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Neco Gce 2019 Mathematics Answer Now Available


 

NECO GCE MATHEMATICS Questions and Answers  2019 NECO GCE

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 Nov/Dec MATHEMATICS  ANSWER

mathematics Paper11 &3 SATURDAY 30/11/19 1.00p.m -5.30p.m

ATTENTION:- PLEASE AFTER SENDING YOUR CARD DO NOT THROW THEM AWAY UNTIL YOU RECEIVE A CONFIRMATION MESSAGE FROM US.

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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