59 questions. Try each one, then open the answer to check yourself and read the explanation.
Which of the following is equal to 72/125?
Answer: A ((23 × 32)/53)
72 = 8 × 9 = 2³ × 3² and 125 = 5³, so 72/125 = (2³ × 3²)/5³.
Express 8.75/0.025 in standard form.
Answer: D (3.5 × 102)
8.75 ÷ 0.025 = 8750 ÷ 25 = 350, and 350 = 3.5 × 10².
Evaluate 271/3/16-1/4
Answer: A (6)
271/3 = 3 and 16-1/4 = 1/2, so the value is 3 ÷ 1/2 = 6.
Simplify: 165/4 × 2-3 × 30
Answer: C (4)
165/4 = 25 = 32, 2-3 = 1/8 and 30 = 1, so 32 × 1/8 × 1 = 4.
Simplify: 2log36 + log312 − log316
Answer: B (3)
Combine the logs: log3(36 × 12 ÷ 16) = log327 = 3.
What is the number whose logarithm to base 10 is 3̄.4771?
Answer: D (0.003)
The mantissa 0.4771 is log 3, and the characteristic bar 3 (that is, −3) puts the first significant figure in the third decimal place, so the number is 0.003.
A house bought for N100,000 was later auctioned for N80,000. Find the loss percent.
Answer: A (20%)
Loss = N20,000, and 20,000/100,000 × 100% = 20%.
A string is 4.8m. A boy measured it to be 4.95m. Find the percentage error.
Answer: D (3⅛%)
Error = 4.95 − 4.8 = 0.15 m. Percentage error = 0.15/4.8 × 100% = 3.125% = 3⅛%.
The sum of the 1st and 2nd terms of an A.P. is 4 and the 10th term is 19. Find the sum of the 5th and 6th terms.
Answer: B (20)
a + (a + d) = 4 and a + 9d = 19 give d = 2 and a = 1. Then T5 + T6 = (1 + 8) + (1 + 10) = 20.
If the 2nd and 5th terms of a G.P. are 6 and 48 respectively, find the sum of the first four terms.
Answer: E (45)
ar = 6 and ar⁴ = 48, so r³ = 8, r = 2 and a = 3. Sum of 4 terms = 3(2⁴ − 1)/(2 − 1) = 45.
Given that 2p − m = 6 and 2p + 4m = 1, find the value of (4p + 3m).
Answer: D (7)
Subtracting the equations gives 5m = −5, so m = −1 and p = 2.5. Then 4p + 3m = 10 − 3 = 7.
Which of the following is a point on the curve y = x2 − 4x + 7?
Answer: B ((1,4))
At x = 1, y = 1 − 4 + 7 = 4, so (1, 4) is on the curve. The other points do not satisfy the equation.
If 8x − 4 = 6x − 10, find the value of 5x.
Answer: B (−15)
8x − 6x = −10 + 4, so 2x = −6 and x = −3. Then 5x = −15.
For what value of x is the expression (x2 + 15x + 50)/(x − 5) not defined?
Answer: D (5)
A fraction is not defined when its denominator is zero: x − 5 = 0 when x = 5.
If x is positive, for what range of values of x is 4 + 3x < 10?
Answer: A (0 < x < 2)
3x < 6 gives x < 2. Since x is positive, x > 0 as well, so 0 < x < 2.
In the graph above, the gradient of the curve at the point P is
Answer: A (−1 1/3)
The gradient at P is the gradient of the tangent drawn at P. The tangent passes through about (−3, 0) and (0, −4), so the gradient = (−4 − 0)/(0 − (−3)) = −4/3 = −1 1/3.
Use the graph to answer this question. What is the equation of the line PQ?
Answer: C (y = x + 5)
P(−2, 3) and Q(2, 7): gradient = (7 − 3)/(2 − (−2)) = 1. Using y = x + c with (2, 7) gives c = 5, so y = x + 5.
Use the graph to answer this question. Which of the following equations can be solved by the points of intersection P and Q of the curve and the line PQ?
Answer: B (x2 − 4 = 0)
The curve is y = x² + x + 1 and the line is y = x + 5. Where they meet, x² + x + 1 = x + 5, which simplifies to x² − 4 = 0.
Which of the following is not a factor of 2p2 − 2?
Answer: E (2p + 1)
2p² − 2 = 2(p − 1)(p + 1), so 2, p − 1, p + 1 and 2(p − 1) = 2p − 2 are all factors; 2p + 1 is not.
