All 23 questions from the West African Examinations Council (WAEC) Mathematics 2014 Theory paper, with the correct answer and a full explanation for each. Free, no signup needed.
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PART I 1(a). Without using tables or calculator, simplify: (0.6×32×0.004)/(1.2×0.008×0.16), leaving the answer in standard form (scientific notation).
Model answer
Numerator = 0.6×32×0.004 = 0.0768 Denominator = 1.2×0.008×0.16 = 0.001536 Result = 0.0768/0.001536 = 50 In standard form: 5×10¹
1(b). In the diagram, line EF is parallel to line GH. If ∠AEF=3x°, ∠ABC=120° and ∠CHG=7x°, find the value of ∠GHB.
Model answer
Draw a line through B parallel to EF and GH. Corresponding angles: the part of ∠ABC nearest EF equals ∠AEF = 3x°. So the remaining part, ∠NBH = 120°−3x° (alternate angle to ∠GHB). Also, by vertically opposite/co-interior relations at H, using ∠CHG=7x°: ∠GHB=180°−7x°. Equating: 180−7x = 120−3x ⟹ 60=4x ⟹ x=15. ∠GHB = 120−3(15) = 120−45 = 75°.
2(a). Simplify 3√75 − √12 + √108, leaving the answer in surd form (radicals).
Model answer
3√75 = 3×5√3 = 15√3 √12 = 2√3 √108 = 6√3 So: 15√3 − 2√3 + 6√3 = 19√3
2(b). If 124ₙ = 232₅, find n.
Model answer
232₅ in base 10 = 2×25 + 3×5 + 2 = 50+15+2 = 67 124ₙ = n²+2n+4 = 67 n²+2n−63=0 (n+9)(n−7)=0 n=7 (taking the positive root)
3(a). Solve the simultaneous equations: 1/x + 1/y = 5, 1/y − 1/x = 1.
Model answer
Let a=1/x, b=1/y. a+b=5 ...(i) b−a=1 ...(ii) Adding (i) and (ii): 2b=6, b=3 From (i): a=5−3=2 So 1/x=2 ⟹ x=1/2 And 1/y=3 ⟹ y=1/3
3(b). A man drives from Ibadan to Oyo, a distance of 48km, in 45 minutes. If he drives at 72km/h where the surface is good and 48km/h where it is bad, find the number of kilometers of good surface.
Model answer
Let d = distance of good surface, so (48−d) = distance of bad surface. Time = distance/speed. Total time = 45min = 0.75h. d/72 + (48−d)/48 = 0.75 Multiply through by 144 (LCM of 72,48): 2d + 3(48−d) = 108 2d+144−3d=108 −d=−36 d=36km
4(a). In the diagram, O is the centre of the circle radius r cm and ∠XOY=90°. If the area of the shaded part is 504cm², calculate the value of r. [Take π=22/7]
Model answer
Sector area (90°) = (90/360)πr² = πr²/4 Triangle XOY area = ½r²sin90° = ½r² Shaded (segment) area = sector − triangle = r²(π/4 − 1/2) Using π=22/7: π/4−1/2 = 11/14 − 7/14 = 2/7 So: r²×(2/7) = 504 r² = 504×7/2 = 1764 r = √1764 = 42cm
4(b). Two isosceles triangles PQR and PQS are drawn on opposite sides of a common base PQ. If ∠PQR=66° and ∠PSQ=109°, calculate the value of ∠RQS.
Model answer
Triangle PQR is isosceles on base PQ, with ∠PQR=∠QPR=66° (base angles equal). Triangle PQS is isosceles on base PQ, with apex ∠PSQ=109°, so base angles ∠PQS=∠QPS=(180−109)/2=35.5°. Since R and S are on opposite sides of PQ, ∠RQS = ∠PQR + ∠PQS = 66°+35.5° = 101.5°
5. A building contractor tendered for two independent contracts, X and Y. The probability that he will win contract X is 0.5, and the probability that he will not win contract Y is 0.3. What is the probability that he will win: (a) both contracts; (b) exactly one of the contracts; (c) neither of the contracts?
