63 questions. Try each one, then open the answer to check yourself and read the explanation.
If {x: 2 ≤ x ≤ 18; x ∈ integer} and 7 + x = 4(mod 9), find the highest value of x.
Answer: C (15)
7 + x ≡ 4 (mod 9) gives x ≡ 6 (mod 9), so x = 6 or 15 in the range. The highest is 15.
The sum of 110112, 111112 and 100002 is 10m10n02. Find the values of m and n.
Answer: C (m = 0, n = 1)
In base ten the sum is 27 + 31 + 16 = 74 = 10010102. Matching 10m10n0 gives m = 0 and n = 1.
A trader bought an engine for $15,000.00 outside Nigeria. If the exchange rate is $0.075 to N1.00, how much did the engine cost in naira?
Answer: B (N200,000.00)
Each naira buys $0.075, so the cost is 15 000 ÷ 0.075 = N200,000.
If (27x × 31−x)/92x = 1, find the value of x.
Answer: B (½)
Write everything as powers of 3: 33x + 1 − x − 4x = 30, so 1 − 2x = 0 and x = ½.
Find the 7th term of the sequence: 2, 5, 10, 17, 26……
Answer: C (50)
The terms are n² + 1, so the 7th term is 49 + 1 = 50 (the differences 3, 5, 7, 9, 11, 13 also give 37 + 13 = 50).
Given that logx 64 = 3, evaluate x log28.
Answer: C (12)
x³ = 64 so x = 4, and log28 = 3, so x log28 = 4 × 3 = 12.
If 2n = y, find 2(2+n/3)
Answer: A (4y1/3)
22 + n/3 = 2² × (2n)1/3 = 4y1/3.
Factorize completely: 6ax − 12by − 9ay + 8bx.
Answer: B ((3a + 4b)(2x − 3y))
Group as 6ax − 9ay + 8bx − 12by = 3a(2x − 3y) + 4b(2x − 3y) = (3a + 4b)(2x − 3y).
Find the equation whose roots are ¾ and −4.
Answer: C (4x2 + 13x − 12 = 0)
(4x − 3)(x + 4) = 0 expands to 4x² + 13x − 12 = 0.
If m = 4, n = 9 and r = 16, evaluate m/n − 17/9 + n/r
Answer: D (−37/48)
4/9 − 16/9 + 9/16 = −4/3 + 9/16 = (−64 + 27)/48 = −37/48.
Adding 42 to a given positive number gives the same result as squaring the number. Find the number.
Answer: C (7)
x² = x + 42 gives (x − 7)(x + 6) = 0; the positive number is 7.
Ada draws the graphs of y = x2 − x − 2 and y = 2x − 1 on the same axes. Which of these equations is she solving?
Answer: B (x2 − 3x − 1 = 0)
Where the graphs meet, x² − x − 2 = 2x − 1, i.e. x² − 3x − 1 = 0.
The volume of a cone of height 3 cm is 38½ cm3. Find the radius of its base. [Take π = 22/7]
Answer: B (3.5 cm)
⅓ × 22/7 × r² × 3 = 38.5 gives r² = 12.25, so r = 3.5 cm.
A sector of a circle with radius 6 cm subtends an angle of 60° at the centre. Calculate its perimeter in terms of π.
Answer: A (2(π + 6) cm)
Arc = 60/360 × 2π × 6 = 2π. Perimeter = 6 + 6 + 2π = 2(π + 6) cm.
The dimensions of a rectangular tank are 2 m by 7 m by 11 m. If its volume is equal to that of a cylindrical tank of height 4cm, calculate the base radius of the cylindrical tank. [Take π = 22/7]
Answer: C (3½ m)
Volume = 2 × 7 × 11 = 154 m³. 22/7 × r² × 4 = 154 gives r² = 12.25, so r = 3½ m (the height must be 4 m).
