WAEC 2018 Mathematics OBJ And Theory Questions And Answers

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👑 2018 MAY/JUNE WAEC MATHEMATICS ANSWERS👑

 

100% VERIFIED MATHEMATICS OBJ ANSWERS

1-10: ACBCDDCBAA
11-20: CDCABCCCAC
21-30: DADBABCDAD
31-40: BADADBCACB
41-50: ADDDBDADCA

100% VERIFIED MATHEMATICS THEORY ANSWERS
Snapped Answers

1. CLICK HERE

2. CLICK HERE

3. CLICK HERE

4. CLICK HERE

6&7. CLICK HERE

6. CLICK HERE

7 & 9. CLICK HERE

8. CLICK HERE

9. CLICK HERE

13. CLICK HERE

13c (Graph).CLICK HERE

Typed

Please Note the below symbols

^ means raise to power
* means multiplication
/ means division

(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
= #299,000
================================

(2a)
Given y-2px^2-p^2x-14
At (3,10)
10=2p(3)^2-p^2(3)-14
3p^2-18p 24=0
p^2-6p 8==0
Using factor method
P^2-2p-4p 8=0
P(p-2)-4(p-2)=0
(p-4)(p-2)=0
p=4 or p=2

(2b)
The lines must be solved simultaneously
3y-2x=21–(i)
4y 5x=5–(ii)
Using elimination method
=>12y-8x=84–(iii)
=>12y 15x=15–(iv)
eq(iv) minus (eqiii)
23x=-69
x=-69/23
x=-3
Put x into eq(i)
3y-2(-3)=21
3y 6=21
3y=21-6
3y=15
y=15/3
Coordiantes of Q is (-3,5)
==============================

(3a)
Using pythagoras theorem
L^2=5.1^2 4.65^2
L^2=26.01 21.6225
L^2=47.6325
L=sqroot(47.6325)

L=6.9cm(1 d.p)
Perimeter of rhombus=4
=4*6.9
=27.6cm

(3b)
Sin x=3/5
DRAW THE TRIANGLE
Using pythagoras tripple the third side=4
therefore cosx=4/5
tanx=3/4
therefore 5cosx-4tanx
=5(4/5)-4(3/4)
=4-3
=1
====================================

(4ai)
Draw the diagram
ai X 90 = 3x 15
90 = 3x – X 15
90 = 2x 15
2x 15 = 90
2x = 90 – 15
X = 75/2
X = 37.5•

(4aii)
(4b)
2N4seven = 15Nnine
Converting both to base 10
2×7 N×7¹ 4×7 = 1×9² 5×9¹ N×9
98 7N 4 = 81 45 N
7N 102 = 126 N
7N – Ń = 126 – 102
6N = 24
Ń = 24/6
Ń = 4
=================================

(5a)
m+n+s+p+q/5=12
m+n+s+p+q=60……(1)
Now;
(m+4)+(n-3)+(5+6)+p-2)+(q+8)/5
=(m+n+s+p+q)+(4-3+3+6-2+8)/5
=60+13/5
=73/5
=14.6

(5b)
75% of 500 = 375 people
Number of people above 65 years = 500-375
=125

25% of 500 = 125
Number of people below 15 years = 125
Number between 15 years and 65 years
=500-(125+125)
=500-250
=250 people
===============================

(6)
Total number of cars on road worthiness = 240
60% passed ie 60/100×240/1 = 144cars.
Number that failed = 240-144 = 96cars

(6a)
Draw the Venn diagram
C = clutch
B = brakes
S = steering

(6b)
From the diagram above
E = 28 12 8 6 x 6 2x
96=60 3x
96-60=3x
36/3 = 3x/3
Therefore X = 12

(i) The no of cars that had faulty brakes
=12 8 6 x (Since X = 12)
=12 8 6 12 = 38

(ii) Only one fault = (no of clutch only) (no of brakes only) (no of steering only)
=28 x 12x = 28 12 24
=64cars
==================================

(7a)
(y-y1)/(x-x1)=(y2-y1)/(x2-x1)
(y-5)/(x-2)=(-7-5)/(-4-2)
(y-5)/(x-2)=-12/-6
(y-5)/(x-2)=2
Cross multiply
y-5=2(x-2)
y-5=2x-4
2x-y-4 5=0
2x-y 1=0

(7bi)
DRAW THE DIAGRAM

(7bii)
(I)
p^2=q r^2-2qrcosP
p^2=8^2 5^2-2*8*5*cos90
p^2=64 25-0
p^2=89
p=sqroot(89)
p=9.4339km
therefore |QR|=9.43km(3 sf)

(II)
q/sinQ=p/sinP
8/sinQ=9.4339/sin90
sinQ=(8*sin90/9.4339
sinq=(8*1)/9.4339 =0.8480
Q=sin^1(0.8480)=57.99 degrees
but Q=30 A
A=Q-30
=57.99-30
A=27.99 degrees
The bearing of R from Q
=180-A
180-27.99
=155.01
=>152 degrees
==================================

(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
================================

(9a)
Draw the diagram
Angles PTR and PSR are similar
|PT|/|PS| = |TQ|/|SR|
In angle PTR
|TQ|²=|PT|² |PQ|²-2|PT||PQ|cos30degrees
=4² 6²-2×4×6×cos30
=16 36-48×0.8660
=52-41.568
=10.432
|TQ|=√10.432 =3.22cm
4/10 = 3.22/|SR|
4|SR| = 10×3.22
|SR| = 32.2/4
|SR| = 8.05cn
Approximately 8cm(to the nearest whole number)

(9b)
Atqrs = AΔPSR – AΔPTR
AΔPTR = 1/2×4×6×sin30
=2×6×0.5
=6cm²
AanglePTQ/AanglePSR = |PT|²/|PS|²
6/AanglePSR = 4²/10²
6/AanglePSR = 16/100
16×AanglePSR = 6×100
AanglePSR = 600/16 = 37.5cm2
ATQRS = 37.5 – 6
=31.5cm2
=32cm2
====================================

(13a)
Frequency=16+x+y
16+x+y=30
x+y=30-16
x+y=14–(eqi)
(900+30x+50y)/30=52
900+30x+50y=52*30
30x+50y=1560-900
30x+50y=660
divide through by 10
3x+5y=66–(eqii)
From (i)
x+y=14
x=14-y–(eqiii)
sub for x in eqii
3(14-y) +5y=66
42-3y+5y=66
2y=66-42
y=24/2
y=12
feom eqiii
x=14-12
x=2

(13b)
TABULATE
Class interval:1-10,11-20,21-30,41,50,51-60,61-70,71-80,81-90
Freq:1,1,2,5,12,1,4,3,1
Class boundary:0.5-10.5,10.5-20.5,20.5-30.5,30.5-40.5,40.5-50.5,50.5-60.5,60.5-70.5,70.5-80.5,80.5-90.5

(13C)
DRAW THE GRAPH (CLICK HERE)

====COMPLETED====

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