Waec 2020 mathematics Answers

WAEC 2020 MATHEMATICS ANSWERS

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MATHS-OBJ 2020
1-10:CBCDACDCCD
11-20:AADBDACBBC
21-30:BDDABDADAD
31-40:CDACCCCCDA
41-50:BBBCDCACDB
SOLVED BY MR ABILITY


MATHS THEORY ANSWERS…………………
(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)
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 = 224πcm²
r:l = 2:5
r/l = 2/5
Cross multiply
2l/2 = 5r/2
L = 5r / 2
Total surface = πrl + πr²
= πr (l + r)
24π/π = πr (5r/2 + r )/ π
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 = 1/2πr²h
= 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) =πrl
Area of hemisphere=2πr²
Total area of structure =πrl + 2πr²
=πr(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
=======================================
(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
============

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