61 questions. Try each one, then open the answer to check yourself and read the explanation.
Evaluate, correct to two decimal places, 75.0785 − 34.624 + 9.83
Answer: C (50.28)
75.0785 − 34.624 = 40.4545, and 40.4545 + 9.83 = 50.2845, which is 50.28 to 2 d.p.
If X = {x : x < 7} and Y = {y : y is a factor of 24} are subsets of μ = {1, 2, 3, … 10}, find X ∩ Y.
Answer: B ({1, 2, 3, 4, 6})
X = {1, 2, 3, 4, 5, 6} and Y = {1, 2, 3, 4, 6, 8}. The common elements are {1, 2, 3, 4, 6}.
Find the least value of x which satisfies the equation 4x = 7 (mod 9)
Answer: D (4)
Try values: 4 × 4 = 16 = 9 + 7, so 16 ≡ 7 (mod 9). Smaller values 1, 2, 3 give 4, 8, 3, so x = 4 is the least.
Express 1 + 2 log103 in the form log10q.
Answer: A (log1090)
1 = log1010 and 2 log103 = log109, so the sum is log10(10 × 9) = log1090.
If 101two + 12y = 23five, find the value of y.
Answer: C (6)
101two = 5 and 23five = 13, so 12y = y + 2 = 8, giving y = 6.
An amount of ₦550,000.00 was realised when a principal, x was saved at 2% simple interest for 5 years. Find the value of x.
Answer: D (₦500,000.00)
Interest = x × 2 × 5/100 = 0.1x, so the amount is 1.1x = 550,000 and x = ₦500,000.
Given that (√3 + √5)/√5 = x + y√15, find the value of (x + y).
Answer: C (1⅕)
(√3 + √5)/√5 = √3/√5 + 1 = √15/5 + 1. So x = 1, y = 1/5 and x + y = 1⅕.
If x = 3 and y = −1, evaluate 2(x2 − y3).
Answer: C (20)
x2 = 9 and y3 = −1, so 2(9 − (−1)) = 2 × 10 = 20.
Solve 3x − 2y = 10 and x + 3y = 7 simultaneously.
Answer: D (x = 4 and y = 1)
From the second equation x = 7 − 3y. Then 3(7 − 3y) − 2y = 10 gives 11y = 11, so y = 1 and x = 4.
The implication x ⇒ y is equivalent to
Answer: A (~y ⇒ ~x)
An implication always has the same truth value as its contrapositive: x ⇒ y is equivalent to ~y ⇒ ~x.
The first term of a Geometric Progression (G.P) is 3 and the 5th term is 48. Find the common ratio.
Answer: A (2)
T5 = ar4, so 3r4 = 48, r4 = 16 and r = 2.
Make m the subject of the relation k = √((m − y)/(m + 1))
Answer: B (m = (y + k2)/(1 − k2))
Square: k2(m + 1) = m − y, so k2m + k2 = m − y. Then y + k2 = m − k2m = m(1 − k2), giving m = (y + k2)/(1 − k2).
Find the quadratic equation whose whose roots are ½ and −⅓.
Answer: D (6x2 − x − 1 = 0)
The factors are (2x − 1) and (3x + 1). Their product is 6x2 − x − 1 = 0.
Given that x is directly proportional to y and inversely proportional to Z, x = 15 when y = 10 and Z = 4, find the equation connecting x, y and Z.
Answer: A (x = 6y/Z)
x = ky/Z. Then 15 = 10k/4, so k = 6 and x = 6y/Z.
Two buses start from the same station at 9.00 a.m. and travel in opposite directions along the same straight road. The first bus travels at a speed of 72 km/h and the second at 48 km/h. At what time will they be 240 km apart?
Answer: C (11:00 am)
Moving apart, the gap grows by 72 + 48 = 120 km each hour. 240 km takes 2 hours, so the time is 11:00 am.
A solid cuboid has length 7 cm, width 5 cm and height 4 cm. Calculate its total surface area.
Answer: B (166 cm2)
Total surface area = 2(lb + bh + lh) = 2(35 + 20 + 28) = 166 cm2.
In the diagram, PQ ∥ SR. Find the value of x.
