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) P lies outside the circle — The squared distance from the origin to P is 8² + 9² = 64 + 81 = 145, which is greater than 100, so P lies outside the circle. Choosing 'P lies inside the circle' comes from adding the coordinates directly (8 + 9 = 17) instead of squaring and adding them, and comparing 17 with 100. Choosing 'P lies on the circle' comes from squaring 9 incorrectly as 36 (as if it were 6²) instead of 81, giving 64 + 36 = 100, equal to the right-hand side. Choosing 'Cannot be determined' wrongly assumes the equation of the line itself is needed, when the radius and the distance from the origin to P are already enough.
- (d) 168 m — Split the area under the graph into two parts. The rising section (0 to 8 s) is a triangle: area = 0.5 × 8 × 12 = 48. The constant section, from 8 s to 18 s (10 s), is a rectangle: area = 12 × 10 = 120. Total distance = 48 + 120 = 168 m. Leaving out the 0.5 doubles the triangle, giving 8 × 12 + 120 = 216 m; ignoring the triangle section altogether gives only 120 m; averaging the start and final speeds over the whole 18 seconds, (0 + 12) ÷ 2 = 6, then 6 × 18 = 108 m, wrongly treats the tram as accelerating the whole time, when the graph is flat for the last 10 seconds.
- (a) 52 — Test successive terms: n=6 gives 6²+3=39, which is not greater than 50. n=7 gives 7²+3=52, which is greater than 50, so the first term greater than 50 is 52. A candidate who stops at n=6, before checking whether 39 actually exceeds 50, would give 39. A candidate who solves n²>50 instead of n²+3>50, ignoring the +3 in the search, would find n=8 is the first value with n²>50 (since 7²=49) and compute 8²+3=67. A candidate who computes n² by doubling n instead of squaring it would compute 2×7+3=17.
- (b) x = 7, y = 3 — Method: one equation contains +y and the other −y, so adding them removes y; the value found is then substituted back to get the other letter. Working: adding x + y = 10 and x − y = 4 gives 2x = 14, so x = 7; substituting into x + y = 10 gives 7 + y = 10, so y = 3. Answer: x = 7, y = 3, and 7 − 3 = 4 as required. The distractors: x = 7, y = 4 comes from finding x correctly and then taking the 4 in x − y = 4 to be the value of y; x = 5, y = 5 comes from splitting the total of 10 equally and never using the difference; x = 14, y = −4 comes from adding the equations to 2x = 14 and forgetting to halve, so that x is taken as 14 and y as 10 − 14.
- (a) They always charge the same, since 3(2n + 4) = 6n + 12. — Expand Advert A's formula by multiplying both terms inside the bracket by 3: 3 × 2n = 6n, and 3 × 4 = 12, giving 3(2n + 4) = 6n + 12, which is identical to Advert B's formula — so the two adverts always charge the same amount, whatever n is. Getting 6n + 4 comes from multiplying the 2n by 3 but leaving the 4 unmultiplied. Getting 2n + 7 comes from adding 3 to the bracket instead of multiplying by it. Saying it depends on n avoids expanding the bracket at all — once expanded, both formulas are identical for every value of n, so the cost can be compared directly.
- (c) x + 2y = 6 — L has gradient 2, so the perpendicular gradient is −1/2. Substituting (4, 1) into y − 1 = −(1/2)(x − 4): y = −(1/2)x + 2 + 1 = −(1/2)x + 3, so 2y = −x + 6, giving x + 2y = 6. Distractor routes: x − 2y = 2 comes from using gradient +1/2 instead of −1/2, forgetting the negative sign the perpendicular rule needs. x + 2y = 9 comes from swapping the point's coordinates, substituting (1, 4) instead of (4, 1). x + 2y = −2 comes from a sign error distributing the negative gradient over the bracket, computing −(1/2)(x − 4) as −(1/2)x − 2 instead of −(1/2)x + 2.
- (d) t = 5 — The maximum height occurs halfway between the two times when the ball is at ground level: the midpoint of t = 1 and t = 9 is (1 + 9) ÷ 2 = 5, so the ball reaches its maximum height at t = 5 seconds. Getting t = 4 comes from halving the DIFFERENCE between the times, 9 − 1 = 8, then 8 ÷ 2 = 4, instead of finding their midpoint. Getting t = 8 uses that difference, 9 − 1 = 8, as if the gap between the two times were itself the time of the maximum. Getting t = 10 adds the two times, 1 + 9 = 10, but forgets to divide by 2.
