O Level Mathematics (4024) — MCQs & Practice Questions

A rational number is any number that can be written in the form p/q where p and q are integers and q ≠ 0. Example: −3/4.

Yes. The real numbers are defined as the rationals together with the irrationals, so every irrational number is real. The converse is false: not every real number is irrational.

(i) Associative property of addition. (ii) Multiplicative identity.

1. Let x = 0.3636… and multiply by 100 because the repeating block has two digits: 100x = 36.3636…the power of 10 must match the length of the repeating block[1]
2. Subtract: 100x − x = 36.3636… − 0.3636… gives 99x = 36every decimal place cancels[1]
3. x = 36/99 = 4/11lowest terms required; 36/99 alone does not get the final mark[1]

4/11

1. Multiply numerator and denominator by the conjugate √7 + √3the same conjugate top and bottom, so the value is unchanged[1]
2. Denominator: (√7 − √3)(√7 + √3) = 7 − 3 = 4difference of two squares[1]
3. Numerator: (√7 + √3)² = 7 + 2√21 + 3 = 10 + 2√21the middle term is 2√(7×3), not √21[1]
4. (10 + 2√21)/4 = (5 + √21)/2cancel the factor 2 for the final mark[1]

(5 + √21) / 2

1. √8 = √(4 × 2) = 2√2extracting the perfect square factor[1]
2. √2 + √8 = √2 + 2√2 = 3√2[1]
3. 3√2 = √(9 × 2) = √18, and √18 ≠ √10or compare decimals: 4.243 against 3.162[1]
4. Correct value 3√2 (accept √18)[1]
5. The student has assumed √a + √b = √(a + b)naming the false rule[1]
6. Radicals distribute over multiplication and division, not over addition and subtraction√(ab) = √a√b is true; √(a+b) = √a + √b is not[1]

(a) √2 + √8 = 3√2 ≈ 4.243 while √10 ≈ 3.162 (b) 3√2 (c) roots do not distribute over addition

A set is a well-defined collection of distinct objects, meaning that for any object it can be decided without dispute whether it belongs. "Tall" has no agreed boundary, so membership cannot be decided, and the collection is not well defined.

P(A) = { ∅, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} } and n(P(A)) = 2³ = 8.

A relation from A to B is any subset of A × B. A function is a relation in which every element of A appears exactly once as a first component — every input is used, and none has more than one output. So every function is a relation, but not every relation is a function.

1. A ∪ B = {1,2,3,4,5,6,7}, so (A ∪ B)′ = {8,9,10}complement taken with respect to U[1]
2. A′ = {6,7,8,9,10}[1]
3. B′ = {1,2,3,8,9,10}, so A′ ∩ B′ = {8,9,10}[1]
4. Both sides equal {8, 9, 10}, therefore the law is verifiedthe concluding statement is a mark in its own right[1]

Both sides equal {8, 9, 10}, so the law holds.

1. Domain = {1, 2, 3, 4}; Range = {2, 4, 6, 8}first components and second components respectively[1]
2. R is a function, because every element of A appears exactly once as a first componenta reason is required, not just the verdict[1]
3. It is one-one, since no two different inputs share an output[1]
4. It is onto, since the range equals B and no element of B is unusedso R is a bijection[1]

(a) Domain {1,2,3,4}, Range {2,4,6,8} (b) yes, a function (c) one-one and onto, so bijective

1. Two overlapping circles inside a rectangle labelled U, with 8 written in the overlapalways fill the intersection first[1]
2. Physics only = 24 − 8 = 16; Chemistry only = 20 − 8 = 12the given totals include the overlap[1]
3. Uses n(A ∪ B) = n(A) + n(B) − n(A ∩ B)the inclusion–exclusion principle[1]
4. = 24 + 20 − 8 = 36 students study at least oneor 16 + 8 + 12 from the diagram[1]
5. Neither = n(U) − n(A ∪ B)[1]
6. = 40 − 36 = 4 students[1]

(b) 36 students (c) 4 students

(x³)⁴ = x¹² by the power law, then x¹² ÷ x⁵ = x⁷ by subtracting the indices.

The first significant figure is 4, so a = 4.02. The point moves four places right to get there, so n = −4: 4.02 × 10⁻⁴.

The cube root of 27 is 3, and 3² = 9.

1. (2a²b)³ = 8a⁶b³ — every factor inside the bracket is cubed, including the 2forgetting to cube the 2 is the usual slip[1]
2. Multiply: 8a⁶b³ × 4a⁻¹b² = 32a⁵b⁵coefficients multiply, indices add[1]
3. Divide: 32 ÷ 8 = 4, and indices subtract[1]
4. = 4a²ba⁵⁻³ = a², b⁵⁻⁴ = b¹[1]

4a²b

1. Distance = speed × time = (3 × 10⁸) × (3.15 × 10⁷)[1]
2. Number parts: 3 × 3.15 = 9.45[1]
3. Indices: 10⁸ × 10⁷ = 10¹⁵add the indices[1]
4. = 9.45 × 10¹⁵ m, already in standard form9.45 lies between 1 and 10, so no adjustment is needed[1]

9.45 × 10¹⁵ metres

1. (1.5 × 10⁻⁵) × (2 × 10⁶) = 3 × 10¹1.5 × 2 = 3 and −5 + 6 = 1[1]
2. = 30 kgaccept either form[1]
3. (1.5 × 10⁻⁵) × (8 × 10¹⁸) = 12 × 10¹³[1]
4. Adjusted to standard form: 1.2 × 10¹⁴ kg12 is not between 1 and 10[1]
5. Divide by 1000, that is by 10³: subtract 3 from the index[1]
6. = 1.2 × 10¹¹ tonnes[1]

(a) 3 × 10¹ kg (b) 1.2 × 10¹⁴ kg (c) 1.2 × 10¹¹ tonnes

Multiply by 1.24: 350 × 1.24 = 434.

Change = 12. Percentage change = (12 ÷ 80) × 100 = 15%, dividing by the original price.

0.375 = 375/1000 = 3/8, and 0.375 × 100 = 37.5%.

