2021 NECO MATHEMATICS ANSWERS

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MATHEMATICS-THEORY

All are 101% Authentic.
Pls 2bi and 4b and 9 and 12 also corrected. We Make Sure Everything
Is 101% Correct. Sorry For The Little
Corrections. Tell Others Pls.
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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
====================================

(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
= 20.7846
= 20.78(approx)

(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
:. 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°
====================================

(8a)
Y = 2x²+ 3

|¹ 2x³/3 + 3x|
|² |

= 2(1)/3 + 3(1) – 2(2)²/3 + 3(2)
=2/3 +3/1 – 16/3 +6
= 2+9/3 – 16+18/3
= 11/3 – 34/3 = -23/3
= -7.67

(b)
(i)
3p +5q = 9500….. (i) *5
5p +10q = 17500….(ii) *3

15p +25q = 47500…..(iii)
15p + 30q = 52500…..(iv)

Subtracting (iii) from (iv)
-5q = -5000
q = -5000/-5
q = 1000

Substituting q =1000 into equation (i)
3p + 5(1000) = 9500
3p + 5000 = 9500
3p = 9500-5000
3p = 4500
p = 4500/3
p = 1500

:. q = ₦1000.00
p = ₦1500.00

(ii)
8 men and 12 women

8p +12p
8(1500) + 12(1000)
= 12000 + 12000
= ₦24000.00

:. ₦24000.00 will be contributed by 8 men and 12 women

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

(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

(9c)
No time difference between Q and S because they are on the same longitude
====================================

(12)
(ai)
TABULATE

Class interval: 1-20| 21-40| 41-60| 61-80| 81-100|

Frequency (f): 7| 13| 5| 4| 1

Class mark(x): 10.5| 30.5| 50.5| 70.5| 90.5|

d=x-A: -20| 0| 20| 40| 60|

Fd: -140| 0| 100| 160| 60|

d²: 400| 0| 400| 1600| 3600|

fd²: 2800| 0| 2000| 6400| 3600|

(ii) Mean= A+∑fd/∑f = 30.5+180/30
= 30.5+6
= 36.5
Standard deviation= √ ∑fd²/∑f – (∑fd/∑f)²
= √14800/30 – (180/30)²
= √493.33 – 36
= √ 457.33
= 21.4

(b) P(above 60 marks) = (4+1/30) = 5/30
= 1/6

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(1&2&3)

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(4)

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(5)

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(6)

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(7)

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(8)

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(9)

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(11)

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Bonus
(12)

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COMPLETED

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