# 2020 Waec General Mathematics Obj & Essay Answer Now Available # Mathematics Questions And Answers WAEC 2020 Images.

MATHS OBJ
1-10: CBCDACDCCD
31-40: CBACCCCCDA

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

[img]https://i.imgur.com/69VZvRZ.jpg[/img]

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[img]https://i.imgur.com/lA1tefu.jpg[/img]

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[img]https://i.imgur.com/69VZvRZ.jpg[/img]

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[img]https://i.imgur.com/lA1tefu.jpg[/img]

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[img]https://i.imgur.com/t7b2on8.jpg[/img]

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[img]https://i.imgur.com/1KXY0z7.jpg[/img]

Exam Time: Monday 17th Aug

Paper II: General Mathematics/ Mathematics (Core) Essay 09:30a.m– 12:00p.m

Paper I: General Mathematics/ Mathematics (Core) Obj 15:00p.m – 16:00p.m

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KEEP REFRESHING THE PAGE

==================================   (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

[img]https://i.imgur.com/69VZvRZ.jpg[/img]

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(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

[img]https://i.imgur.com/lA1tefu.jpg[/img]

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(3a)

Diagram

CBD = CDB (base angles an scales D)

BC

(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
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

href=”https://i.imgur.com/t7b2on8.jpg“>src=”https://i.imgur.com/t7b2on8.jpg“>

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(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

href=”https://i.imgur.com/eaiedbu.jpg “>src=”https://i.imgur.com/eaiedbu.jpg “>

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(5)
href=” https://i.imgur.com/3Z69sIv.jpg “>src=” https://i.imgur.com/3Z69sIv.jpg “>

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(7)
href=” https://i.imgur.com/smln85b.jpg “>src=” https://i.imgur.com/smln85b.jpg “>

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(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

href=” https://i.imgur.com/k2XM7QM.jpg “>src=” https://i.imgur.com/k2XM7QM.jpg “>
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(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°

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