1. The stated price is 115% of the pre-tax priceidentifying this is the key step[1]
2. So divide by the multiplier: 1 495 000 ÷ 1.15not subtract 15%[1]
3. = Rs 1 300 000[1]
4. Check: 1 300 000 × 1.15 = 1 495 000 ✓a verification is expected in reverse percentage questions[1]

Rs 1 300 000

1. Uses A = P(1 + R/100)ᵀ with P = 50 000, R = 8, T = 3[1]
2. A = 50 000 × 1.08³the multiplier raised to the power of the years[1]
3. = 50 000 × 1.259712 = Rs 62 985.60[1]
4. Interest = 62 985.60 − 50 000 = Rs 12 985.60the formula gives the total, so the principal must be subtracted[1]

Amount Rs 62 985.60; interest Rs 12 985.60

1. Marked price = 30 000 × 1.40[1]
2. = Rs 42 000[1]
3. Sale price = 42 000 × 0.75a 25% reduction is a multiplier of 0.75[1]
4. = Rs 31 500[1]
5. Profit = 1500, so percentage profit = (1500 ÷ 30 000) × 100 = 5%divided by the cost price, the original[1]
6. No — the combined multiplier is 1.40 × 0.75 = 1.05, a net 5% gain, because the 25% is taken from the larger marked price rather than from the costthe explanation is the mark[1]

(a) Rs 42 000 (b) Rs 31 500 (c) 5% profit — the percentages act on different amounts

The first significant figure is the 4, so the three are 4, 0 and 7. The next digit is 8, which rounds up: 0.0408.

Half a centimetre either side: 23.5 ≤ length < 24.5 cm.

Rounding to 1 s.f.: (0.5 × 400) ÷ 20 = 200 ÷ 20 = 10.

1. Bounds of the length: 11.5 ≤ l < 12.5[1]
2. Bounds of the width: 7.5 ≤ w < 8.5[1]
3. Maximum area = 12.5 × 8.5 = 106.25 cm²both upper bounds, since multiplying[1]
4. Minimum area = 11.5 × 7.5 = 86.25 cm²[1]

86.25 cm² ≤ area < 106.25 cm²

1. Distance bounds: 145 ≤ d < 155 mhalf of 10 either side[1]
2. Time bounds: 11.5 ≤ t < 12.5 shalf of 1 either side[1]
3. Maximum speed uses the largest distance and the smallest timethis reversal is the point of the question[1]
4. 155 ÷ 11.5 = 13.5 m/s (3 s.f.)[1]

13.5 m/s

1. 14.5 ≤ s < 15.5 cm[1]
2. Minimum area = 14.5² = 210.25 cm²[1]
3. Maximum area = 15.5² = 240.25 cm²[1]
4. The floor area is 60 000 cm², so the number of tiles lies between 60 000 ÷ 240.25 ≈ 250 and 60 000 ÷ 210.25 ≈ 285note the reversal — larger tiles mean fewer needed[1]
5. The exact number cannot be found because the true tile size is unknown within that range[1]
6. A builder should order for the smallest tile size, that is 285, since ordering too few would leave the floor unfinishedthe practical judgement is the mark[1]

(a) 14.5 ≤ s < 15.5 (b) 210.25 to 240.25 cm² (c) between 250 and 285 tiles; order 285

The highest common factor is 15, so dividing each part gives 3 : 4 : 5.

There are 3 + 5 = 8 parts altogether, so the first part is 3/8 of the total.

2 hours 30 minutes is 2.5 hours, so speed = 150 ÷ 2.5 = 60 km/h.

1. The difference between Bilal and Ali is 8 − 5 = 3 partswork with the difference in parts, not the shares[1]
2. 3 parts = 2400, so one part = 800[1]
3. Total parts = 5 + 7 + 8 = 20[1]
4. Total sum = 20 × 800 = Rs 16 000check: shares 4000, 5600, 6400 differ by 2400 ✓[1]

Rs 16 000

1. Inverse proportion means y = k/x, so xy = k[1]
2. k = 4 × 15 = 60find k from the given pair[1]
3. When x = 10: y = 60 ÷ 10 = 6x went up, y went down ✓[1]
4. When y = 12: x = 60 ÷ 12 = 5[1]

y = 6; x = 5

1. Rice is in direct proportion to the number of people: 300 ÷ 4 = 75 g each[1]
2. 75 × 10 = 750 g[1]
3. No — cooking time is not proportional to quantity[1]
4. Rice cooks by absorbing water at a rate that does not depend on how much is in the pot, so a larger quantity takes only a little longer, not two and a half times as longthe reason is the mark[1]
5. Painting is inverse: 8 × 6 = 48 worker-days, so 48 ÷ 3 = 16 days[1]
6. Assumption: all workers paint at the same rate and do not obstruct one another[1]

(a) 750 g (b) no — cooking time is not proportional to quantity (c) 16 days, assuming equal and independent working rates

Profit = 150. Percentage profit = (150 ÷ 600) × 100 = 25%, dividing by the cost price.

Rupees to pounds divides: 91 000 ÷ 350 = £260.

45 ÷ 60 = 0.75, so the time is 3.75 hours.

1. Basic pay = 40 × 950 = Rs 38 000[1]
2. Overtime hours = 46 − 40 = 6[1]
3. Overtime rate = 950 × 1.5 = Rs 1425 per hour"time and a half" means the rate is multiplied by 1.5[1]
4. Total = 38 000 + (6 × 1425) = 38 000 + 8550 = Rs 46 550[1]

Rs 46 550

1. Multiplier = 1.03 per year[1]
2. P = 24 000 × 1.03⁵the multiplier raised to the number of years[1]
3. = 24 000 × 1.159274 = 27 823 (nearest whole)a population must be a whole number[1]
4. Increase = 27 823 − 24 000 = 3823 people[1]

27 823 people, an increase of 3823

1. Total cost = 200 × 850 = Rs 170 000[1]
2. Marked-up selling price = 850 × 1.40 = Rs 1190, so 150 × 1190 = Rs 178 500[1]
3. Clearance price = 850 × 0.90 = Rs 765, so 50 × 765 = Rs 38 250a 10% loss is a multiplier of 0.9[1]
4. Total revenue = 178 500 + 38 250 = Rs 216 750[1]
5. Profit = 216 750 − 170 000 = Rs 46 750[1]
6. Percentage profit = (46 750 ÷ 170 000) × 100 = 27.5%of the total cost price[1]

(a) Rs 170 000 (b) Rs 216 750 (c) 27.5% profit

p terms: 7p − 2p = 5p. q terms: −3q + 5q = 2q. So the answer is 5p + 2q.

