2020/2021 WAEC/NECO Answers: Nabteb Gce 2020. Maths answers



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IRS OBJECTIVE

1-10: AAEDCEBBDA

11-20: DACDACECAE

21-30: ACEEDBBECB

31-40: AACEEABCDC

41-50: AAADEBECCC

51-60: CEDDAEACCE

This is actually the Nabteb Gce 2019. Maths answers Answers as questions sent to our subscribers on WHATSAPP, if you are looking for 2020 WAEC/NECO Answers midnight before the exam, try contacting us.



*SECTION A*

*_(Answer ALL QUESTIONS From This Section)_*


(1a)

6^n+1*9^n*4^2n/18^n*2^n*12^2n

=(2*3)^n+1*(3*3)^n*(2*2)^2n/(2*3*3)^n*2^n*(2*2*3)^2n


=2^n+1*2^n+1*3^n*3^n*2^2n*2^2n/2^n*3^n*3^n*2^n*2^n*2^n*3^2n


= 2^n*2^1*2^n*2^1*3^n*3^n*2^n*2^n*2^n*2^n/2^n*3^n*3^n*2^n*2^n*2^n*2^n*2^n*3^n*3^n


= 2^6n*3^2n*2*2/2^6n*3^4n


= (2^6n / 2^6n) *(3^2n / 3^4n)*2*2


=2^6n - 6n * 3^2n - 4n * 4


= 2^0 * 3^-2n * 4


= 4*3^-2n


= 4 * 1/3^2n


= 4/3^2n


(1b)

Log 243/log 27

= log 3^5/log 3^3

= 5log3/3log3 = 5/3 = 1 whole no 2/3


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



(2a) 

3t - 2p = 8×2 ...(I) 

2t - 3p = 14×2...(II) 


6t - 4p = 16 ...(III) 

6t - 9p = 42 ...(IV) 

  9P - 4P = -42 + 16

  5p = -26

P = -26/5 = -5 1/5


Put p = -26/5 into eqn (I) 

3t - 2(-26/5) = 8

3t + 52/5 = 8

3t = 8 - 52/5

3t = 40 - 52/5 = -12/5

Divide both sides by 3

36/3 = -12/5 / 3

t = -12/5 × 1/3

= -12/15 = -4/5


(2b)

3/√3(2/√3 - √12/6)

2/√3 × 3/√3 - 3/√3 × √12/6

6/√9 - 3√12/6/√3

6/3 - 3√12/6√3

2 - √12/2√3

2 × 2√3 - √12/2√3

4√3 - (√4 × 3)/2√3

= 4√3 - 2√3/2√3

= 2√3/2√3 = 1


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



(3)

P(k) = 2/3

P(Y) = 5/8

P(I) = 3/4

P(k fail) = 1- 2/3 = 1/3

P(Y fail) 1 - 5/8 = 3/8

P(I fail) = 1 - 3/4 = 1/4


(a) 2/3*5/8*3/4 = 5/16

(b) 1/3*3/8*1/4 = 1/32

(c) 2/3*5/8*1*4 = 5/48


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



(4)

Draw the triangle 

<GFN = 180 - ( 68+90)

Sum of angles in a Δ

= 180 - 158 = 22°

Sin 22/|GN| = Sin 68/8

|GN|= 8sin22/sin68

= 8×0.3746/0.9272

= 2.9969/0.9272

= 3.2322cm


|GT| = |TN| = 3.2322

= 1.616≈ 1.62cm

Let tita = <FTG


(i) tan tita = 8/1.6161 = 4.9502

tita = tan^-¹(4.9502)

tita = 78.6°


FTN = 180 - tita

= 180 - 78.6

(sum of angles on a straight line) = 101.4


(ii) >TFN 

= 180 - (68+101.4)

Sum of angles in a Δ

180 - 169.4

=10.6°


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




(5a)

