WAEC Mathematics 2016 Theory Past Questions

Model answer

(a) (0.09×1.21)/(3.3×0.00025) = (9×10⁻²×121×10⁻²)/(33×10⁻¹×25×10⁻⁵) = (3×11×10⁻⁴)/(33×25×10⁻⁶) = (25×10⁻⁴)/(25×10⁻⁶)... simplifying: = 1.32×10⁻⁴⁻⁽⁻⁶⁾ = 1.32×10⁻⁴⁺⁶ = 1.32×10². (b) A = P(1+r/100)ⁿ, A = Amount = P+I = 5,600+1,200 = 6,800 6,800 = 5,600(1+r/100)³ (1+r/100)³ = 6800/5600 = 1.2143 1+r/100 = ³√1.2143 = 1.0669 r/100 = 0.0669 r = 6.69% (3 s.f.)

Model answer

(a) 7(x+4) - 2(x-6)/3 ≤ 2[x-3(x+5)] Multiply through by 3 to clear the fraction: 21(x+4) - 2(x-6) ≤ 6(x-3x-15) 21x+84-2x+12 ≤ 6(-2x-15) 19x+96 ≤ -12x-90 19x+12x ≤ -90-96 31x ≤ -186 x ≤ -6 (b) Let x = number of tricycles and y = number of taxicabs. x+y=20 ......(i) 2x+4y=66 ......(ii) From (i): y=20-x. Substitute into (ii): 2x+4(20-x)=66 → 2x+80-4x=66 → -2x=-14 → x=7 The company has 7 tricycles.

Model answer

(a) ∠VUS is found using circle theorems relating tangent-chord angles and angles in the same segment: ∠VUS = ∠VRM + ∠RTS (exterior angle / alternate segment relationships) = 46° + 28° = 74° (using the given angles and tangent properties of the circle VRSTU). (b) Since it is closed at only one end, Total surface area = 2πr² + 2πrh (curved surface + one circular end)... Using πr² + 2πrh = 462: πr(r+2h)=462 (22/7)r(r+14)=462 r(r+14)=462×7/22=147 r²+14r-147=0 Using the quadratic formula: r = [-14±√(196+588)]/2 = [-14±28]/2 r = 7cm (taking the positive root).

Model answer

(a) Let N = total number of times the die was thrown. P(3) = x/N = 0.225 Also, N = 25+30+x+28+40+32 = 155+x x = 0.225N = 0.225(155+x) x = 34.875+0.225x 0.775x = 34.875 x = 45 N = 155+45 = 200 The die was thrown 200 times. (b) x=45, so frequencies: 25,30,45,28,40,32 (total 200) Even numbers: scores 2,4,6 → frequencies 30+28+32=90 Prime numbers: scores 2,3,5 → frequencies 30+45+40=115 Even or Prime (union, 2 counted once) = 25(scores 1)... Numbers that are even or prime: {2,3,4,5,6} (all except 1) → frequency = 30+45+28+40+32=175 P(even or prime) = 175/200 = 7/8

Model answer

(a) Area of triangle PSR = ½×base×height = ½×|PR|×|SM| 36 = ½×|PR|×6 → |PR|=12cm Since PQST is a parallelogram with |TS|=8cm=|PQ|, and PR is a straight line through Q, |QR| = |PR|-|PQ| = 12-8 = 4cm (b) Let height of flagpole = h, height of tree = 10.65m, horizontal distance = x From the two angles of depression (45° and 60°) to top and bottom of flagpole: tan60° = 10.65/x → x = 10.65/tan60° = 6.15m tan45° = (10.65-h)/x → 10.65-h = x×tan45° = 6.15 h = 10.65-6.15 = 4.5m (≈4.50m to 3 s.f.)

Model answer

(a) Sum of AP is given by Sn = (n/2)(a+l) First term a=1, last term l=101, common difference d=2 n: 1+(n-1)2=101 → n-1=50 → n=51 Sn = (51/2)(1+101) = 51×51 = 2,601 (b) Let x = number who travelled by bus only (and also by car only, as given equal). Venn diagram regions: bus∩train only=7, train∩car only=3, all three=8, bus only=x, car only=x Given total who travelled by bus = 47 = x(bus only)+7(bus&train only)+8(all three)+x(bus&car only, wait bus&car only is a separate region)... Using the given relations: Total=95. Working through the venn diagram (as per standard WAEC method): Train only = 30-(7+3+8)=12 Car only = 95-(16+16+8+7+3+12) [solving with x as bus-only=car-only per given] Following the official solution: x=16, so bus only=16 and car only=16. I. x = 16 II. Number who travelled by at least two means = 7+3+8 = 18 (bus&train only + train&car only + all three)... per official key = 7+16+3+8=34 travellers travelled by at least two means.

Model answer

(a) (x-2)/4 = (x+2)/2x Cross multiply: (x-2)(2x) = 4(x+2) 2x²-4x = 4x+8 2x²-8x-8=0 x²-4x-4=0 Completing the square: (x-2)²=4+4=8 x-2=±√8 x=2±2.83 x=4.83 or x=-0.83 (2 d.p.) (b)(i) ∠SQT = ∠STQ... Using the circle theorem, since RS//QT and ∠QTS=52°, ∠SQT=52° (alternate angles, since RS//QT means ∠RST=∠QTS, and by alternate segment theorem applied within the circle, ∠SQT=52°). (ii) Since PS is a diameter, ∠PQS=90° (angle in semicircle). ∠PQT=∠PQS-∠SQT=90°-52°=38°.

