10th Class Mathematics — MCQs & Practice Questions

A complex number is any number of the form z = a + bi, where a and b are real and i = √(−1). The real part is a and the imaginary part is b.

The powers of i repeat every four. 103 = 4 × 25 + 3, so i¹⁰³ = i³ = −i.

|z| = √((−5)² + 12²) = √(25 + 144) = √169 = 13.

1. Multiplies numerator and denominator by the conjugate 1 + 4i[1]
2. Denominator: (1 − 4i)(1 + 4i) = 1 + 16 = 17a² + b², because −16i² = +16[1]
3. Numerator: (3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i² = −5 + 14i[1]
4. = −5/17 + (14/17)imust be split into a + bi form[1]

−5/17 + (14/17)i

1. z₁z₂ = (2 + 3i)(4 − i) = 8 − 2i + 12i − 3i² = 11 + 10i−3i² = +3, which combines with the 8[1]
2. |z₁z₂| = √(121 + 100) = √221[1]
3. |z₁| = √13 and |z₂| = √17[1]
4. |z₁||z₂| = √13 × √17 = √221, equal to |z₁z₂|, so the result is verifiedthe concluding comparison is required[1]

z₁z₂ = 11 + 10i, and both sides equal √221.

1. Discriminant = (−6)² − 4(1)(25) = 36 − 100 = −64[1]
2. The discriminant is negative, so there are no real rootsthe conclusion must be stated, not just the number[1]
3. x = [6 ± √(−64)] / 2 with √(−64) = 8i[1]
4. x = 3 ± 4i[1]
5. Sum = (3 + 4i) + (3 − 4i) = 6 ✓the imaginary parts cancel[1]
6. Product = (3 + 4i)(3 − 4i) = 9 − 16i² = 9 + 16 = 25 ✓difference of two squares, with i² = −1[1]

(a) discriminant = −64 < 0 (b) x = 3 ± 4i (c) sum 6, product 25

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)

3(x² − 4) = 3(x − 2)(x + 2)

3V = πr²h → r² = 3V/(πh) → r = √(3V/(πh))

8x⁹y⁶ ÷ 4x⁴y = 2x⁵y⁵

1. Multiplies to match one coefficient, e.g. first ×3 and second ×2[1]
2. 9x + 6y = 48 and 10x − 6y = 18[1]
3. Adds to eliminate y: 19x = 66[1]
4. x = 66/19not every set of coefficients gives whole numbers[1]
5. Substitutes back into either original equation[1]
6. y = (16 − 3x)/2 = 53/19[1]

x = 66/19 ≈ 3.47, y = 53/19 ≈ 2.79

1. Length is w + 3[1]
2. w(w + 3) = 88, so w² + 3w − 88 = 0[1]
3. Factorises: (w + 11)(w − 8) = 0[1]
4. w = −11 or w = 8[1]
5. Width = 8 m[1]
6. Length = 11 m[1]
7. A width cannot be negative, so w = −11 is rejected[1]

8 m by 11 m

1. Factorises the denominator: x² − 16 = (x − 4)(x + 4)[1]
2. Cancels the common factor (x + 4)[1]
3. = 1/(x − 4)[1]
4. x = −4cancelling removed a genuine restriction[1]
5. 1/(x − 4) = 1/2 → x − 4 = 2[1]
6. x = 6[1]

(a) 1/(x − 4) (b) x = −4 (c) x = 6

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

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

A scalar has magnitude only — for example a mass of 5 kg. A vector has both magnitude and direction — for example a displacement of 5 m due east.

|u| = √(9 + 16) = 5, and the unit vector is ⟨3/5, −4/5⟩.

BA = ⟨−4, 2⟩. It is the same length but in the exactly opposite direction, so BA = −AB.

