WAEC 2018 Mathematics OBJ and Theory Answer May/June Expo

WAEC 2018 Mathematics OBJ and Theory Answer May/June Expo Lawrence Victor 5 of 5

Verified WAEC 2018 Mathematics Answer WAEC 2018 Mathematics Essay/Theory and Objective (OBJ) QUESTIONS and Answer Runs Verified WAEC 2018 Ma...

Verified WAEC 2018 Mathematics Answer

WAEC 2018 Mathematics Essay/Theory and Objective (OBJ) QUESTIONS and Answer Runs

Verified WAEC 2018 Mathematics ANSWER

Maths Obj

1-10: ACBCDDCBAA

11-20: CDCABCCCAC

21-30: DADBABCDAD

31-40: BADADBCACB

41-50: ADDDBDADCA

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Maths Theory

2018 Waec Mathematics portal 


1-10 Answers Posted 


(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


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(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)

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(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

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(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

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(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
(b)
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

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(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

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(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 
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(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

(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

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