Find the quadratic equation whose roots are 3 and 2/3.
Answer: E (3x2 − 11x + 6 = 0)
Sum of roots = 11/3 and product = 2, so x² − (11/3)x + 2 = 0. Multiply by 3: 3x² − 11x + 6 = 0.
Which of the following is a root of the equation x2 + 6x = 0?
Answer: A (0)
x² + 6x = x(x + 6) = 0, so x = 0 or x = −6. Only 0 is listed.
Factorise: 6x2 + 7xy − 5y2
Answer: D ((3x + 5y)(2x − y))
(3x + 5y)(2x − y) = 6x² − 3xy + 10xy − 5y² = 6x² + 7xy − 5y².
Solve the equation log8x − 4log8x = 2
Answer: A (1/4)
−3log8x = 2, so log8x = −2/3 and x = 8-2/3 = 1/4.
The angle subtended at the centre by a chord of a circle radius 6cm is 120°. Find the length of the chord.
Answer: E (6√3cm)
Half the chord = 6 sin 60° = 3√3 cm, so the chord = 6√3 cm.
A cuboid of base 12.5cm by 20cm holds exactly 1 litre of water. What is the height of the cuboid? (1 litre = 1000cm3)
Answer: B (4cm)
Base area = 12.5 × 20 = 250 cm², so height = 1000 ÷ 250 = 4 cm.
Two ships on the equator are on longitudes 45°W and 45°E respectively. How far are they apart along the equator, correct to 2 significant figures? (Take the radius of earth = 6400km and π = 22/7)
Answer: B (10,000km)
The difference in longitude is 90°. Distance = 90/360 × 2 × 22/7 × 6400 ≈ 10 057 km ≈ 10,000 km (2 s.f.).
Calculate, correct to 2 significant figures, the length of the arc of a circle of radius 3.5cm which subtends an angle of 75° at the centre of the circle. (Take π = 22/7).
Answer: B (4.6cm)
Arc = 75/360 × 2 × 22/7 × 3.5 = 75/360 × 22 ≈ 4.58 cm ≈ 4.6 cm.
A circle has radius x cm. What is the area of a sector of the circle with angle 135°, leaving the answer in terms of x and π?
Answer: B (3πx2/8)
Area = 135/360 × πx² = 3/8 × πx² = 3πx²/8.
In the diagram, PQRS is a trapezium with PQ||SR. STR is a straight line, PS||QT, PT||QR. If |PQ| = 12cm and the height of the trapezium is 9cm, find its area.
Answer: D (162cm2)
PQTS and PQRT are parallelograms, so ST = TR = PQ = 12 cm and SR = 24 cm. Area = ½(12 + 24) × 9 = 162 cm².
Two solid spheres have volumes 250cm3 and 128cm3 respectively. Find the ratio of their radii.
Answer: E (5 : 4)
Volumes are in the ratio of the cubes of the radii: R³ : r³ = 250 : 128 = 125 : 64, so R : r = 5 : 4.
In the diagram ST and QR are parallel. |PS| = 6cm, |SQ| = 8cm and |PR| = 18 2/3. Find |PT|
Answer: B (8cm)
Triangles PST and PQR are similar, so PT/PR = PS/PQ = 6/14. PT = 6/14 × 56/3 = 8 cm.
In the diagram, PQ||RS and the angles are as shown. Find x.
Answer: E (165°)
Draw a line through T parallel to PQ. It splits the 55° angle into 180° − 140° = 40° and 180° − x. So 40° + (180° − x) = 55°, giving x = 165°.
If the exterior angles of a quadrilateral are y°, (2y + 5)°, (y + 15)° and (3y − 10)°, find y°.
Answer: E (50.)
Exterior angles add up to 360°: 7y + 10 = 360, so y = 50.
The sum of the angles of a regular polygon is 2520°. How many sides does the polygon have?
Answer: D (16)
(n − 2) × 180° = 2520°, so n − 2 = 14 and n = 16.
Which of the following is not a property of all parallelograms?
Answer: C (The diagonals cut at right angles)
Diagonals cut at right angles only in special parallelograms such as the rhombus and square, not in every parallelogram.
In the diagram, PQ is the tangent to the circle RST at T. |ST| = SR and ∠RTQ = 68°. Find ∠PTS.