In the diagram, O is the centre of the circle. If PQ//RS and ∠ONS = 140°, find the size of ∠POM.
Answer: A (40°)
∠ONM = 180° − 140° = 40°. OM = ON (radii), so ∠OMN = 40°, and ∠POM = ∠OMN = 40° (alternate angles, PQ//RS).
In the diagram, PTR is a tangent to the circle centre O. If angle TON = 108°, calculate the size of angle PTN.
Answer: B (126°)
Triangle TON is isosceles, so ∠OTN = (180° − 108°)/2 = 36°. The radius OT meets the tangent at 90°, so ∠PTN = 90° + 36° = 126°.
In the diagram, PQ//RT, QR//SU, ∠PQR = 48° and ∠RTS = x. Find the value of x.
Answer: B (132°)
∠QRT = ∠PQR = 48° (alternate angles, PQ//RT). Since QR//SU, ∠RTS and ∠QRT are co-interior, so x = 180° − 48° = 132°.
In the diagram, O is the centre of the circle, RT is a diameter, ∠PQT = 33° and ∠TOS = 76°. Calculate the value of angle PTR.
Answer: C (57°)
∠TPR = 90° (angle in a semicircle) and ∠PRT = ∠PQT = 33° (same segment), so ∠PTR = 180° − 90° − 33° = 57°.
In the diagram, O is the centre of the circle, RT is a diameter, ∠PQT = 33° and ∠TOS = 76°. Find the size of angle PRS.
Answer: B (71°)
∠PRT = ∠PQT = 33° (same segment) and ∠TRS = ½ × ∠TOS = 38° (angle at centre is twice angle at circumference), so ∠PRS = 33° + 38° = 71°.
The table shows some values for a linear graph.
x: 0, 1¼, 2, 4
y: 3, 5½, __, __
Find the gradient of the line.
Answer: B (2)
Gradient = (5½ − 3)/(1¼ − 0) = 2½ ÷ 1¼ = 2.
The table shows some values for a linear graph.
x: 0, 1¼, 2, 4
y: 3, 5½, __, __
What is the value of y when x = 2?
Answer: B (7)
The line has gradient 2 and y-intercept 3, so y = 2x + 3 and at x = 2, y = 7.
Given that tan x = ⅔ where 0° ≤ x ≤ 90°, find the value of 2sin x.
Answer: C (4√13/13)
Opposite 2, adjacent 3, hypotenuse √13. sin x = 2/√13, so 2 sin x = 4/√13 = 4√13/13.
PQRS is a square. If X is the midpoint of PQ, calculate, correct to the nearest degree, ∠SXP.
Answer: C (63°)
In right triangle SPX, PS = side and PX = half the side, so tan∠SXP = 2 and ∠SXP ≈ 63°.
The angle of elevation of an aircraft from a point K on the horizontal ground is 30° if the aircraft is 800m above the ground, how far is it from K?
Answer: D (1,600.00m)
sin 30° = 800/d, so d = 800 ÷ 0.5 = 1600 m.
The population of students in a school is 810. If this is represented on a pie chart, calculate the sectoral angle for a class of 72 students.
Answer: A (32°)
72/810 × 360° = 32°.
The scores of twenty students in a test are as follows: 44, 47, 48, 49, 50, 51, 52, 53, 53, 54, 58, 59, 60, 61, 63, 65, 67, 70, 73, 75. Find the third quartile.
Answer: B (63)
The scores are already in order; the third quartile is the ¾ × 20 = 15th score, which is 63.
The table shows the distribution of the scores of some students in a test.
Scores: 0 − 4, 5 − 9, 10 − 14
Frequency: 2, 1, 2
Calculate the mean score.
Answer: D (7.0)
Class midpoints are 2, 7 and 12: mean = (2×2 + 1×7 + 2×12)/5 = 35/5 = 7.0.
The probabilities that Kebba, Ebou and Omar will hit a target are ⅔, ¾ and ⅘ respectively. Find the probability that only Kebba will hit the target.
