# WAEC GCE 2019 (1ST SERIES) – MATHS ANSWERS

Exam Time: Friday, 1st February, 2019
Saturday, 2nd February, 2019
General Mathematics 2 (Essay) – 9.30 am – 12.00 noon
General Mathematics 1 (Objective) – 2.00 pm – 3.30 pm
++++++++++++++++++++++++++++++++++++

=============================
MATHS OBJ:
11-20: BCABACDBAC
21-30: ABCDCACBCA
41-50: DCCAACBDCC

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

(1a)
1+4x/2 – 5+2x/7 < x-2/1
Multiply through by 14
14(1+4x)/2 – 14(5+2x)/7 <_ 14(x-2)

7(1+4x) – 2(5+2x) <_ 14(x-2)
7+28x – 10 – 4x<_14x – 28
28x – 4x – 14x <_ -28 – 7 +10
10x/10 <_ -25/10
X <_ -2.5

(1b)
X = y = 3 : 5
X = 3, y = 5
2x² – y²/y² – x²
=2(3)² – 5²/5² – 3²
= 18 – 25/25 – 9
= -7/16=

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

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

(2a)
Common difference,=> (x+1) -(x-1)
d=x+1-x-1
d=2

(2b)
Common difference,d=2
T3 = a+2d
7=a+2(2)
7=a+4
7-4=a
a=3

(2c)
Common difference,d=>(x+1)-(x-1)=7-(x+1)
X+1-x-1=7-x-1
2=6-x
X=6-2
X=4

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

(3a)
Log8base2 + Log16base2 – 4log2base2/log16 base4
= log2³base2 + log2³base2 – 4log2base2/log4²base4
= 3log2base2 + 4log2bae2 – 4log2base2
= 3(1) + 4(1) – 4(1)/2(1)
= 3+4-4/2
= 3/2
= 1½

(3b)
1342five – 241five = Xten
[(1×5³)+(3×5²)+(4×5¹)+(2×5^0)] – [(2×5²)+(4×5¹)+(1×5^0)] = Xten
[125+75+20+2] – [50+20+1] = Xten
[222 – 71]ten = Xten
151ten = Xten
X = 151

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

(4a)
If log10^a=1.3010
a=10^1.3010———–(1)

If log10^b=1.4771
b=10^1.4771

ab=10^1.3010 * 10^1.4771
ab=10^2.7781
ab = 599.9
(from antilog tables)

(4bi)
ABC + CBE=180( angles of a straight line)
ABC + 62=180
ABC=180-62=118°
Reflex AOC=2*118(angle center = 2*angle at circle)
=236degree
AOC + Reflex AOC=360°(angle at a point)
Obtuse AOC+236=360°
Obtuse AOC=360-236
Obtuse AOC =124°
:. Interior Angle AOC =124°

(4bii)
BAC+ACB=CBE(Ext. angle = sum of two opp. Int. angle )
BAC+39=62
BAC=62-39
BAC=23d°

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

(5)
Let the constant be a and b
C = a + bN
740 = a + 20b —–(1)
960 = a + 30b ——(2)
equation 2 minus equation 1
960 – 740 = 30b – 20b
220 = 10b
b = 220/10 = 22

Put b = 22 into eqn 1
740 = a+20(22)
740 = a+440
a = 740 – 440
a = 300

Relationship is C = 300 + 22N
When N = 15
C = 300 + 22(15)
C = 300 + 330
C = 630
When only 15 tourist were present, cost was \$630

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

(6a)
Cost price of a car =\$5600
Selling price of car=90% of cost price
Selling price =\$5600*90/100
=560*9=5040
Selling price =\$5040
He spent 1,310 of \$5040

Remainder =5040-1310
Remainder which he invested =\$3730

I =PRT/100
I=?, P=\$3730, R=6%, T=3yrs

I=3730*3*6/100
I=3730*18/100

I=\$671.4

(6b)
2^x(4^-1)=2
3^x(9^2y)=3
2^x(4^-7)=2—-(1)
3^-x(9^2y)=3—(2)
2^x[2^2(-7)]=2^1
2^x(2^-14)=2^1
2^x+(-14)=2^1
2^x-14=2^1
Equating the process
x-14=1
x=1+14=15
Put x=15 into the equation (II) to get the value of y
3^-x[3^2(2y)]=3^1
3^-x(3^4y)=3^1
3^-x+4y=3^1
Equating the powers :
-x+4+y=1
Substitute x=15
-15+4y=1
4y=1+15
4y=16
y=16/4=4