WAEC 2018 MATHEMATICS ESSAY AND THEORY ANSWERS – MAY/JUNE – wakagist.com

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

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

NO2)
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)

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

Number 3 ( Image )

(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

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

Number 4 ( image )

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)

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<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
=====================

Number 5

(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

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

Number 6 ( Image )

(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)
(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

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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

Number 10( image )

(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= 12×0.3843

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= 12×0.60000

y= 7.2m

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