, which is wrong because the step from each term to the next is a multiplication by √3, not a pattern in the coefficients alone. Doubling the previous term instead of multiplying by the surd ratio √3 gives 6√3 × 2 = 12√3, which is wrong because the common ratio is √3, not 2. Using 3 instead of √3 as the common ratio — squaring the true ratio by mistake — gives 6√3 × 3 = 18√3, which is wrong because 3 is the SQUARE of the common ratio, not the ratio itself.
- (a) 18 — Method: substitute the value, apply the index before the multiplication, and remember that a negative number multiplied by itself gives a positive result. Working: x² = (−3) × (−3) = 9, and then 2 × 9 = 18. Answer: 18. The distractors: −18 comes from squaring only the 3 and leaving the minus sign outside the index, giving 2 × (−9); 36 comes from multiplying 2 by −3 first and squaring afterwards, giving (−6)²; −12 comes from reading x² as 2x, so that the calculation becomes 2 × 2 × (−3).
- (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.
- (c) y = 200 − 8x — Method: in a linear model the amount present at the start is the constant term and the steady rate of change is the gradient, which is negative because the amount is falling. Working: at x = 0 minutes there are 200 litres, so the constant term is 200; 8 litres are lost every minute, so after x minutes 8x litres have gone and the amount left is y = 200 − 8x. Answer: y = 200 − 8x. The distractors: y = 8x + 200 treats the draining as filling, so the butt would gain 8 litres a minute; y = 200x − 8 swaps the two numbers over, using the starting 200 litres as the rate per minute and the 8 litres per minute as the starting amount; y = −8x − 200 makes the starting amount negative as well as the rate, so the butt would begin 200 litres in deficit.
- (a) 2 — The common ratio is 54 ÷ 162 = 1/3, confirmed by 18 ÷ 54 = 1/3 and 6 ÷ 18 = 1/3. The next term is 6 × 1/3 = 2. A candidate who wrongly treats the sequence as arithmetic and subtracts the last difference (18 − 6 = 12) again gets 6 − 12 = −6. A candidate who multiplies by 3 instead of dividing, effectively reversing the sequence, gets 18. A candidate who guesses the ratio is 1/2 gets 6 × 1/2 = 3.
- (a) 18 cm² — Method: substitute the width into the formula, applying the index to the letter before multiplying by the 2 in front of it. Working: x² = 3 × 3 = 9, and then A = 2 × 9 = 18, so the area is 18 cm². Answer: 18 cm². The distractors: 36 cm² comes from multiplying 2 by 3 first and squaring afterwards, giving (2 × 3)²; 12 cm² comes from doubling instead of squaring, so that x² is replaced by 2x and the calculation becomes 2 × 2 × 3; 6 cm² comes from working out 2 × 3 and never applying the index at all.
- (b) (3x + 1)(2x − 3) — To factorise 6x² − 7x − 3, look for two numbers that multiply to 6 × (−3) = −18 and add to −7: these are −9 and 2. Splitting the middle term gives 6x² − 9x + 2x − 3, which groups to 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3). Writing (3x − 1)(2x + 3) comes from using the pair 9 and −2 instead — the right product but the wrong signs, giving +7x instead of −7x. Writing (6x − 1)(x + 3) comes from picking the factor pair 18 and −1: it multiplies to −18 correctly, but it adds to 17, not −7, so checking only the product splits the middle term as 6x² + 18x − x − 3 and pairs the wrong factors of 6 and 3 together. Writing (x − 3)(6x + 1) comes from that same unchecked pair used the other way round, splitting the middle term as 6x² + x − 18x − 3.
- (a) 3x/2 — Factorise top and bottom first: 6x³ − 15x² = 3x²(2x − 5), and 4x² − 10x = 2x(2x − 5). The bracket (2x − 5) is common to both, so it cancels, leaving 3x²/(2x); dividing the power of x, 3x² ÷ x = 3x, gives 3x/2. Writing 3x²/2 cancels the (2x − 5) correctly and removes the denominator's x, but never reduces the power of x left in the numerator — 3x² ÷ x should give 3x, not stay as 3x². Writing 3/2 cancels the (2x − 5) correctly but then drops the x from the numerator altogether, treating 3x² over x as if it cancelled completely to 3 instead of reducing to 3x. Writing −3x/2 comes from factorising the denominator with the wrong sign, as 2x(5 − 2x) instead of 2x(2x − 5); cancelling (5 − 2x) against the numerator's (2x − 5) then needs an extra minus sign, which flips the answer to −3x/2.
