Neco 2020/2021 Mathematics Obj & Theory Question and Answers Verified Expo

 


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2020 NECO MATHEMATICS ANSWERS

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MATH OBJECTIVE

1-10: ACBAECABAE
11-20: CCABBCACEE
21-30: BBEBBBACBD
31-40: CCAEBAAABD
41-50: DABBBCCBBD
51-60: DACEADAEEA

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

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

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.

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

(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.

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

(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

 

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(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

 

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.

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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=√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) 

[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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