Answer: C (1/30)
Kebba hits and the other two miss: ⅔ × ¼ × ⅕ = 2/60 = 1/30.
In the diagram, VW//YZ, |WX| = 6cm, |XY| = 16 cm, |YZ| = 20cm and |ZX| = 12cm. Calculate |VX|
Answer: D (8cm)
Triangles XVW and XYZ are similar (VW//YZ), so VX/XY = WX/XZ: VX = 16 × 6/12 = 8 cm.
Tom will be 25 years old in n years' time. If he is 5 years younger than Bade, find Bade's present age.
Answer: A ((30 − n) years)
Tom is now 25 − n, so Bade is 25 − n + 5 = (30 − n) years.
If (√2 + √3)/√3 is simplified as m + n√6, find the value of (m + n).
Answer: C (1⅓)
Multiply top and bottom by √3: (√6 + 3)/3 = 1 + ⅓√6, so m + n = 1 + ⅓ = 1⅓.
In the diagram, QT and PR are straight lines, ∠ROS = (3n − 20)°, ∠POT = 2m, ∠SOT = n, ∠POL = m and ∠QOL is a right angle. Find the value of n.
Answer: A (35°)
On line QT: 90° + m + 2m = 180°, so m = 30°. On line PR: 2m + n + (3n − 20) = 180°, so 4n = 140° and n = 35°.
Make k the subject of the relation T = √((Tk − H)/(k − H))
Answer: A (k = H(T2 − 1)/(T2 − T))
Square: T²(k − H) = Tk − H, so k(T² − T) = T²H − H and k = H(T² − 1)/(T² − T).
Which of the following is used to determine the mode of a grouped data
Answer: D (Histogram)
The mode of grouped data is estimated from the tallest bar of a histogram (by drawing the crossing lines).
The area of a rhombus is 110 cm2. If the diagonals are 20 cm and (2x + 1) cm long, find the value of x.
Answer: A (5.0)
Area = ½ × d₁ × d₂: ½ × 20 × (2x + 1) = 110, so 2x + 1 = 11 and x = 5.
Simplify: (3x − y)/xy − (2x + 3y)/2xy + ½
Answer: D ((4x − 5y + xy)/2xy)
Over 2xy: (6x − 2y − 2x − 3y + xy)/2xy = (4x − 5y + xy)/2xy.
Illustrate graphically, the solution of y/2 + 1/8 > 5y/8.
Answer: A (Number line −3 to 4: open circle at 1, arrow to the left)
Multiply by 8: 4y + 1 > 5y, so y < 1: an open circle at 1 with the arrow pointing left.
A farmer uses ⅖ of his land to grow cassava, ⅓ of the remainder for yam and the rest for maize. Find the part of the land used for maize.
Answer: B (⅖)
Remainder after cassava = ⅗; yam = ⅓ × ⅗ = ⅕; maize = 1 − ⅖ − ⅕ = ⅖.
The rate of consumption of petrol by a vehicle varies directly as the square of the distance covered. If 4 litres of petrol is consumed on a distance of 15 km, how far would the vehicle go on 9 litres of petrol?
Answer: A (22½ km)
R = kD²: 4 = k × 225. Then 9 = (4/225)D², so D² = 506.25 and D = 22½ km.
A trader bought 100 oranges at 5 for N40.00 and sold them at 20 for N120.00. Find the profit or loss per cent.
Answer: D (25 % loss)
Cost per orange = N8, selling price = N6, so there is a loss of N2 on N8, i.e. 25% loss.
Describe the shaded portion in the diagram.
Answer: A (P' ∩ Q ∩ R')
The shaded part is inside Q but outside both P and R, i.e. P' ∩ Q ∩ R' = (P ∪ R)' ∩ Q.
Find the value of p if ¼p + 3q = 10 and 2p − ⅓q = 7.
