2018/2019 Waec Mathematics Question & Answers/Runs | WAEC 2018 Maths Expo

How to Get Real 2018 Waec Mathematics Expo Questions and Answers 

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Confimed Maths Obj
1-10: ACBCDDCBAA
11-20: CDCABCCCAC
21-30: DADBABCDAD
31-40: BADADBCACB

41-50: ADDDBDADCA

1) 
On February 28th 2012, value = (100-30/100) * #900,00.00

= 70/100 * #900,00

= #630,000.00

On february 28th 2013, value = (100-22/1000 * #630,00

= 78/100 8 #630,000

= #491,400

On february 28th 2014, value = 78/100 8 #491,400

=383,292

On february 28th 2015, value = 78/100 * #383,292

= #298,967.76

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

2b) The lines must be solved simultenously

3y – 2x = 21 ——- (1)

4y + 5x = 5 ——-(2)

using elimination method,

(4)  3y – 2x = 21

(30 4y + 5x = 5

12y – 8y = 84 ——— (3)

12y + 15x = 15 ——-(4)

equ (4) minus equ(3)

23x = -69

x = -69/23

x = -3

Put this into equation (1)

3y -2(-3) = 21

3y = 6 = 21

3y = 21 -6

3y = 15

y =15/3

y = 5

coordinates of Q is (-3, 5)

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

(3a)
The diagonal = 10.2m and 9.3cm
Using Pythagoras theory 
Ac² = 10.2² + 9-3²
Ac² = 104.04 + 86.49
Ac² = 190.53
Ac² = √190.53
Ac² = 13.80

(3b)
DRAW THE DIAGRAM
Using Pythagoras theory 
5² = 3² + x²
x² = 5² - 3²
X²= 25 - 9
X² = √16
X= 4cm
CosX = adjacent/hyp
= 4/5
Tan X = opp/adj. = 3/4
5cos x - 4tan x
5(4/5)- 4(3/4)
20/5 - 12/4
4-3= 1

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

4ai)
sum of angle in a D =180degree
xdegree + 90degree + 180degree - (3x+15)=180degree
xdegree + 90degree + 180degree - 3x+15=180degree
-2x=180degree - 255
+2x/2=+75/2
x=37.5


4aii)
=180-(112.5 + 15)
=180 - 127.5

4b)
2N4seven =15Nnine
2*7^2+N*7^1+4*7degree =1*9^2 + 5*9^1+N*9degree
9*49+N*7+4*1=1*81+5*9+N*1
98+7N+4=81+45+N
7N+102=126+N
7N-N=126-102
6N/6 =24/6
N=4

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

5a)
m + n + s + p + q / 5 = 12
m + n + s + p + q = 60 ----> equ 1

Now;

(m+4) + (m-3) + (5+6) + (p-2) + (q+8) / 5
= (m + n + s + p + q) + (4 - 3 + 6 - 2 + 8) / 5
= 60 + 13 / 5
= 73 / 5
= 14.6

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

(6a)
Draw the Venn diagram
Let the number of cars with faults in brakes only be x

(6b)
Number that passed = 60% × 240 = 144
Number that failed =
240 - 144 = 96
Therefore; 28+2x+x+14+6+6-x+8 = 96
2x + 62 = 96
2x = 96 - 62
2x = 34
X = 34/2
X = 17
(i) faulty brakes cars = 8+6+x+6-x
= 8+6+6
=20
(ii) only one fault = 28+x+2x
=28+3x
=28+3(19)
=28+51
= 79

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

(7a) Given the points (2,5) and (-4,-7)
Gradient (m)= -7-5/-4-2= -12/-6 = 2
Let (x, y) be a point on the line and (2,5) on the line
2/1 = y-5/x-2
y-5= 2(x-2)=2x-4
y=2x-4+5
y= 2x+1
(7bi) (QR) ^2= 8^2 + 5^2
= 64+25
= 89
QR= sqroot of 89
= 9.43km
(7bii) 1. Distance between Q and R
9.43/sin90 = 5/sinR
5sin90= 9.43SinR
Sin R= 5Sin90/9.43
Sin R= 5/9.43
= 0.5302
R= Sin^-1 0.5302
=32
2. Bearing of R from Q
= 32+90
=122degrees

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

(8a)
Cost price for Lami= #300.00
Profit made by lami = x%
Ie selling price for lami=(100+x/100)×#300
=#3(100+x)
=#(300+3x)

Bola's cost price = #3(100+x)
Profit made by bola =x%
Selling price for bola =(100+x/100)×#3(100+x)
=#3/100(100+x)²

James cost price =#3/100(100+x)²=300+(6x+3/4)
expanding;
3/100(10000+200+x²) = 300+3/4+6x
3(10000+200x+x²)=30000+75+600x
30000+600x+3x²=30000+75+600x
3x²=75
X² = 75/3
X² = 25
X = square root 25
X = 5

(8b)
3x-2<10+x<2+5x
3x-2<10+x & 10+x<2+5x
3x-x<10+2 & 10-2<5x-x
2x<12 8<4x
X<12/2 4x>8
X<6 x>8/4
X>2

Also; 3x-2<2+5x
-4<2x
2x > -4
X > -2
Therefore; Range is -2
====================

(10a) Using Pythagoras theorem from SPQ
|SQ|^2 = 12^2 + 5^2
= 144+25
=169
SQ= sqroot of 169
= 13cm
Sin tita= 5/13 = 0.3846
Tita= Sin^-1(0.3846)
= 22.6degrees
From PRQ
Sin tita= |PR|/12
Sin 22.6 = PR/12
Sin 22.6= PR/12
PR= 12xsin 22.6
PR= 12x0.3843
PR= 4.61cm
(10bii)Let the height at which m touches the wall= y
Cos x^degrees= 8/10= 0.8
x^degrees= Cos^-1(0.8)
= 36.87degrees
Sin x^degrees = y/12
Sin 36.87= y/12
y= 12xsin36.87
y= 12x0.60000
y= 7.2m