4 1/2 - 3(y - 2) = 2y + 1/3

= 9/2 - 3y + 6 = 2y + 1/3

= -3y + 6 + 9/2 = 2y + 1/3

6 + 9/2 = 2y + 1/3 + 3y

12 + 9/2 = 2y + 1/3 + 3y

21/2 = 2y + 1 + 9y/3

21 × 3 = 2(2y + 1 + 9y)

63 = 4y + 2 + 18y

63 = 22y + 2

22y = 63 - 2

22y = 61

y = 61/22 = 2 17/22


(5b) 

Let the number be y 

7 - 2y >_ 16

-2y >_16 - 7

-2y >_ 9

y _< _9/2 = 4 1/2

The greatest possible values of y are;  -4, -3, -2, -1, 0


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



*SECTION B*

_(All Candidates should Answer FOUR while SECRETARIAL & BUSINESS CANDIDATES Should Answers Only TWO from this Section)_



(6a)

Let the water melon be x

Let the mango be y


12x + 24y = 432 ...(i) × 24

24x + 12y = 360...(ii) × 12


288x + 576y = 10368 ...(iii) 

288x + 144y = 4320 ... (iv) 

-144y + 576y = -4320+10368

432y = 6048

y = 6018/432 = 14


Put y = 14 into eqn (i) 


12x + 24(14) = 432

12x + 336 = 432

12x = 432 - 336

12x = 96

X = 96/12 = 8


(i) Water melon per kg = 8

(ii) Mango per kg = 14

(iii) (3 × 8) + (2 × 14)

= 24 + 28

= 52


(6b) 

123x = 83ten

1*X² + 2*X¹ + 3*X^0

= 83


X² + 2x + 3 = 83

X² + 2x + 3 - 83 = 0

X² + 2x - 80 = 0

X² + 10x -8x - 80 = 0

X(x+10) - 8(x+10) = 0

Then either 

X - 8 = 0 or X + 10 = 0

X = 8 or X = -10


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




(8a) 

Distance /xy/ = tita/360 × 2πr

tita = 45° - 15° = 30°

r = Rcosα

r = 6400 × cos40°

r = 6400 × 0.7600

r = 4902.4km

Distance /xy/ = 30/360 × 2 × 22/7 × 4902.4

= 30 × 2 × 22 × 4902.4/2520

=6471168/2520 = 2567.9238

= 2600km


(8bi) 

Time taken to fly from X to y

Speed = Total distance/Total time taken 

850 = 2600/t

t = 2600/850 = 3.0588hrs

t = 3hours


Time taken to fly to Y from X

Time = Distance/Speed

= 6700/850 = 7.8824hrs = 8hrs

Total time = 3hrs + 8hrs = 11hrs


(8bii) 

Latitude of Z

Distance /yz/ = tita/360 × 2πr

6700 = Z-40/360×2×22/7×6400

6700×360×7 =(Z - 40)281600

16884000 = 281600Z - 11264000

281600Z = 16884000 + 11264000

281600Z = 28148000

Z = 28148000/281600

Z = 99.9574 = 100°


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



(9a)

DRAW THE TRIANGLE DAIGRAM

|YX| = x + 620

From ∆ YFP

Tan56 = R/620

R = 620 x tan56

= 620 x 1.4826

h = 919.19m

To find x

Tan 20 = h/x

X = h/tan20 = 919.19/0.3639

X = | XF | = 2525.45m

|YX| = |XF| + |FY|

|YX| = 2525.45 + 620

= 3145.5.45m

Hence the value of |YX| correct to the four significant figure is 3145m

|XY| = 3145m


(9b)

DRAW THE DAIGRAM

|YX|² = |OY|² + |OX|² - 2|OY| |OX| cosØ

(6)² = (5)² + (5)² - 2 (5) (5) cosØ

36 = 25 +25 – 50cosØ

36 = 50 – 50cos Ø

50cosØ = 50 – 36

50cos = 14

cosØ = 14/50 = 0.28

Ø = cos-¹ (0.028) = 73.34º

Therefore <XZY = ½ (73.74) (Angle at the center twice the angle at the circumference)

<XZY = ½ (73.74) = 36.87º

<XZY = 37º (nearest degree)


