60 questions. Try each one, then open the answer to check yourself and read the explanation.
Multiply 2.7 × 10−4 by 6.3 × 106 and leave your answer in standard form.
Answer: C (1.701 × 103)
2.7 × 6.3 = 17.01 and 10−4 × 106 = 102, so the product is 17.01 × 102 = 1.701 × 103.
If 9(2−x) = 3, find x.
Answer: B (3/2)
32(2 − x) = 31, so 4 − 2x = 1 and x = 3/2.
In what number base is the addition 465 + 24 + 225 = 1050?
Answer: D (Seven.)
In base seven: 5 + 4 + 5 = 14 = 2 sevens + 0; 6 + 2 + 2 + 2 = 12 = 1 seven + 5; 4 + 2 + 1 = 7 = 10. This gives 1050.
Simplify: (1 7/8 × 2 2/5)/(6¾ ÷ ¾)
Answer: D (½)
15/8 × 12/5 = 9/2 and 27/4 ÷ 3/4 = 9. So (9/2) ÷ 9 = ½.
If Un = n(n2 + 1), evaluate U5 − U4.
Answer: C (62)
U5 = 5 × 26 = 130 and U4 = 4 × 17 = 68. 130 − 68 = 62.
If √50 − K√8 = 2/√2, find K.
Answer: D (2)
5√2 − 2K√2 = √2, so 5 − 2K = 1 and K = 2.
A sales boy gave a change of ₦68 instead of ₦72. Calculate his percentage error.
Answer: B (5 5/9 %)
Error = 72 − 68 = ₦4. Percentage error = error ÷ true value × 100% = 4/72 × 100% = 5 5/9 %.
Four oranges sell for ₦x and three mangoes sell for ₦y. Olu bought 24 oranges and 12 mangoes. How much did he pay in terms of x and y?
Answer: B (₦(6x + 4y))
24 oranges are 6 lots of 4, costing 6x. 12 mangoes are 4 lots of 3, costing 4y. Total ₦(6x + 4y).
Simplify: (x2 − y2)/(x + y)2 ÷ (x − y)2/(3x + 3y)
Answer: C (3/(x − y))
(x − y)(x + y)/(x + y)2 = (x − y)/(x + y). Dividing by (x − y)2/3(x + y) gives (x − y)/(x + y) × 3(x + y)/(x − y)2 = 3/(x − y).
Solve the inequality: (2x − 5)/2 < (2 − x).
Answer: D (x < 2¼)
2x − 5 < 4 − 2x, so 4x < 9 and x < 9/4 = 2¼.
If x = 64 and y = 27, calculate: (x½ − y⅓)/(y − x⅔)
Answer: C (5/11)
x½ = 8, y⅓ = 3, x⅔ = 16. (8 − 3)/(27 − 16) = 5/11.
Which of the following lines represents the solution of the inequality 7x < 9x − 4?
Answer: A (Diagram A)
7x − 9x < −4 gives −2x < −4, so x > 2. That is an open circle at 2 with the arrow pointing right, as in A.
If ½x + 2y = 3 and (3/2)x − 2y = 1, find (x + y).
Answer: A (3)
Adding: 2x = 4, so x = 2. Then 1 + 2y = 3 gives y = 1. x + y = 3.
Given that p⅓ = ∛q/r, make q the subject of the equation.
Answer: C (q = pr3)
∛q = rp⅓. Cube both sides: q = r3p = pr3.
In the diagram, PRST is a square. If |PQ| = 24cm, |QR| = 10cm and ∠PQR = 90°; find the perimeter of the polygon PQRST.
Answer: A (112cm)
|PR| = √(242 + 102) = 26 cm, the side of the square. Perimeter = 24 + 10 + 3 × 26 = 112 cm.
In the diagram, the height of a flagpole (TF) and the length of its shadow (FL) are in the ratio 6:8. Using K as a constant of proportionality, find the shortest distance between T and L.
Answer: B (10K units)
TF = 6K and FL = 8K with a right angle at F. TL = √(36K2 + 64K2) = 10K.
A chord is 2cm from the centre of a circle. If the radius of the circle is 5cm, find the length of the chord.
Answer: A (2√21 cm)
The perpendicular from the centre bisects the chord. Half-chord = √(25 − 4) = √21, so the chord = 2√21 cm.
A cube and a cuboid have the same base area. The volume of the cube is 64 cm3 while that of the cuboid is 80 cm3. Find the height of the cuboid.
Answer: C (5 cm)
The cube has side 4 cm, so the base area is 16 cm2. Cuboid height = 80 ÷ 16 = 5 cm.
