## CRK-OBj

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1DABDBABCDA

11DDCADBCCDD

21ACADBBCBDC

31BDDCBDBBDB

41BCCBDDCACB

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(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

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(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

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(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

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(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

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(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

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(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

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(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

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(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%

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(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

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(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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You are required to get at least a credit pass (50%) in your core subjects. Under the new WAEC grading system, A1 is Excellent, B2 Very Good, B3 Good, C4, C5 and C6 are interpreted as Credit, D7 and D8 are interpreted as Pass, while F9 is Fail.

A1: 1 x 4 = 4 B2: 3 x 3.6 = 10.8 C4: 1 x 2.8 = 2.8 Total points: 4 + 10.8 + 2.8 = 17.6 Now, we divide the UTME by 8 i.e 280 ? 8 = 35 So, the aggregate score will be 17.6 + 35 + Post-UTME score.

waec expo runs is real. WAEC Answers, NECO ANSWERS and GCE answers comes a night before exam or Hours before exam

waec expo runs is real. WAEC Answers, NECO ANSWERS and GCE answers comes a night before exam or Hours before exam

A1: 1 x 4 = 4 B2: 3 x 3.6 = 10.8 C4: 1 x 2.8 = 2.8 Total points: 4 + 10.8 + 2.8 = 17.6 Now, we divide the UTME by 8 i.e 280 ? 8 = 35 So, the aggregate score will be 17.6 + 35 + Post-UTME score.

A1: 1 x 4 = 4 B2: 3 x 3.6 = 10.8 C4: 1 x 2.8 = 2.8 Total points: 4 + 10.8 + 2.8 = 17.6 Now, we divide the UTME by 8 i.e 280 ? 8 = 35 So, the aggregate score will be 17.6 + 35 + Post-UTME score.

Under the new WAEC grading system, A1 is Excellent, B2 Very Good, B3 Good, C4, C5 and C6 are interpreted as Credit, D7 and D8 are interpreted as Pass, while F9 is Fail. A1 and B2 in the WASSCE means Excellent, B3 is B (Very Good), C4 is C (Good), C5 and C6 are D (Credit), D7 and E8 are E (Pass) and F9 is F (Fail).

waec expo runs is real. WAEC Answers, NECO ANSWERS and GCE answers comes a night before exam or Hours before exam