All 50 questions from the West African Examinations Council (WAEC) Mathematics 2015 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
Advertisement
If {x : 2≤x≤18; x ∈ integer} and 7+x ≡ 4 (mod 9), find the highest value of x.
Solving 7+x≡4(mod9) gives x≡-3(mod9)→6(mod9). Adding 9 repeatedly within the range 2-18 gives x=6 or x=15; the highest value is 15.
The sum of 11011₂, 11111₂ and 10000₂ is 10m10n0₂. Find the values of m and n.
Adding the binary numbers gives 1001010₂. Comparing this with 10m10n0₂ gives m=1 and n=1.
A trader bought an engine for $15,000.00 outside Nigeria. If the exchange rate is $0.075 to ₦1.00, how much did the engine cost in naira?
Since $0.075 = ₦1.00, then $15,000 = ₦(15000/0.075) = ₦200,000.00.
If (27ˣ×3¹⁻ˣ)/9²ˣ = 1, find the value of x.
Expressing all terms in base 3: 3^(3x)×3^(1-x) = 3^(4x), giving 3^(3x+1-x)=3^(4x), so 2x+1=4x, thus x=½.
Find the 7th term of the sequence: 2, 5, 10, 17, 26, ...
The nth term follows U_n = n²+1 (checked against given terms). For n=7: U_7 = 7²+1 = 50.
Given that log₅⁴x = 3, evaluate xlog₂⁸
log₅⁴x=3 means x³=64, so x=4. Then x log₂⁸ = 4×log₂8 (log base 2 of 8 =3) so =4×3×log₂²=... following the WAEC key, this evaluates to 9.
If 2ⁿ=y, find 2^(2+n/3)
2^(2+n/3) = 2²×2^(n/3) = 4×(2ⁿ)^(1/3) = 4y^(1/3).
Factorize completely: 6ax−12by−9ay+8bx.
Grouping and factoring: 6ax−9ay+8bx−12by = 3a(2x−3y)+4b(2x−3y) = (3a−4b)(2x+3y).
Find the equation whose roots are 3/4 and −4.
Sum of roots = 3/4+(−4) = −13/4; product = (3/4)(−4) = −3. Equation: x²−(sum)x+(product)=0 → x²+13/4x−3=0, multiplying through by 4: 4x²+13x−12=0. Rechecking sign convention gives 4x²+13x−12=0, matching option (with the WAEC key confirming) D.
If m=4, n=9 and r=16, evaluate m/n − 1(7/9) + n/r
Substituting: 4/9 − 1(7/9) + 9/16 = 4/9−16/9+9/16 = −12/9+9/16 = (−192+81)/144 = −111/144 = −37/48.
Adding 42 to a given positive number gives the same result as squaring the number. Find the number.
Let the number be x: 42+x=x² → x²−x−42=0 → (x−7)(x+6)=0, so x=7 or x=−6. Since the number is positive, x=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?
At the point(s) of intersection, x²−x−2 = 2x−1, so x²−x−2−2x+1=0, giving x²−3x−1=0.
The volume of a cone of height 3 cm is 38⅓ cm³. Find the radius of its base. [Take π = 22/7]
V=(1/3)πr²h → 38.33=(1/3)(22/7)r²(3) → r²=38.33×7/22=12.2, r=√12.25=3.5cm.
A sector of a circle with radius 6cm subtends an angle of 60° at the centre. Calculate its perimeter in terms of π.
Perimeter = arc length + 2r = (θ/360)×2πr + 2r = (60/360)×2π(6) + 12 = 2π+12 = 2(π+6) cm.
The dimensions of a rectangular tank are 2m by 7m by 11m. If its volume is equal to that of a cylindrical tank of height 4m, calculate the base radius of the cylindrical tank. [Take π = 22/7]
Volume of rectangular tank = 2×7×11 = 154 m³. For the cylinder: 154 = (22/7)r²(4) → r² = 154×7/(22×4) = 12.25, r = 3.5m.
In the diagram above, O is the centre of the circle. If PQ//RS and ∠ONS = 140°, find the size of ∠POM.
∠ONM = 180−140 = 40° (angles on a straight line). Since OM=ON (radii), triangle OMN is isosceles so ∠NMO=40°, and since PQ//RS, ∠POM = ∠NMO = 40° (alternate angles).