Answer: D (56°)
By the alternate segment theorem, ∠TSR = ∠RTQ = 68°. Triangle STR is isosceles, so ∠STR = (180° − 68°)/2 = 56°. Then ∠PTS = 180° − 68° − 56° = 56°.
The diagram below shows a circle PQRS in which ∠PRQ = 54° and ∠SPQ = 97°, find ∠PQS
Answer: E (29°)
∠PSQ = ∠PRQ = 54° (angles in the same segment). In triangle PQS, ∠PQS = 180° − 97° − 54° = 29°.
In the diagram, |PS| is a diameter of circle PQRS. |PQ| = |QR| and ∠RSP = 74°. Find ∠QPS.
Answer: D (53°)
∠PQR = 180° − 74° = 106° (cyclic quadrilateral), so ∠QPR = (180° − 106°)/2 = 37°. ∠PRS = 90° (angle in a semicircle), so ∠RPS = 90° − 74° = 16°. ∠QPS = 37° + 16° = 53°.
If sinθ = k, find tanθ, 0° ≤ θ ≤ 90°
Answer: C (k/√(1 − k2))
Take opposite = k and hypotenuse = 1; then adjacent = √(1 − k²), so tanθ = k/√(1 − k²).
A ladder 6m long leans against a vertical wall so that it makes an angle of 60° with the wall. Calculate the distance of the foot of the ladder from the wall.
Answer: D (3√3m)
The distance from the wall is opposite the 60° angle at the wall: 6 sin 60° = 6 × √3/2 = 3√3 m.
The bearing of Q from P is 122°, what is the bearing of P from Q?
Answer: E (302°)
The back bearing is 122° + 180° = 302°.
Cos 57° has the same value as
Answer: C (sin 147°)
cos 57° = sin 33°, and sin 147° = sin(180° − 33°) = sin 33°. The other options are all negative.
The table shows the masses to the nearest kilogram, of 40 students in a university.
Mass(kg): 65, 66, 67, 68, 69, 70
No. of Students: 4, 5, 11, 8, 7, 5
The percentage of the students with mass less than 69kg is
Answer: D (70%)
Students under 69 kg: 4 + 5 + 11 + 8 = 28. 28/40 × 100% = 70%.
The mean heights of the three groups of students consisting of 20, 16 and 14 students are 1.67m, 1.50m and 1.40m respectively. Find the mean height of all the students.
Answer: C (1.54m)
Total height = 20 × 1.67 + 16 × 1.50 + 14 × 1.40 = 77 m for 50 students, so the mean = 77 ÷ 50 = 1.54 m.
Two cards are drawn one after the other with replacement from a well-shuffled ordinary pack of 52 cards containing four Aces. Find the probability that they are both Aces.
Answer: B (1/169)
With replacement each draw has probability 4/52 = 1/13, so both Aces = 1/13 × 1/13 = 1/169.
Find the probability that a number selected from the numbers 30 to 50 inclusive is a prime.
Answer: B (5/21)
There are 21 numbers from 30 to 50. The primes are 31, 37, 41, 43 and 47, so the probability is 5/21.
The length of an exercise book is given as 20cm correct to the nearest centimetre. In which of the following ranges of possible measurements does the actual length lie?
Answer: A (19.5 − 20.5cm)
To the nearest centimetre, the length can be up to 0.5 cm either side of 20 cm: 19.5 cm to 20.5 cm.
(a) Given that 3 × 91+x = 27-x, find x.
(b) Evaluate log10√35 + log10√2 − log10√7.
(a) Write everything as powers of 3: 3 × 32(1+x) = 3-3x
So 1 + 2 + 2x = −3x, 5x = −3, x = −3/5
(b) Combine: log10(√35 × √2 ÷ √7) = log10√10
= ½ log1010 = ½
Use a common base for indices, and the product/quotient laws for logs.
(a) Evaluate, without using mathematical tables, 17.572 − 12.432.
(b) Prove that angles in the same segment of a circle are equal.
(a) Difference of two squares: (17.57 − 12.43)(17.57 + 12.43) = 5.14 × 30 = 154.2
(b) Diagram: circle centre O, chord AB, points P and Q on the same arc; join OA and OB.
Given: ∠APB and ∠AQB in the same segment. To prove: ∠APB = ∠AQB.
∠AOB = 2∠APB (angle at centre is twice angle at circumference)
∠AOB = 2∠AQB (same reason)
So ∠APB = ∠AQB.
For (b), the angle-at-the-centre theorem does all the work; state the reason for each step.