1. PQ = ⟨5 − 1, 6 − 3⟩ = ⟨4, 3⟩head minus tail[1]
2. QR = ⟨9 − 5, 9 − 6⟩ = ⟨4, 3⟩[1]
3. PQ = QR, so PQ is a scalar multiple of QR with k = 1, hence they are parallel[1]
4. They share the point Q, so P, Q and R lie on a single straight line and are collinearthe shared point is essential — parallel alone is not enough[1]

PQ = QR = ⟨4, 3⟩ and they share Q, so the three points are collinear (with Q the midpoint of PR).

1. 2a = ⟨4, 10⟩both components multiplied[1]
2. 3b = ⟨−9, 3⟩[1]
3. 2a − 3b = ⟨4 − (−9), 10 − 3⟩ = ⟨13, 7⟩subtracting a negative is where marks are lost[1]
4. |2a − 3b| = √(169 + 49) = √218 ≈ 14.8exact surd or 3 s.f. both accepted[1]

⟨13, 7⟩, magnitude √218 ≈ 14.8

1. AB = ⟨5 − 1, 2 − 1⟩ = ⟨4, 1⟩[1]
2. DC = ⟨6 − 2, 6 − 5⟩ = ⟨4, 1⟩take the vertices in the order that makes DC correspond to AB[1]
3. AB = DC, so AB and DC are equal in length and parallel[1]
4. One pair of opposite sides equal and parallel is sufficient for a parallelogramthe reason must be stated, not just the equality[1]
5. The diagonals of a parallelogram bisect each other, so they meet at the midpoint of ACor the midpoint of BD, which gives the same point[1]
6. Midpoint = (a + c)/2 = (⟨1,1⟩ + ⟨6,6⟩)/2 = ⟨3.5, 3.5⟩checking with BD: (⟨5,2⟩ + ⟨2,5⟩)/2 = ⟨3.5, 3.5⟩ ✓[1]

(a) AB = DC = ⟨4, 1⟩ (b) equal and parallel, so a parallelogram (c) ⟨3.5, 3.5⟩

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°

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

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

Join the points to form two chords and construct the perpendicular bisector of each. They meet at the circumcentre, which is equidistant from all three points.

At the incentre, where the three angle bisectors meet. A point equidistant from two lines lies on the bisector of the angle between them, so the intersection of two bisectors is equidistant from all three sides.

Join the centre O to T, then construct the perpendicular to OT at T. That perpendicular is the required tangent, since a tangent is always perpendicular to the radius at the point of contact.

1. Triangle constructed accurately by SSS, with arcs visibledraw the longest side first[1]
2. Perpendicular bisector of one side constructed with equal arcs[1]
3. Perpendicular bisector of a second side, meeting the first at the circumcentre O[1]
4. Circle drawn through all three vertices; radius approximately 3.6 cmaccept 3.5–3.7 cm[1]

Circumradius ≈ 3.6 cm

1. Circle and point P marked with OP = 10 cmP is outside since 10 > 6[1]
2. Midpoint M of OP found by perpendicular bisectorarcs must be visible[1]
3. Circle centre M radius 5 cm drawn, cutting the given circle at the two points of contactthe angle in a semicircle guarantees the right angle[1]
4. Tangent length = √(10² − 6²) = √64 = 8 cmmeasurement should agree to within 1 mm[1]

Each tangent is 8 cm long.

1. 60° angle constructed at Q with compasses, not measured with a protractorthe equilateral-triangle construction[1]
2. PQ = 8 cm and QR = 6 cm marked off, and PR joined[1]
3. Angle bisector of one angle constructed with arcs[1]
4. Angle bisector of a second angle, meeting the first at the incentre I[1]
5. Perpendicular dropped from I to a side to obtain the radius, and the circle drawn touching all three sidesmeasuring from I to a vertex is the standard error here[1]
6. An angle bisector always lies inside the triangle, so the bisectors must meet inside; a perpendicular bisector is not confined to the interior, so in an obtuse triangle the circumcentre falls outside[1]

(c) angle bisectors are interior lines, so their intersection is interior; perpendicular bisectors are not, so the circumcentre can lie outside.

A bar chart displays discrete categories with gaps between the bars, and only the heights are meaningful. A histogram displays continuous data with no gaps, and it is the area of each bar that represents the frequency.

