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WAEC GCE 2019 mathematics Questions And Answers – Jan/Feb EXPO

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How to Get 2019 WAEC GCE Jan/Feb Mathematics Question Paper.

Follow the steps below to get your gce jan/feb mathematics questions and answers 1 hours before your exam time.

Section A:-

(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

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

3) typing

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

(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=118degree
Reflex AOC=2*118(angle center = 2*angle at circle)
=236degree
AOC + Reflex AOC=360degree(angle at a point)
Obtuse AOC+236=360degree
Obtuse AOC=360-236
Obtuse AOC =124degree
:. Interior Angle AOC =124degree

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

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

5) Let the constant be a and b
C=a+bN
740=a+20b———(1)
960=a+30b———(2)

Equation (2) minus equ(1)
960-740=30b-20b
220=10b
b=220/10=22

Put b=22 into equ(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 where present , cost was $630

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

Section B:-

(13a)
The curved surface area of a cone is given by A = πrl
Where l = slant height
Let A1 = A2 therefore
πr1L1 = πr2L2
r1 = 5cm, L1 = 12cm, L2 = ?
r2 = 6cm
Therefore π×5×12 = π×6×L2
L2 = 5×12/6 = 5×2 = 10cm

Draw the right angle triangle
h² = 10² – 6²
h² = 100 – 36
h² = 64
h = √64 = 8cm

(13bi)
A = [2 5] B = [3 -2]
[-1 3] [4 1]

BA
=[3 -2][2 5] = [6+2 15+6]
[4 1][-1 3] [8+-1 20+-3]
=[8 21]
[7 17]

(13bii)
|8 21|
|7 17|
= 8×17 – 7×21
= 136 – 147
= 11

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