2019 Neco Gce Mathematics Obj And Theory Answers

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MATHS OBJ
01-10: BCBEDDCCDC
11-20: DEBCBBDDEC
21-30: BCBCCBAEED
31-40: EBABEDCDBB
41-50: DDCECDCBAB
51-60: DDBCEACBBE

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MATHS Theory

(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

(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

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(3b)
RS= √(X1X2)^2 (y1y2)^2
√(-2-3)^2 + (-6-5)^2
√(-5)^2 + (-11)^2
√25 + 121
√146
= 12.08

(4b)
1+x² u/1-x2 v
= Sy/su =V Sy/su -U su/sx
U = 1+x², V = 1-x²
Sy/sx =2x
sr/sx = -2x
Sy/sx 1-x2)2x-(4x³)-2x/(1-x²)²
= 2x-2X³-(-2x-2x³)/(1-x²)²
=2x-2x³+2x+2x³/ (1-x³)(1-2x)
= 4x/(1-x²)³

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

(5ai) the same colour

BB + RR
(6/16×5/15) + (10/16 × 9/15)

30/240 + 90/240

=30+90/240
=120/240
=1/2

(5ii) different color
BR+RB

6/16×10/15) + (10/16×/15)

= 60/240 + 60/240
= 60+60/240
= 120/240
= 1/2

(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

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(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

(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

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(7bii)
Total cost of 10kg of maize and 5kg of meat
=10x+5y
=10(120)+5(850)
1200+4250
=N5450

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