Because the cumulative total for a class counts every observation up to and including the top of that class. The count "36 or fewer" is only complete at the upper boundary, so plotting at the midpoint would claim the total was reached earlier than it was.

IQR = Q₃ − Q₁, the spread of the middle 50% of the data. Unlike the range it is unaffected by extreme values, since it depends only on the quartiles and ignores the top and bottom quarters entirely.

1. Midpoints 2.5, 7.5, 12.5, 17.5, 22.5[1]
2. fx values 15, 105, 225, 157.5, 67.5 with Σfx = 570[1]
3. x̄ = 570 / 50 = 11.4 kgaccept 11.4 kg[1]
4. Modal class 10–15 kg, having the greatest frequency of 18a class, not a single value[1]

Estimated mean 11.4 kg; modal class 10–15 kg

1. IQR = 68 − 42 = 26 marks[1]
2. Semi-IQR = 26 / 2 = 13 marks[1]
3. Q₃ is the value below which three quarters of the data lie, so 3/4 × 80 = 60 students scored 68 or less[1]
4. 80 − 60 = 20 students scored above 68a quarter of the data lies above Q₃, by definition[1]

IQR 26 marks; semi-IQR 13 marks; 20 students above 68

1. Cumulative frequencies 10, 35, 70, 90, 100the last value must equal 100[1]
2. Points plotted at the upper boundaries 200, 400, 600, 800, 1000, starting from (0, 0)[1]
3. Draw a horizontal line at N/2 = 50 to the curve and read down to the horizontal axisthe median is about 490 hours[1]
4. Midpoints 100, 300, 500, 700, 900 with Σfx = 1000 + 7500 + 17500 + 14000 + 9000 = 49 000[1]
5. x̄ = 49 000 / 100 = 490 hours[1]
6. The mean and median are almost identical, which indicates that the distribution is roughly symmetrical with no strong skewthe comment is required, not just the two numbers[1]

(a) 10, 35, 70, 90, 100 (b) read across at 50 and down to the axis (c) mean ≈ 490 h, almost equal to the median, so the data is nearly symmetrical

Mutually exclusive events cannot occur together, so P(A ∩ B) = 0. Independent events can occur together, but one occurring does not affect the probability of the other, so P(A ∩ B) = P(A) × P(B).

1/2. Successive tosses are independent, so the coin has no memory of previous results and the probability is unchanged. Expecting a tail to "even things out" is the gambler's fallacy.

No. Removing the first ball changes both the number of balls left and the composition of the bag, so the probability for the second draw depends on what the first was. The draws are dependent, and the second-stage probabilities are conditional.

1. P(heart) = 13/52 and P(face) = 12/52twelve face cards: three in each of four suits[1]
2. The events are not mutually exclusive; there are 3 cards that are bothjack, queen and king of hearts[1]
3. Uses P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 13/52 + 12/52 − 3/52[1]
4. = 22/52 = 11/26[1]

11/26

1. Independent, so P(both) = 0.8 × 0.6[1]
2. = 0.48[1]
3. P(neither) = 0.2 × 0.4 = 0.08multiply the two failure probabilities[1]
4. P(at least one) = 1 − 0.08 = 0.92accept the direct route 0.48 + 0.32 + 0.12 = 0.92[1]

(i) 0.48 (ii) 0.92

1. First stage: P(good) = 6/10, P(defective) = 4/10the two branches must sum to 1[1]
2. Second stage after a good bulb: 5/9 good, 4/9 defective; after a defective: 6/9 good, 3/9 defectivedenominator 9 throughout, since one bulb has gone[1]
3. P(both good) = (6/10) × (5/9)multiply along the branch[1]
4. = 30/90 = 1/3[1]
5. Uses the complement: P(at least one defective) = 1 − P(both good)"at least one defective" is the exact opposite of "both good"[1]
6. = 1 − 1/3 = 2/3[1]

(b) 1/3 (c) 2/3