(a) A tower and a building stand on the same horizontal level. From the point P at the bottom of the building, the angle of elevation of the top T, of the tower is 65°. From the top Q of the building, the angle of elevation of the point T is 25°. If the building is 20m high, calculate the distance PT.
(b) Hence or otherwise, calculate the height of the tower. (Give your answers correct to 3 significant figures.)
(a) In triangle PQT: ∠QPT = 90° − 65° = 25°, ∠PQT = 90° + 25° = 115°, so ∠PTQ = 40°
Sine rule: PT/sin 115° = 20/sin 40°
PT = 20 × 0.9063/0.6428 = 28.2 m
(b) Height of tower = PT sin 65° = 28.2 × 0.9063 = 25.6 m
Sketch the building PQ and tower first; the key is finding the angles of triangle PQT.
(a) Divide 11111111two by 101two
(b) A sector of a circle 6cm has an angle of 105° at the centre. Calculate its (i) perimeter (ii) area (Take π = 22/7)
(a) 11111111two = 255 and 101two = 5; 255 ÷ 5 = 51 = 110011two (or by long division in base two)
(b)(i) Arc = 105/360 × 2 × 22/7 × 6 = 11 cm; perimeter = 11 + 6 + 6 = 23 cm
(ii) Area = 105/360 × 22/7 × 62 = 33 cm2
The perimeter of a sector includes the two radii as well as the arc.
The table below gives the frequency distributions of marks obtained by some students in a scholarship examination.
Scores (x): 15, 25, 35, 45, 55, 65, 75
Freq. (f): 1, 4, 12, 24, 18, 8, 3
(a) Calculate, correct to 3 significant figures, the mean mark.
(b) Find the: (i) mode, (ii) range of the distribution.
(a) Σf = 70; Σfx = 15 + 100 + 420 + 1080 + 990 + 520 + 225 = 3350
Mean = 3350/70 = 47.86 ≈ 47.9
(b)(i) Mode = 45 (highest frequency, 24)
(ii) Range = 75 − 15 = 60
Set out a column of fx; the mean is Σfx ÷ Σf.
(a) if log10(3x − 1) − log102 = 3, find the value x.
(b) Use logarithm tables to evaluate √(0.897 × 3.536)/0.00249, correct to 3 significant figures
(a) log10((3x − 1)/2) = 3, so (3x − 1)/2 = 1000
3x − 1 = 2000, x = 2001/3 = 667
(b) log 0.897 = 1̄.9528, log 3.536 = 0.5485; sum = 0.5013
÷ 2 for the square root: 0.2507 (antilog ≈ 1.781)
Subtract log 0.00249 = 3̄.3962: 0.2507 − 3̄.3962 = 2.8545
Antilog = 715 (3 s.f.)
Take the square root by halving the log before dividing by 0.00249.
A bag contains 12 white balls and 8 black balls, another contains 10 white balls and 15 black balls. If two balls are drawn, without replacement, from each bag, find the probability that (a) all four balls are black (b) exactly one of the four balls is white.
(a) Bag 1 both black: 8/20 × 7/19 = 14/95; Bag 2 both black: 15/25 × 14/24 = 7/20
P(all black) = 14/95 × 7/20 = 49/950
(b) P(exactly one white from bag 1) = 2 × 12/20 × 8/19 = 48/95; P(exactly one white from bag 2) = 2 × 10/25 × 15/24 = 1/2
P(exactly one white) = P(1 white from bag 1, both black from bag 2) + P(both black from bag 1, 1 white from bag 2)
= 48/95 × 7/20 + 14/95 × 1/2 = 336/1900 + 140/1900 = 476/1900 = 119/475 (≈ 0.251)
Treat each bag separately, then multiply; for 'exactly one' add the two ways it can happen.
(a) Using a ruler and a pair of compasses only, construct: (i) a traingle XYZ in which |YZ| = 8cm, XYZ = 60° and XZY = 75°. Measure |XY|; (ii) the locus l1 of points equidistant from Y and Z; (iii) the locus l2 of points equidistant from YX and YZ.
(b) Measure QY where Q is the point of intersection of l1 and l2
(a)(i) Draw YZ = 8 cm; construct 60° at Y and 75° at Z (60° + bisected 30°, i.e. 60° + 15°); the arms meet at X.
|XY| ≈ 10.9 cm (check: 8 sin 75°/sin 45° = 10.93)
(ii) l1 is the perpendicular bisector of YZ.