Answer: B (46)
The angle at the middle vertex inside the bend is 360° − 246° = 114°. For parallel lines this angle equals the sum of the other two: x + 68 = 114, so x = 46.
Find the equation of the line parallel to 2y = 3(x − 2) and passes through the point (2, 3).
Answer: C (y = (3/2)x)
2y = 3x − 6 has gradient 3/2. Through (2, 3): y − 3 = (3/2)(x − 2), so y = (3/2)x.
The expression (5x + 3)/(6x(x + 1)) will be undefined when x equals.
Answer: B ({0, −1})
A fraction is undefined when its denominator is zero: 6x(x + 1) = 0 when x = 0 or x = −1.
A man is five times as old as his son. In four years time, the product of their ages would be 340. If the son's age is y, express the product of their ages in terms of y.
Answer: D (5y2 + 24y − 324 = 0)
In four years the ages are 5y + 4 and y + 4. (5y + 4)(y + 4) = 5y2 + 24y + 16 = 340, so 5y2 + 24y − 324 = 0.
Simplify: a/b − b/a − c/b.
Answer: D ((a2 − b2 − ac)/(ab))
Use the common denominator ab: (a·a − b·b − c·a)/(ab) = (a2 − b2 − ac)/(ab).
In the diagram, XYZ is an equilateral triangle of side 6 cm and T is the midpoint of XY. Find tan (∠XZT).
Answer: A (1/√3)
ZT bisects the 60° angle at Z, so ∠XZT = 30° and tan 30° = 1/√3.
A fence 2.4 m tall, is 10 m away from a tree of height 16 m. Calculate the angle of elevation of the top of the tree from the top of the fence.
Answer: B (53.67°)
Above the fence top the tree rises 16 − 2.4 = 13.6 m over 10 m. tan θ = 1.36, so θ ≈ 53.67°.
Fati buys milk at ₦x per tin and sells each at a profit of ₦y. If she sells 10 tins of milk, how much does she receive from the sales?
Answer: D (₦10(x + y))
Each tin sells for ₦(x + y), so 10 tins bring ₦10(x + y).
If tan y is positive and sin y is negative, in which quadrant would y lie?
Answer: C (Third only)
sin is negative in the third and fourth quadrants, and tan is positive in the first and third. Both hold only in the third quadrant.
The dimensions of a rectangular base of a right pyramid are 9 cm by 5 cm. If the volume of the pyramid is 105 cm3, how high is the pyramid?
Answer: D (7 cm)
Volume = ⅓ × base area × height: 105 = ⅓ × 45 × h = 15h, so h = 7 cm.
Each interior angle of a regular polygon is 168°. Find the number of sides of the polygon.
Answer: A (30)
Each exterior angle is 180° − 168° = 12°, so the number of sides is 360 ÷ 12 = 30.
In the diagram, MN ∥ PQ, ∠MNP = 2x, and ∠NPQ = (3x − 50°). Find the value of ∠NPQ.
Answer: D (100°)
∠MNP and ∠NPQ are alternate angles, so 2x = 3x − 50°, giving x = 50°. Then ∠NPQ = 3(50°) − 50° = 100°.
The length of an arc of a circle of radius 3.5 cm is 1 19/36 cm. Calculate, correct to the nearest degree, the angle subtended by the arc at the centre of the circle. [Take π = 22/7]
Answer: C (25°)
The circumference is 2 × (22/7) × 3.5 = 22 cm. θ = (55/36) ÷ 22 × 360° = 25°.
In the diagram, PU ∥ SR, PS ∥ TR, QS ∥ UR, UR = 15 cm, SR = 8 cm, PS = 10 cm and area of ΔSUR = 24 cm2. Calculate the area of PTRS.
Answer: B (48 cm2)
The height between the parallel lines comes from ΔSUR: ½ × 8 × h = 24, so h = 6 cm. PTRS is a parallelogram on base SR = 8 cm with that height, so its area is 8 × 6 = 48 cm2.
In the diagram, PQR is a circle with centre O. If ∠OPQ = 48°, find the value of m.
Answer: A (96°)
OP = OQ (radii), so ∠OQP = 48°. Angle m is an exterior angle of triangle OPQ, so m = 48° + 48° = 96°.