- (a) 33 — Substitute n=4 into 2n²+1: 2×4²+1=2×16+1=33. A candidate who computes n² as 2×n instead of n×n would compute 2×(2×4)+1=2×8+1=17. A candidate who correctly finds 2×16 but forgets to add the constant 1 would stop at 32. A candidate who squares the whole term 2n, rather than squaring n before multiplying by 2, would compute (2×4)²+1=64+1=65.
- (c) The object was stationary (not moving) — On a distance-time graph, the gradient at any point represents the speed at that point. A gradient of 0 means distance is not changing over time, so the object is stationary. A straight, sloped line (not flat) shows constant nonzero speed; a curve bending one way shows acceleration or deceleration; a flat section is not a maximum speed — it is no speed at all. Read the shape of the graph, not just how steep it looks.
- (b) 3 — Method: a term is any part of an expression separated from the rest by a + or − sign, including a number on its own. Working: 8 − 3x + 5x² splits at the + and − signs into 8, −3x and 5x² — that is three separate terms. Answer: 3. Ryan's answer of 2 comes from wrongly excluding the number 8, thinking a term must contain a letter. 5 comes from miscounting by treating the coefficients and powers as separate terms as well as the letters. 1 comes from treating the whole expression as a single term because it is written without brackets.
- (a) −3/4 — Method: the tangent at a point on a circle is perpendicular to the radius drawn to that point, so find the gradient of the radius and then take its negative reciprocal. Working: the radius joins (0, 0) to (6, 8), so its gradient is 8 ÷ 6, which cancels to 4/3. Turning 4/3 upside down gives 3/4, and changing the sign gives −3/4. Answer: the gradient of the tangent is −3/4. The distractors: 4/3 is the gradient of the radius itself, quoted without taking the perpendicular at all; −4/3 changes the sign but leaves the fraction the same way up, so the two gradients do not multiply to give −1; 3/4 turns the fraction upside down but keeps it positive, which is the other half of the same rule left undone.
- (a) 5x − 3 = 12 — 5x − 3 = 12 is an equation with exactly one solution: adding 3 and dividing by 5 gives x = 3, and no other value works. 5x − 3 = 5x − 3 is true for every value of x, since both sides are identical — it has infinitely many solutions, not one. 5x − 3 > 12 is an inequality: any value of x greater than 3 satisfies it, so it has a whole range of solutions, not a single one. 5x − 3 = 5x + 2 has no solution at all, since subtracting 5x from both sides leaves −3 = 2, which is never true. The equation with exactly one solution is 5x − 3 = 12.
- (d) 7n − 2 — Method: find how much the total cost rises each month, then find the constant that fits the cost for one month. Working: the cost rises by £7 for each extra month (12 − 5 = 7, 19 − 12 = 7, 26 − 19 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 5, so c = −2. Answer: the total cost in pounds is 7n − 2. The value 7n comes from leaving out the constant. The value 7n + 5 comes from using the cost of one month as the constant directly, without subtracting the monthly rise first. The value 5n + 7 comes from swapping the roles of the cost of one month, £5, and the monthly rise, £7 — using the cost of one month as the coefficient of n and the rise as the constant.
- (b) y = f(−x) — y = f(−x) reflects the graph of y = f(x) in the y-axis: every x-coordinate changes sign. The x-intercepts −3 and 7 become 3 and −7, matching the second graph's intercepts −7 and 3. A point already on the y-axis is unaffected, since −0 = 0, so the y-intercept (0, 21) stays exactly where it is — matching the second graph as well. y = −f(x) leaves the x-intercepts unchanged at −3 and 7, since f(x) = 0 exactly where −f(x) = 0, which does not match; it also sends the y-intercept to (0, −21), a second mismatch. y = −f(−x) does send the x-intercepts to the right places, −7 and 3, but it sends the y-intercept to (0, −21) instead of (0, 21), so it fails the second clue. y = f(x) − 4 moves every point down 4, sending the y-intercept to (0, 17) instead of (0, 21), so it fails the y-axis clue. Test each option against BOTH clues — the pair of x-intercepts and the point on the y-axis — because more than one option gets only one of the two right.
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