- (d) 110 — Method: form the equation 15 + 0.08m = 23.80, where m is the number of minutes, then solve for m. Working: subtract the fixed fee: 0.08m = 23.80 − 15 = 8.80. Divide by the cost per minute: m = 8.80 ÷ 0.08 = 110. Answer: 110 minutes. 1.1 comes from using 8 instead of 0.08 as the cost per minute, forgetting to convert pence to pounds. 297.5 comes from dividing the whole bill by the cost per minute without subtracting the fixed fee first, 23.80 ÷ 0.08. 485 comes from adding the fixed fee to the bill instead of subtracting it, before dividing by the cost per minute, (23.80 + 15) ÷ 0.08.
- (d) An equation, true for one value of x — 2x + 5 = 17 is only true when x = 6, so it is an equation. A candidate who ignores the equals sign and focuses only on the x terms and numbers present picks the expression option. A candidate who sees an equals sign and assumes it must relate two different quantities, as a formula does, picks that option, missing that there is only one quantity, x, involved here. A candidate who wrongly checks whether both sides look similar in structure rather than testing specific values picks the identity option, missing that the two sides are not equal for every value of x.
- (b) $y = x^3 - 4x$ — Method: count how many times the curve crosses the x-axis and check whether it is a cubic (an S-shape with up to three crossing points) rather than a lower power, and note which way it runs overall from bottom-left to top-right or the reverse. Working: the curve crosses the x-axis at three points, x = −2, 0 and 2, and runs from bottom-left to top-right, which matches y = x³ − 4x = x(x − 2)(x + 2). Answer: y = x³ − 4x. Distractor refutation: y = −x³ + 4x comes from a sign error on every term, which would flip the curve so it ran from top-left to bottom-right instead. y = x³ + 4x comes from a sign error on the x term only, which removes two of the three crossing points, since x(x² + 4) has only x = 0 as a real root. y = x² − 4x comes from dropping the cubic term altogether, giving a parabola with only two crossing points instead of an S-shaped curve with three.
- (b) −3 — Method: the gradient of a straight line is the change in y divided by the change in x, with the two coordinates taken in the same order in the numerator as in the denominator. Working: going from (−1, 5) to (3, −7), the change in y is −7 − 5 = −12 and the change in x is 3 − (−1) = 4, so the gradient is −12 ÷ 4 = −3. Answer: −3. The distractors: 3 comes from subtracting the y-coordinates in one order and the x-coordinates in the other, giving 12 ÷ 4; −1/3 comes from dividing the change in x by the change in y instead of the other way round, giving 4 ÷ (−12); −6 comes from working out 3 − (−1) as 3 − 1 = 2, so that the change in y is divided by 2 rather than by 4.
- (d) 13 — Method: the numbers of matchsticks form a sequence with a term-to-term rule, so count the first square in full and then add the repeated amount once for every extra square. Working: one square uses 4 matchsticks; a row of 4 squares has 3 extra squares after the first, and each of those adds 3 matchsticks, giving 3 × 3 = 9 to add on to the 4. Answer: 13. The distractors: 16 comes from counting each square as a separate set of 4 matchsticks, 4 × 4, and ignoring the shared sides; 12 comes from using 3 matchsticks for all four squares, 3 × 4, and forgetting that the first square needs a fourth side; 10 comes from adding the 3 only twice, as though a row of four squares had two extra squares rather than three.
- (b) 1215 — Method: generate the terms one at a time with the term-to-term rule and compare each with 1000 as you go, stopping at the first one that passes it. Working: the terms are 5, then 5 × 3 = 15, then 45, then 135, then 405, and 405 × 3 = 1215; 405 is still below 1000 while 1215 is above it. Answer: 1215. The distractors: 405 comes from stopping at the last term that is still below 1000 instead of giving the first one above it; 3645 comes from carrying on one term too far, past the first term that passes 1000; 2187 comes from using the multiplier 3 as the first term as well, generating 3, 9, 27, 81, 243, 729, 2187 instead of the sequence described.
- (a) 5p + 7q — Tom's total spend combines like terms: (2p + 6q) + (3p + q) = 5p + 7q, adding the notebook terms (2p + 3p = 5p) and the pencil terms (6q + q = 7q) separately. Answering 12pq adds every coefficient together (2 + 6 + 3 + 1 = 12) and multiplies the letters, combining unlike terms as though notebooks and pencils were the same item. Answering 5p + 6q correctly combines the notebook terms but forgets to add Tuesday's extra pencil to the 6q. Answering 2p + 7q correctly combines the pencil terms but forgets to add Tuesday's 3 extra notebooks to the 2p. The total amount Tom has spent is 5p + 7q pence.
- (c) x + 3 — Expand the bracket: 0.5(4x + 6) = 2x + 3. Then subtract the x: 2x + 3 − x = x + 3. The option 2x + 3 comes from expanding the bracket correctly but then forgetting to subtract the x at all. The option x + 6 comes from forgetting to multiply the 6 inside the bracket by 0.5 (treating it as 2x + 6), then subtracting x. The option 3x + 3 comes from adding the x instead of subtracting it: 2x + 3 + x = 3x + 3.
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