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WAEC 2020 GENERAL MATHEMATICS OBJ & THEORY ANSWERSMATHS OBJ

MATHS-OBJ

1 -10-CBCDACDCCD

11 – 20-AADBDACBBC

21-30-BDDABDADAD

31-40-CDACCCCCDA

41-50-BBBCDCACDB

COMPLETED

==================================

**(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)?**

**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)**

**Diagram**

**CBD = CDB (base angles an scales D)**

**BCD+CBD+CDB=180° (Sum of < in a D)**

**2CDB+BCD=180°**

**2CDB+108°=180°**

**2CDB=180°-108°=72°**

**CDB=72/2=36°**

**BDE=90°(Angle in semi circle)**

**CDE=CDB+BDE**

**=36°+90**

**=126**

**(3b)**

**(Cosx)² – Sinx given**

**(Sinx)² + Cosx**

**Using Pythagoras theory thrid side of triangle**

**y²= 1²+?3**

**y²= 1+ 3=4**

**y=?4=2**

**(Cosx)² – sinx/(sinx)² + cosx**

**(1/2)² – ?3/2/**

**(?3/2)² + 1/2 = 1/4 – ?3/2 = 1-2?3/4**

**3/4+1/2 = 3+2/4**

**=1-2?3/4 * 4/5**

**=1-2?3/5**

**=======================================**

**(4a)**

**Total Surface Area = 224?cm²**

**r:l = 2:5**

**r/l = 2/5**

**Cross multiply**

**2l/2 = 5r/2**

**L = 5r / 2**

**Total surface = ?rl + ?r²**

**= ?r (l + r)**

**24?/? = ?r (5r/2 + r )/ ?**

**224 = 5r²/2 + r²/1**

**L.c.m = 2**

**448 = 5r² + 2r²**

**448 / 7= 7r²/7**

**r² = 64**

**r = ?64 = 8cm**

**L = 5*8/2 = 20cm**

**(4b)**

**Volume = 1/2?r²h**

**= 1/3 * 22/7 * 8 * 8 * 18.33**

**= 1228.98cm³**

**L² = h² + r ²**

**20² = h² + 8²**

**400 – 64 = h²**

**h² = 336**

**h = ? 336**

**h = 18.33cm**

**=======================================**

**(5a)**

**Total income = 32+m+25+40+28+45**

**=170+m**

**PR(²)=m/170+m = 0.15/1**

**M=0.15(170+m)**

**M=25.5+0.15m**

**0.85m/0.85=25.5/0.85**

**M=30**

**(5b)**

**Total outcome = 170 + 30 = 200**

**(5c)**

**PR(even numbers) = 30+40+50/200**

**=115/200 = 23/40**

**=======================================**

**(7a)**

**Diagram**

**Using Pythagoras theorem, l²=48² + 14²**

**l²=2304 + 196**

**l²=2500**

**l=?2500**

**l=50m**

**Area of Cone(Curved) =?rl**

**Area of hemisphere=2?r²**

**Total area of structure =?rl + 2?r²**

**=?r(l + 2r)**

**=22/7 * 14 [50 + 2(14)]**

**=22/7 * 14 * 78**

**=3432cm²**

**~3430cm² (3 S.F)**

**(7b)**

**let the percentage of Musa be x**

**Let the percentage of sesay be y**

**x + y=100 ——————-1**

**(x – 5)=2(y – 5)**

**x – 5=2y – 10**

**x – 2y=-5 ——————-2**

**Equ (1) minus equ (2)**

**y – (-2y)=100 – (-5)**

**3y=105**

**y=105/3**

**y=35**

**Sesay’s present age is 35years**

**=======================================**

**(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**

**============================**

** **

(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

================================================

(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)

Diagram

(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

**Solving/Typing…..[Keep Refreshing]**

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