All 23 questions from the West African Examinations Council (WAEC) Mathematics 2018 Theory paper, with the correct answer and a full explanation for each. Free, no signup needed.
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1.(a) Evaluate without using a calculator: [¼×9⅑ + ⅕(⅔+¾)] ÷ (⅕−¼)
Model answer
9⅑ = 64/7, so ¼×64/7 = 16/7. ⅔+¾ = 17/12, so ⅕×17/12 = 17/30. 16/7+17/30 = (480+119)/210 = 599/210. ⅕−¼ = 3/20. Dividing: 599/210 ÷ 3/20 = 599/210 × 20/3 = 1198/63 = 19¹⁄₆₃.
1.(b) A hunter walked 250m from point P to Q on a bearing of 042°. Calculate, correct to the nearest metre, the vertical distance he has moved.
Model answer
The vertical (northward) component of the displacement is 250cos42° = 250×0.7431 ≈ 186m.
2.(a) Musa is three years older than Manya. Seven years ago, Musa was twice as old as Manya. How old are they now?
Model answer
Let Musa=x, Manya=y. x=y+3 …(1). x−7=2(y−7) → x=2y−7 …(2). Equating: y+3=2y−7 → y=10, x=13. Musa is 13 years old, Manya is 10 years old.
2.(b) When will the sum of their ages be 45?
Model answer
(13+n)+(10+n)=45 → 23+2n=45 → 2n=22 → n=11. The sum of their ages will be 45 in 11 years.
3.(a) The diagram above shows an athletics track with two parallel sides and two semi-circular ends. Each of the parallel sides is 60 metres long and the diameter of each semi-circular end is 120 metres long. Calculate the distance covered by an athlete who runs the track two times. [Take π = 22/7]
Model answer
Perimeter = 2(60) + π(120) = 120 + (22/7×120) = 120+377.14 = 497.14m. Running the track twice: 2×497.14 ≈ 994.3m.
3.(b) If the athlete spends 200 seconds for the race, calculate the speed in km/h.
Model answer
Speed = 994.28m ÷ 200s = 4.9714m/s. Converting: 4.9714×3600/1000 ≈ 17.9km/h.
4.(a) A man was charged 2 kobo per month for every ₦1.00 he borrowed from a bank. At what rate per annum was the interest charged?
Model answer
2 kobo/month per ₦1.00 = 2×12 = 24 kobo per annum per ₦1.00, i.e. a rate of 24% per annum.
4.(b) In the diagram above, |WX| = |XY| = |YZ| and ∠WXY = 80°. What is the size of ∠XWZ?
Model answer
Since WX=XY=YZ, the triangles formed are isosceles. Working through the base angles (each equal to b°=50°), ∠XWZ = b°+b° = 100°.
5. The table shows the distribution of scores obtained when a fair die was rolled 50 times. (a) Draw a bar chart for the distribution. (b) Calculate the mean score of the distribution.
Model answer
(a) A bar chart should show bars of heights 2, 5, 13, 11, 9 and 10 for scores 1–6 respectively. (b) Mean = Σfx/Σf = (1×2+2×5+3×13+4×11+5×9+6×10)/50 = 200/50 = 4.
6.(a) If tan x = 5/12, 0° < x < 90°, evaluate without using Mathematical tables or calculator: sin x / [(sin x)² + cos x]
Model answer
For a 5-12-13 triangle: sinx=5/13, cosx=12/13. sinx/(sin²x+cosx) = (5/13) ÷ [(25/169)+(12/13)] = (5/13) ÷ (181/169) = 65/181.
6.(b) The diagram above shows a rectangular lawn measuring 14m by 11m. A path of uniform width x metres surrounds it. If the total area of the path is 186m², how wide is the path?
Model answer
(14+2x)(11+2x) − (14×11) = 186 → 4x²+50x−186=0 → 2x²+25x−93=0. Solving: x = (−25+√1369)/4 = (−25+37)/4 = 3. The path is 3m wide.
7. A shop had two reduction sales during which prices of all items were reduced by 40% in the first sales and 30% in the second. (a) If a shirt was sold at GH¢3,500 during the second reduction sales, find the price before the first sale. (b) If the price of an article before the first reduction sales was GH¢180.00, find the total: (i) Reduction in the price due to the sales; (ii) Percentage reduction in the price of the article.
Model answer
(a) Let the original price be x. Price after 1st reduction = 0.6x = GH¢50 (price before 2nd reduction). Solving 0.7 of this = 3500 (pesewas) gives the price before the second reduction as GH¢50, so before the first sale the shirt cost approximately GH¢83.33. (b) Reduction due to first sales = 40% of 180 = GH¢72. Price after first sale = GH¢108. Reduction due to second sales = 30% of 108 = GH¢32.40. Total reduction = GH¢104.40. Percentage reduction of the article = (104.40/180)×100 ≈ 58%.
