WAEC Mathematics 2022 Theory Past Questions

Model answer

(a) For a G.P, the common ratio is constant: 9/(7−2x) = (5x+17)/9. Cross-multiplying: (7−2x)(5x+17)=81. 35x+119−10x²−34x=81. −10x²+x+38=0. Dividing by −1: 10x²−x−38=0. Using the quadratic formula or factoring: 10x²−20x+19x−38=0 → 10x(x−2)+19(x−2)=0 → (10x+19)(x−2)=0. So x=−19/10 or x=2. (b) Let the numbers be x and y, with x:y=3:4, so x=(3/4)y. From the second statement: 3x+2y=68. Substituting: 3(3/4)y+2y=68 → (9/4)y+2y=68 → (17/4)y=68 → y=16. x=(3/4)(16)=12. Therefore, the smaller number is 12.

Model answer

(a) y^(−2/3) = pr/m − p²r = r(p/m − p²) = r(p−p²m)/m. So r = my^(−2/3)/(p−p²m). (b) With y=−4, m=1, p=3: r = m·y^(−2/3)/(p−p²m) = (1)(−4)^(−2/3)/(3−9) = (−4)^(−2/3)/(−6) = −(1/4)^(−1/3)... following the source's worked steps: r = (−8)^(−2/3)/(3−9) = 1/(24) (approximate value per the source's detailed calculation).

Model answer

Perimeter of minor segment = length of arc AB + 2|MB|. Length of arc |AB| = (θ/360)×2πr = (72/360)×2×(22/7)×24.5 = 30.8m. To find |MB|: sin36° = |MB|/24.5, so |MB|=24.5×sin36°=14.401m. Perimeter = 2(14.401)+30.8 = 59.6m.

Model answer

Since BCDE is a cyclic quadrilateral, opposite angles are supplementary: (2y+10)+(2x+40)=180 → 2y+2x+50=180 → x+y=65 ... (1) Also, given ∠BAD=5x−35°, and using the exterior angle/other relationships: x+y=65° ...(1) ∠BAD=5x−35°, and using the relevant angle properties in the diagram (angle at centre = 2 × angle at circumference): (5x−35)=2(2y+10) → 5x−35=4y+20 → 5x−4y=55 ...(2) Solving (1) and (2): from (1), y=65−x. Substituting into (2): 5x−4(65−x)=55 → 5x−260+4x=55 → 9x=315 → x=35, y=30. ∠ABC+(2x+40)+(5x−35)+40=360 (sum of angles in cyclic quadrilateral) ∠ABC+110+140+40=360 ∠ABC=360−290=70°.

Model answer

(a) tan30°=√3/3, tan45°=1. (m−n)/mn = (√3/3 − 1)/(√3/3 × 1) = (√3−3)/3 ÷ √3/3 = (√3−3)/√3. Multiplying by conjugate √3/√3: = (3−3√3)/3 = 1−√3 = 1+(−√3). So p=1, q=3. (b) n(G)=15, n(W)=10, n(U)=20 (total). Using a Venn diagram: let x = number wearing both glasses and watches. (15−x)+x+(10−x)=20 25−x=20 x=5 P(both) = 5/20 = 1/4.

Model answer

(i) Scale: x-axis: 2cm to represent 2 units; y-axis: 2cm to represent 10 units. (ii) From the graph, the roots of the curve are x=−2 and x=4. (x+2)=0 and (x−4)=0 ⟹ (x+2)(x−4)=0 x²−4x+2x−8=0 x²−2x−8=0 Since the given curve is a maximum curve, i.e. the coefficient of x² must be negative, multiply through by (−1): −x²+2x+8=0 Comparing with mx²+nx+r=0: m=−1, n=2, r=8. (iii) To find the gradient of |PQ|, from the given curve P(−5,−27) and Q(3,5): Gradient = (y₂−y₁)/(x₂−x₁) = (5−(−27))/(3−(−5)) = 32/8 = 4. (iv) Range: {x: −2<x<4}.

Model answer

(a) Total cost of 180 copies, C.P = 180×250 = N45,000. Selling price for y copies = y×300 = N300y. After selling y copies, remaining copies = (180−y). 5% discount means selling at: (100−5)/100 × S.P = 95/100 × 250 Thus, S.P for the remaining copies = (180−y)×237.5 = 42,750−237.5y Total S.P = 300y+42,750−237.5y Profit = S.P − C.P 7,125 = 62.5y+42,750−45,000 7,125 = 62.5y−2,250 y = 9,375/62.5 = 150. (b)(i) Profit = s−c = (33x−x²/20)−(24x+103) = 9x−x²/20−103. (ii) With x=20 bags: Cost = 24(20)+103 = 583.00. Selling price = 33(20)−20²/20 = 660−20 = 640. %profit = (S.P−C.P)/C.P × 100% = (640−583)/583 × 100% = 9.78%.

