NECO Mathematics Questions and Answers 2020 | Theory and Objectives

NECO 2020 Mathematics Essay and OBJ
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Today’s Math Essay Answers:
No (1a)
Equation:
y – y1/x – x1=y2 – y1/x2 – x1
y – 5/x – 6 = 7 – 5/-2 – 6
y – 5/x – 6 = 2/-8 = 1/-4
y – 5 = 1/4(x – 6)
y – 5 = -1/4x + 6/4
y = -1/4x + 6/4 + 5
y = -1/4x + 3/2 + 5
y = -1/4x + 13/2
OR 4y = -x+26
X + 4y = 26
No (1b)
2
S(2x + 9)dx
-1
= 2X¹+¹/1+1 + 9x]2
-1
= x²+9x]2
-1
=[2²+9(2)] – [(-1)² + 9(-1)]
=[4 + 18] – [1 – 9]
= 22 + 8
= 30
(2a)
No that traveled by bus
= u+18+5+12 = 52
=u+35 = 52
u = 52 – 35
u = 17
No that traveled by train
= 6+5+12+v = 35
V + 23 = 35
V = 35 – 23
V = 12
Total no of tourist
=u+v+w+18+12+6+5 = 100
17+12+w+18+12+6+5 = 100
W + 70 = 100
W = 100 – 70
W = 30
(2b)
No who travelled by at least two means
of transportation
= 18 + 6 + 12 + 5
= 41
(3ai)
EF = 50
Median = 25th + 26th/2
= 6+6/2 = 6
(3aii)
Range = 10 – 3 = 7
Median + Range = 6 + 7 = 13
(3b)
Prob(Olu passes) = 2/5
Prob(Tony passes) = 3/4
Prob(Olu fails) = 3/5
Prob(Tony fails) = 1/4
Prob(both passes) =
2/5 × 3/4 = 3/10
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(4a)
X² + 3x – 28 = 0
X² +7x – 4x – 28 = 0
X(x+7)-4(x+7) = 0
(X – 4)(X + 7) = 0
X – 4 = 0 OR x + 7 = 0
X = 4 OR x = -7
(4b)
8x/9 – 3x/2 = 5/6 – x
Multiply by 18
18(8x/9) -18(3x/2) = 18(5/6) -18x
16x – 27x = 15 – 18x
18x + 16x – 27x = 15
7x = 15
X = 15/7
X = 2 whole 1/7
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(5a)
d/dx(4x³ – 2x + 4)⁵
= 5(4x³-2x+4)⁴ × (12x² – 2)
= 10(6x² – 1)(4x³ – 2x + 4)⁴
(5b)
5/3(2-x) – (1-x)/(2-x) = 2/3
Multiply through with 3(2-x)
3(2-x)[5/3(2-x)] -3(2-x)[1-x/2-x] =
3(2-x)(2/3)
5 – 3(1-x) = (2 – x)(2)
5 – 3 + 3x = 4 – 2x
3x + 2x = 4 + 3 – 5
5x = 2
X = 2/5
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(6ai)
Volume of sphere = 9 ⅓ ×(its surface
area)
4/3πr³ = 28/3×4πr²
r = 28 units
Surface area = 4πr²
= 4×22/7×28×28
= 9856 units squared.
(6aii)
Volume = 9 ⅓ × 9856
= 28/3 × 9856
= 91989.33
=91989 cubic units
(6b)
log10(3x – 5)² – log10(4x -3)² = log10
25
Log10(3x – 5/4x – 3)² = log10 25
(3x – 5/4x – 3)² = 25
Square root of both sides gives
3x-5/4x-3 = ±5
3x-5 = 5(4x-3) OR 3x-5 = 5(4x-3)
3x-5 = 20x-15 OR 3x-5 = 20x+15
3x-20x = 5-15 OR 3x+20x = 15+5
-17x = -10 OR 23x = 20
X = 10/17 OR x = 20/23
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(7a)
Given: 3x + 5y = 10
5y = -3x + 10 OR y = -3/5x+ 2
Gradient of the straight line = -1 ÷
(-2/5)
= 5/3
Equation of line:y-2/x-3 = 5/3
3y-6 = 5x – 15
3y = 5x – 15 + 6
3y = 5x – 9
y = 5/3x – 3
Intercept of line = -3
(7b)
Amount = P(1+R/100)³
=8000(1+5/100)³
=8000(1.05)³ OR 8000(1.05/100)³
=8000×1.157625
=#9261
Compound interest = Amount – principal
= 9261 – 8000
= #1261
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(10a)
Let the woman’s age be W
Let the daughter’s age be d
Given: w = 4d—–(1)
Givenw+5)²=(d+5)²+120–(2)
Put eqn(1)into(2)
(4d+5)² = (d+5)² + 120
(4d+5)² – (d+5)² = 120
[4d+5+d+5][4d+5-d-5] = 120
[5d+10][3d] = 120
5(d+2)(3d) = 120
15(d+2)(d) = 120
d(d+2) = 8
d² + 2d – 8 = 0
d² + 4d – 2d – 8 = 0
d(d+4) -2(d+4) = 0
(d – 2)(d + 4) = 0
d – 2 = 0 (only)
d = 2
Daughter is 2 years old
(10b)
t = w + wy²/PZ
Multiply through by PZ
Pat = PWZ + wy²
Wy² = ptz – pwz
y² = Pz(t – w)/w
y = ±√PZ(t – w)/w
If P = 5, Z = 10, t = 9, w=3
y = ±√(5)(10) (9-3)/3
y = ±√(5)(10)(6)/3
y = ±√100
y = ±10
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(11)
|Score|5-9|10-14|15-19|20-24|25-29|
30-34|
|Frequency|3|10-T|3T|8|2T+2|T+2|
(a)Ʃ+=50
3+10-T+3T+8+2T+2+T+2=50
3+10+8+2+2-T+3T+2T+T=50
25+5T=50
5T=50-20
5T/5=25/5
T=5
(11b)
|Score|5-9|10-14|15-19|20-24|25-29|
30-34|
|Frequency|3|5|15|8|12|7|
The frequency of the modal class is 15.
(11c)
Tabulate.
|Score|5-9|10-14|15-19|20-24|25-29|
30-34|
|F|3|5|1|5|8|1|2|7|50
|X|7|12|17|22|27|32|
|FX|21|60|255|176|324|224|1060
|X-x̅|-14.2|-9.2|-4.2|0.8|5.8|10.8|
|(X-x̅)²|201.64|84.64|17.64|0.64|33.64|
116.64|
F|(X-x̅)|604.92|423.2|264.6|5.12|
403.68|816.48|2518
Mean (x̅) =Ʃfx/Ʃf=1060/50=21.2
Variance =Ʃf(x-x̅)²/Ʃf
=2518/50
=50.36
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Today’s Maths OBJ:
1-10: ACBAECABAE
11-20: CCABBCACEE
21-30: BBEBBBACBD
31-40: CCAEBAAABD
41-50: DABBBCCBBD
51-60: DACEADAEEA