8.(a) Using ruler and a pair of compasses only, construct: (i) a trapezium PQRS such that |PQ| = 6.8cm, ∠PQR = 120°, QR//PS, |PS| = 10.6cm and |PR| = 9.3cm; (ii) locus l₁ of points equidistant from P and R; (iii) locus l₂ of points equidistant from Q and R. 8.(b) Measure: (i) |QR|; (ii) ∠PSR; (iii) |QY|, where Y is the point of intersection of l₁ and l₂.
Model answer
This is a construction exercise to be carried out with ruler and compasses. l₁ is the perpendicular bisector of PR; l₂ is the perpendicular bisector of QR; Y is their intersection point. Measured values of |QR|, ∠PSR and |QY| should be read directly from the accurate scale drawing.
9.(a) A donkey is tied with a rope to a post which is 15m from a fence. If the length of the rope between the donkey and the post is 17m, calculate the length of the fence within the reach of the donkey.
Model answer
By Pythagoras' theorem: 17² = x²+15² → x²=289−225=64 → x=8m. Length of fence accessible = 2x = 16m.
9.(b) The base of a right pyramid with vertex V is a square PQRS of side 15cm. If the slant edge is 32cm long: (i) Represent the information in a diagram. (ii) Calculate its Height (correct to 1 decimal place) and Volume (correct to the nearest cm³).
Model answer
Half-diagonal of the square base = (15√2)/2 ≈ 10.6cm. Height² = 32²−10.6² = 1024−112.36 = 911.64, so Height ≈ 30.2cm. Volume = ⅓ × base area × height = ⅓ × 225 × 30.2 ≈ 2265cm³.
10.(a) Copy and complete the table of values for y = 2cos x − sin x, 0° ≤ x ≤ 300°. 10.(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 = 2cos x − sin x for 0° ≤ x ≤ 300°. 10.(c) Use the graph to find the value(s) of x for which: (i) 2cos x − sin x = 1; (ii) tan x = 2.
Model answer
(a) Computed values: x=0°:2.00, 30°:1.23, 60°:0.13, 90°:−1.00, 120°:−1.87, 150°:−2.23, 180°:−2.00, 210°:−1.23, 240°:−0.13, 270°:1.00, 300°:1.87. (b) Plot these points and draw a smooth curve. (c)(i) Reading where y=1 on the graph gives x≈37° and x=270°. (ii) tan x = 2 corresponds to x≈63° and x≈243° (where the curve crosses the tan x=2 line).
11.(a) If logₐ(y+2) = 1 + logₐx, find x in terms of y.
Model answer
logₐ(y+2) − logₐx = 1 → logₐ[(y+2)/x] = 1 → (y+2)/x = a → x = (y+2)/a.
11.(b) The table below shows the distribution of timber production in five communities in a certain year. (i) Draw a pie chart to represent the information. (ii) What percentage of timber produced that year was from Amenfi? (iii) If a tonne of timber is sold at $560.00, how much more revenue would Oda community receive than Bibiani?
Model answer
Total production = 600+900+1800+1500+2400 = 7200 tonnes. Pie chart sectors (×360°/7200): Bibiani 30°, Amenfi 45°, Oda 90°, Wiawso 75°, Sankore 120°. (ii) % from Amenfi = 900/7200×100 = 12.5%. (iii) Oda revenue = 1800×560 = $1,008,000. Bibiani revenue = 600×560 = $336,000. Difference = $672,000.
12.(a) Mr. John paid ₦8,400.00 in ₦1.00 ordinary shares of a company which sold at ₦2.50 per share. If dividend was declared at 25k per share, how much dividend did he get?
Model answer
Number of shares bought = 8400/2.50 = 3,360 shares. Dividend = 3360×0.25 = ₦840.00.
12.(b) Using the method of completing the square, solve: (1−x)/x + x/(1−x) = 5/2
Model answer
Let u = x/(1−x); then 1/u + u = 5/2 → 2u²−5u+2=0 → u=2 or u=½. Solving x/(1−x)=2 gives x=2/3; solving x/(1−x)=½ gives x=1/3. So x = 1/3 or 2/3.
13.(a) The points R(3,−6), S(6,−2) and T(p,q) are on the x–y plane. If ⅓OR + OS + OT = RS, find the coordinates of T.
Model answer
⅓(3,−6)+(6,−2)+(p,q) = (6−3,−2+6) = (3,4). (1,−2)+(6,−2)+(p,q)=(3,4) → (7+p, −4+q)=(3,4) → p=−4, q=8. T = (−4, 8).
13.(b) If 2(3,m) + n(1,2) = (8,4), find the values of m and n.
Model answer
6+n=8 → n=2. 2m+2n=4 → 2m+4=4 → m=0. So m=0, n=2.
13.(c) Given that w*u = w+u+1, and (y*4)*y = 12, find the value of y.
Model answer
y*4 = y+4+1 = y+5. (y+5)*y = (y+5)+y+1 = 2y+6 = 12 → 2y=6 → y=3.
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