Neco Maths Expo 2021 | Neco Mathematics Obj and Essay Runz Questions & Answers Free

neco Maths Expo 2021

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2020 NECO MATHEMATICS EXPO Obj and Essay Runz Questions & Answers

NECO 2020 OCT/NOV ANSWERS (SSCE)

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NECO 2020 MATHEMATICS OBJ AND THEORY ANSWERS FROM EXAMSPOT.NET

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MATHS OBJ

1-10: ACBAECABAE

11-20: CCABBCAAEC

21-30: BBEBBBACBD

31-40: CCADBAEABD

41-50: DEBBBCCBBD

51-60: DACEADAEEA

NECO MATHS THEORY 2021 ANSWERS

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(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

(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

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(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

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(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)

Given:(w+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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(8a)

T9 = a+8d = 50 —(1)

T12 = a+11d = 65 —(2)

Eqn(2)minus eqn(1) gives

3d = 15

d = 15/3 = 5

Put d = 5 into eqn (1)

a+8(5) = 50

a+40 = 50

a = 50 – 40

a = 10

Sn = n/2[2a+(n-1)d]

S10 = 70/2[2(10)+(70-1)5]

= 35[20+345]

= 35 × 365

= 12,775

(8b)

Given v = t² – 3t + 2

S = svdt

=t³/3 – 3t²/2 + 2t +S0

Given S = 40, when t = 6

40 =6³/3 – 3(6)²/2 + 2(6)+ S0

40 = 72 – 54 + 12 + S0

40 = 30 + S0

S0 = 40 – 30 = 10

Therefore,

S = t³/3 – 3t²/2 + 2t + 10

 

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Completed!!

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2021 NECO MATHEMATICS EXPO Obj and Essay Runz Questions & Answers

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