**Verified Waec 2018 Mathematics Answers**

**2018 WAEC mathematics OBJ AND THEORY ANSWERS**

Free waec 2018 mathematics answers

KEPPR REFRESHING………

**Maths OBJ:**

**1ABBCDDBBAA**

**11ADAACCDCCC**

**21DADBABCDAD**

**31BADADBCACB**

**41ADDDBDADCA**

==> 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

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

(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

(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

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

(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

(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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