In the diagram, VX is a tangent to the circle UYW at W. If WY//UV, ∠UYW = 95° and ∠UWY = 46°, find ∠UVW.
Answer: C (39°)
WY ∥ UV gives ∠VUW = ∠UWY = 46° (alternate). The angle between tangent WV and chord WU equals the angle in the alternate segment: ∠UWV = 95°. So ∠UVW = 180° − 46° − 95° = 39°.
In the diagrams, |XZ| = |MN|, |ZY| = |MO| and |XY| = |NO|. Which of the following statements is true?
Answer: D (ΔXYZ ≡ ΔNOM)
Matching the equal sides: X ↔ N, Y ↔ O, Z ↔ M. So ΔXYZ ≡ ΔNOM (SSS).
In the diagram, PQRS is a rhombus and ∠PSQ = 35°. Calculate the size of ∠PRQ.
Answer: B (55°)
The diagonals cross at right angles, so in the small triangle at P, ∠SPR = 180° − 90° − 35° = 55°. PS ∥ QR, so ∠PRQ = ∠SPR = 55° (alternate angles).
Find the value of m in the diagram.
Answer: D (17°)
The two vertical lines are parallel. The bend of 153° turns the line by 180° − 153° = 27°, and the angles with the parallel lines give 7m + 2m + 27° = 180°. So 9m = 153° and m = 17°.
In the diagram, O is the centre of the circle, OM||XZ and ∠ZOM = 25°. Calculate ∠XYZ.
Answer: D (65°)
OM ∥ XZ gives ∠OZX = ∠ZOM = 25°. OX = OZ, so ∠OXZ = 25° and ∠XOZ = 130°. The angle at the circumference is half: ∠XYZ = 65°.
If sin x = 5/13 and 0° ≤ x ≤ 90°, find the value of (cos x − tan x).
Answer: C (79/156)
cos x = 12/13 and tan x = 5/12. 12/13 − 5/12 = (144 − 65)/156 = 79/156.
An object is 6 m away from the base of a mast. The angle of depression of the object from the top of the mast is 50°. Find, correct to 2 decimal places, the height of the mast.
Answer: C (7.15 m)
The angle of elevation of the top from the object is 50°: height = 6 tan 50° = 6 × 1.1918 ≈ 7.15 m.
The bearing of Y from X is 060° and the bearing of Z from Y = 060°. Find the bearing of X from Z.
Answer: B (240°)
X, Y and Z lie on one straight line in the direction 060°, so X is in the opposite direction from Z: 060° + 180° = 240°.
Which of the following is not a probability of Mary scoring 85% in a mathematics test?
Answer: D (1.01)
A probability cannot be greater than 1, so 1.01 is impossible.
Use the histogram to answer this question. Estimate the mode of the distribution.
Answer: C (53.3)
The tallest bar is 50.5–60.5. The lines joining its top corners to the neighbouring bars cross at about 53.3.
Use the histogram to answer this question. What is the median class?
Answer: B (50.5 − 60.5)
Cumulative frequencies: 10 (to 40.5), 22 (to 50.5), 37 (to 60.5). The middle students (24th and 25th of 48) are in the class 50.5 − 60.5.
If 2logx(3⅜) = 6, find the value of x.
Answer: A (3/2)
logx(27/8) = 3, so x3 = 27/8 and x = 3/2.
If P = {y: 2y ≥ 6} and Q = {y: y − 3 ≤ 4}, where y is an integer, find P∩Q.
Answer: C ({3, 4, 5, 6, 7})
P: y ≥ 3 and Q: y ≤ 7. The integers in both are 3, 4, 5, 6, 7.
Find the values of k in the equation 6k2 = 5k + 6.
Answer: B ({−2/3, 3/2})
6k2 − 5k − 6 = 0 factorises as (3k + 2)(2k − 3) = 0, so k = −2/3 or 3/2.
If y varies directly as the square root of (x + 1) and y = 6 when x = 3, find x when y = 9.
Answer: A (8)
y = k√(x + 1): 6 = 2k, so k = 3. 9 = 3√(x + 1) gives x + 1 = 9 and x = 8.
The graph of the relation y = x2 + 2x + k passes through the point (2, 0). Find the value of k.
Answer: D (−8)
0 = 4 + 4 + k, so k = −8.
The pie chart shows the distribution of 600 Mathematics textbooks for Arts, Business, Science and Technical classes. How many textbooks are for the Technical class?
Answer: D (250)
Technical angle = 360° − 90° − 75° − 45° = 150°. 150/360 × 600 = 250.
