NECO GCE 2019 MATHS ANSWER

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SUBJECT: NECO GCE 2019 - MATHS ANSWERS

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AYOSTUFF - 30th, November 2019
==> Exam Time: – Saturday 30th Nov.2019
General Mathematics Paper III (Objective) – 1.00 pm – 2.45 pm
General Mathematics Paper II (Essay) – 3.00 pm – 5.30 pm
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MATHS OBJ
1-10: BCBEDDCCDC
11-20: DEBCBBDDEC
21-30: BCBCCBBEBD
31-40: EBABEDCDEB
41-50: DDAECACBAB
51-60: DDBCEACBBE



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THEORY
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(1)
√((0.0024)×35000)/0.0105

No | Log
0.024 |_3.3802
35000 |4.5441
- |=1.9243
0.0105|_2.0212 -
|3.9031 ÷2
|1.9516

√((0.0024)×35000)/0.0105


Antilog of .9516 =89.45

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(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]

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(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

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(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

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(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

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(6a)
27^(2x+1) × 3^-x = 81^(x-2)/9^(x+2)
= 3^3(2x+1) × 3^-x = 3^4(x-2)/3^2(x+2)
=3^6x+3-x = 3^4x-8-2x-4
5x+3 = 2x - 12
5x - 2x = -12-3
3x = -15
3x/3 = -15/3
X = -5

(6b)
X/x+101 = 11/1000
Since all the members are in binary, convert all to denary (base 10)
Xbase2 = Xbase10
101base2 = (1×2^2)+(1×2^0) = 4+1 = 5base10
11base2 = (1×2¹)+(1×2raise to power 0) = 2+1 = 3base10

1000base2 = 1×2^3 = 8base10
X/X+101 = 11/1000 --> X/X+5 = 3/8
3(x+5) = 8(x)
3x+15 = 8x
15 = 8x - 3x
15 = 5x
15/5 = 5/5
X = 3
Convert X=3 to base 10 to base 2
2|3
2|1R1
|0R1
.:. x = 11

(6c)
Given that log5 base 10 = 0.699

and log3 base 10 = 0.477
10
Log75 base 10 = log(3×5×5) base 10

=log(3×5²) base 10

=log3 base 10 + 2log5 base 10

=0.477 + 2(0.699)
= 0.477 + 1.398
= 1.875
log75 base 10 = 1.875

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(7a)
Given that Y=2x2+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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