**National Examination Council, NECO GCE mathematics Questions and Answers for 2019 NECO GCE Examination is now available, subscribe now and get the questions and already solved answers by Gidirunz.**

# Neco Gce 2019 Mathematics Answer Now Available

Maths OBJ:

1-10: BCBEDDCCDC

11-20: DEBCBBDDEC

21-30: BCBCCBBEBD

31-40: EBABEDCDEB

41-50: DACECDCBAB

51-60: DECCEABBBE

===

(2a)

Given that the roots of the equation are

X = -2/3 and X = -3/2

3x = -2 and 2x = -3

3x + 2 = 0 and 2x + 3 =0

(3x+2)(2x+3) = 0

6x² + 9x + 4x + 6 = 0

6x² + 13x + 6 = 0

(2b)

R = [3 4 0 ]

[2 0 3 ]

[1 2 2]

(2ci)

2/3R

=2/3[3 4 0]

[2 0 3]

[1 2 2]

= [⅔(3) ⅔(4) ⅔(0)]

[⅔(2) ⅔(0) ⅔(3)]

[⅔(1) ⅔(4) ⅔(2)]

= [2 8/3 0 ]

[4/3 0 2 ]

[2/3 4/3 4/3]

(2cii)

|R|

= |3 4 0|

|2 0 3|

|1 2 2|

=3|0 3| -4|2 3| +0|2 0|

|2 2| |1 2| |1 2|

=3(2×0-3×2)-4(2×2-3×1)

+0(2×2-0×1)

=3(0 – 6)-4(4 – 3) +0(4 – 0)

=3(-6) -4(1) +0(4)

= -18 – 4 + 0 = -22

(2ciii)

The transpose of R

= [3 2 1]

[4 0 2]

[0 3 2]

***

(3a)

Given : R(3,5) and S(-2, -6)

equation to line through them is :

y-5/x-3 = -6-5/-2-3

y-5/x-3 = -11/-5

y-5/x-3 = 11/5

5(y-5)= 11(x-3)

5y-25 = 11x-33

5y-11x = 25-33

5y-11x = -8 or 11x-5y = 8

(3b)

RS= √(X1X2)^2 (y1y2)^2

√(-2-3)^2 + (-6-5)^2

√(-5)^2 + (-11)^2

√25 + 121

√146

= 12.08

*****

(4a)

Given: curve; y x² – 3x

gradient ; dy/dx = 2x – 3

At (2-2), gradient = 2(2) – 3

4-3 = 1

(4b)

Given; y= 1+x²/1-x²

dy/dx = (1-x²)(2x) – (1+x²)(-2x)/(1-x²)²

= (1-x²)(2x) – (1+x²)(2x)/(1-x²)²

= 2x(1-x² + 1+x²)/(1-x²)²

= 2x(2)/(1-x²)²

= 4x/(1-x²)²

*****

(5)

No of blue balls = 6

No of red balls = 10

(i)

Prob (2 balls of some colour)

= BB or RR

Total no of balls = 6+10 = 16

BB or RR

(6/16 × 5/15) + (10/16 × 9/15)

=30/240 + 90/240

=120/240 = 1/2

(ii)

Prob (2 balls of different colours)

= BR or RB

= (6/16 × 10/15) or (10/16 × 6/15)

= 60/240 + 60/240 = 120/240

=1/2

*****

(7a)

Given that Y=2×2+7x-6

To find the gradient of the curve at the point x=3

dy/dx=4x+7 (at x=3)

dy/dx=4(3)+7

dy/dx=19

(7bi)

Ade bought 7kg of maize + 4kg of meat=N4240

Kemi bought 3kg of maize + 5kg meat =N4610

Let x = 1kg of maize and y=1kg of meat

Therefore 7x+4y=4240—–(eq1)

3x+5y=4610———(eq2)

Substituting simultaneously

Multiply eq1 by 3 and eq2 by 7

21x+12y=12720—eq3

21x+35y=32270—eq4

Substract eq3 from eq4

23y=19550

23y/23=19550/23=850

y=850

Substitute y=850 in1

7x+4y=4240

7x+4(850)=4240

7x=4240-3400

x=840/7

=120

Hence the total cost price per Kg of maize is N120.00 while the total cost per kg of meat is N850.00

(7bii)

Total cost of 10kg of maize and 5kg of meat

=10x+5y

=10(120)+5(850)

1200+4250

=N5450

++++++

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