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WAEC 2018 Mathematics Obj And Essay Answer – May/June Expo

WAEC Mathematics Obj And Essay/Theory Solution Questions and Answer – May/June 2018 Expo Runz.

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Maths OBJ:
1ABBCDDBBAA
11ADAACCDCCC
21DADBABCDAD
31BADADBCACB
41ADDDBDADCA


9a)



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



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



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



7a)
X1-X/Y1-Y = X2-X1/Y2-Y1
2-X/5-Y = -4-2/-7-5
2-X/5-Y= -6/-12
-12(2-X)=-6(5-Y)
-24+12X=-30+6Y
6Y-12X=30+24
6Y-12X=-6
6y-12x+6=0
y-2x+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 



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



CLICK HERE FOR NO.2 nd 8 co. ANSWER
CLICK HERE FOR NO.9 ANSWER
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



2)
Given that y = 2pxˆ² – p² x – 14
AT (3, 10)
10 = 2p(3)² – p² (3) – 14
10 = 18p – 3p² – 14
3p² – 18p + 24 = 0
p² – 6p + 8 = 0
using factor method,
p² – 2p -4p + 8 = 0
p(p-2) – 4(p-2) = 0
(p-4)(p-2) = 0
p-4 = 0 or p-4 = 0
p= 4 or p =2

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)
<RsQ =180 - (3x+15)
<RsQ =180-(3*37.5+15)
=180-(112.5 + 15)
=180 - 127.5
<RsQ= 52.5degree

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



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



CLICK HERE FOR NO.3 AND 4 ANSWERS



CLICK HERE FOR NO.1 ANSWER


CLICK HERE FOR NO.3 ANSWER


CLICK HERE FOR NO.3b ANSWER


CLICK HERE FOR NO.4 ANSWER


CLICK HERE FOR NO.6a ANSWER



CLICK HERE FOR NO.7 TABLE




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