ID 80

## CRK-OBj

This is actually the WAEC 2019 - MATHS ANSWERSAnswers as questions sent to our subscribers on WHATSAPP, if you are looking for 2022 WAEC/NECO Answers midnight before the exam, try contacting us.

1DABDBABCDA

11DDCADBCCDD

21ACADBBCBDC

31BDDCBDBBDB

41BCCBDDCACB

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

(1)

110â‚“ = 40â‚…

change all to base ten.

110â‚“ = 40â‚…

1x²+1x ¹+0xâ° = 4Ã—5¹+0Ã—5â°

x² + x + 0 = 20

x² + x â€“ 20 = 0

Solve quadratically

x² + 5x â€“ 4x â€“ 20 = 0

x(x+5) â€“ 4(x+5) = 0

(x-4)(x+5) = 0

Either x-4 = 0 or x + 5 = 0

x = -5 or 4

therefore, x= 4

1b)

15/âˆš75 + âˆš108 + âˆš432

= 15/âˆš253 + âˆš363 + âˆš144Ã—3

= 15/5âˆš3 + 6âˆš3 + 12âˆš3

= 3/âˆš3 + 18âˆš3

= 3âˆš3/3 = 18âˆš3

= âˆš3 + 18âˆš3

=19âˆš3

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

(2)

(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

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

(3a)

Total ratio =3+5+8=16

Ali's share =3/16 * 420,000/1

=210,000

Son's of Ali and Yusuf's share = N78,750+ N210,000

= N288,750

(3b)

2(1/8)^{x}=32^{x}-1

2^{1}(2^{-3})^{x}=2^{5}(x-1)

2^{1}* 2^{-3}x=2^{5}(x-1)

2^{-3}x+1=2^{5}(x-1)

-3x+1=5(x-1)

-3x+1=5x-5

-5x-3x= -5-1

-8x/-8= -6/-8

x=3/4

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

(4)

Since

|PR|² = |PQ|² + |QR|²

|PR|² = 3² + 4²

|PR|² = 9 + 16

|PR|² = 25 PR = âˆš25

|PR| = 5cm

Considering PRS

|PS|² = |PR|²+|SR|²

13² = 5² + |SR|²

169 = 25 + |SR|²

|SR|² = 169 - 25

|SR|² = 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/d²

F = KM1M2/d²

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

20 = k(25)(10)/5²

250k = 500

k = 500/250 = 2

Expression is

F = 2M1M2/d²

(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— Tan52° = 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)

ar² = 1/4 ......(1)

ar^5= 1/32 .....(2)

Divide eqn (2) by eqn(1)

ar^5/ar² = 1/32Ã·1/4

r³ = 1/32 Ã— 4/1

r³= 1/8

r³ = 2-³

r = 2-¹

r = 1/2

Common ratio = 1/2

Put this into eqn (1)

a(1/2)² = 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

(11b)

Given : X = 2 and X = -3

(X - 2)(X + 3) = 0

X² + 3x - 2x - 6 , 0

X² + x - 6 = 0

Comparing with ax²+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

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