(iii) l2 is the bisector of angle XYZ.
(b) Q lies on the bisector (30° to YZ) above the midpoint of YZ: QY ≈ 4.6 cm (4/cos 30° = 4.62)
Leave all construction arcs visible; a locus equidistant from two lines is an angle bisector.
The table below gives the ages, to the nearest 5 years, of 50 people.
Age in years: 10, 15, 20, 25, 30
No. of People: 8, 19, 10, 7, 6
(a) Construct a cumumlative frequency table for the distribution. (b) Draw a cumulative frequency curve (ogive). (c) From your ogive, find the: (i) median age: (ii) number of people who are at most 15 years of age. (iii) number of people who are between 20 and 25 years of age.
(a) Upper class boundaries 12.5, 17.5, 22.5, 27.5, 32.5 with cumulative frequencies 8, 27, 37, 44, 50
(b) Plot cumulative frequency against upper class boundary (start at (7.5, 0)) and join with a smooth S-shaped curve.
(c)(i) Median at the 25th person: about 17 years (12.5 + 17/19 × 5 ≈ 17.0)
(ii) At most 15 years (read at 17.5): 27 people
(iii) Read the curve at 20 and 25 years (about 32 and 40): about 8 people
Ages 'to the nearest 5 years' mean each class spans 2.5 years either side, so plot at upper class boundaries.
(a) Copy and complete the following table of values for y = 2x2 − 9x − 1.
x: −1, 0, 1, 2, 3, 4, 5, 6
y: __, −1, −8, −11, __, __, __, 17
(b) Using a scale of 2cm to represent 1 unit on the x-axis and 2cm to represent 5 unit on the y-axis, draw the graph of y = 2x2 − 9x − 1 for −1 ≤ x ≤ 6.
(c) Use your graph to find the: (i) roots of the equation 2x2 − 9x = 4, correct to one decimal place; (ii) gradient of the curve y = 2x2 − 9x − 1 at x = 3.
(a) x = −1: y = 10; x = 3: y = −10; x = 4: y = −5; x = 5: y = 4
(b) Plot the points with the given scales and join with a smooth U-shaped curve.
(c)(i) 2x2 − 9x = 4 means 2x2 − 9x − 1 = 3: draw y = 3 and read x ≈ −0.4 and x ≈ 4.9
(ii) Draw the tangent at x = 3 and find rise ÷ run: gradient ≈ 3
Rewrite the new equation so its left side matches the graph; the right side is the horizontal line to draw.
(a) The fourth term of an A.P. is 37 and the 6th term is 12 more than the fouth term. Find the first and seventh terms
(b) If P = (1,2,3,4) and Q = (3,5,6), find: (i) P∩Q; (ii) P∪Q (iii) (P∩Q)∪Q; (iv) (P∪Q)∩P.
(a) a + 3d = 37 and (a + 5d) − (a + 3d) = 12, so 2d = 12, d = 6
a = 37 − 18 = 19 (first term); seventh term = 19 + 6 × 6 = 55
(b)(i) P∩Q = {3}
(ii) P∪Q = {1, 2, 3, 4, 5, 6}
(iii) (P∩Q)∪Q = {3, 5, 6}
(iv) (P∪Q)∩P = {1, 2, 3, 4}
nth term of an A.P. is a + (n − 1)d.
(a) Two points X(32°N, 47°W) and Y(32°N, 25°E), are on the earth's surface. If it takes an aeroplane 11 hours to fly from X to Y along the parallel of latitude, calculate its speed, correct to the nearest kilometre per hour (Take the radius of the earth = 6400km and π = 22/7).
(b) Two observers P and Q, 15 metres apart observe a kite (K) in the same vertical plane and from the same side of the kite. The angles of elevation of the kite from P and Q 35° and 45° respectively. Find the height of the kite to the nearest metre.
(a) Difference in longitude = 47° + 25° = 72°
Distance = 72/360 × 2 × 22/7 × 6400 × cos 32° ≈ 8045.7 × 0.8480 ≈ 6823 km
Speed = 6823 ÷ 11 ≈ 620 km/h
(b) Let height = h. From Q (45°), horizontal distance = h; from P, distance = h + 15
tan 35° = h/(h + 15): 0.7002(h + 15) = h
0.2998h = 10.50, h ≈ 35 m
Along a parallel of latitude the radius is R cos(latitude).