6x − 3 − 2x + 8 = 4x + 5.

Subtract u: v − u = at. Divide by a: t = (v − u)/a.

1. (2x + 3)(x − 5) = 2x² − 10x + 3x − 15all four products[1]
2. = 2x² − 7x − 15the middle terms combine[1]
3. (3y − 2)² = (3y − 2)(3y − 2) = 9y² − 6y − 6y + 4writing the bracket twice prevents the missing middle term[1]
4. = 9y² − 12y + 4not 9y² + 4[1]

2x² − 7x − 15; 9y² − 12y + 4

1. 2a²b = 2 × (−2)² × 3 = 2 × 4 × 3 = 24the square of −2 is +4[1]
2. −3bc = −3 × 3 × (−1) = +9two negatives make a positive[1]
3. c³ = (−1)³ = −1an odd power keeps the sign negative[1]
4. Total = 24 + 9 − 1 = 32[1]

32

1. Perimeter = 2[(3x + 2) + (x − 1)] = 2(4x + 1)[1]
2. = 8x + 2the 2 multiplies both terms[1]
3. Area = (3x + 2)(x − 1) = 3x² − 3x + 2x − 2all four products[1]
4. = 3x² − x − 2[1]
5. 8x + 2 = 36, so 8x = 34 and x = 4.25[1]
6. Area = 3(4.25)² − 4.25 − 2 = 54.19 − 6.25 = 48.1 cm² (3 s.f.)check: length 14.75, width 3.25, product 47.9 — accept either from rounding[1]

(a) 8x + 2 (b) 3x² − x − 2 (c) x = 4.25, area ≈ 48 cm²

Both terms are perfect squares with a minus between them, so this is a difference of two squares: (5x)² − (7y)² = (5x − 7y)(5x + 7y).

Using a² + b² = (a + b)² − 2ab = 81 − 40 = 41.

This is a difference of two cubes with b = 2: x³ − 2³ = (x − 2)(x² + 2x + 4).

1. Take out the common factor 3x: 3x(x² − 9)both the number and the variable come out[1]
2. Recognises x² − 9 as a difference of two squares[1]
3. x² − 9 = (x − 3)(x + 3)[1]
4. Complete factorisation 3x(x − 3)(x + 3)stopping at 3x(x² − 9) does not earn the final mark[1]

3x(x − 3)(x + 3)

1. Groups the first three terms: (4x² − 12x + 9) − y²a 3-and-1 grouping, not 2-and-2 — the three terms form a perfect square[1]
2. 4x² − 12x + 9 = (2x − 3)²check the middle term: 2 × 2x × 3 = 12x ✓[1]
3. Now a difference of two squares: (2x − 3)² − y²[1]
4. = (2x − 3 − y)(2x − 3 + y)[1]

(2x − 3 − y)(2x − 3 + y)

1. p = (x − 2)(x − 3)two numbers multiplying to 6 and adding to −5[1]
2. q = (x − 2)(x + 2)difference of two squares[1]
3. HCF = (x − 2), the only common factor[1]
4. LCM = (x − 2)(x − 3)(x + 2)every distinct factor at its highest power[1]
5. HCF × LCM = (x − 2)² (x − 3)(x + 2)[1]
6. p × q = (x − 2)(x − 3) × (x − 2)(x + 2) = (x − 2)²(x − 3)(x + 2), the same expression, so the identity is verifiedthe concluding statement is required[1]

(a) p = (x−2)(x−3), q = (x−2)(x+2) (b) HCF = x − 2, LCM = (x−2)(x−3)(x+2) (c) both sides equal (x−2)²(x−3)(x+2)

Factorising, (x − 4)(x + 4)/(x + 4) = x − 4, provided x ≠ −4.

The x values are terms inside sums, not factors of the whole numerator and denominator. Cancelling is only permitted for factors that multiply the entire expression, and here neither the numerator nor the denominator factorises further.

Factorising the denominator gives (x − 2)(x − 3), which is zero at x = 2 and x = 3. The expression is undefined at those two values.

1. Common denominator (x − 3)(x + 2)no shared factor, so the LCM is the product[1]
2. Numerator = 2(x + 2) + 3(x − 3)each numerator multiplied by the missing factor[1]
3. = 2x + 4 + 3x − 9 = 5x − 5[1]
4. = 5(x − 1) / [(x − 3)(x + 2)], with x ≠ 3, −2factorising the numerator shows nothing further cancels[1]

5(x − 1) / [(x − 3)(x + 2)], x ≠ 3, −2

1. x² − 1 = (x − 1)(x + 1)difference of two squares[1]
2. x² + 4x + 3 = (x + 1)(x + 3)[1]
3. Cancels (x + 1), (x + 3) and (x − 1) across the two fractionsin a multiplication you may cancel a numerator factor against either denominator[1]
4. Result 1, with x ≠ 1, −1, −3the restrictions are needed for the mark[1]

1, provided x ≠ 1, −1, −3

1. Common denominator (x − 2)(x + 2) = x² − 4; numerator = (x + 2) − (x − 2) = 4the x terms cancel — that is the design of the question[1]
2. First bracket = 4/(x² − 4)[1]
3. Division becomes multiplication by the reciprocal: [4/(x² − 4)] × [(x² − 4)/4][1]
4. E = 1[1]
5. Denominators vanish at x = 2 and x = −2[1]
6. Those are the only exclusions; the divisor 4/(x² − 4) is never zero, since its numerator is the constant 4checking whether the divisor can be zero is the step being tested[1]

(a) 4/(x² − 4) (b) E = 1 (c) undefined only at x = 2 and x = −2

Expanding, 5x − 10 = 3x + 4. Collecting: 2x = 14, so x = 7.

