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# WAEC 2020 General Mathematics Essay & Obj Question And Answer

## How to Get 2020 WAEC  MATHEMATICS  ESSAY AND OBJECTIVE Question Paper.

Monday 17th August 2020
General mathematics(core) 2(Essay) – 9:30am – 12:00noon (2hrs 30mins)
General mathematics(core) 2(objective) – 3:00pm – 4:30pm
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MATHS OBJ

MATHS-OBJ
1-10:CBCDACDCCD
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 = \$16.00Amount 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
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

====

(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
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

======

(12a)
BCD=ABC=40°(alternate D)

DDE=2*BCD(<at centre = twice < at circle)

DDE = 2*40 = 80°
OD3=OED(base < of I sealed D ODE)
ODE + OED + DOE= 180°(sum of < is in D)
2ODE+DOE=180°
2ODE+80°=180
2ODE+180=180
2ODE+100°
ODE+100/2=50°

(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

======

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### OR

1. How to get the math obj

by Ogunnubi oluwaTosin on Aug 17, 2020 at Reply

2. Hi

by Gabriel rotimi julius on Aug 16, 2020 at Reply

3. Help me to waec anwer

by Aminu on Aug 16, 2020 at Reply