WAEC 2019- MATHEMATICS

Thursday 16th May, 2019

Mathematics 2 (Essay) â€“ 09:30 a.m. â€“ 12:00pm

Mathematics 1 (Objective) â€“ 3:00 p.m â€“ 4.30 p.m.

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MATHS OBJ

1-10: CABDBADCBC

11-20: BCBCBBAACD

21-30: BCCBCCABBA

31-40: AABDDCDDCC

41-50: BBCDCCACCD

(2a)

The equation of the line through the points

A(-2,7) and B(2,-3)

Using the equation Y=mx+b

Where m=slope of the gradient, b=the intercept at the vertical axis

Hence slope=Change in Y/Change in X

=>Y2-Y1/X2-X1=Y-y1/X-x1

=(-3-7)/(2–2)=Y-7/X+2

=-10/4=(Y-7)/(X+2)

=4(y-2)=-10(x+2)

4y-28=-10x-20

4y=-10x+8

5x+2y=4

(2b)

(5b-a)/(8b+3a)=1/5

5(5b-a)=1(8b+3a)

25b-5a=8b+3a

25b-8b=3a+5a

17b=8a

Therefore a/b=17/8

(4)

Since <PQR = <PRS = 90Â°

Using Pythagoras theorem

|PR|Â² = |PQ|Â² + |QR|Â²

|PR|Â² = 3Â² + 4Â²

|PR|Â² = 9 + 16

|PR|Â² = 25 PR = âˆš25

|PR| = 5cm

Considering PRS

|PS|Â² = |PR|Â²+|SR|Â²

13Â² = 5Â² + |SR|Â²

169 = 25 + |SR|Â²

|SR|Â² = 169 – 25

|SR|Â² = 144

|SR| = âˆš144 = 12cm

Hence the area of the quadrilateral = Area of triangle PQR + area of PRS

= 1/2bh + 1/2bh

= 1/2Ã—4Ã—3 + 1/2Ã—12Ã—5

= 6+30 = 36cm

(5a)

No of red balls = 3

No of green balls = 5

No of blue balls = x

Prob.(red ball) = no of total outcome/no of possible outcome

Pr(red) = 3/3+5+x = 1/6

3/8+x = 1/6

6(3) = 1(8+x)

18 = 8 + x

X = 18 – 8 = 10

Therefore the no of blue ball = 10

(5b)

Probability of picking a green ball

P(g) = no of green balls/no of possible outcome

P(g) = 5/3+5+10 = 5/18

=5/18

(6ai)

F Î± M1M2/dÂ²

F = KM1M2/dÂ²

Given F = 20N, M1= 25kg, M2 = 10kg and d = 5m

20 = k(25)(10)/5Â²

250k = 500

k = 500/250 = 2

Expression is

F = 2M1M2/dÂ²

(6aii)

Making d subject

d = âˆš2M1M2/F

d = âˆš2 Ã—7.5Ã—4/30

d = âˆš60/30 = âˆš2

d = âˆš2m or 1.41m

(6b)

Draw the diagram

X+X+60+X+80+X+40+X+20 = 540(sum of angles in a Pentagon)

5x + 200 = 540

5x = 540 – 200

5x = 340

X = 340/5

X = 68

(8a)

1/3x – 1/4(x+2)>_ 3x -1â…“

1/3x – 1/4(x+2)>*3x – 4/3
Multiply through by the L. C. M(12), we have
4x – 3(x + 2)>_36x – 16
4x – 3x – 6 >* 36x – 16

-6+16 >*36x + 3x – 4x
10 >* 35x

35x _< 10

X = 10/35

X = 2/7

(8bi)

Draw the triangle

|AB|/66 = sin35

|AB| = 66sin35 = 66Ã—0.5736 = 37.8576

Draw the right angled triangle

|AD|/|AB| = Tan52

|AD| = 37.8576 Ã— Tan52Â° = 37.8576 Ã— 1.2799 = 48.45m

Height of tower = 48.45m

(10)

130kg of tomatoes for #52,000

Half of the tomatoes

130/2 = 65kg sold at 30%

Profit = #52,000/2 = 26,000

# 26,000 = 100%

X = 130%

X = 26000 Ã— 130/100

= #33,800

Then 65kg was then sold at reduction of 12% per kg

Recall that the initial cost price = 52000/130

=400kg

65kg sold at = 33,000/65

=#520/kg

Then for 12% reduction

520 Ã— 88/100 = 457.6/kg

(a)

The new selling price per kg = #457.6/kg

(b) 65kg – 5kg = 60kg

(60kgÃ—457.6kg)+33,800

= #61,256.00

# profit = selling price /cost price Ã— 1000/1

=61256/52000Ã—100/1= 117.8

= 17.8%

(11ai)

arÂ² = 1/4 â€¦â€¦(1)

ar^5= 1/32 â€¦..(2)

Divide eqn (2) by eqn(1)

ar^5/arÂ² = 1/32Ã·1/4

rÂ³ = 1/32 Ã— 4/1

rÂ³= 1/8

rÂ³ = 2-Â³

r = 2-Â¹

r = 1/2

Common ratio = 1/2

Put this into eqn (1)

a(1/2)Â² = 1/4

a(1/4) = 1/4

a = (1/4)/(1/4) = 1

First term, a = 1

(11aii)

Seventh term, T7 = ar^6

=(1)(1/2)^6

=1/64

(1b)

Given : X = 2 and X = -3

(X – 2)(X + 3) = 0

XÂ² + 3x – 2x – 6 , 0

XÂ² + x – 6 = 0

Comparing with axÂ²+bx+c = 0

a = 1

b = 1

C = -6

(12a)

Given : siny = 8/17

Draw the right angle

From Pythagorean triple, third side is 15

Draw the right angle triangle

tan y = 8/15

tan y/1+2tany = 8/15/1+2(8/15) = 8/15/1+16/15

tany /1+2tan y = 8/31

(12b)

Amount shared = #300,000

Otobo’s share = #60,000

Ade’s share = 5/12 Ã— #(300,000-60,000)

= 5/12 Ã— #240,000

=#100,000

Adeobi’s share = #300,000 – (#60,000 + #100,000)

= 300,000 – 160,000

=#140,000

Ratio : Otobo : Ade : Adeola

60,000 : 100,000 : 140,000

60 : 100 : 140

6 : 10 : 14

3 : 5 : 7

(13a)

RT^0=RS^0=90Â° (radius o targets)

T^0S=2TU^0 (angle at centre = 2 time angle at circumstance)

TOs=2*68Â°=136Â°

Now RT^0+RS^0+T^0S+SRT^0=360Â°(sum of angle in a quadrilateral)

90Â° + 90Â° + 136Â° + x = 360Â°

X+316Â°=360Â°

X=360Â°-316

X=44Â°

(13b)

Let tank B hold x litres

; Tank A hold (x+600)literally

3(x-100)=(x+600-100)

3x-300=x+500

3x-x=500+300

2x=800

X=800/2=400

Tank B holds 400 litres

Tank A holds (400+600)=1000litres

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