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Under: 2020 NECO MATHEMATICS ANSWERS
OBJECTIVE
*MATHS OBJ:*
1-10: ACBAECABAE
11-20: CCDBBCAAEE
21-30: BBEBBBACBD
31-40: CCADBAEAAB
41-50: DDBBBCCBBD
51-60: DACEADAEEA
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https://codedfans.com.ng
THEORY(1a)
x¹ y¹. x¹ y¹
A (6,5). B ( -2 ,7)
The equation of a straight line
y² – y¹/x² – x¹ = y-y/x¹-x¹
7-5/-2-6 = y-5/x-6
2/-8 = y-5/x-6
-1/4 = y-5/x-6
4(y-5)= -1(x-6)
4y-20= -x+6
4y+x=6+20
4y+x=26
(1b)
∫² (2x + 9)dx
-²
2x²/2+9x + c|²¹ x² + 9x + c|²¹
(2² + 9(2) + C) – ((-1)² + 9 (-1)+ c)
(4+18+c) – (1-9+C)
(22 + C) – (-8 + C)
22+C + 8 – C
=30.
(2a)
To calculate u; 18+5+12+u=52
35+u=52
u=52-32=17
u=17
To calculate v
6+5+12+v=35
23+v=35
v=35-23=12
v=12
To calculate w
w+18+17+6+5+12+12
=100
w+70=100
w=100+70=30
w=30
(2b)
at least two means =18+6+5+12=41
; 41 tourist travelled by at least two means of transportation.
(2a)
To calculate u; 18+5+12+u=52
35+u=52
u=52-32=17
u=17
To calculate v
6+5+12+v=35
23+v=35
v=35-23=12
v=12
To calculate w
w+18+17+6+5+12+12
=100
w+70=100
w=100+70=30
w=30
(2b)
at least two means =18+6+5+12=41
; 41 tourist travelled by at least two means of transportation.
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(3a)
(i)median =6+6/2=12/2=6
(ii)median + range = 6+7=13
Range = highest value – lowest value = 10-3=7
(3b)
Pr(both pass)
=2/5*3/4=6/20=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=0+4 or x=0-7
x=4 or -7
(4b)
8x/9 – 3x/2 =5/6 – x/1
L.c.m=54
68x-273x=95-54x
48x-81x=45-54x
48x-81x+54x=45
21x/21=45/21
x=45/21=15/7=2⅐
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(5a)
Let u = 4x³ – 2x + 4 ,
y = u³ , du/dx = 12x²-2,
dy/du=3u²
dy/dx= dy/dudu/dx =3u²12x²-2
=(36x² – 6) u²
Recall u=4x³ – 2x + 4
dy/dx= (36x² – 6) (4x³-2x + 4)²
(5b)
5/3(2-x) – (1-x)/2-x = 2/3
L.c.m= 3(2-x)
5-3(1-x) = 2(2-x)
5-3+3x=4-2x
2+3x=4-2x
3x+2x=4-2
5x/5=2/5
x=2/5
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(6a)
Volume of a sphere = 9⅓ of its surface area 4/3 πr³=28/34πr² 4πr³/3112πr²/3
34πr³=3112πr²
12πr³=336πr²
Divide through by 12πr²
12πr³/12πr² = 336πr²/12πr²
r=28cm
(i)surface area =4πr²
=422/728cm*28cm
=98cm²
(ii) volume =4/3πr³
=4/322/728cm28cm28cm
=91989.33cm³ ≅ 91989cm³
(6b)
Log10 (3x-5)² – Log10 (4x-3)² =Log²⁵,¹⁰
(3x-5/4x-3)²=25
Square root both side
3x-5/4x-3=5/1
5(4x-3)=3x-5
20x-15=3x-5
20x-3x=-5+15
17x=10=x = 10/17
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(7a)
(y,x)
(3,2)perpendicular to the line 3x + 5y = 10
3x + 5y =10
5y/5= -3x/5 +10/5
y=3x/5 + 2
m=3/5
The equation of a line y= y¹=1/m(x-x¹)
y-2= 1/⅗(x-3)
y-2=-5/3(x-3)
3(y-2)=-5(x-3)
3y-6= -5x + 15
3y= – 5x + 15 + 6
3y/3= -5x/3 + 21/3
y= -5x/3 + 7
Gradient = -5/3 intercept = 7
(7b)
Compound interest = p(1+R/100)n-p
=8000(1+5/100)³ – 8000
=8000(1+0.05)³ – 8000
=9261 – 8000
Compound interest= ₦1261
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(8a)
Ta; a+8d=50 ——–(1)
T12; a+11d=65———(2)
Subtract equation (1) from (2)
a+11d-(a+8d)=65-50
a+11d-a-8d=15
11d-8d=15
3d/3=15/3
d=5
Substitute for d=5 in each equation (1)
a+8d=50
a+8(5)=50
a+40=50
a=50-40=10
Sn=n/2(2a+(n-1)d)
S70=70/2(210+(70-1)5) =35(20+695)
=35*365=12,775
(8b)
V=t²-3t+2
d³/dt=v
d³/dt=t²-3t+2
v=6²-3(6)+2
=36-18+2
v=20mls
Recall; ds/dt=v
ds=vdt
∫ds=∫vdt
S=∫vdt
S=∫20dt
S=20t+c
Where c is constant.
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(10a)
Let daughter = x
Woman = 4x
In 5 years time ;
daughter = x + 5
Woman = 4x + 5
(4x + 5)² = (x + 5)² + 120.
(4x + 5 ) (4x + 5) = (x + 5) (x + 5)+ 120
16x² + 20x + 20x + 25= x²+5x+5x+25+120
16x²+40x+25= x²+10x+145
16x²-x²+40x-10x+25-145=0
15x²+30x-120=0
Divide through by 15;
15x²/15+30x/15-120/15=0/15
x² + 2x – 8= 0
x² + 4x – 2x – 8= 0
x(x+4) – 2 (x + 4) = 0
(x-2) (x+4)=0
x – 2= 0 or x+ 4= 0
x=2 or x=-4
x=2
The daughters age is 2years
(10b)
t=w+wy²/pz
t-w/1 * wy²/pz
10/y²/10 = pz(t-w)/w
y=√pz(t-w)/w
y=√510(9-3)/3 y=√510*6/2
=√100=10
;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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(12a)
y=²-2x-3
Tabulate
|x|-2|-1|0|1|2|3|
|y|5|0|-3|-4|-3|0|
12b)
[Check the Diagram below]
(12c)
y=1-3x
|y|-2|-1|0|1|2|3|
|x|7|4|1|-2|-5|-8|
(12d)
(i)the root of the equation x² – 2x – 3= 1-3x are -3 and 1.6
(ii) the minimum value of y is – 4 and the corresponding value of x is 1
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