Model answer
P(X)=0.5, so P(X')=0.5. P(Y')=0.3, so P(Y)=0.7. (a) P(both) = P(X)×P(Y) = 0.5×0.7 = 0.35 (b) P(exactly one) = P(X)P(Y') + P(X')P(Y) = (0.5×0.3)+(0.5×0.7) = 0.15+0.35 = 0.50 (c) P(neither) = P(X')×P(Y') = 0.5×0.3 = 0.15
PART II — Answer five questions from this part. All questions carry equal marks. 6(a). If 3/(2p−½) = (1/3)/(¼p+1), find p.
Model answer
Cross-multiplying: 3×(¼p+1) = (1/3)×(2p−½) (3p/4)+3 = (2p/3)−(1/6) Multiply through by 12: 9p+36 = 8p−2 9p−8p = −2−36 p = −38
6(b). A television set was marked for sale at GH¢760.00 in order to make a profit of 20%. The television set was actually sold at a discount of 5%. Calculate, correct to 2 significant figures, the actual percentage profit.
Model answer
Marked price = Cost price × 1.20, so Cost Price = 760/1.20 = GH¢633.33 Selling price (after 5% discount) = 760×0.95 = GH¢722.00 Actual profit = SP−CP = 722.00−633.33 = GH¢88.67 Percentage profit = (88.67/633.33)×100% ≈ 14%
7(a). Copy and complete the table of values for the relation y=2sinx+1 for x=0° to 270° in steps of 30°. 7(b). Using scales of 2cm to 30° on the x-axis and 2cm to 1 unit on the y-axis, draw the graph of y=2sinx+1 for 0°≤x≤270°. 7(c). Use the graph to find the values of x for which sinx=1/4.
Model answer
(a) Computed table: x: 0°,30°,60°,90°,120°,150°,180°,210°,240°,270° y: 1.0, 2.0, 2.7, 3.0, 2.7, 2.0, 1.0, 0.0, −0.7, −1.0 (b) Plot these points on the given scale and draw a smooth sine curve through them. (c) sinx=1/4 corresponds to y=2(1/4)+1=1.5. Reading from the graph where the curve crosses y=1.5 gives x≈14.5° and x≈165.5°.
8(a). Copy and complete the following table for multiplication modulo 11 (values: 1, 5, 9, 10). Use the table to: (i) evaluate (9⊗5)⊗(10⊗10); (ii) find the truth set of I. 10⊗m=2 II. n⊗n=4.
Model answer
Completed table (mod 11): 5⊗5=3, 5⊗9=1, 5⊗10=6, 9⊗5=1, 9⊗9=4, 9⊗10=2, 10⊗5=6, 10⊗9=2, 10⊗10=1 (i) (9⊗5)⊗(10⊗10) = 1⊗1 = 1 (ii) I. 10⊗m=2 ⟹ from the table, m=9. II. n⊗n=4 ⟹ n=9.
8(b). When a fraction is reduced to its lowest term, it is equal to ¾. The numerator of the fraction when doubled would be 34 greater than the denominator. Find the fraction.
Model answer
Let the fraction be 3k/4k (lowest terms 3:4). Doubling numerator: 2(3k) = 6k. This is 34 greater than the denominator: 6k = 4k+34 2k=34 k=17 Fraction = 3(17)/4(17) = 51/68
9(a). In the Venn diagram, P, Q and R are subsets of the universal set U. If n(U)=125, find: (i) the value of x; (ii) n(P∩Q∩R′).
Model answer
Sum of all regions = 125: (16−2x)+5x+(6+x)+8x+4x+7x+(19−3x)+4 = 125 Constants: 16+6+19+4=45. x-terms: −2x+5x+x+8x+4x+7x−3x=20x 45+20x=125 ⟹ 20x=80 ⟹ x=4 (ii) n(P∩Q∩R′) = the region in both P and Q but not R = 5x = 5(4) = 20
9(b). In the diagram, O is the centre of the circle. If WX is parallel to YZ and ∠WXY=50°, find the value of: (i) ∠WYZ; (ii) ∠YEZ.
Model answer
(i) Since WX∥YZ, with XY as transversal, ∠WYZ = ∠WXY = 50° (alternate angles). (ii) In triangle EYZ (E is the intersection of the diagonals), by the isosceles symmetry of the trapezium, the base angles at Y and Z are each 50°. Sum of angles in triangle EYZ: ∠YEZ = 180°−50°−50° = 80°
10(a). Solve: (x−2)(x−3)=12.