The pie chart shows the population of men, women and children in a city. If the population of the city is 1,800,000, how many men are in the city?
Answer: C (355,000)
The men's sector is 360° − (120° + 169°) = 71°. Men = 71/360 × 1,800,000 = 355,000.
The mean of the numbers 15, 21, 17, 26, 18 and 29 is 21. Calculate the standard deviation.
Answer: C (5)
Squared deviations from 21: 36, 0, 16, 25, 9, 64, total 150. Variance = 150/6 = 25, so the standard deviation is 5.
In the diagram, O is the centre of the circle. SOQ is the diameter and ∠SRP = 37°. Find ∠PSQ.
Answer: C (53°)
∠SQP = ∠SRP = 37° (angles in the same segment) and ∠SPQ = 90° (angle in a semicircle). So ∠PSQ = 180° − 90° − 37° = 53°.
Find the sum of the interior angles of a pentagom.
Answer: C (540°)
Sum of interior angles = (n − 2) × 180° = 3 × 180° = 540°.
The diameter of a sphere is 12 cm. Calculate, correct to the nearest cm3, the volume of the sphere. [Take π = 22/7]
Answer: C (905 cm3)
r = 6 cm. V = (4/3) × (22/7) × 216 ≈ 905.1, so 905 cm3.
A box contains 12 identical balls of which 5 are red, 4 blue and the rest are green. If a ball is selected at random from the box, what is the probability that it is green?
Answer: D (1/4)
There are 12 − 5 − 4 = 3 green balls, so P(green) = 3/12 = 1/4.
A box contains 12 identical balls of which 5 are red, 4 blue and the rest are green. If two balls are selected at random one after the other with replacement, what is the probability that both are red?
Answer: A (25/144)
With replacement the chance stays 5/12 each time: (5/12) × (5/12) = 25/144.
In the diagram, PQ is a straight line. If m = ½(x + y + z), find value of m.
Answer: B (60°)
Angles on a straight line add to 180°: x + y + z + m = 180°. Since x + y + z = 2m, 3m = 180° and m = 60°.
The points on a linear graph are as shown in the table. Find the gradient (slope) of the line.
x: 6.20, 6.85, 7.50
y: 3.90, 5.20, 6.50
Answer: B (2)
Gradient = change in y ÷ change in x = (5.20 − 3.90)/(6.85 − 6.20) = 1.30/0.65 = 2.
In the diagram, O is the centre of the circle, PQ and RS are tangents to the circle. Find the value of (m + n).
Answer: B (90°)
The tangents touch at the ends of a diameter, and the angle at the third point is 90° (angle in a semicircle). Each of m and n is 90° minus the triangle's angle beside it, so m + n = 180° − 90° = 90°.
In the diagram, O is the centre of the circle. If ∠NLM = 74°, ∠LMN = 39° and ∠LOM = x, find the value of x.
Answer: A (134°)
∠LNM = 180° − 74° − 39° = 67°. The angle at the centre is twice that: x = 134°.
Which of the following in not a sufficient condition for two triangles to be congruent?
Answer: D (SSA)
Two sides and a non-included angle (SSA) can fit two different triangles, so it does not prove congruence.
A woman received a discount of 20% on a piece of cloth she purchased from a shop. If she paid $525.00 what was the original price?
Answer: C ($656.25)
She paid 80% of the price: 525 ÷ 0.8 = $656.25.
The interquartile range of a distribution is 7. If the 25th percentile is 16, find the upper quartile.
Answer: C (23)
Interquartile range = Q3 − Q1, and the 25th percentile is Q1. Q3 = 16 + 7 = 23.
The graphs of the equations y = 2x + 5 and y = 2x2 + x − 1 are shown. Find the points of intersection of the two graphs.
Answer: A ((2.0, 9.0) and (−1.5, 2.0))
Set 2x2 + x − 1 = 2x + 5: 2x2 − x − 6 = 0, (2x + 3)(x − 2) = 0. x = 2 gives y = 9 and x = −1.5 gives y = 2.
The graphs of the equations y = 2x + 5 and y = 2x2 + x − 1 are shown. If x = −2.5, what is the value of y on the curve?
