 iBlogTV: WAEC 2020 Free Mathematics (Obj & Essay) Answers - May/June Expo

## Pages # WAEC 2020 Free Mathematics (Obj & Essay) Answers - May/June Expo

## Aug 16, 2020

### WAEC 2020 Free Mathematics (Obj & Essay) Answers - May/June Expo

(1a) Given A={2,4,6,8,...} B={3,6,9,12,...} C={1,2,3,6} U= {1,2,3,4,5,6,7,8,9,10}

A' = {1,3,5,7,9} B' = {1,2,4,5,7,8,10} C' = {4,5,7,8,9,10} A'nB'nC' = {5, 7}

(1b) Cost of each premiere ticket = \$18.50 At bulk purchase, cost of each = \$80.00/50 = \$16.00

Amount saved = \$18.50 - \$16.00 =\$2.50 =================================================

(2ai) P = (rk/Q - ms)⅔ P^3/2 = rk/Q - ms rk/Q = P^3/2 + ms Q= rk/P^3/2 + ms

(2aii) When P =3, m=15, s=0.2, k=4 and r=10 Q = rk/p^3/2 + ms = 10(4)/(3)^3/2 + (15)(0.2) = 40/8.196 = 4.88(1dp)

(2b) x + 2y/5 = x - 2y Divide both sides by y X/y + 2/5 = x/y - 2 Cross multiply 5(x/y) - 10 = x/y + 2 5(x/y) - x/y = 2 + 10 4x/y = 12 X/y = 3 X : y = 3 : 1 =================================================
(3a) Draw the diagram

CBD = CDB(Base angles of an issoceles triangle) BCD + CBD + CDB = 180°(sum of angles in a triangle) 2CDB + BCD = 180° 2CDB + 108° = 180° 2CDB = 180° - 108° =72° CDB = 72/2 = 36° BDE = 90°(angle in a semi-circle) CDE = CDB + BDE = 36° + 90° = 126°

(3b)
(CosX)² - SinX/(SinX)²+ CosX Using Pythagoras theorem, third side of triangle y² = 1² + |3² y² = 1 + 3 = 4 y = square root e = 2 Cos X = 1/2(adj/hyp) Sin X = root 3/2(opp/hyp) (CosX)² - SinX/(SinX)² + CosX = (1/2)² - root3/2 / (root3/2)² + 1/2 = 1/4 - root3/2 / 3/4 + 1/2 = 1 - 2root3/4 / 3+2/4 = 1-2root3/5 =================================================

(4a) Given: r : l = 2 : 5 (ie l = 5/2r) Total surface area of cone =πr² + url 224π = π(r² + r(5/2r)) 224 = r² + 5/2r² 224 = 7/2r² 7r² = 448 r² = 448/7 = 64 r = root 64 = 8.0cm

(4b) L = 5/2r = 5/2 × 8 = 20cm Using Pythagoras theorem L² = r² + h² h² = l² - r² h² = 20² - 8² h² = (20 + 8)(20 - 8) h² = 28 × 12 h = root28×12 h = 18.33cm

Volume of cone = 1/3πr²h = 1/3 × 22 × 7 × 8² × 18.33 =1229cm³ ================================================

(5a) Prob(2) = no of 2s/Total outcomes 0.15 = m/32+m+25+40+28+45 0.15 = m/m + 170 m = 0.15m + 25.5 m - 0.15m = 25.5 0.85m = 25.5 m = 25.5/0.85 = 30

(5b) Number of times dice was rolled = m + 170 = 30 + 70 = 200

(5c) Prob(even number) = no of even numbers/Total outcome = m+40+45/200 =30+40+45/200 =115/200 = 23/40 = 0.575 ================================================
(7a) Total surface area = url + 2πr² =πr(l + 2r)

Draw the diagram From pythagoras theorem Hyp² = Adj² + Opp² L² = 14² + 48² L² = 196 + 2304 L² = 2500 L = /2500 = 50m

=πr(L + 2r) = 22/7 ×14(50 + 2(14)) = 44(50 + 28) = 3432m² Total surface area = 3432m² ~3430m²(to 3s.f)

(7b) Five years ago, Let Musa's age = x Let Sesay's age = y X - 5 = 2(Y - 5) X - 5 = 2y - 10 X - 2y = 5 - 10 X - 2y = -5 ..... (1) -X + y = 100 ..... (2) -3y = -105 Subtracting eqn 2 from 1 -3y/3 = -105/-3 y = 35 Sesay's present age = 35 years ================================================
(8a) Let Ms Maureen's Income = Nx 1/4x = shopping mall 1/3x = at an open market

Hence shopping mall and open market = 1/4x + 1/3x = 3x + 4x/12 = 7/12x

Hence the remaining amount = X-7/12x = 12x-7x/12 =5x/12
Then 2/5(5x/12) = mechanic workshop = 2x/12 = x/6 Amount left = N225,000 Total expenses = 7/12x + X/6 + 225000 = Nx

7x+2x+2,700,000/12 =Nx 9x + 2,700,000 = 12x 2,700,000 = 12x - 9x 2,700,000/3 = 3x/3 X = N900,000

(ii) Amount spent on open market = 1/3X = 1/3 × 900,000 = N300,000

(8b) T3 = a + 2d = 4m - 2n T9 = a + 8d = 2m - 8n -6d = 4m - 2m - 2n + 8n -6d = 2m + 6n -6d/-6 = 2m+6n/-6 d = -m/3 - n d = -1/3m - n =========================================

(9a) Draw the triangle
(9b) (i)Using cosine formulae q² = x² + y² - 2xycosQ q² = 9² + 5² - 2×9×5cos90° q² = 81 + 25 - 90 × 0 q² = 106 q = square root 106 q = 10.30 = 10km/h Distance = 10 × 2 = 20km

(ii) Using sine formula y/sin Y = q/sin Q 5/sin Y = 10.30/sin 90° Sin Y = 5 × sin90°/10.30 Sin Y = 5 × 1/10.30 Sin Y = 0.4854 Y = sin‐¹(0.4854), Y = 29.04

Bearing of cyclist X from y = 90° + 19.96° = 109.96° = 110°

(9c) Speed = 20/4, average speed = 5km/h ===============================================

(11a) (11b) Given 8y+4x=24 8y=-4x + 24 y=4/8x + 24/8 y=-1/2x +3 Gradient = -1/2 Using m = y-y/x-x¹ and given (x¹=-8) (y¹=12) -1/2=y-12/x+8 2(y-12)=-x-8 2y-24=-x-8 2y+x=24-8 2y+x=16 ===============================================

(12a) BCD=ABC=40°(alternate D)
DDE=2*BCD(
(12bi) Digram

(12bii) Area of parallelogram = absin =5*7*sin125° =35*sin55° =35*0.8192 =28.67 =28.7cm²(1dp)

(12c) Given x=1/2(1-√2) 2x²-2x=2[1/2(1-√2]²-2(1/2(1-√2)} =2[1-2√2+2/4]-(1-√2) =(3-2√2/2)-(1-√2) =3-2√2-2+2√2/2=1/2 OR

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