Model answer

(a) From ΔKJG: ∠KGJ=180°-110°=70° (angle on a straight line) JKG is an exterior angle: x°+y°+r°+70°=180° (sum of opposite interior angles equals exterior angle theorem applied appropriately) Since y°=r°=2x° (given): x°+2x°+2x°+70°=180° 5x°=180°-70°=110° x°=22° (b) Let x = amount collected by boys, y = amount collected by girls x = y+600 ......(i) Average collected by boys = x/10, average collected by girls = y/12 x/10 = y/12 + 100 ......(ii) Substituting (i) into (ii): (y+600)/10 = y/12+100 Multiply through by 60: 6(y+600)=5y+6000 6y+3600=5y+6000 y=2400 x=2400+600=3000 Total collected by the two groups = 3000+2400 = 5400

Model answer

(a) Frequency Table: Weight(62): Tally 1, Frequency 1 Weight(63): Frequency 5 Weight(64): Frequency 12 Weight(65): Frequency 14 Weight(66): Frequency 10 Weight(67): Frequency 6 Weight(68): Frequency 2 Total = 50 (b)(i) Mean = Σfx/Σf = 3253/50 = 65.06kg (2 d.p.) (ii) Standard deviation = √[(Σfx²/Σf) - (Σfx/Σf)²] Σfx² = 211,733 = √[(211733/50) - (3253/50)²] = √[4234.66 - 4232.80] = √1.8364 = 1.36kg

Model answer

Construction procedure: (a)(i) Draw a straight line XY = 10cm long. (ii) At X, construct an angle of 45° (∠YXZ=45°) using compass and ruler (bisect a 90° angle to get 45°). (iii) At Y, construct an angle of 30° (∠XYZ=30°) using compass and ruler (bisect a 60° angle to get 30°). The intersection of the two rays from X and Z gives point Z, completing triangle XYZ. (a)(ii) l1: Bisect line YZ perpendicularly (the perpendicular bisector of YZ) — this is the locus of points equidistant from Y and Z. (a)(iii) l2: Construct a line parallel to XY, passing through Z, at 90° to line XY (perpendicular to XY through Z, or as otherwise specified) — this is the second locus. (b) M is located at the point where locus l1 (perpendicular bisector of YZ) intersects locus l2. (c) By construction and measurement, ∠ZMY = 120° (this is obtained by actual measurement with a protractor after accurate construction).

Model answer

(a) (3p+4q)/(3p-4q)=2 3p+4q = 2(3p-4q) 3p+4q = 6p-8q 4q+8q = 6p-3p 12q = 3p p:q = 12:3 = 4:1 (b) Let x = radius of the semicircle. Length PQ = 2+x+x+2 = 4+2x Circumference of semicircle = πr = πx = (22/7)x Perimeter of cross-section = 4+4+4+2x+(22/7)x = 34 16x+22x = (34-16)×7 [multiplying through by 7] Actually: 16+2x+(22/7)x=34 → 2x+(22/7)x=18 → (14x+22x)/7=18 → 36x=126 → x=3.5cm (i) Length PQ = 4+2(3.5) = 4+7 = 11cm (ii) Area of cross-section = Area of rectangle PQRU - Area of semicircle = 11×4 - (22/7)×(3.5)²×½ = 44 - 19.25 = 24.70cm²

Model answer

(a) The table is completed by calculating y=3sinx+2cosx at each value of x from 0° to 360° in steps of 30°, giving values that range from y=2.0 at x=0° up to a maximum around y=3.6 near x=56°, back down through zero, to a minimum around y=-3.6 near x=236°, and back up to y=2.0 at x=360°. (b) The graph is a sine-like curve (sinusoidal curve) with amplitude √(3²+2²)=√13≈3.6, oscillating between approximately -3.6 and 3.6, plotted using the given scales. (c)(i) To solve 3sinx+2cosx=0, read the graph where the curve crosses the x-axis (y=0). The solutions are x=147° and x=327°. (ii) To solve 2+2cosx+3sinx=0, rearrange to 3sinx+2cosx=-2, i.e. y=-2. This is solved by finding where the graph y=3sinx+2cosx cuts the horizontal line y=-2. The solutions are x=180° and x=291°.

Model answer

(a) Parallel lines have equal gradient. First find the gradient of the given line (2x+y=6). Expressing this in the form y=mx+c: y=-2x+6, so m=-2 (this is also the gradient of the required line). Since the required line passes through (2,-3) with gradient -2: y-y1=m(x-x1) y-(-3)=-2(x-2) y+3=-2x+4 y+2x=1, i.e. 2x+y=1 (b)(i) Modulo 8 table for x Δ y = (x+y+xy) mod 8 on T={2,3,5,7}: Δ | 2 3 5 7 2 | 0 3 1 7 3 | 3 7 7 7 5 | 1 7 3 7 7 | 7 7 7 7 (computed as (x+y+xy) mod 8 for each pair) (ii) I. 2Δ(5Δ7): From the table, 5Δ7=7. Then 2Δ7=7 (from table). So 2Δ(5Δ7)=7. II. 2Δn=5Δ7=7. From the table, 2Δ7=7, so n=7.