Answer: C (y = 9.0)
On the curve y = 2(−2.5)2 + (−2.5) − 1 = 12.5 − 3.5 = 9.0.
If (x + 2) is a factor of x2 + Px − 10, find the value of P.
Answer: B (−3)
If (x + 2) is a factor, x = −2 makes the expression zero: 4 − 2P − 10 = 0, so P = −3.
(a) If A = {multiples of 2}, B = {multiples of 3} and C = {factors of 6} are subsets of μ = {x : 1 ≤ x ≤ 10}, find A′ ∩ B′ ∩ C′.
(b) Tickets for a movie premiere cost $18.50 each while the bulk purchase price for 5 tickets is $80.00. If 4 gentlemen decide to get a fifth person to join them so that they can share the bulk purchase price equally, how much would each person save?
(a) A = {2, 4, 6, 8, 10}, B = {3, 6, 9}, C = {1, 2, 3, 6}
A′ = {1, 3, 5, 7, 9}, B′ = {1, 2, 4, 5, 7, 8, 10}, C′ = {4, 5, 7, 8, 9, 10}
A′ ∩ B′ ∩ C′ = {5, 7}
(b) Bulk share = $80 ÷ 5 = $16 each
Saving per person = $18.50 − $16.00 = $2.50
List μ first, then each complement; only elements in all three complements remain.
(a) Given that P = (rk/Q − ms)2/3, (i) make Q the subject of the relation; (ii) find, correct to two decimal places, the value of Q when P = 3, m = 15, s = 0.2, k = 4 and r = 10.
(b) Given that (x + 2y)/5 = x − 2y, find x : y.
(a)(i) Raise both sides to the power 3/2: P3/2 = rk/Q − ms
rk/Q = P3/2 + ms → Q = rk/(P3/2 + ms)
(ii) Q = 40/(33/2 + 3) = 40/(5.196 + 3) = 40/8.196 ≈ 4.88
(b) x + 2y = 5x − 10y → 12y = 4x
x/y = 3, so x : y = 3 : 1
Undo the power 2/3 by raising both sides to 3/2.
(a) In the diagram, O is the centre of the circle ABCDE, |BC| = |CD| and ∠BCD = 108°. Find ∠CDE.
(b) Given that tan x = √3, 0° ≤ x ≤ 90°, evaluate ((cos x)2 − sin x)/((sin x)2 + cos x).
(a) Triangle BCD is isosceles: ∠CDB = (180° − 108°)/2 = 36°
BE is a diameter (passes through O), so ∠BDE = 90° (angle in a semicircle)
∠CDE = 36° + 90° = 126°
(b) x = 60°: cos x = ½, sin x = √3/2
Numerator = ¼ − √3/2 = (1 − 2√3)/4; denominator = ¾ + ½ = 5/4
Value = (1 − 2√3)/5 ≈ −0.49
In (a), look for the diameter through O that gives a right angle.
The total surface area of a cone of slant height l cm and base radius r cm is 224π cm2. If r : l = 2 : 5, find: (a) correct to one decimal place, the value of r; (b) correct to the nearest whole number, the volume of the cone [Take π = 22/7]
(a) l = 5r/2; TSA = πrl + πr2 = π(5r2/2 + r2) = (7/2)πr2
(7/2)r2 = 224 → r2 = 64 → r = 8.0 cm
(b) l = 20 cm; h = √(202 − 82) = √336 ≈ 18.33 cm
V = ⅓ × (22/7) × 64 × 18.33 ≈ 1229 cm3
Total surface area of a cone includes the base: πrl + πr2.
A die was rolled a number of times. The outcomes are as shown in the table.
Number: 1, 2, 3, 4, 5, 6
Outcomes: 32, m, 25, 40, 28, 45
If the probability of obtaining 2 is 0.15, find the: (a) value of m; (b) number of times the die was rolled; (c) probability of obtaining an even number.
(a) m/(170 + m) = 0.15 → m = 25.5 + 0.15m → 0.85m = 25.5 → m = 30
(b) Total = 170 + 30 = 200 rolls
(c) Even outcomes = 30 + 40 + 45 = 115
P(even) = 115/200 = 23/40
The known outcomes add up to 170; the total is 170 + m.