Model answer
Expand: x²−5x+6=12 x²−5x−6=0 (x−6)(x+1)=0 x=6 or x=−1
10(b). In the diagram, M and N are the centres of two circles of equal radii 7cm. The circles intersect at P and Q. If ∠PMQ=∠PNQ=60°, calculate, correct to the nearest whole number, the area of the shaded portion. [Take π=22/7]
Model answer
Sector area (60°, r=7) = (60/360)×(22/7)×7² = (1/6)×154 = 25.67cm² Triangle MPQ is equilateral (MP=MQ=7=radius, angle 60°): Area = (√3/4)×7² ≈ 21.22cm² Segment area (one side) = 25.67−21.22 = 4.45cm² Total shaded (lens) area = 2×4.45 ≈ 8.9cm² ≈ 9cm²
11. The table shows the distribution of outcomes when a die is thrown 50 times (Scores 1–6; Frequencies 2,5,13,11,9,10). Calculate the: (a) mean deviation of the distribution; (b) probability that a score selected at random is at least 4.
Model answer
Mean = Σfx/Σf = (1×2+2×5+3×13+4×11+5×9+6×10)/50 = (2+10+39+44+45+60)/50 = 200/50 = 4 (a) Mean deviation = Σf|x−mean|/Σf = [2(3)+5(2)+13(1)+11(0)+9(1)+10(2)]/50 = [6+10+13+0+9+20]/50 = 58/50 = 1.16 (b) P(score≥4) = (11+9+10)/50 = 30/50 = 3/5 = 0.6
12(a). Given that 5cos(x+8.5)°−1=0, 0°≤x°≤90°, calculate, correct to the nearest degree, the value of x.
Model answer
5cos(x+8.5)°=1 cos(x+8.5)° = 0.2 x+8.5° = cos⁻¹(0.2) = 78.46° x = 78.46−8.5 ≈ 70° (to the nearest degree)
12(b). The bearing of Q from P is 150° and the bearing of P from R is 015°. If Q and R are 24km and 32km respectively from P: (i) represent this information in a diagram; (ii) calculate the distance between Q and R, correct to two decimal places; (iii) find the bearing of R from Q, correct to the nearest degree.
Model answer
(i) Sketch P with PQ=24km at bearing 150°, and PR=32km at bearing 195° (since bearing of P from R is 015°, bearing of R from P = 015+180=195°). (ii) Angle QPR = 195°−150° = 45°. By cosine rule: QR² = 24²+32²−2(24)(32)cos45° = 576+1024−1536×0.7071 ≈ 513.9 QR ≈ 22.67km (iii) By sine rule: sin(PQR)/32 = sin45°/22.67 ⟹ sin(PQR) ≈ 0.998 ⟹ ∠PQR ≈ 86.6° Bearing of P from Q = 150+180 = 330°. Bearing of R from Q = 330°−86.6° ≈ 243° (to the nearest degree).
13(a). Two functions, f and g, are defined by f:x→2x²−1 and g:x→3x+2, where x is a real number. (i) If f(x−1)−7=0, find the values of x; (ii) Evaluate f(−½)·g(3) / (f(4)−g(5)).
Model answer
(i) f(x−1) = 2(x−1)²−1 = 7 ⟹ (x−1)²=4 ⟹ x−1=±2 ⟹ x=3 or x=−1 (ii) f(−½)=2(¼)−1=−0.5; g(3)=3(3)+2=11; f(4)=2(16)−1=31; g(5)=3(5)+2=17 f(4)−g(5)=31−17=14 Result = (−0.5×11)/14 = −5.5/14 ≈ −0.39 (= −11/28)
13(b). An operation (*) is defined on the set R of real numbers by m(*)n = −n/(m²+1), where m,n∈R. If −3,−10∈R, show whether or not (*) is commutative.
Model answer
m*n = −n/(m²+1). Test with m=−3, n=−10: m*n = −(−10)/((−3)²+1) = 10/10 = 1 n*m = −(−3)/((−10)²+1) = 3/101 ≈ 0.0297 Since m*n ≠ n*m (1 ≠ 0.0297), the operation (*) is NOT commutative.
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