The pie chart shows the distribution of 600 Mathematics textbooks for Arts, Business, Science and Technical classes. What percentage of the total number of textbooks belongs to science?
Answer: A (12½%)
Science has 45° of 360°: 45/360 × 100% = 12½%.
In the diagram, PQRST is a regular polygon with sides QR and TS produced to meet at V. Find the size of ∠RVS.
Answer: A (36°)
Each exterior angle of a regular pentagon is 360° ÷ 5 = 72°. In triangle RVS, ∠VRS = ∠VSR = 72°, so ∠RVS = 180° − 144° = 36°.
What is the locus of the point X which moves relative to two fixed points P and M on a plane such that ∠PXM = 30°?
Answer: B (An arc of a circle with PM as a chord)
All points from which PM is seen at the same angle lie on an arc of a circle through P and M (angles in the same segment are equal).
In the diagram, PQ is a straight line. Calculate the value of the angle labelled 2y.
Answer: D (100°)
Below PQ: 80° + 50° + (x + 10)° = 180°, so x = 40. Above PQ: 2y + 2x = 180°, so 2y = 180° − 80° = 100°.
When a number is subtracted from 2, the result equals 4 less than one-fifth of the number. Find the number.
Answer: C (5)
2 − n = n/5 − 4, so 6 = 6n/5 and n = 5.
Express 2/(x + 3) − 1/(x − 2) as a simple fraction.
Answer: A ((x − 7)/(x2 + x − 6))
[2(x − 2) − (x + 3)]/[(x + 3)(x − 2)] = (x − 7)/(x2 + x − 6).
An interior angle of a regular polygon is 5 times each exterior angle. How many sides has the polygon?
Answer: B (12)
Exterior + 5 × exterior = 180°, so the exterior angle is 30° and n = 360 ÷ 30 = 12.
In the diagram, ST//PQ, reflex angle SRQ = 198° and ∠RQP = 72°. Find the value of y.
Answer: B (54°)
Draw a line through R parallel to ST and PQ. It splits the reflex angle into (180° − y), co-interior with y at S, and 72°, alternate to ∠RQP. So (180° − y) + 72° = 198°, giving y = 54°.
Using the Venn diagram, find n(X∩Y′).
Answer: A (2)
X∩Y′ is the part of X outside Y: {a, b}, which has 2 elements.
A pyramid has a rectangular base with dimensions 12 m by 8 m. If its height is 14 m, calculate the volume.
Answer: B (448 m3)
V = ⅓ × 12 × 8 × 14 = 448 m3.
The slant height of a cone is 5 cm and the radius of its base is 3 cm. Find, correct to the nearest whole number, the volume of the cone. [Take π = 22/7]
Answer: C (38 cm3)
Height = √(52 − 32) = 4 cm. V = ⅓ × 22/7 × 32 × 4 ≈ 37.7 ≈ 38 cm3.
The distance between two towns is 50 km. It is represented on a map by 5 cm. Find the scale used.
Answer: A (1 : 1,000,000)
50 km = 5,000,000 cm. 5 cm : 5,000,000 cm = 1 : 1,000,000.
An open cone with base radius 28 cm and perpendicular height 96 cm was stretched to form a sector of a circle. Calculate the area of the sector. [Take π = 22/7]
Answer: A (8800 cm2)
Slant height = √(282 + 962) = 100 cm. The sector is the cone's curved surface: πrl = 22/7 × 28 × 100 = 8800 cm2.
(a) Simplify, without using tables or calculator: ¾(3⅜ + 1⅝)/(2⅛ − 1½)
(b) Given that log102 = 0.3010 and log103 = 0.4771, evaluate correct to 2 significant figures and without using tables or calculator, log101.125.
(a) 3⅜ + 1⅝ = 5, so the numerator is ¾ × 5 = 15/4. Denominator: 17/8 − 12/8 = 5/8.
15/4 ÷ 5/8 = 15/4 × 8/5 = 6.
(b) 1.125 = 9/8 = 32/23.
log 1.125 = 2(0.4771) − 3(0.3010) = 0.9542 − 0.9030 = 0.0512 ≈ 0.051.
Write 1.125 as a fraction made of powers of 2 and 3.
(a) Solve: 7x + 4 < ½(4x + 3).
(b) Salem, Sunday and Shaka shared a sum of ₦1,100.00. For every ₦2.00 that Salem gets, Sunday gets 50 kobo and for every ₦4.00 Sunday gets, Shaka gets ₦2.00. Find Shaka's share.
(a) 14x + 8 < 4x + 3 → 10x < −5 → x < −½.