Answer: A (4)
From the second equation q = 6p − 21. Then ¼p + 18p − 63 = 10, so 73p/4 = 73 and p = 4.
Calculate the mean deviation of 5, 3, 0, 7, 2, 1.
Answer: B (2.0)
Mean = 18/6 = 3. Deviations 2, 0, 3, 4, 1, 2 add to 12, so mean deviation = 12/6 = 2.0.
In the diagram, the shaded part is a carpet laid in a room with dimensions 3.5m by 2.2m leaving a margin of 0.5m round it. Find the area of the margin.
Answer: A (4.7 m2)
The carpet measures (3.5 − 1) × (2.2 − 1) = 2.5 × 1.2 = 3.0 m². Margin = 3.5 × 2.2 − 3.0 = 7.7 − 3.0 = 4.7 m².
Two angles of a pentagon are in the ratio 2:3. The others are 60° each. Calculate the smaller of the two angles.
Answer: D (144°)
Angles of a pentagon add to 540°. 540° − 3 × 60° = 360°, shared 2:3 gives 144° and 216°; the smaller is 144°.
A letter is selected from the letters of the English alphabet. What is the probability that the letter selected is from the word MATHEMATICS?
Answer: C (4/13)
MATHEMATICS uses 8 different letters (M, A, T, H, E, I, C, S), so the probability is 8/26 = 4/13.
In a circle radius r cm, a chord 16√3 cm long is 10 cm from the centre of the circle. Find, correct to the nearest cm, the value of r.
Answer: B (17cm)
Half the chord is 8√3. r² = 10² + (8√3)² = 100 + 192 = 292, so r ≈ 17.1 ≈ 17 cm.
Consider the following statements:
X: Locally manufactured tyres are attractive.
Y: Many locally manufactured tyres do not last long.
Denoting locally manufactured tyres by M, attractive tyres by R and long lasting tyres by L, which of these Venn diagrams illustrates the statements?
Answer: A (Diagram A)
M must lie wholly inside R (all are attractive) and only partly inside L (many do not last long). Only diagram A shows M inside R and overlapping L.
In the diagram, OX bisects ∠YXZ and OZ bisects ∠YZX. If ∠XYZ = 68°, calculate the value of ∠XOZ.
Answer: D (124°)
∠YXZ + ∠YZX = 112°, so half of it is 56°. In triangle XOZ, ∠XOZ = 180° − 56° = 124°.
(a) Without using Mathematical tables or calculators, simplify: 34/9 ÷ (5⅓ − 2¾) + 59/10
(b) A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. Find the probability that the sum of the two numbers is greater than 3 and less than 7.
(a) 5⅓ − 2¾ = 16/3 − 11/4 = 31/12
31/9 ÷ 31/12 = 31/9 × 12/31 = 4/3
4/3 + 59/10 = 40/30 + 177/30 = 217/30 = 77/30
(b) There are 3 × 3 = 9 equally likely pairs; sums: 3, 5, 7, 4, 6, 8, 5, 7, 9
Sums greater than 3 and less than 7: 5, 4, 6, 5 → 4 outcomes
Probability = 4/9
Do the bracket first, then the division, then the addition.
(a) Solve the inequality: 4 + ¾(x + 2) ≤ ⅜x + 1.
(b) The diagram shows a rectangle PQRS from which a square of side x cm has been cut. If the area of the shaded portion is 484 cm2, find the values of x.
(a) Multiply through by 8: 32 + 6(x + 2) ≤ 3x + 8
6x + 44 ≤ 3x + 8 → 3x ≤ −36 → x ≤ −12
(b) The rectangle is 20 cm high and (10 + x + 10) = (20 + x) cm long.
Shaded area = 20(20 + x) − x² = 484
400 + 20x − x² = 484 → x² − 20x + 84 = 0 → (x − 6)(x − 14) = 0
x = 6 cm or x = 14 cm
In (b), the length of the rectangle includes the cut-out width x.
