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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- (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).
- (b) y = 2x² + 14x + 19 — A translation by the vector (−2, 0) moves the graph 2 units in the negative x-direction, which means replacing x with (x + 2): 2(x + 2)² + 6(x + 2) − 1. Expanding 2(x + 2)² gives 2x² + 8x + 8, and 6(x + 2) gives 6x + 12; collecting terms 2x² + 8x + 8 + 6x + 12 − 1 simplifies to 2x² + 14x + 19. Substituting (x − 2) instead of (x + 2) — translating in the wrong direction — gives 2(x−2)² + 6(x−2) − 1, which simplifies to 2x² − 2x − 5. Distributing the outer 2 only onto the x² term of the expansion, instead of onto every term of (x + 2)², gives 2x² + 10x + 15. Dropping the −1 when collecting the constant terms, treating 8 + 12 as 20 rather than 8 + 12 − 1 as 19, gives 2x² + 14x + 20.
- (d) Below y = x + 1, below x + y = 5, above y = 0 — For y ≤ x + 1, R lies on or below the line y = x + 1. For x + y ≤ 5 (that is, y ≤ 5 − x), R lies on or below that line too. For y ≥ 0, R lies on or above the x-axis. Combining all three: R is below y = x + 1, below x + y = 5, and above y = 0. Distractor routes: "Above y = x + 1" flips the first inequality, describing the wrong side of that line. "Above x + y = 5" flips the second inequality, describing the wrong side of that line. "Below y = 0" flips the third inequality, describing the wrong side of the x-axis.
- (d) 20 km/h — Method: total distance = 9 + 21 = 30 km. Total time = 15 + 45 + 30 = 90 minutes = 1.5 hours. Average speed = total distance ÷ total time = 30 ÷ 1.5 = 20 km/h. Distractor origins: 40 km/h ignores the 45-minute stop, dividing the total distance by the travel time only (30 km ÷ 45 minutes = 40 km/h); 39 km/h averages the two section speeds, 36 km/h and 42 km/h, instead of using total distance over total time; 36 km/h uses only the first section's speed (9 km in 15 minutes), ignoring the second section and the stop.
- (b) x = y/3 − 2 — The bracket containing x has been multiplied by 3, so divide both sides by 3 first, which gives y/3 = x + 2. Subtracting 2 from both sides then leaves y/3 − 2 = x, so x = y/3 − 2. Taking the 2 away before dividing gives (y − 2)/3, which divides the 2 by 3 as well, although the 2 was never divided in the original formula. Adding 2 rather than subtracting it gives y/3 + 2, and multiplying by 3 instead of dividing gives 3y − 2.
- (a) Outside: OQ² = 221 > r² = 169 — OQ² = 10² + 11² = 100 + 121 = 221. Comparing this with r² = 169: since 221 > 169, OQ > r, so Q lies outside the circle — this is the correct verdict AND the correct working. 'Outside: OQ = 21 (10 + 11) > r = 13' reaches the same Outside verdict, but by invalid working: it adds the coordinates instead of squaring them (10 + 11 = 21, rather than 10² + 11² = 221), so the stated 'OQ' of 21 is not a distance at all — the verdict happens to match, but the method is wrong. 'Inside: OQ ≈ 14.87 < r² = 169' correctly finds the distance OQ = √221 ≈ 14.87, but then compares that DISTANCE with r² = 169 instead of with r = 13 — comparing two different kinds of quantity gives a meaningless, and here wrong, verdict. 'Inside: OQ² = 221 < (2r)² = 676' confuses the radius with the diameter: it compares OQ² with the diameter squared, (2 × 13)² = 676, instead of with r² = 169.
- (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) £14.50 — Method: the first 200 minutes are already covered by the £12, so only 250 − 200 = 50 minutes are charged extra. Extra charge = 50 × £0.05 = £2.50. Total cost = £12 + £2.50 = £14.50. Distractor origins: £24.50 charges 5p for all 250 minutes instead of only the 50 minutes over 200 (250 × £0.05 = £12.50, plus £12 = £24.50); £12.50 makes that same slip of charging all 250 minutes but then forgets to add the £12 monthly fee; £13.50 works out 250 − 200 wrongly as 30 extra minutes instead of 50 (30 × £0.05 = £1.50, plus £12 = £13.50).