Multiplying or dividing by a negative number reflects every point in the number line about zero, so the left-to-right order of the two values is reversed. For example 3 < 5, but −3 > −5.

No solution. The modulus of any expression is a distance from zero and can never be negative, so no value of x can make |2x + 1| equal to −4.

1. Multiply every term by the LCM 12: 3(x + 2) − 2(x − 3) = 12the right-hand side must be multiplied too[1]
2. Expand: 3x + 6 − 2x + 6 = 12the minus sign in front of the second bracket changes both signs inside it[1]
3. Collect: x + 12 = 12[1]
4. x = 0zero is a perfectly good answer — students often assume they have made an error[1]

x = 0

1. Uses the rule |a| < k ⟺ −k < a < k: −3 < x − 4 < 3one interval because the sign is less-than[1]
2. Add 4 to all three partsthe same operation on every part[1]
3. 1 < x < 7[1]
4. Number line with open circles at 1 and 7 and the segment between them shadedcircles must be open because the inequality is strict[1]

1 < x < 7

1. Perimeter = 2[(2x + 3) + (x − 1)] = 2(3x + 2) = 6x + 4perimeter is twice the sum of length and width[1]
2. 6x + 4 = 34 so 6x = 30 and x = 5[1]
3. Length = 2(5) + 3 = 13 cm[1]
4. Width = 5 − 1 = 4 cm; check 2(13 + 4) = 34 ✓[1]
5. x − 1 > 4 gives x > 5[1]
6. Values with x ≤ 1 make the width zero or negative, which is impossible for a real rectanglethe physical restriction must be stated[1]

(a) x = 5 (b) 13 cm by 4 cm (c) x > 5; values x ≤ 1 give a non-positive width

The discriminant is b² − 4ac. It gives the number of real roots: two if positive, one repeated if zero, none if negative.

b² − 4ac = 0 → k² − 36 = 0 → k = ±6

(3, 5), a minimum, because the coefficient of the squared term is positive.

1. Factorising: (2x − 1)(x − 3) = 0[1]
2. x = 1/2 or x = 3[1]
3. Formula: identifies a = 2, b = −7, c = 3[1]
4. x = [7 ± √(49 − 24)] / 4[1]
5. = (7 ± 5)/4[1]
6. x = 3 or x = 1/2, the same pair[1]

x = 1/2 and x = 3

1. Takes out the factor: h = −5(t² − 4t)[1]
2. Completes the square inside: t² − 4t = (t − 2)² − 4[1]
3. h = −5[(t − 2)² − 4] = 20 − 5(t − 2)²[1]
4. Greatest height = 20 mthe squared term is zero there[1]
5. At t = 2 s[1]
6. Sets h = 0: t(20 − 5t) = 0[1]
7. t = 4 st = 0 is the moment of throwing[1]

(a) 20 − 5(t − 2)² (b) 20 m at t = 2 s (c) 4 s

1. Equates the two expressions: x + k = x² + 3x + 4[1]
2. Rearranges to x² + 2x + 4 − k = 0[1]
3. A tangent means one repeated root, so b² − 4ac = 0[1]
4. 4 − 4(4 − k) = 0[1]
5. 4 − 16 + 4k = 0 → k = 3[1]
6. x² + 2x + 1 = 0 → x = −1, so the point is (−1, 2)[1]

k = 3, touching at (−1, 2)

A sequence is an ordered list of terms, such as 2, 5, 8, 11. A series is the sum of the terms of a sequence, such as 2 + 5 + 8 + 11 = 26.

a = 7 and d = 4, so a₂₀ = 7 + 19(4) = 7 + 76 = 83.

The sum to infinity exists only when |r| < 1, and then S∞ = a / (1 − r).

1. S₁₀ = 5[2a + 9d] = 155, so 2a + 9d = 31divide through by 5 immediately to keep the numbers small[1]
2. S₂₀ = 10[2a + 19d] = 610, so 2a + 19d = 61[1]
3. Subtracting: 10d = 30, so d = 3the 2a terms cancel[1]
4. 2a + 27 = 31, so a = 2check: S₁₀ = 5(4 + 27) = 155 ✓[1]

a = 2, d = 3

1. Writes it as 0.47 + 0.0047 + 0.000047 + …each block of two digits is one term[1]
2. Identifies a = 0.47 and r = 0.01each term is the previous one divided by 100[1]
3. |r| < 1, so S∞ = 0.47 / (1 − 0.01) = 0.47 / 0.99the condition should be stated[1]
4. = 47/99[1]

47/99

1. Heights form a GP with a = 8 and r = 0.75; after the 4th bounce the height is 8(0.75)⁴the drop height is the term before the first bounce[1]
2. = 2.53 m (3 s.f.)[1]
3. Total distance = the initial 8 m drop + twice each subsequent rise, since each bounce goes up and comes back downthe factor of 2 is the mark most often missed[1]
4. Rises form a GP with first term 8(0.75) = 6 and r = 0.75, so their sum is 6/(1 − 0.75) = 24[1]
5. Total = 8 + 2(24) = 56 m[1]
6. Because |r| < 1, the heights shrink geometrically towards zero fast enough for the infinite sum to converge to a finite limitthe explanation must refer to |r| < 1[1]

(a) 2.53 m (b) 56 m (c) the heights form a convergent GP since |r| < 1

(i) The object is stationary — its distance from the start is not changing. (ii) The object is moving at a constant speed — the speed is not changing, but it is still moving.

The distance travelled. Speed multiplied by time gives distance — metres per second times seconds gives metres — so the area, which is a product of the two axes, has the units of distance.

It is a fixed charge of Rs 150 applied before any distance is travelled — a flag-fall or booking fee. The gradient of the line then gives the additional cost per kilometre.