(9bii)

Let |XZ| = |YZ| = y

|XY|² = |YZ|² + |XZ|² - 2 |YZ| |XZ| Cos〆

(6)² = y² + y² - 2y² cos 36.87

36 = 0.4y²

y² = 36/0.4 = 60

y = √60 = 7.75m

hence, |XZ| = 7.8m (1dp)


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



(11a)

In a tabular form 


Under scores(x)

2, 3, 4, 5, 6, 7


Under frequency(f) 

2, 4, 5, 3, 4, 2 Ef, = 20


Under fx 

4, 12, 20, 15, 24, 14 Efx = 89


(i) Modal score = 4

(ii) Median = n1+n2 = 4+4/2 = 8/2 = 4

(iii) Mean X = Efx/Ef = 89/20 = 4.45


(11b) 

A = P(1+r/100)n 

A = amount in compound interest.

P = principal, r = rates, n = no of years the compound interest is charged. 

A = 53000(1+7/100)^5

= 53000(1+0.07)^5 = 53000(1.07)^5


In a tabular form 

  No | Log

53000| 4.7243 =  4.7243

     | 0.0294×5 = 0.1469

                  4.8712

Antilog of 4.8712 = ₦74436.14

Compound interest = 74436.14 - 53000 = ₦21336.14


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




*SECTION C*

_(Only For SECRETARIAL & BUSINESS Candidates. They shoud answers only TWO from this section)_


(12a) 

If passengers paid ₦8,400.00 per trip 

Commission on each trip = 5/100 × 8400 = ₦420.00


For 52 trips, his commission would be 52 × ₦420 = ₦21,840.00


His take home for the month = ₦6000.00 + ₦21,840.00 = ₦27,840.00


(12bi) 

Gross pay = ₦125018.35 per month. 

5% income tax = 5/100 × 125018.35 = 6250.9175


2% Union dues = 2/100 × 125018.35 = 2500.3670


1% housing fund = 1/100 × 125018.35 = 1250.1835


7.5% pension scheme = 7.5/100 × 125018.35 = 9376.3763


2.5% health insurance scheme = 2.5/100 × 125018.35 = 3125.4588


10% cooperative contribution = 10/100 × 125018.35 = 12501.8350


Total deductions = 6250.9175+2500.3670+1250 1835+9376.3763+3125.4588+12501.8350 = 35005.14


(12bii)

His net pay = ₦125018.35 - 35005.14 = ₦90013.21


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



(13a)

4 - point moving average. 

16, 18, 20, 27, 28, 31, 32


16+11+20+27/4 = 81/4 = 20.25


18+20+27+28/4 = 93/4 = 23.25


20+27+28+31/4 = 106/4 = 23.25


20+27+28+31/4 = 106/4 = 26.5


27+28+31+32/4 = 118/4 =

 29.5


(13b)

First 20weeks, she earns ₦400×20 = ₦8000


The next 20weeks she earns ₦720×20 = ₦14400


Total earns for 40weeks = ₦8000 + ₦14400 = ₦22400


Total earns for the whole

 year = ₦52×600 = ₦31200


Amount earns for the remaining 12months = ₦31200 - ₦22400 = ₦8800


Average weekly earns for the last 12 months = 8800/12 = ₦733.33


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



(15a)

Rent = 15/100 * ₦365,500 = ₦54,825

Personal allowance = 12/100 ×₦365,500 = ₦43,860

Annual tax free allowance 

= ₦54825 + ₦43860 = ₦98685


(15b) 

Taxable income = ₦(365500 - 98685) = ₦266,815

Tax calculation 

The first ₦65,000 = 0% = ₦0


Next ₦100,000 = 10/100 × 100,000 = ₦10,000


Next 100,000 = 15/100 × 100,000 = ₦15,000


The remaining ₦1815 = 20/100 × 1815 = ₦363


Annual income tax = ₦(0 + 10,000 + 15,000 + 363)

=₦25,363


(15c) 

Percentage of his salary paid as tax = 25,363/365,500 × 100 = 6.94%


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


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