- (a) n² + 4n + 4 — The whole patio-plus-border square has side length (n + 2), so the total is (n + 2)². Expanding this bracket correctly gives n² + 4n + 4, which matches 9, 16, 25, 36 for n = 1, 2, 3, 4. Expanding (n + 2)² by squaring each term separately, as if (a + b)² = a² + b², gives n² + 4, which is already wrong at n = 1 (it gives 5, not 9). Multiplying out (n + 2)(n + 2) as n² + 2n + 2n but forgetting the final 2 × 2 gives n² + 4n, which is 5 short at every value of n. Counting the patio twice — once as the n² inner square and again inside the (n + 2)² total — gives n² + (n² + 4n + 4) = 2n² + 4n + 4.
- (d) 8 — Method: write the nth term of the sequence, 2000 + 800(n − 1), and find the smallest whole n for which it is greater than 7000. Working: 2000 + 800(n − 1) > 7000, so 800(n − 1) > 5000, giving n − 1 > 6.25. Since n − 1 must be a whole number, the smallest value is 7, so n = 8. Check: year 8's total is 2000 + 800 × 7 = 7600, which is more than £7000, while year 7's total is 2000 + 800 × 6 = 6800, which is not. Answer: year 8. 7 comes from rounding 6.25 to the nearest whole number, 6, and then adding 1, instead of rounding up to the next whole number before adding 1. 6 comes from using 6.25 rounded down to 6 as the year number directly, without adding the 1 needed to convert from the number of increases to the year number. 9 comes from adding one extra year beyond the year that already satisfies the condition.
- (d) x + y = 7 — The midpoint of AB is (−1 + 5)/2, (2 + 8)/2, which is (2, 5). The gradient of AB is (8 − 2)/(5 − (−1)) = 6/6 = 1. The perpendicular bisector has gradient −1 and passes through (2, 5): y − 5 = −(x − 2), which rearranges to x + y = 7. Distractor routes: x − y = −3 uses the gradient of AB itself, 1, rather than its negative reciprocal, giving a line PARALLEL to AB through the midpoint instead of perpendicular to it. x + y = 12 comes from adding the y-coordinates, 2 + 8 = 10, but forgetting to divide by 2, using the midpoint (2, 10) instead of (2, 5). x + y = 4 comes from using A's x-coordinate, −1, directly instead of averaging it with B's, giving the point (−1, 5) instead of the true midpoint (2, 5).
- (a) x = 2, x = 7; y-intercept cannot be found here — Translating y = f(x) to y = f(x − 3) shifts the graph 3 units to the right, so each root increases by 3: x = −1 becomes x = 2, and x = 4 becomes x = 7. The y-intercept is the value at x = 0, which for this new graph is f(0 − 3) = f(−3) — and f(−3) is not one of the values given, so the new y-intercept cannot be worked out from the information given. Writing 'y-intercept stays at (0, −8)' wrongly assumes a horizontal translation leaves the y-intercept unchanged — it generally does not, since it moves the whole graph sideways, including the point that used to sit on the y-axis. Writing roots at x = −4 and x = 1 comes from translating 3 units to the LEFT instead of to the right — f(x − 3) shifts the graph in the positive x-direction, not the negative direction.
- (d) x = 4 or x = −4 — Method: divide both sides by 2 to get x² = 16, then take the square root of both sides: x = 4 or x = −4. Distractor origins: x = 4 forgets the negative root; x = 8 comes from halving 16 instead of taking its square root; x = 16 stops at x² = 16 without ever taking the square root.
- (b) x = −3, y = −7 — x² + 6x + 2 = (x + 3)² − 3² + 2 = (x + 3)² − 7. Substituting x = −3: (−3)² = 9, 6 × (−3) = −18, so 9 − 18 + 2 = −7, confirming the minimum value −7 at x = −3: turning point (−3, −7). Using 6 instead of half of 6 inside the bracket gives (x + 6)² − 34, turning point (−6, −34) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (3, −7) — wrong, because (x + 3)² is zero at x = −3, not x = 3. Computing 9 − 2 = 7 instead of 2 − 9 = −7 flips the sign of the constant, giving (−3, 7) — wrong, since the completed square's constant must be evaluated as 2 minus 9, not 9 minus 2. Always check a turning point by substituting its x-value back into the original equation.
- (a) 40 — From 9am to 12 noon is 3 hours, so the population doubles three times: 5 × 2³ = 40. A candidate who counts the elapsed time as 4 hours (an off-by-one counting error) would reach 5 × 2⁴ = 80. A candidate who counts it as only 2 hours would reach 5 × 2² = 20. A candidate who misreads 'doubles' as 'increases by 2' each hour would compute 5 + 3 × 2 = 11.
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