(b) Salem : Sunday = 2 : 0.5 = 4 : 1; Sunday : Shaka = 4 : 2 = 2 : 1.
So Salem : Sunday : Shaka = 8 : 2 : 1 (11 parts).
Shaka's share = 1/11 × 1100 = ₦100.00.
Make Sunday's part the same in both ratios before combining.
(a) The present ages of a father and his son are in the ratio 10 : 3. If the son is 15 years old now, in how many years will the ratio of their ages be 2 : 1?
(b) The arithmetic mean of x, y and z is 6 while that of x, y, z, t, u, v and w is 9. Calculate the arithmetic mean of t, u, v, and w.
(a) Father = 15 × 10/3 = 50. (50 + n)/(15 + n) = 2 → 50 + n = 30 + 2n → n = 20 years.
(b) x + y + z = 18 and the seven numbers total 7 × 9 = 63. t + u + v + w = 45, so their mean = 45/4 = 11.25.
Turn each mean back into a total first.
The area of a circle is 154 cm2. It is divided into three sectors such that two of the sectors are equal in size and the third sector is three times the size of the other two put together. Calculate the perimeter of the third sector. [Take π = 22/7]
22/7 × r2 = 154 → r2 = 49 → r = 7 cm.
Angles a, a and 3(2a) = 6a: 8a = 360° → a = 45°, so the third sector is 270°.
Arc = 270/360 × 2 × 22/7 × 7 = 33 cm.
Perimeter = 33 + 7 + 7 = 47 cm.
The perimeter of a sector is its arc plus two radii.
A boy 1.2 m tall, stands 6 m away from the foot of a vertical lamp pole 4.2 m long. If the lamp is at the tip of the pole, (a) represent this information in a diagram. (b) calculate the: (i) length of the shadow of the boy cast by the lamp; (ii) angle of elevation of the lamp from the boy, correct to the nearest degree.
(a) Sketch: pole 4.2 m and boy 1.2 m standing 6 m apart on level ground; a line from the lamp over the boy's head to the ground marks the end of the shadow s.
(b)(i) Similar triangles: 1.2/s = 4.2/(6 + s) → 7.2 + 1.2s = 4.2s → s = 2.4 m.
(ii) From the top of the boy's head: tan θ = (4.2 − 1.2)/6 = 0.5 → θ ≈ 27°.
(Measured from the boy's feet: tan θ = 4.2/6 → θ ≈ 35°.)
The lamp, the top of the boy's head and the tip of the shadow lie on one straight line.
(a) Two positive whole numbers P and q are such that P is greater than q and their sum is equal to three times their difference. (i) Express P in terms of q. (ii) Hence, evaluate (P2 + q2)/Pq.
(b) A man sold 100 articles at 25 for ₦66.00 and made a gain of 32%. Calculate his gain or loss per cent if he sold them at 20 for ₦50.00.
(a)(i) P + q = 3(P − q) → 4q = 2P → P = 2q.
(ii) (4q2 + q2)/(2q × q) = 5q2/2q2 = 5/2.
(b) Selling price = 4 × 66 = ₦264 = 132% of the cost, so cost = ₦200.
New selling price = 5 × 50 = ₦250. Gain = ₦50 = 50/200 × 100% = 25% gain.
Find the cost price first from the 32% gain.
(a) If (3 − x), 6, (7 − 5x) are consecutive terms of a geometric progression (G.P.) with constant ratio r > 0; find the: (i) values of x; (ii) constant ratio.
(b) In the diagram, |AB| = 3 cm, |BC| = 4 cm, |CD| = 6 cm and |DA| = 7cm. Calculate ∠ADC, correct to the nearest degree.
(a)(i) 62 = (3 − x)(7 − 5x) → 36 = 21 − 22x + 5x2 → 5x2 − 22x − 15 = 0 → (5x + 3)(x − 5) = 0 → x = −3/5 or 5.
x = 5 gives −2, 6, −18 (ratio −3, not allowed). So x = −3/5.
(ii) Terms 18/5, 6, 10: r = 6 ÷ 18/5 = 5/3.
(b) ∠ABC = 90°, so |AC| = √(32 + 42) = 5 cm.
In ΔACD: cos ∠ADC = (72 + 62 − 52)/(2 × 7 × 6) = 60/84 ≈ 0.714, so ∠ADC ≈ 44°.
In a GP the middle term squared equals the product of its neighbours.