1. A straight line rising from (0, 0) to (20, 6)time in minutes on the horizontal axis[1]
2. A horizontal section from (20, 6) to (30, 6) for the restdistance unchanged means stationary[1]
3. Outward speed = 6 km ÷ (1/3) h = 18 km/h20 minutes is one third of an hour[1]
4. A line falling from (30, 6) to (60, 0); return speed = 6 ÷ 0.5 = 12 km/hthe downward slope means returning towards the start[1]

Rise, horizontal, then fall to zero. Out at 18 km/h, back at 12 km/h.

1. Recognises that the distance is the area under the speed–time graphthis is the mark most often missed[1]
2. Acceleration triangle: ½ × 15 × 30 = 225 m[1]
3. Constant rectangle: 45 × 30 = 1350 m[1]
4. Deceleration triangle: ½ × 30 × 30 = 450 m; total = 2025 m[1]

2025 m

1. The first part is a straight line rising steadilyconstant cross-section means constant rate of rise[1]
2. The second part is a curve that becomes progressively shallowerstill rising, but more slowly[1]
3. The rate of water entering is constant, so equal volumes arrive in equal times[1]
4. As the tank widens, that same volume spreads over a larger area and so raises the level by less — the depth increases more slowlythe explanation must connect area to rate of rise[1]
5. A narrowing tank gives the opposite: the curve becomes progressively STEEPER[1]
6. because the same volume is confined to a smaller area and therefore raises the level more[1]

(a) straight, then a flattening curve (b) a wider section spreads the same volume over more area (c) a narrowing tank steepens instead

Divide by 2 to reach y = 3x − 4. Gradient 3, y-intercept −4.

The roots are the x-coordinates of the points where the curve crosses the x-axis, because y = 0 at every point on that axis. Here they are x = 2 and x = 3.

The lines never intersect, so the pair of equations has no solution. Algebraically this shows up as a contradiction such as 0 = 5 during elimination.

1. y values: 5, 0, −3, −4, −3, 0, 5one wrong value is tolerated; two is not[1]
2. Roots at x = −1 and x = 3the two x values where y = 0[1]
3. Turning point at x = 1 by symmetry, midway between the roots[1]
4. Minimum point (1, −4); it is a minimum because the coefficient of x² is positivethe reason is required for the mark[1]

Roots at x = −1 and x = 3; minimum at (1, −4).

1. First line through (0, 2) with gradient 1two correct points are enough to draw a line[1]
2. Second line through (0, 8) with gradient −2[1]
3. Lines cross at (2, 4)[1]
4. Check by substitution: 4 = 2 + 2 ✓ and 4 = −4 + 8 ✓a substitution check is expected in a graphical solution[1]

x = 2, y = 4

1. y values: −2, 2, 0, −2, 2e.g. at x = −1: (−1)³ − 3(−1) = −1 + 3 = 2[1]
2. A cubic curve rising, turning down, then rising againtwo turning points[1]
3. The ends go in opposite directions because the highest power is odd[1]
4. It crosses the x-axis three timesat x = 0, √3 and −√3[1]
5. Draw the horizontal line y = 1 on the same axesthe left-hand side is already the drawn curve[1]
6. The x-coordinates of the three intersections are the solutionsthe number of solutions must be stated as three[1]

(a) −2, 2, 0, −2, 2 (b) a cubic with two turning points, crossing the x-axis three times (c) draw y = 1 and read the x values where it meets the curve

The denominator is zero when x² = 9, that is at x = 3 and x = −3. The domain is therefore all real numbers except 3 and −3.

It is not one-one: f(3) = f(−3) = 9, so the output 9 does not identify a unique input. An inverse would have to assign two values to 9, which no function may do. Restricting the domain to x ≥ 0 makes it one-one and the inverse √x then exists.

Since x² ≥ 0 for every real x, the smallest value of f is 3, reached at x = 0. The range is f(x) ≥ 3.

1. fg(x) = f(x² − 3) = 2(x² − 3) + 1 = 2x² − 5substitute the whole of g into f[1]
2. gf(x) = g(2x + 1) = (2x + 1)² − 3 = 4x² + 4x − 2expand the bracket fully[1]
3. Set them equal: 2x² − 5 = 4x² + 4x − 2, so 2x² + 4x + 3 = 0[1]
4. Discriminant = 16 − 24 = −8 < 0, so there is no real value of x for which they are equala reasoned "no solution" is the answer, not an omission[1]

fg(x) = 2x² − 5; gf(x) = 4x² + 4x − 2; no real x makes them equal

1. Let y = (2x − 3)/(x + 1) and multiply up: y(x + 1) = 2x − 3clear the fraction before rearranging[1]
2. xy + y = 2x − 3, so xy − 2x = −3 − ygather every term containing x on one side[1]
3. x(y − 2) = −(3 + y), so x = −(y + 3)/(y − 2)factorising out x is the key step[1]
4. f⁻¹(x) = −(x + 3)/(x − 2), or equivalently (x + 3)/(2 − x)either form accepted[1]

f⁻¹(x) = (x + 3)/(2 − x), x ≠ 2

1. Need x − 4 ≥ 0, so the domain is x ≥ 4a square root requires a non-negative argument[1]
2. A square root is never negative, so the range is f(x) ≥ 0[1]
3. y = √(x − 4) gives y² = x − 4, so x = y² + 4squaring is safe here because y ≥ 0[1]
4. f⁻¹(x) = x² + 4[1]
5. Domain of f⁻¹ is x ≥ 0not all of ℝ — it must match the range of f[1]
6. The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f — the inverse swaps the two sets[1]

(a) domain x ≥ 4, range f(x) ≥ 0 (b) f⁻¹(x) = x² + 4 with domain x ≥ 0 (c) inverting swaps domain and range

|AB| = √[(4 −(−2))² + (−5 − 3)²] = √(36 + 64) = √100 = 10 units.

The negative reciprocal: m = −4/3, since (3/4) × (−4/3) = −1.

A gradient of 0 means the rise is zero, so the line is horizontal. An undefined gradient means the run is zero, so the formula divides by zero and the line is vertical.

