9th Class Mathematics — 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

If ax = y, where a > 0 and a ≠ 1, then x is called the logarithm of y to the base a, written loga y = x. It is the power to which the base must be raised to give the number.

Because 1x = 1 for every value of x. No power of 1 can produce any number other than 1, so log₁ y would have no value for y ≠ 1 and infinitely many for y = 1.

In standard form 0.00456 = 4.56 × 10⁻³, so the characteristic is −3, written 3̄. The characteristic equals the power of 10 in standard form.

1. log₂ 64 = 6 because 2⁶ = 64[1]
2. log₃ 81 = 4 because 3⁴ = 81[1]
3. log₅ 125 = 3, so the value is 6 + 4 − 3 = 7final answer required[1]

7

1. Combine using the product law: log₂[(x+3)(x−3)] = 4sum of logs with the same base becomes the log of a product[1]
2. Convert to exponential form: (x+3)(x−3) = 2⁴ = 16[1]
3. x² − 9 = 16, so x² = 25 and x = ±5difference of two squares[1]
4. Reject x = −5 because it makes x − 3 negative and the log undefined; x = 5the rejection must be stated with a reason[1]

x = 5

1. N = 500 × 2⁶ = 500 × 64[1]
2. N = 32 000 bacteria[1]
3. 32 000 = 500 × 2ᵗ so 2ᵗ = 64divide by the initial amount first[1]
4. Take logs: t log 2 = log 64power law brings t down[1]
5. t = log 64 / log 2 = 1.8062 / 0.3010 = 6.00 hoursaccept t = 6 from inspection with working shown[1]
6. In (a) the exponent is known and the answer is a direct calculation; in (b) the unknown is the exponent itself, and logarithms are the only way to bring it down to where it can be solvedthe explanation must mention that the unknown is in the exponent[1]

(a) 32 000 (b) t = 6.00 hours (c) in (b) the unknown is the exponent

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

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)

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

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°

|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 statement is a declarative sentence that is either true or false, but not both. "Open the window" is not a statement, because a command cannot be assigned a truth value.

Converse: if a number is divisible by 3, then it is divisible by 6. Contrapositive: if a number is not divisible by 3, then it is not divisible by 6.

Only when p is true and q is false. In all three other combinations the conditional is true, including both cases where p is false.

1. Four rows listed in a systematic order: TT, TF, FT, FFa missing row loses this mark even if the rest is right[1]
2. Column for ~(p ∧ q): F, T, T, Tp ∧ q is T only in the first row, so its negation is F only there[1]
3. Column for ~p ∨ ~q: F, T, T, T[1]
4. The two columns are identical, so the statements are logically equivalent — this is De Morgan's lawnaming the law is not required, but the equivalence statement is[1]

Both columns read F, T, T, T, so ~(p ∧ q) ≡ ~p ∨ ~q.

1. Four rows, with the column for p ∧ q: T, F, F, F[1]
2. Row 1: T → T = T[1]
3. Rows 2–4 all have a false hypothesis, so each conditional is Tthis is where vacuous truth is being tested[1]
4. The final column is T, T, T, T, so the statement is a tautology[1]

A tautology — the final column is all T.

1. Let p: the quadrilateral is a square; q: the quadrilateral is a rectangle. Statement: p → qboth p and q must be defined[1]
2. Converse: q → p — if a quadrilateral is a rectangle then it is a square[1]
3. The converse is false[1]
4. Counterexample: a 2 cm by 5 cm rectangle is not a squarea counterexample is required, not just the verdict[1]
5. Contrapositive: ~q → ~p — if a quadrilateral is not a rectangle then it is not a square[1]
6. A conditional and its contrapositive have identical truth tables, so they are logically equivalent and must always agreeaccept an argument from the truth table[1]

(a) p → q (b) q → p, false — a 2 × 5 rectangle is not a square (c) ~q → ~p, equivalent to the original

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

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

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

Class limits are the values written in the table, such as 20–29. Class boundaries are the true dividing values, 19.5 and 29.5, obtained by going half a unit beyond each limit so that consecutive classes meet with no gap.

The mean uses every value in its calculation, so one very large value pulls it far above the bulk of the data and it no longer represents a typical member of the set. The median should be used instead, because it depends only on position and is unaffected by extremes.

The mode, which is size 8. The manager needs to know which size sells most often; a mean of 8.14 is not a size that can be ordered.

1. Σx = 12 + 15 + 11 + 15 + 18 + 14 + 15 + 20 = 120, n = 8[1]
2. Mean = 120/8 = 15[1]
3. In order: 11, 12, 14, 15, 15, 15, 18, 20. Median is the mean of the 4th and 5th values = (15 + 15)/2 = 15the data must be ordered first — this is where marks are lost[1]
4. Mode = 15, occurring three times[1]

Mean = 15, median = 15, mode = 15

1. Midpoints 12, 17, 22, 27, 32midpoint of 10–14 is (10 + 14)/2 = 12[1]
2. fx values 48, 119, 242, 162, 64, so Σfx = 635[1]
3. x̄ = 635 / 30 = 21.2 cm (1 d.p.)accept 21.17[1]
4. Modal class is 20–24, the class with the highest frequencythe class is required, not a single value[1]

Estimated mean ≈ 21.2 cm; modal class 20–24

1. Ali: Σx = 300, mean = 300/5 = 60[1]
2. Bilal: Σx = 300, mean = 300/5 = 60the means are identical, which is the point of the question[1]
3. Ali's range = 62 − 58 = 4[1]
4. Bilal's range = 80 − 40 = 40[1]
5. Ali is more consistent[1]
6. Both have the same mean, so the mean cannot separate them; Ali's much smaller range shows his marks are clustered close together, while Bilal's vary widelythe justification must refer to the equal means as well as the ranges[1]

(a) both 60 (b) Ali 4, Bilal 40 (c) Ali — same mean but far smaller spread

The sample space is the set of all possible outcomes of the experiment. For two coins, S = {HH, HT, TH, TT}, so n(S) = 4.

Two events are mutually exclusive if they cannot occur at the same time, so P(A and B) = 0. Example: on a single roll of a die, "getting a 2" and "getting a 5".

P(no rain) = 1 − 0.35 = 0.65, using the complement rule P(E′) = 1 − P(E).

1. Total n(S) = 5 + 4 + 3 = 12the total is the denominator throughout[1]
2. P(green) = 4/12 = 1/3lowest terms expected[1]
3. P(not blue) = 1 − 3/12 = 9/12 = 3/4accept counting 5 + 4 = 9 favourable outcomes directly[1]

(i) 1/3 (ii) 3/4

1. n(S) = 6 × 6 = 36ordered pairs, so (2,5) and (5,2) both count[1]
2. Total 7 arises from (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — six ways, so P = 6/36 = 1/67 is the most likely total on two dice[1]
3. At least 10 means 10, 11 or 12: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6) — six wayslisting them is safer than trying to count in your head[1]
4. P = 6/36 = 1/6[1]

(i) 1/6 (ii) 1/6

1. First-stage branches 7/10 white and 3/10 black[1]
2. Second-stage branches with denominator 9 throughout: 6/9, 3/9 after white; 7/9, 2/9 after blackone ball fewer, so the denominator drops to 9[1]
3. P(both white) = (7/10) × (6/9)multiply along the branch[1]
4. = 42/90 = 7/15[1]
5. Different colours = white then black, or black then white: (7/10)(3/9) + (3/10)(7/9)both orders are needed[1]
6. = 21/90 + 21/90 = 42/90 = 7/15[1]

(b) 7/15 (c) 7/15