(a) The ratio of the interior angle to the exterior angle of a regular polygon is 5 : 2. Find the number of sides of the polygon.
(b) The diagram shows a circle PQRS with centre O, ∠UQR = 68°, ∠TPS = 74° and ∠QSR = 40°. Calculate the value of ∠PRS.
(a) Interior + exterior = 180°, so exterior angle = 2/7 × 180° = 360/7°
Number of sides = 360° ÷ 360/7° = 7
(b) TPQU is a straight line: ∠PQR = 180° − 68° = 112° and ∠SPQ = 180° − 74° = 106°
∠PSR = 180° − 112° = 68° (opposite angles of cyclic quadrilateral PQRS)
∠QPR = ∠QSR = 40° (same segment), so ∠SPR = 106° − 40° = 66°
In triangle PRS: ∠PRS = 180° − 68° − 66° = 46°
Use opposite angles of a cyclic quadrilateral and angles in the same segment.
(a) By how much is the sum of 3⅔ and 2⅕ less than 7?
(b) The height, h m, of a dock above sea level is given by h = 6 + 4cos(15p)°, 0 < p < 6. Find: (i) the value of h when p = 4; (ii) correct to two significant figures, the value of p when h = 9m.
(a) 3⅔ + 2⅕ = 55/15 + 33/15 = 88/15 = 513/15
7 − 513/15 = 12/15
(b)(i) h = 6 + 4cos 60° = 6 + 4 × 0.5 = 8 m
(ii) 9 = 6 + 4cos(15p)° → cos(15p)° = 0.75 → 15p = 41.41° → p = 2.76 ≈ 2.8
In (b)(ii), make cos(15p)° the subject before using inverse cosine.
A trapezium PQRS is such that PQ//RS and the perpendicular from P to RS is 40 cm. If |PQ| = 20cm, |SP| = 50 cm and |SR| = 60 cm, calculate, correct to 2 significant figures, the: (a) area of the trapezium; (b) ∠QRS.
(a) Area = ½(20 + 60) × 40 = 1600 cm² ≈ 1.6 × 10³ cm²
(b) Let the perpendicular from P meet SR at N: SN = √(50² − 40²) = 30 cm
The perpendicular from Q meets SR at M with SM = 30 + 20 = 50 cm, so MR = 60 − 50 = 10 cm
tan∠QRS = 40/10 = 4 → ∠QRS ≈ 76°
Drop perpendiculars from P and Q to SR and use right-angled triangles.
(a) (i) Illustrate the following statement in a Venn diagram: All good Literature students in a school are in the General Arts class. (ii) Use the diagram to determine whether or not the following are valid conclusions from the given statement: I. Vivian is in the General Arts class therefore she is a good Literature student; II. Audu is not a good Literature student therefore he is not in the General Arts class; III. Kweku is not in the General Arts class therefore he is not a good Literature student.
(b) The cost (c) of producing n bricks is the sum of a fixed amount h, and a variable amount y, where y varies directly as n. If it costs GH¢ 950.00 to produce 600 bricks and GH¢ 1,030.00 to produce 1000 bricks, (i) find the relationship between c, h and n; (ii) calculate the cost of producing 500 bricks.
(a)(i) Universal set = students in the school; circle L (good Literature students) drawn wholly inside circle G (General Arts class).
(ii) I: not valid (Vivian may be in G but outside L). II: not valid (Audu may be in G but outside L). III: valid (outside G means outside L).
(b)(i) c = h + kn
950 = h + 600k and 1030 = h + 1000k → 400k = 80 → k = 0.2, h = 830
c = 830 + 0.2n (i.e. c = h + n/5 with h = 830)
(ii) c = 830 + 0.2 × 500 = GH¢ 930.00
A conclusion is valid only if it must be true in the diagram.
