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» » WAEC 2018 VERIFIED MATHEMATICS ANSWERS

Wednesday, 18 April 2018

WAEC 2018 VERIFIED MATHEMATICS ANSWERS





GENERAL MATHS LOADING….

OBJ - MATHS
1-10ABBCDDBBAA
11 -20ADAACCDCCC
21 -30DADBABCDAD
31 -40BADADBCACB
41 -50ADDDBDADCA

         WORKING SOLUTIONS PICTURE           

















NO1) 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
4a) Rate = 2/100 * N0.02 per month Rate per annum = 0.02 * 12 = 0.24 per annum
*4A ii*
RSQ=(x+90)°
RSQ=37.5°+90°
RSQ=127.5°
(4b) Draw the Diagram
(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
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


6a) Draw it your self

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


(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

(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
6 bi ) faulty brakes cars = 8 +6 + x+ 6- x
= 8 + 6 + 6
= 20
bii ) Only one fault = 28 + x + 2 x
= 28 + 3 x
= 28 + 3 ( 17)
= 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

NO9) Using cosine rule,

|TQ|ˆ² = 4ˆ² + 6 ˆ² – 2(4)(6) cos30°

|TQ|ˆ² = 16 + 36 – 48(0.8660)

|TQ|ˆ² = 52 – 41.568

|TQ|ˆ² = 10.432

TQ = √10.432

TQ = 3.23CM

From similar triangles;

|PT|/|TQ| = |PS|/|SR|

4/3.23 = 10/|SR|

4|SR| = 32.3

|SR| = 32.3/4

|SR| = 8CM (nearest whole number)

9b) Now: |PT|/|PQ| = |PS|/|PR|

4/6 = 10/|PR|

4|PR| = 60

|PR| = 60/4

= 15CM

Therefore, |QR| = 15 – 6 = 9cm

Area of TQRS = Area of ΔPRS – Area of ΔPQT

=1/2(PS)(PR)sin30° – 1/2(PT)(PQ)sin30°

= 1/2sin30°(|PS| (PR) – |PT| |PQ|)

=1/2 * 1/2((10)(15) – (4)(6))

=1/4(150 – 24)

=1/4 * 126

=31.5 apx 32cm²

=======
5a) m+n+s+p+q/5 = 12

m+n+s+p+q =60

Now; (m+4) + (n-3) + (s+6) + (p-2) + (q+8)

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

= 60+13/5

= 73/5 = 14.6


( 8 a)
Cost price for Lami= # 300 .00
Profit made by lami = x %
Ie selling price for lami=( 100 + x/ 100 )×# 300
= # 3 (100 + x)
= # ( 300+ 3 x)
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 +6 x
3 ( 10000 +200 x+ x ²)= 30000+ 75+ 600 x
30000 + 600 x+ 3x ² =30000 + 75+ 600x
3 x² = 75
X² = 75/ 3
X² = 25
X = square root 25
X = 5

( 8 b)
3 x- 2< 10+ x<2 + 5 x
3 x- 2< 10+ x & 10 +x <2 +5 x
3 x- x< 10+2 & 10 -2 <5 x- x
2 x< 12 8 <4 x
X< 12/ 2 4x >8
X< 6 x >8 / 4
X> 2
Also; 3x -2 <2 + 5x
- 4< 2x 2 x > -4
X > -2
Therefore ; Range is - 2


5b) 75% of 500 = 375 people

Number of people above 65 yrs = 500 – 375 =125

25% of 500 = 125

Number of people below 15 yrs = 125

Number between 15yrs and 65yrs

= 500 – (125 + 125)

= 500 – 250

= 250 people

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 NIGERIA QUESTIONS AND ANSWERS
Note that 15 question to answer 10 don’t answer more Dan dat cos u re seeing how answers re coming for betament of results. 10Q




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