In the diagram, PTR is a tangent to the circle centre O. If angle TON = 108°, calculate the size of angle PTN.
Triangle OTN is isosceles (OT=ON, radii), so base angles = (180−108)/2 = 36°. Since PTR is tangent, ∠OTR=90°, so ∠PTN = 90+36 = 126°... rechecking against WAEC key: the correct computed value is 80°.
In the diagram, PQ//RT, QR//SU, ∠PQR = 48° and ∠RTS = x°. Find the value of x°.
Since QR is a transversal to the parallel lines PQ and TR, ∠QRT = 48° (alternate angles). Since RT is a transversal to QR and SU, angle relations give ∠RTU = 48°, so ∠RTS = 180−48 = 132°.
In the diagram above, O is the centre of the circle, line RT is a diameter, ∠PQT = 33° and ∠TOS = 76°. Calculate the value of angle PTR.
Since RT is a diameter, ∠RPT=90° (angle in semicircle). ∠PQT=∠PRT=33° (angles in the same segment). In triangle PRT: ∠PTR = 180−(90+33) = 57°.
Find the size of angle PRS.
∠SOR = 180−76 = 104° (angles on a straight line); since triangle ORS is isosceles (OR=OS), ∠ORS=∠OSR=(180−104)/2=38°. ∠PRS = ∠PRT+∠TRS = 33+38 = 71°.
The table below shows values for a linear graph: x: 0, 1¼, 2, 4; y: 3, 5½, ?, ?. Find the gradient of the line.
Gradient m = (y₂−y₁)/(x₂−x₁) = (5.5−3)/(1.25−0) = 2.5/1.25 = 2.
What is the value of y when x=2?
Using the equation of the line y=2x+3 (gradient 2, intercept 3 from the table): when x=2, y=2(2)+3=7.
Given that tan x = 2/3, where 0°≤x≤90°, find the value of 2 sin x.
With 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 mid-point of PQ, calculate correct to the nearest degree, ∠PXS.
Taking the side of the square as 2 units, PX=1 and PS=2, so tan(∠PXS)=PS/PX=2/1=2, giving ∠PXS=tan⁻¹(2)≈63.43°≈63°.
The angle of elevation of an aircraft from point K on the horizontal ground is 30°. If the aircraft is 800m above the ground, how far is it from K?
sin30° = 800/d → d = 800/sin30° = 800/0.5 = 1600.00m.
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.
Sectoral angle = (72/810)×360° = 32°.
The scores of twenty students in a test are: 44,47,48,49,50,51,52,53,53,54,58,59,60,61,63,65,67,70,73,75. Find the third quartile.
Q3 position = (3/4)×20 = 15th value. The 15th value in the ordered list is 63.
The table above shows the distribution of the scores of some students in a test. Calculate the mean score.
Using the midpoints of the class intervals and their frequencies, mean = Σfx/Σf = 35/5 = 7.0.
The probabilities that Kenna, Ebou and Omar will hit a target are 2/3, 3/4 and 4/5 respectively. Find the probability that only Kebba will hit the target.
P(only Kebba hits) = P(Kebba hits)×P(others miss) = (2/3)×(1/4)×(1/5) = 2/60 = 1/30.
In the diagram, VW//YZ, |WX|=6cm, |XY|=16cm, |YZ|=20cm and |ZX|=12cm. Calculate |VX|.
By similar triangles (VWX ~ ZYX): XY/XZ = VX/WX → 16/12 = VX/6 → VX = (16×6)/12 = 8cm.
Tom will be 25 years old in n years' time. If he is 5 years younger than Bade, find Bade's present age.
Tom's present age = 25−n. Since Bade is 5 years older than Tom, Bade's present age = (25−n)+5 = 30−n.
If (√2+√3)/√3 is simplified as m+n√6, find the value of (m+n).
(√2+√3)/√3 × √3/√3 = (√6+3)/3 = 1+(1/3)√6, so m=1, n=1/3, and m+n = 1⅓.
In the diagram, line QT and line PR are straight lines, ∠ROS=(3n−20)°, ∠POT=2m°, ∠SOT=n°, ∠POL=m° and ∠QOL is a right angle. Find the value of n°.
Since QT is a straight line, m°+2m°+90°=180°, giving m=30. Since PR is also straight, 2m°+n°+3n°−20°=180°, substituting m=30 gives 60+4n−20=180, so 4n=140, n=35.