1. The y-coordinate of M gives (k + 8)/2 = 5the x-coordinates already check out: (3 + (−1))/2 = 1 ✓[1]
2. k + 8 = 10, so k = 2[1]
3. Uses the distance formula with A(3, 2) and B(−1, 8)[1]
4. |AB| = √[(−1 − 3)² + (8 − 2)²] = √(16 + 36) = √52 = 2√13simplified surd required[1]

k = 2; |AB| = 2√13 ≈ 7.21 units

1. Gradient of PQ = (5 − 1)/(3 − 1) = 4/2 = 2[1]
2. Gradient of QR = (11 − 5)/(6 − 3) = 6/3 = 2[1]
3. The two gradients are equal[1]
4. PQ and QR share the point Q, so the three points lie on one straight line and are collinearthe shared point must be mentioned — equal gradients alone only prove the segments are parallel[1]

Gradient PQ = gradient QR = 2, and they share Q, so P, Q and R are collinear.

1. AB: (0 − 0)/(6 − 0) = 0; CD: (4 − 4)/(2 − 8) = 0both horizontal[1]
2. BC: (4 − 0)/(8 − 6) = 2; DA: (0 − 4)/(0 − 2) = 2[1]
3. AB is parallel to CD and BC is parallel to DA, since each pair has equal gradients[1]
4. Both pairs of opposite sides are parallel, so ABCD is a parallelogramadjacent gradients 0 and 2 do not multiply to −1, so it is not a rectangle[1]
5. Uses the distance formula on A(0,0) and C(8,4)[1]
6. |AC| = √(64 + 16) = √80 = 4√5[1]

(a) 0, 2, 0, 2 (b) a parallelogram — both pairs of opposite sides are parallel (c) 4√5 ≈ 8.94 units

A decagon has 10 sides, so the sum is (10 − 2) × 180 = 8 × 180 = 1440°.

The exterior angles total 360°, so n = 360 ÷ 30 = 12 sides.

180 − 115 = 65°, because co-interior angles add to 180°.

1. Exterior angle = 180 − 162 = 18°interior and exterior lie on a straight line[1]
2. n = 360 ÷ 18 = 20 sidesthe exterior angles total 360°[1]
3. Interior sum = (20 − 2) × 180[1]
4. = 3240°; check 3240 ÷ 20 = 162 ✓[1]

20 sides; interior angles total 3240°

1. Angles in a triangle sum to 180°: 3x + (2x + 10) + (x + 50) = 180the reason should be stated[1]
2. 6x + 60 = 180collecting terms[1]
3. 6x = 120, so x = 20[1]
4. Angles are 60°, 50° and 70°, so the largest is 70°; check they total 180 ✓the largest is C, not A — worth checking rather than assuming[1]

x = 20; the largest angle is 70°

1. A regular pentagon has interior angle (5 − 2) × 180 ÷ 5 = 108°[1]
2. 360 ÷ 108 = 3.33, not a whole number, so pentagons cannot fit round a point without a gap or an overlapthe whole-number requirement is the key idea[1]
3. A regular hexagon has interior angle (6 − 2) × 180 ÷ 6 = 120°[1]
4. 360 ÷ 120 = 3 exactly, so three hexagons meet at each point — which is why honeycomb is hexagonal[1]
5. A regular octagon has interior angle (8 − 2) × 180 ÷ 8 = 135°[1]
6. Two octagons and one square give 135 + 135 + 90 = 360° exactly, so yes — this is a common floor tiling[1]

(a) 360 ÷ 108 is not a whole number (b) 120° × 3 = 360° (c) yes — two octagons and a square give exactly 360°

A locus is the set of all points that satisfy a given condition. Example: the locus of points 5 m from a fixed post is a circle of radius 5 m centred on the post — the path traced by a goat at the end of a taut 5 m rope.

The perpendicular bisector of AB. Construct it by drawing arcs of equal radius (greater than half AB) from A and from B, above and below the line, and joining the two intersections.

The marks in a construction question are awarded for the method, which is shown by the arcs. A line drawn accurately by measurement, with no arcs, earns no marks even if it is in exactly the right place.

1. Draws AB = 7 cm accurately as the basestart with the longest side; it makes the arcs easier to cross[1]
2. Arc of radius 6 cm from A and arc of radius 5 cm from B, crossing at Cboth arcs must be visible[1]
3. Completes triangle ABC with straight lineswithin 2 mm tolerance[1]
4. Perpendicular bisector of AB drawn with arcs from A and from B of equal radiusthe bisector must pass through the midpoint at right angles[1]

Triangle constructed by SSS, with the bisector of AB drawn using equal arcs from A and B.

1. Circle centred P with radius 6 km, and the required region is inside itscale drawing, e.g. 1 cm to 1 km[1]
2. Perpendicular bisector of PQ constructed with arcs[1]
3. Identifies the side of the bisector containing Q as "nearer to Q"[1]
4. Shades the overlap: inside the circle and on Q's side of the bisectorshading the union instead of the intersection loses this mark[1]

The region inside the 6 km circle about P and on Q's side of the perpendicular bisector of PQ.

1. A line parallel to AB at a distance of 3 m from itinside the garden only, since the tree must be in the garden[1]
2. The required region is on the far side of that line from AB"at least" makes this a region, not just a line[1]
3. The perpendicular bisector of AD[1]
4. Which, since AD is a side of the rectangle, runs parallel to AB through the midpoint of AD[1]
5. The tree lies where the bisector meets the allowed region[1]
6. That is the part of the perpendicular bisector of AD that is 3 m or more from AB, i.e. a segment of it inside the gardenaccept a clearly drawn and labelled answer[1]

(a) a parallel line 3 m from AB, with the region beyond it (b) the perpendicular bisector of AD (c) the part of the bisector lying at least 3 m from AB

Congruent figures have the same shape and the same size, so corresponding sides are equal. Similar figures have the same shape only; corresponding sides are in a constant ratio k, and congruence is the case k = 1.

Areas scale by k², so k² = 9/25 and k = 3/5. The sides are in the ratio 3 : 5.

A circle is determined completely by its radius, so any circle is an enlargement of any other. A rectangle needs two independent lengths, so two rectangles are similar only if their length-to-width ratios match — a 2 × 5 and a 3 × 5 rectangle are not similar.