(a) Copy and complete the table of values for the relation y = 3 sin 2x.
x: 0°, 15°, 30°, 45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°
y: 0.0, __, __, __, __, 1.5, __, __, __, __, −2.6
(b) Using a scale of 2 cm to 15° on the x-axis and 2 cm to 1 unit on the y-axis, draw the graph of y = 3 sin 2x for 0° ≤ x ≤ 150°.
(c) Use the graph to find the truth set of: (i) 3 sin 2x + 2 = 0; (ii) (3/2) sin 2x = 0.25.
(a) y = 0.0, 1.5, 2.6, 3.0, 2.6, 1.5, 0.0, −1.5, −2.6, −3.0, −2.6
(b) Plot the points and join with a smooth curve
(c)(i) 3 sin 2x = −2: read where the curve meets y = −2, x ≈ 111°
(ii) Doubling gives 3 sin 2x = 0.5: read where the curve meets y = 0.5, x ≈ 5° and x ≈ 85°
Rearrange each equation into the form 3 sin 2x = k, then draw the line y = k.
(a) The diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is 48 m and the base radius is 14 m. Caculate, correct to three significant figures, the surface area of the structure. [Take π = 22/7]
(b) Five years ago, Musah was twice as old as Sesay. If the sum of their ages is 100, find Sesay's present age.
(a) Slant height l = √(482 + 142) = √2500 = 50 m
Curved surface of cone = πrl = (22/7) × 14 × 50 = 2200 m2
Curved surface of hemisphere = 2πr2 = 2 × (22/7) × 196 = 1232 m2
Total = 3432 ≈ 3430 m2 (3 s.f.)
(b) M + S = 100 and M − 5 = 2(S − 5) → M = 2S − 5
3S − 5 = 100 → S = 35
Sesay is 35 years old
The flat circle where the cone sits on the hemisphere is hidden, so it is not part of the surface.
(a) Ms. Maureen spent ¼ of her monthly income at a shopping mall, ⅓ at an open market and 2/5 of the remaining amount at a Mechanic workshop. If she had ₦225,000.00 left, find: (i) her monthly income; (ii) the amount spent at the open market.
(b) The third term of an Arithmetic Progression (A.P.) is 4m − 2n. If the ninth term of the progression is 2m − 8n, find the common difference in terms of m and n.
(a) Let income = x. Mall + market = x/4 + x/3 = 7x/12; remainder = 5x/12
Mechanic = 2/5 × 5x/12 = x/6; left = 5x/12 − x/6 = x/4
(i) x/4 = 225,000 → x = ₦900,000
(ii) Open market = ⅓ × 900,000 = ₦300,000
(b) a + 2d = 4m − 2n and a + 8d = 2m − 8n
Subtract: 6d = −2m − 6n
d = −(m + 3n)/3 = −m/3 − n
In (a) the ¼ and ⅓ are fractions of the whole income; only the 2/5 is of the remainder.
Two cyclists X and Y leave town Q at the same time. Cyclist X travels at the rate of 5 km/h on a bearing of 049° and cyclist Y travels at the rate of 9 km/h on a bearing of 319°.
(a) Illustrate the information on a diagram.
(b) After travelling for two hours, calculate, correct to the nearest whole number, the: (i) distance between cyclists X and Y; (ii) bearing of cyclist X from Y.
(c) Find the average speed at which cyclist X will get to Y in 4 hours.
(a) Diagram from Q: QX on bearing 049°, QY on bearing 319°; the angle XQY = 360° − 319° + 49° = 90°
(b) After 2 h: QX = 10 km, QY = 18 km
(i) XY = √(102 + 182) = √424 ≈ 20.6 ≈ 21 km
(ii) ∠QYX: tan = 10/18 → 29.1°; bearing of Q from Y = 139°, so bearing of X from Y = 139° − 29.1° ≈ 110°
(c) Speed = 20.6 km ÷ 4 h ≈ 5.1 km/h (about 5 km/h)
The two bearings differ by 90°, so the triangle QXY is right-angled at Q.
