NECO MATHS ANSWERS 2020

NECO MATHEMATICS ANSWERS 2020

THEORY ANSWERS

Here is the confirmed maths answers for 2020. Please do well to confirm each questions in your question paper before answering. Goodluck,  just share to others.

(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

2x²/2+9x + c|²_¹
x² + 9x + c|²_¹
(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.
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.

(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
6*8x-27*3x=9*5-54*x
48x-81x=45-54x
48x-81x+54x=45
21x/21=45/21
x=45/21=15/7=2⅐

(5a)
Let u = 4x³ – 2x + 4 ,
y = u³ , du/dx = 12x²-2,
dy/du=3u²

dy/dx= dy/du*du/dx
=3u²*12x²-2
=(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

(6a)
Volume of a sphere = 9⅓ of its surface area 4/3 πr³=28/3*4πr²
4πr³/3*112πr²/3
3*4πr³=3*112πr²
12πr³=336πr²
Divide through by 12πr²
12πr³/12πr² = 336πr²/12πr²

r=28cm

(i)surface area =4πr²
=4*22/7*28cm*28cm
=98cm²

(ii) volume =4/3πr³
=4/3*22/7*28cm*28cm*28cm
=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

(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

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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

(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(2*10+(70-1)5)
=35(20+69*5)
=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.

(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

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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=√5*10(9-3)/3
y=√5*10*6/2
=√100=10
;y=10

(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

(12a)
y=²-2x-3
Tabulate
|x|-2|-1|0|1|2|3|
|y|5|0|-3|-4|-3|0|

(12b)
Graph.

(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

8 Comments

  1. Samuel November 9, 2020
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