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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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