(a) Using ruler and a pair of compasses only, construct: (i) a trapezium WXYZ such that |WX| = 10.2 cm, |XY| = 5.6 cm, |YZ| = 5.8 cm, ∠WXY = 60° and WX is parallel to YZ; (ii) a perpendicular from Z to meet WX at N.
(b) Measure: (i) |WZ|; (ii) |ZN|.
(a)(i) Draw WX = 10.2 cm; construct 60° at X and mark XY = 5.6 cm. Through Y draw a line parallel to WX (e.g. construct 120° at Y) and mark YZ = 5.8 cm towards W. Join WZ.
(ii) From Z construct the perpendicular to WX, meeting it at N.
(b)(i) |WZ| ≈ 5.1 cm. (ii) |ZN| ≈ 4.8 cm (= 5.6 sin 60° ≈ 4.85 cm).
Co-interior angles: if ∠X = 60°, the angle at Y between YX and YZ is 120°.
(a) A segment of a circle is cut off from a rectangular board as shown in the diagram. If the radius of the circle is 1½ times the length of the chord, calculate, correct to 2 decimal places, the perimeter of the remaining portion. [Take π = 22/7]
(b) Evaluate without using calculators or tables, (3/√3)(2/√3 − √12/3).
(a) Chord = 22 − 5 − 3 = 14 cm, so the radius = 21 cm.
sin(½θ) = 7/21 → ½θ ≈ 19.47° → θ ≈ 38.94°.
Arc = 38.94/360 × 2 × 22/7 × 21 ≈ 14.28 cm.
Perimeter = 22 + 12 + 12 + 5 + 3 + 14.28 = 68.28 cm.
(b) 3/√3 = √3 and √12/3 = 2√3/3. √3 × 2/√3 − √3 × 2√3/3 = 2 − 2 = 0.
The curved edge replaces the 14 cm chord in the perimeter.
The frequency distribution table shows the marks obtained by 100 students in a Mathematics test.
Marks (%): 1-10, 11-20, 21-30, 31-40, 41-50, 51-60, 61-70, 71-80, 81-90, 91-100
Frequency: 2, 3, 5, 13, 19, 31, 13, 9, 4, 1
(a) Draw a cumulative frequency curve for the distribution. (b) Use the graph to find the: (i) 60th percentile; (ii) probability that a student passed the test if the pass mark was fixed at 35%.
(a) Upper boundaries 10.5, 20.5, …, 100.5 with cumulative frequencies 2, 5, 10, 23, 42, 73, 86, 95, 99, 100; plot and draw a smooth S-curve.
(b)(i) 60th student: about 50.5 + (60 − 42)/31 × 10 ≈ 56.3 marks.
(ii) Reading at 35: about 16 students scored below 35, so about 84 passed. Probability ≈ 84/100 = 0.84.
(Accept readings close to these from the curve.)
Plot each cumulative frequency at the upper class boundary.
An aeroplane flies due north from a town T on the equator at a speed of 950 km per hour for 4 hours to another town P. It then flies eastwards to town Q on longitude 65° E. If the longitude of T is 15° E, (a) represent this information in a diagram. (b) calculate the: (i) latitude of P, correct to the nearest degree; (ii) distance between P and Q, correct to 4 significant figures. [Take π = 22/7, Radius of the earth = 6400 km]
(a) Sketch of the earth with the equator, the meridian 15°E through T and P, and the parallel through P to Q on 65°E.
(b)(i) T to P = 950 × 4 = 3800 km. θ/360 × 2 × 22/7 × 6400 = 3800 → θ ≈ 34.0°, so P is about 34°N.
(ii) Longitude difference = 50°. PQ = 50/360 × 2 × 22/7 × 6400 × cos 34° ≈ 5587.3 × 0.8290 ≈ 4632 km.
Along a parallel of latitude the radius is R cos(latitude).
When one end of a ladder, LM, is placed against a vertical wall at a point 5 metres above the ground, the ladder makes an angle of 37° with the horizontal ground. (a) Represent this information in a diagram. (b) Calculate, correct to 3 significant figures, the length of the ladder. (c) If the foot of the ladder is pushed towards the wall by 2 metres, calculate, correct to the nearest degree, the angle which the ladder now makes with the ground.
(a) Right triangle: wall 5 m, ladder LM as hypotenuse, 37° between ladder and ground.
(b) |LM| = 5/sin 37° = 5/0.6018 ≈ 8.31 m.
(c) Original foot distance = 5/tan 37° ≈ 6.64 m; new distance = 4.64 m. The ladder length stays 8.31 m: cos θ = 4.64/8.31 ≈ 0.558, so θ ≈ 56°.
When the foot moves, the ladder slides up the wall; its length does not change.