Neco Maths answers 2021
Here is neco mathematics answers for 2021, enjoy and share to your friends.
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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
:. 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
(9c)
No time difference between Q and S because they are on the same longitude
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