The table is for the relation y = px2 − 5x + q.
x: −3, −2, −1, 0, 1, 2, 3, 4, 5
y: 21, 6, __, −12, __, __, __, 0, 13
(a)(i) Use the table to find the values of p and q. (ii) Copy and complete the table.
(b) Using scales of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of the relation for −3 ≤ x ≤ 5.
(c) Use the graph to find: (i) y when x = 1.8; (ii) x when y = −8.
(a)(i) At x = 0, y = −12, so q = −12. At x = 4: 16p − 20 − 12 = 0 → p = 2. So y = 2x² − 5x − 12.
(ii) x = −1: y = −5; x = 1: y = −15; x = 2: y = −14; x = 3: y = −9
(b) Plot the nine points with the given scales and join them with a smooth U-shaped curve (minimum near x = 1.25, y ≈ −15.1).
(c)(i) y ≈ −14.5 (exact −14.52)
(ii) x ≈ −0.6 or x ≈ 3.1 (roots of 2x² − 5x − 4 = 0)
Find q first from the point where x = 0.
(a) Using ruler and a pair of compasses only, construct a: (i) trapezium WXYZ such that |WX| = 8 cm, |XY| = 5.5 cm, |XZ| = 8.3 cm, ∠WXY = 60° and WX//ZY; (ii) rectangle PQYZ where P and Q are on WX.
(b) Measure: (i) |QX| (ii) ∠XWZ.
(a)(i) Draw WX = 8 cm; construct 60° at X and mark Y with XY = 5.5 cm; through Y construct a line parallel to WX; with centre X and radius 8.3 cm cut this line at Z; join WZ.
(ii) Drop perpendiculars from Z and Y to WX, meeting it at P and Q; PQYZ is the rectangle.
(b)(i) |QX| ≈ 2.8 cm (± 0.1) (ii) ∠XWZ ≈ 76° (± 1°)
Keep construction arcs visible; measured answers within ± 0.1 cm and ± 1° are accepted.
(a) The first term of an Arithmetic Progression (AP) is −8. If the ratio of the 7th term to the 9th term is 5 : 8, find the common difference of the AP.
(b) A trader bought 30 baskets of pawpaw and 100 baskets of mangoes for N2,450.00. She sold the pawpaw at a profit of 40% and the mangoes at a profit of 30%. If her profit on the entire transaction was N855.00, find the: (i) cost price of a basket of pawpaw; (ii) selling price of the 100 baskets of mangoes.
(a) (−8 + 6d)/(−8 + 8d) = 5/8 → −64 + 48d = −40 + 40d → 8d = 24 → d = 3
(b) Let a basket of pawpaw cost x and a basket of mangoes cost y.
30x + 100y = 2450 … (1)
Profit: 0.4(30x) + 0.3(100y) = 855 → 12x + 30y = 855 … (2)
Solving: x = N40.00, y = N12.50
(i) Cost price of a basket of pawpaw = N40.00
(ii) Selling price of 100 baskets of mangoes = 1.3 × 100 × 12.50 = N1,625.00
Write one equation for the total cost and one for the total profit.
(a) Without using Mathematical tables or calculators, simplify: (2 tan 60° + cos 30°)/sin 60°.
(b) From an aeroplane in the air and at a horizontal distance of 1050 m, the angle of depression of the top and base of a control tower at an instance are 36° and 41° respectively. Calculate, correct to the nearest metre, the: (i) height of the control tower; (ii) shortest distance between the aeroplane and the base of the control tower.
(a) tan 60° = √3, cos 30° = sin 60° = √3/2
(2√3 + √3/2) ÷ √3/2 = (5√3/2) × (2/√3) = 5
(b)(i) Height of plane above the base = 1050 tan 41° ≈ 912.8 m
Height of plane above the top = 1050 tan 36° ≈ 762.9 m
Height of tower ≈ 912.8 − 762.9 ≈ 150 m
(ii) Distance = 1050/cos 41° ≈ 1391 m
The tower's height is the difference of two vertical distances from the plane.