1. By the intercept theorem AD/DB = AE/EC, so 4/6 = 5/ECpieces compared with pieces[1]
2. 4 × EC = 30, so EC = 7.5 cm[1]
3. For the ratio of DE to BC use whole sides: AD/AB = 4/10 = 2/5AB = AD + DB = 10 cm — this is the step most often skipped[1]
4. DE : BC = 2 : 5accept 0.4[1]

EC = 7.5 cm; DE : BC = 2 : 5

1. Area scale factor = 50² = 2500areas scale by the square of the length ratio[1]
2. Real wing area = 120 × 2500 = 300 000 cm²accept 30 m²[1]
3. Volume scale factor = 50³ = 125 000[1]
4. Real volume = 400 × 125 000 = 5 × 10⁷ cm³accept 50 000 000 cm³ or 50 m³[1]

(a) 300 000 cm² (b) 5 × 10⁷ cm³

1. k = 1.8 / 1.2 = 1.5from smaller to larger, so k > 1[1]
2. Area factor = k² = 2.25[1]
3. Larger surface area = 4.8 × 2.25 = 10.8 m²[1]
4. Volume factor = k³ = 3.375[1]
5. Going from larger to smaller, divide: 2430 / 3.375direction of the scale factor matters here[1]
6. = 720 litres[1]

(a) k = 1.5 (b) 10.8 m² (c) 720 litres

The angle of rotation, the direction (clockwise or anticlockwise), and the centre of rotation.

The area is multiplied by 3² = 9, since area involves two lengths and each has been tripled.

No lines of symmetry, but rotational symmetry of order 2 — a half turn maps it onto itself.

1. With centre at the origin, each coordinate is multiplied by the scale factorthis shortcut only works when the centre is (0, 0)[1]
2. Q has vertices (4,2), (10,2) and (4,6)[1]
3. Areas scale by k² = 2² = 4[1]
4. Ratio = 4 : 1; check by calculating both areas — P is 3 and Q is 12 ✓[1]

(a) (4,2), (10,2), (4,6) (b) 4 : 1

1. Same size and same orientation with every point moving equally means a translation[1]
2. Column vector (3, −2)right is positive, down is negative[1]
3. If B were upside down as well, it could not be a translation, since a translation never changes orientation[1]
4. It would be a rotation of 180° about some centre, which produces the same displacement with the orientation reversedaccept an enlargement with k = −1[1]

A translation by (3, −2); if inverted, a 180° rotation instead.

1. 6 lines of symmetrythree through opposite vertices, three through opposite edge midpoints[1]
2. Rotational symmetry of order 6 — a regular n-gon always has n of each[1]
3. Perimeter is a length, so it is multiplied by 1.5[1]
4. Area is multiplied by 1.5² = 2.25[1]
5. A scale factor of −1 keeps every distance from the centre the same but reverses the directionmagnitude 1 means no change of size[1]
6. Reversing the direction of every point about a centre is exactly a half turn, so the two produce identical images[1]

(a) 6 and order 6 (b) perimeter × 1.5, area × 2.25 (c) k = −1 reverses direction without changing distance, which is a 180° rotation

The tangent is perpendicular to the radius drawn to the point of contact — they meet at exactly 90°.

PA = PB, the two tangents from an external point are equal in length; and OP bisects ∠APB. Also, both ∠OAP and ∠OBP are right angles.

The diameter subtends a straight angle of 180° at the centre. By the angle-at-the-centre theorem, the angle it subtends at the circumference is half of that, namely 90°, wherever the point is chosen on the arc.

1. Angle at the centre is twice the angle at the circumference: ∠BAC = 130°/2A is on the major arc, so the non-reflex central angle is the right one[1]
2. ∠BAC = 65°[1]
3. Triangle OBC is isosceles with OB = OC = 9 and included angle 130°; using the cosine rule or dropping the perpendicular: BC = 2 × 9 × sin 65°the perpendicular from O bisects both BC and the 130° angle[1]
4. BC = 18 × 0.9063 = 16.3 cm[1]

∠BAC = 65°; BC ≈ 16.3 cm

1. P and R are opposite angles of a cyclic quadrilateral, so they sum to 180°naming the theorem earns this mark[1]
2. (3x + 10) + (2x + 20) = 180, so 5x + 30 = 180[1]
3. 5x = 150, giving x = 30[1]
4. ∠P = 100° and ∠R = 80°; check 100 + 80 = 180 ✓[1]

x = 30; ∠P = 100° and ∠R = 80°

1. The angle between a tangent and a chord equals the angle subtended by that chord in the alternate segment[1]
2. ∠ACB is in the alternate segment to ∠TABC is on the far side of the chord AB from the angle TAB[1]
3. ∠ACB = 58°[1]
4. Uses the angle at the centre being twice the angle at the circumference[1]
5. ∠AOB = 2 × 58° = 116°[1]
6. Alternative check: ∠OAT = 90° since the tangent is perpendicular to the radius, so ∠OAB = 90 − 58 = 32°; triangle OAB is isosceles, giving ∠AOB = 180 − 2(32) = 116° ✓either route is accepted[1]

(b) 58°, by the alternate segment theorem (c) 116°, twice the angle at the circumference

A chord is a straight line segment whose two endpoints lie on the circle. The longest chord is the diameter, which passes through the centre.

The perpendicular from the centre to a chord bisects the chord. Conversely, the line from the centre to the midpoint of a chord is perpendicular to it.

The perpendicular from the centre bisects each chord, forming a right-angled triangle with the radius as hypotenuse and half the chord as one leg. Equal chords give equal halves, and the radii are equal, so by Pythagoras the third sides — the distances from the centre — must also be equal.