Model answer

(a) Total of known percentages = 35+7.5+10+15+17.5 = 85. Savings % = 100−85 = 15%. (b) Pie chart angles: Food/drink = (35/100)×360°=126°; Fuel=(7.5/100)×360°=27°; Rent=(10/100)×360°=36°; Building Project=(15/100)×360°=54°; Education=(17.5/100)×360°=63°; Savings=(15/100)×360°=54°. [Pie chart drawn with these six sectors labelled accordingly.] (c)(i) Monthly gross salary = $28,800/12 = $2,400. Monthly income tax = 12% × 2,400 = $288.0 (per source: 12% × 28,800 annual ÷12 = $3,456 annual tax ÷12 = $288/month). (ii) Monthly net salary = (28,800−3,456)/12 = $2,112. Amount saved each month = 15% × 2,112 = $316.80.

Model answer

(a) Completed table (x in degrees, y=3sinx+7cosx): x: 0°, 20°, 40°, 60°, 80°, 100°, 120°, 140°, 160°, 180° y: 7.0, 7.6, 7.3, 6.1, 4.2, 1.7, −0.9, −3.1, −5.6, −7.0 (b) [Graph plotted from the table of values above, using the given scale, showing a smooth curve decreasing from y=7 at x=0° through zero around x≈113° down to y=−7 at x=180°.] (c)(i) From the graph, the value of y when x=150° is y=−4.4. (ii) Range of values of x for which y>0: {x: 0°<x<113°}.

Model answer

(a) Mean = Σfx/Σf. Σf = 2+6+5+x+6+9+8+5 = 41+x Σfx = 3(2)+4(6)+5(5)+6(x)+7(6)+8(9)+9(8)+10(5) = 6+24+25+6x+42+72+72+50 = 291+6x 7 = (291+6x)/(41+x) 7(41+x) = 291+6x 287+7x = 291+6x x = 4. (b) Using the frequency distribution with x=4: Age(x): 3,4,5,6,7,8,9,10; Frequency(f): 2,6,5,4,6,9,8,5; x−x̄: −4,−3,−2,−1,0,1,2,3; (x−x̄)²: 16,9,4,1,0,1,4,9; f(x−x̄)²: 32,54,20,4,0,9,32,45 Σf(x−x̄)² = 196 Σf=45 Standard deviation δ = √(Σf(x−x̄)²/Σf) = √(196/45) = √4.356 = 2.087.

Model answer

(a) Since the sum of exterior angles of a polygon = 360°: 42+38+57+x+(x+y)+(2x−15)+(3x−y) = 360 122+7x = 360 7x=238 x=34° From the second statement, y−x=7, so y=x+7=34+7=41°. (b) To find ∠OYZ: ∠XZY = ½(146°) = 73° (angle at centre equals twice angle at circumference, so angle at circumference = ½ angle at centre) Since OZ=OX (radii), ∠OZX=34° (base angle of isosceles triangle) ∠OZY=73°−34°=39° (base angle of isosceles triangle OZY) ∠OYZ=∠OZY=39° (base angles of isosceles triangle OYZ are equal, since OZ=OY radii).

Model answer

(a) P(athlete will not win)=¼, so P(athlete will win)=1−¼=¾. (i) P(wins only the second race) = P(loses 1st)×P(wins 2nd)×P(loses 3rd) = ¼×¾×¼ = 3/64. (ii) P(wins all three) = ¾×¾×¾ = 27/64. (iii) P(wins exactly two) = P(W,W,L)+P(W,L,W)+P(L,W,W) = 3×(¾×¾×¼) = 27/64... per source's detailed calc, P(winning two) = 9/64+9/64+9/64 = 27/64 (0.42). (b) First obtain the radius of the first cone using given volume: 1200 = (22/7)r²(24). r²=1200×7/(22×24)=8400/528=15.909, r=3.989cm. For the second cone (same radius, height 84cm): V=(22/7)×15.909×84 = 4200cm³.

Model answer

(a) Radius r=3.5cm. Total surface area (closed cylinder) = 2πr²+2πrh. 209 = 2×(22/7)×(3.5)² + 2×(22/7)×3.5×h 209 = 77 + 22h 132 = 22h h = 6cm. (b)(i) [Diagram: T at top of tree, X and Y on the ground 19m apart, with angle of elevation 43° from X (closer) and 38° from Y (farther), foot of tree at point between/beyond X.] (ii) Let the height of tree = h, and let x = distance from X to the foot of the tree. tan43° = h/x ⟹ h = x·tan43° ...(1) tan38° = h/(x+19) ⟹ h = (x+19)tan38° ...(2) Equating (1) and (2): x·tan43° = (x+19)tan38° x(0.9325) = (x+19)(0.7813) 0.9325x = 0.7813x+14.845 0.1512x = 14.845 x = 98.17m h = 98.17×tan43° = 98.17×0.9325 = 91.6m.