(a) Make m the subject of the relation h = mt/(d(m + p))
(b) In the diagram, WY and WZ are straight lines, O is the centre of circle WXM and ∠XWM = 48°. Calculate the value of ∠WYZ.
(c) An operation ⊗ is defined on the set X = {1, 3, 5, 6} by m ⊗ n = m + n + 2 (mod 7), where m, n ∈ X. (i) Draw a table for the operation. (ii) Using the table, find the truth set of: I. 3 ⊗ n = 3; II. n ⊗ n = 3.
(a) hd(m + p) = mt → hdm + hdp = mt → m(t − hd) = hdp → m = hdp/(t − hd)
(b) WM is a diameter, so ∠WXM = 90°; ∠XMW = 180° − 90° − 48° = 42°
∠XMZ = 180° − 42° = 138° (WMZ straight)
XYZM is cyclic, so ∠XYZ = 180° − 138° = 42°; ∠WYZ = 42°
(c)(i) Rows m = 1, 3, 5, 6; columns n = 1, 3, 5, 6:
1: 4, 6, 1, 2
3: 6, 1, 3, 4
5: 1, 3, 5, 6
6: 2, 4, 6, 0
(ii) I. 3 ⊗ n = 3 → n = 5: {5}
II. n ⊗ n = 3: the diagonal is 4, 1, 5, 0, so the truth set is { } (empty)
In (b), use the angle in a semicircle and then opposite angles of a cyclic quadrilateral.
A water reservoir in the form of a cone mounted on a hemisphere is built such that the plane face of the hemisphere fits exactly to the base of the cone and the height of the cone is 6 times the radius of its base.
(a) Illustrate this information in a diagram.
(b) If the volume of the reservoir is 333⅓ π m3 calculate, correct to the nearest whole number, the: (i) volume of the hemisphere; (ii) total surface area of the reservoir. [Take π = 22/7]
(a) A cone of base radius r and height 6r standing on the flat face of a hemisphere of radius r.
(b) Volume = ⅔πr³ + ⅓πr²(6r) = (8/3)πr³ = (1000/3)π → r³ = 125 → r = 5 m
(i) Hemisphere = ⅔ × 22/7 × 125 ≈ 261.9 ≈ 262 m³
(ii) Slant height l = √(5² + 30²) = √925 ≈ 30.41 m
Curved surface of hemisphere = 2πr² = 2 × 22/7 × 25 ≈ 157.1 m²
Curved surface of cone = πrl = 22/7 × 5 × 30.41 ≈ 477.9 m²
Total surface area ≈ 635 m²
Find r first from the total volume.
The table shows the distribution of marks of candidates.
Marks %: 0−9, 10−19, 20−29, 30−39, 40−49, 50−59, 60−69, 70−79, 80−89, 90−99
Frequency: 7, 11, 17, 20, 29, 34, 30, 25, 21, 6
(a) Construct a cumulative frequency table for the distribution and draw a cumulative frequency curve.
(b) Use the curve to estimate, correct to one decimal place, the: (i) lowest mark for distinction if 5 % of the candidates passed with distinction; (ii) probability of selecting a candidate who scored at most 45%.
(a) Upper class boundaries 9.5, 19.5, …, 99.5; cumulative frequencies 7, 18, 35, 55, 84, 118, 148, 173, 194, 200. Plot cumulative frequency against upper class boundary and join with a smooth S-shaped curve.
(b)(i) 5% of 200 = 10, so read the mark at cumulative frequency 190: ≈ 79.5 + (17/21) × 10 ≈ 87.6
(ii) At 45 marks the cumulative frequency ≈ 71, so probability ≈ 71/200 ≈ 0.36 (accept readings from the curve)
Plot against upper class boundaries, not class marks.