Neco Maths answers 2021

Neco Maths answers 2021

Here is neco mathematics answers for 2021, enjoy and share to your friends.

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11BCADDDCADB

21BBDCEBADBA

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Maths-THEORY

All are 101% Authentic.

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(1a)

Log¹⁰6+ Log¹⁰45 – Log¹⁰27

Log¹⁰(6*45/27) = Log(270/27)

= Log¹⁰10 = 1

(1b)

8^x = 32

2^3x = 2⁵

:. 3x = 5

x=5/3

(1c) 81⅙/27-⅑ = 3⁴*⅙/3³*-⅑ = 3⅔/3-⅓

:. 3⅔-(-⅓) = 3⅔+⅓ = 3³/³

= 3¹

= 3

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(2a)

(i) Gradient (m) = y¹ – y²/x¹- x²

m = -1-0/2-3 = +1/+1=1

(ii) y – y¹= m(x-x¹)

y-0 = 1(x-3)

y = x-3

y-x+3=0

(2b)

(i) Area of ∆ABC = ½absinϴ

= ½*6*8*sin60

= 3*8*sin60

= 1.73

(ii) Area of parallelogram = absinϴ

= 6*8*sin60

= 48*sin60

= 48*0.8660

= 41.6

(3a)

|PQ| = 12km, |QP| = 12km

Speed from P to Q = 6km/h

Speed from Q to P = (6+x)km/h

Total time taken 3hrs20mins

Speed = distance/time

From P to Q = 6/1*12/t

6t = 12

t = 12/6 = 2hrs

:. Time left = 3hrs 20mins – 2hrs

= 1hr 20mins

From Q to P , Speed = distance/time

6+x = 12/ 1²⁰/⁶⁰

6+x = 12/⁴/³

6+x = 12*3/4

6+x = 9

x = 9-6 = 3

(3b)

x² – (sum)x + product = 0

Sum = ⅔ + ¾= 8+9/12 = 17/12

Products = ⅔*¾ = ½

x² – 17x/12 + ½ = 0

12x² – 17x + 6 = 0

(4a)

Y = x²/ 1+x²

U=x² , V= 1+x²

du/dx = 2x , dv/dx = 2x

dy/dx = (vdu/dx – udv/dx)/v²

= (1+x²)2x – x² * 2x/(1+x²)²

= 2x+2x³-2x³/(1+x²)²

dy/dx = 2x/(1+x²)²

(4b)

⅔(3x +2) = ¾(2x -3

6x +4/3 = 6x -9/4

4(6x +4)= 3(6x-9)

24x +16 = 18x-27

24x-18x = -27-16

6x/6 = -43/6

 

x = -7⅙

(5)

TABULATE

Mass (kg): 31-40| 41-50| 51-60| 61-70| 71-80| 81-90

F: 3| 10| 15| 12| 6| 4

x: 35.5| 45.5| 55.5| 65.5| 75.5| 85.5

Fx: 106.5| 455| 832.5| 786| 453| 342

Class boundaries: 30.5-40.5| 40.5-50.5| 59.5-60.5| 60.5-70.5| 70.5-80.5| 80.5-90.5

(i) Mean (x-bar) = ∑fx/∑f = 2975/50 = 59.5

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:. Mean = 60kg

(ii) Mode = L¹ + ( fm-fa/2fm-fa-fb) c

= 50.5. + (15-10/2*15-10-12)10

= 50.5 +(5/8)10

= 50.5+6.25

:. Mode = 56.75

Mode = 57kg(approx)

(6a)

T² = ar = 6 …… (T¹)

T⁴ = ar³ = 54….. (T²)

Common ratio = T²/T¹

ar³/at = 54/6

r²= 9

r = ± √9 = ±3

r =3

Subtract r=3 in equation T¹

ar= 6

3a=6

a = 6/3 = 2

:. a = 2 , r =3

(i) 1st term is 2

(ii) 5th term T⁵=ar⁴

T⁵ = 2*3⁴

= 2*81

= 162

(6b)

(i)

Let pencil be x

Let pens be y

Let Ruler be z

U= 160

n(x) = 75

n(y) = 87

n(z) = 93

n(xny) =25

n(xnz) = 30

n(ynz) = 47

pd n(xnynz) = x

n(xnynz¹) = 25-x

n(xnzny¹) = 30-x

n(ynznx¹) = 47-x

n(xnynz¹) = 75-(25-x+x+30-x)

= 75 -(55-x)

= 75-55+x

= 20+x

n(ynx¹nz¹) = 87-(25-x+x+47-x)

= 87-(72-x)

= 15+x

n(znx¹ny¹) = 93-(30-x+x+47-x)

= 93-77+x

= 16+x

:. 20+x+25-x+x+30-x+15+x+47-x+16+x=160

= 153+x =160

x = 160-153

x = 7

(ii)

n(xny¹nz¹) = 20+7

= 27

:. 27 pupils has pencils only

(7a)

∆XAB = ∆ABC (corresponding angle)

:. ∆BAD + ADC + ∆ACD = 180( sum of angles at triangle)

∆CAD + 83+47= 180

∆CAD = 180-83-47

CAD = 50

:. ∆ADY = CAD ( parallel to each other)

x = 50°

(7b)

Using cosine rule

c² = a²+b²- 2abcosC

x² = 6²+8²-2(6)(8)cos120

= 36+64- 96cos120

= 36+64+48

x² = 148

x = √148

x = 12km

Using sine rule

a/sinA = b/sinB

12/sin120 = 6/sinϴ

12sinϴ = 6sin120

Sinϴ = 6sin120/12

Sinϴ = 0.4330

ϴ = sin-¹ 0.4330

ϴ = 25.66

= 26°(approx)

:. The bearing of the boat from its starting point is

360 -(26+80)

360 – 106

= 254°

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(9a)

Distance PQ = ϴ/360 * 2πRcosα

Where ϴ = 11+11 = 22° and α = 12°

PQ = 22/360*2*3.142*6400*cos12°

PQ = 884*787.2*0.9781/360

PQ = 2403.9

Distance QS= ϴ/360 *2πR

Where ϴ = 44-12 = 32°

= 32/360 *2*3.142*6400

= 1286963.2/360 = 3574.9

Total distance= 2403.9+3574.9 = 5978.8

= 5980km( 3 s.f)

 

(9b)

Average speed = Total distance/Total time = 5978.8/8 = 747.35

= 747km/hr

 

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(9c)

No time difference between Q and S because they are on the same longitude

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