# Neco Maths answers 2021

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

1CBBADCCBEC

11BCADDDCADB

21BBDCEBADBA

31DCAABCCCDC

41CBDCEDECCE

51BECBCCDDAD

Maths-THEORY

All are 101% Authentic.

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

(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

= 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°

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

(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

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

More coming like Palliative Rice.

For a clearer picture of the photos down there, You can chat me on Whatsapp with MTN CARD 200.

- Enjoy

Can u send me the answers through facebook

No sir

Tankiu

Welcome

Link for telegram

Ok

Clearer picture

Please your number

I tried joining the WhatsApp group but I was informed that it was full

Try now

Send your WhatsApp details

Join the Whatsapp group on top of the site and chat the admin

Please drop the telegram link. Thanks

Link for telegram, please.

Check the site well

Your WhatsApp number

Join the group

Comment Text*HELP ME THEORY MATHS ANSWER

Please send your WhatsApp details

The screenshots are not clear pls

Ok