The table shows the distribution of marks obtained by students in an examination.
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.
(b) Draw the cumulative frequency curve for the distribution.
(c) Using the curve, find correct to one decimal place, the: (i) median mark; (ii) lowest mark for distinction if 5% of the students passed with distinction.
(a) Upper class boundaries 9.5, 19.5, …, 99.5; cumulative frequencies 7, 18, 35, 55, 84, 118, 148, 173, 194, 200
(b) Plot cumulative frequency against upper class boundary and draw a smooth S-curve
(c)(i) Median at the 100th student: ≈ 54.2 (49.5 + 16/34 × 10)
(ii) Top 5% = 10 students, so read the mark at cumulative frequency 190: ≈ 87.6 (79.5 + 17/21 × 10)
For the distinction mark, count down from the top: 200 − 10 = 190.
(a) In the diagram, MNPQ is a circle with centre O, |MN| = |NP| and ∠OMN = 50°. Find: (i) ∠MNP; (ii) ∠POQ.
(b) Find the equation of the line which has the same gradient as 8y + 4x = 24 and passes through the point (−8, 12).
(a)(i) MQ is a diameter, so ∠MPQ = 90°. In cyclic quadrilateral MNPQ, ∠NPQ = 180° − 50° = 130°, so ∠MPN = 130° − 90° = 40°
MN = NP, so ∠NMP = 40° and ∠MNP = 180° − 80° = 100°
(ii) ∠PMQ = 50° − 40° = 10°; angle at centre ∠POQ = 2 × 10° = 20°
(b) 8y = −4x + 24 → gradient = −½
y − 12 = −½(x + 8) → 2y − 24 = −x − 8
x + 2y − 16 = 0 (or y = −½x + 8)
Use the angle in a semicircle and opposite angles of a cyclic quadrilateral.
(a) In the diagram, AB is a tangent to the circle with centre O, and COB is a straight line. If CD ∥ AB and ∠ABE = 40°, find: ∠ODE.
(b) ABCD is a parallelogram in which |CD| = 7 cm, |AD| = 5 cm and ∠ADC = 125°. (i) Illustrate the information in a diagram. (ii) Find, correct to one decimal place, the area of the parallelogram.
(c) If x = ½(1 − √2), evaluate (2x2 − 2x).
(a) ∠DCO = ∠ABO = 40° (alternate angles, CD ∥ AB)
OC = OD, so ∠ODC = 40° and ∠COD = 100°; ∠DOE = 180° − 100° = 80°
OD = OE, so ∠ODE = (180° − 80°)/2 = 50°
(b)(i) Sketch parallelogram ABCD with CD = 7 cm, AD = 5 cm, ∠ADC = 125°
(ii) Area = 7 × 5 × sin 125° ≈ 28.7 cm2
(c) 2x2 − 2x = 2x(x − 1); x − 1 = −½(1 + √2)
x(x − 1) = −¼(1 − √2)(1 + √2) = −¼(1 − 2) = ¼
2x2 − 2x = ½
In (c), factorising first makes the surds cancel neatly.
(a) Using a ruler and a pair of compasses only, construct: (i) (α) ΔABC with |AB| = 7.5 cm, |AC| = 13.5 cm and ∠ABC = 120°; (β) locus l1 of points equidistant from A and B; (γ) locus, l2 of points equidistant from B and C. (ii) Using N, the point of intersection of l1 and l2 as centre, draw a circle to pass through points A, B, and C.
(b) Using the method of completing the square, solve, 4x2 − 4√3x + 3 = 0, leaving the answer in surd form.
(a)(i)(α) Draw AB = 7.5 cm; construct 120° at B (60° + 60°); with centre A, radius 13.5 cm, cut the arm at C (BC ≈ 8.1 cm)
(β) l1 = perpendicular bisector of AB; (γ) l2 = perpendicular bisector of BC
(ii) With centre N and radius NA (≈ 7.8 cm) draw the circle through A, B and C
(b) x2 − √3x = −¾
x2 − √3x + ¾ = 0 → (x − √3/2)2 = 0
x = √3/2 (twice)
Construction arcs must be left visible; the circle through A, B and C is the circumcircle.