1. Half the chord is 12 cm; r² = 5² + 12² = 25 + 144 = 169perpendicular from the centre bisects the chord[1]
2. r = 13 cm[1]
3. For the second chord, (c/2)² = 13² − 12² = 169 − 144 = 25, so c/2 = 5same circle, so the same radius[1]
4. Second chord = 10 cm[1]

r = 13 cm; the second chord is 10 cm long

1. Fraction = 90/360 = 1/4[1]
2. Arc = ¼ × 2 × (22/7) × 14 = ¼ × 88 = 22 cm[1]
3. Area = ¼ × (22/7) × 196 = ¼ × 616 = 154 cm²[1]
4. Perimeter = arc + two radii = 22 + 14 + 14 = 50 cmthe two radii are part of the boundary and are often forgotten[1]

Arc 22 cm; area 154 cm²; perimeter 50 cm

1. Triangle OAB is right-angled at O with OA = OB = 10 cmboth are radii[1]
2. AB = √(100 + 100) = √200 = 14.1 cmaccept 10√2[1]
3. Sector area = (90/360) × 3.14 × 100a quarter of the circle[1]
4. = 78.5 cm²[1]
5. Triangle area = ½ × 10 × 10 = 50 cm²the two radii are perpendicular, so they are the base and the height[1]
6. Segment = 78.5 − 50 = 28.5 cm²segment = sector minus triangle[1]

(a) 14.1 cm (b) 78.5 cm² (c) 28.5 cm²

A = ½(8 + 12) × 5 = ½ × 20 × 5 = 50 cm².

1 m² = 10 000 cm², so 3.5 × 10 000 = 35 000 cm².

C = 2πr = 2 × (22/7) × 7 = 44 cm.

1. Rectangle area = 20 × 14 = 280 m²[1]
2. Pond area = π × 3² = 3.142 × 9[1]
3. = 28.278 m²[1]
4. Grass = 280 − 28.278 = 251.7 m² (1 d.p.)surround and subtract[1]

251.7 m²

1. The semicircle has radius 3 cm and its arc length is ½ × 2π × 3 = 9.426 cmhalf the circumference[1]
2. The 6 cm side where the semicircle joins is not part of the perimeterthis is the key mark[1]
3. Remaining straight edges: 10 + 6 + 10 = 26 cmtwo long sides and one short side[1]
4. Perimeter = 26 + 9.426 = 35.4 cm (1 d.p.)[1]

35.4 cm

1. Field area = 120 × 85 = 10 200 m²[1]
2. 1 hectare = 10 000 m², so the field is 1.02 hectares[1]
3. The path takes 2 m from each side, so 4 m from each dimensionthe doubling is the mark most often missed[1]
4. Inner region = 116 m by 81 m[1]
5. Inner area = 116 × 81 = 9396 m²[1]
6. Path area = 10 200 − 9396 = 804 m²subtracting is far quicker than adding four strips[1]

(a) 1.02 hectares (b) 116 m by 81 m (c) 804 m²

Volume = 8 × 5 × 3 = 120 cm³. Surface area = 2(40 + 24 + 15) = 158 cm².

The vertical height is the perpendicular distance from the centre of the base to the apex, used in the volume. The slant height is measured along the sloping surface from the rim to the apex, used in the curved surface area. They are related by l² = r² + h².

V = 4⁄3 πr³ = 4⁄3 × 3.142 × 216 = 904.9 cm³.

1. V = πr²h = 3.142 × 1.5² × 4square the radius, do not double it[1]
2. = 3.142 × 2.25 × 4 = 28.278 m³[1]
3. Capacity = 28.278 × 1000[1]
4. = 28 278 litres, about 28 300 litres to 3 s.f.[1]

28.3 m³, about 28 300 litres

1. By Pythagoras h² = l² − r² = 289 − 64 = 225the slant height is the hypotenuse, so it is squared and the radius subtracted[1]
2. h = 15 cman 8-15-17 triple[1]
3. V = ⅓πr²h = ⅓ × 3.142 × 64 × 15the volume uses the vertical height[1]
4. = 1005.4 cm³[1]

Height 15 cm; volume 1005 cm³

1. Cylinder = πr²h = 3.142 × 9 × 10 = 282.78 m³[1]
2. Hemisphere = ½ × 4⁄3 πr³ = ⅔ × 3.142 × 27 = 56.56 m³; total = 339.3 m³half a sphere[1]
3. Curved cylinder = 2πrh = 2 × 3.142 × 3 × 10 = 188.5 m²[1]
4. Hemisphere = ½ × 4πr² = 2 × 3.142 × 9 = 56.6 m²; total = 245.1 m²the circle where they join is internal and excluded[1]
5. Surface area ratio = (1/10)² = 1 : 100[1]
6. Volume ratio = (1/10)³ = 1 : 1000[1]

(a) 339.3 m³ (b) 245.1 m² (c) areas 1 : 100, volumes 1 : 1000

a/sin A = b/sin B = c/sin C. Use it when you have a matched pair — a side and the angle opposite it — plus one more piece of information.

sin 30° = 1/2, cos 60° = 1/2, tan 45° = 1

1 − cos²θ = sin²θ, so the expression is sin²θ/(sin θ cos θ) = sin θ/cos θ = tan θ.

1. Uses the cosine rule a² = b² + c² − 2bc cos Atwo sides and the included angle[1]
2. BC² = 49 + 81 − 2(7)(9)cos 52°[1]
3. BC = 7.29 cm[1]
4. Uses ½ab sin C = ½(7)(9)sin 52°[1]
5. = 24.8 cm²[1]
6. Sine rule: sin B = 9 sin 52° / 7.29 → B = 76.6°[1]

(a) 7.29 cm (b) 24.8 cm² (c) 76.6°

1. Diagram with the mast vertical, both angles marked at ground level and PQ = 45 m[1]
2. Lets the height be h and the distance from Q be d[1]
3. From Q: h = d tan 58°[1]
4. From P: h = (d + 45) tan 34°[1]
5. Equates: d tan 58° = (d + 45) tan 34°[1]
6. Solves: d(1.600 − 0.6745) = 30.35 → d = 32.8 m[1]
7. h = 32.8 × tan 58° = 52.5 m[1]

height ≈ 52.5 m, Q is ≈ 32.8 m from the foot

1. sin θ = 1/2[1]
2. θ = 30°[1]
3. Sine is also positive in the second quadrant[1]
4. θ = 180° − 30° = 150°[1]
5. No further solutions in the range, so θ = 30° and 150°[1]

θ = 30° and 150°