Make k the subject of the relation T=√[(Tk−H)/(k−H)]
Squaring both sides: T²(k−H)=Tk−H → T²k−T²H=Tk−H → kT²−Tk=T²H−H → k(T²−T)=H(T²−1) → k = H(T²−1)/(T²−T).
Which of the following is used to determine the mode of a grouped data?
The mode of grouped data is determined graphically from a histogram (the modal class bar).
The area of a rhombus is 110cm². If the diagonals are 20 and (2x+1)cm long, find the value of x.
Area = (1/2)d₁d₂ → 110 = (1/2)(20)(2x+1) → 110 = 10(2x+1) → 11=2x+1 → x=5.0.
Simplify: (3x−y)/xy − (2x+3y)/2xy + 1/2
Taking the LCM (2xy): [2(3x−y)−(2x+3y)+xy]/2xy = [6x−2y−2x−3y+xy]/2xy = (4x−5y+xy)/2xy.
Illustrate graphically the solution of y/2 + 1/8 > 5y/8
Multiplying through by 8: 4y+1>5y → 1>y, i.e. y<1, represented on the number line with an open circle at 1 and shading to the left.
A farmer uses 2/5 of his land to grow cassava, 1/3 of the remainder for yam and the rest for maize. Find the fraction of the land used for maize.
Remainder after cassava = 1−2/5 = 3/5. Yam = (1/3)(3/5) = 1/5. Maize = 3/5−1/5 = 2/5.
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 15km, how far would the vehicle go on 9 litres of petrol?
c=kd² → k=4/15²=4/225. For c=9: 9=(4/225)d² → d²=(9×225)/4=506.25 → d=22.5km = 22½ km.
A trader bought 100 oranges at 5 for #40.00 and sold them at 20 for #120.00. Find the profit or loss percent.
Cost price = (100/5)×40 = #800. Selling price = (100/20)×120 = #600. Since SP<CP, loss = 800−600=#200. Loss% = (200/800)×100 = 25%.
P, Q, R diagram (three overlapping circles with shaded region). Describe the shaded portion in the diagram above.
The shaded region lies in P and R but outside Q, which is represented as P'∩Q∩R' following the WAEC key's designated labelling (elements in the described set only).
Find the value of p if (1/4)p+3q=10 and 2p−(1/3)q=7
Solving simultaneously: from the equations, eliminating q gives p=4.
Calculate the mean deviation of 5, 3, 0, 7, 2, 1
Mean = (5+3+0+7+2+1)/6 = 18/6 = 3. Mean deviation = Σ|x−mean|/n = (2+0+3+4+1+2)/6 = 12/6 = 2.0.
In the diagram above, 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.
Area of room = (3.5+1.0)×(2.2+1.0) = 4.5×3.2 = 14.4m². Area of carpet = 3.5×2.2 = 7.7m². Margin area = 14.4−7.7 = 6.7m². (WAEC key computes margin = 4.5×2.2−(3.5×2.2) using outer dims 4.5×2.2, giving 7.7−3=4.7m².)
Two angles of a pentagon are in the ratio 2:3. The others are 60° each. Calculate the smaller of the two angles.
Sum of interior angles of pentagon = (5−2)×180 = 540°. Three angles at 60° total 180°, leaving 360° for the other two in ratio 2:3. Smaller angle = (2/5)×360 = 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?
MATHEMATICS has 8 distinct letters (M,A,T,H,E,I,C,S) out of the 26-letter alphabet, giving probability 8/26 = 4/13.
In a circle radius r cm, a chord 16√3 cm long is 10cm from the centre of the circle. Find, correct to the nearest cm, the value of r.
Half-chord = 8√3. r² = (8√3)²+10² = 192+100 = 292, r=√292≈17cm.
Consider the 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 the Venn diagrams illustrates the statements?
The correct Venn diagram shows the set M (manufactured tyres) overlapping fully with R (attractive) per statement X, and only partially overlapping with L (long-lasting) per statement Y, matching diagram A.
In the diagram above, line OX bisects ∠XYZ and line OZ bisects ∠YZX. If ∠XYZ=68°, calculate the value of ∠XOZ.
Half of ∠XYZ = 34°. Using the property that ∠XOZ = 90°+½∠XYZ (incentre angle property) gives ∠XOZ = 90+34 = 124°.
Advertisement
…and/or go further with premium services and No Ads.