2020/2021 WAEC/NECO Answers: Neco Gce 2021. Mathematics answers



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OBJ

1-10: DABBABBBAD

11-20: CDBDCCCBBA

21-30: CDABDACCCC

31-40: BCBCABABBC

41-50: DACCCCDBDB

This is actually the Neco Gce 2018. Mathematics 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.


(10)

Draw the triangle 


Hence we have A+B+C = 180

50+80+C = 180

130+C = 180

C = 180 - 130

C = 50


(i)

The bearing of B from C = 90 - 50 = 40°


(ii) 

Bearing of A from B

90+50(alternate angle to A) 

= 140°


(iii) 

Distance between B and C

Using sine rule 

C/sinC = a/SinA

40/sin50 = a/sin50

Cross multiply 

asin50 = 40sin50

a = 40km

Hence distance between B and C = 40km


(iv) 

Using cosine rule 

b²= a²+c² - 2acCosB

where a = 40, c= 40 and B=80

b² = 40²+40² -2*40*40cos80

b² = 1600+1600 - 3200(0.1736)

b² = 3200 - 555.52

b² = 2644.48

b = √2644.48

b = 51.42km


(v)

Height of ΔABC

Draw the triangle 


Area of the triangle 

=1/2acSinB

=1/2*40*40sin80

=800(0.9848)

=787.84km²

Hence 1/2bh=787.84

bh=2*787.84

h=2*787.84/b

h=2*787.84/51.42

h=30.64

=31






(12)

TABULATE:

Marks | frq | c.f | boundary

11-20 | 8 | 8 | 10.5 – 20.5

21-30 | 6 | 14 | 20.5 – 30. 5

31-40 | 10 | 24 | 30.5 – 40.5

41-50 | 12 | 36 | 40.5 – 50.5

51-60 | 8 | 44 | 50.5-60.5

61-70 | 6 | 50 | 60.5 – 70.5


(12b)

DRAW THE CUMULATIVE FREQUENCE CURVE


(12ci)

Medium mark = N + ½

= 50 + 1/2 = 51/2

= 25. 5

= 48


(12cii)

Lower quartile Q1

= ¼ x 50

= 12.5

= 28


(12ciii)

Upper quartile Q2

= ¾ x 50

= 37.5

=53


(12civ)

Both percentile 

70/100 x 50 

= 35

= 50

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

(1) 

TABULATE:

No| Log

3081| 3.4887

0.775 | 1.8893

0.456 | 1. 6589

| UNDER Log

| 3.4887

| 1. 5482

| 3.9405

| 0.9851

Square root of 4/3081/0.775.0456

Antilog = 9663

= 9.663

= 2.9






(2a)

1101â‚‚ = 2x + 1

1 x 2² + 0 x 2¹+ 1 x 2º = 2x + 1

8 + 4 + 1 = 2x +1

13 – 2x

X = 12/2 = 6


(2b)

Sin x = 12/13 = 0.9231

Z = sin-¹ 0.9231 = 67.4º

Therefore 3 sin x + ½ cos x

= 3 sin 67.4 + ½ cos 67.4

= 3 x 0.9232 + ½ x 0.3843

= 7.7696 + 0.1921






(3a)

PQ x 90º  [angle in a semicircle]

QPO = 90º - 62 [ angle in a triangle]

= 28º

Therefore POZ = 28º [alternative angles]

(ii)

PXZ = ½ x 28º = 14º [angle at center is twice angle at cirumfeence]


(3b)

6/5 + 3/x+3 – 9/5(x+3)

= 6 (x+3) + 3 (5) – 9/5(x+3)

= 6x + 18 + 15 – 9/5 (x+3)

= 6x + 24/5(x+3) 

= 6(x+4)/5(x+3)






(4a)

1/3(y-1)+2>1/2(2y-1)+1

2y-2+12>6y-3+1

2y-6y>-2-10

-4y>-12

y<3


(4bi)

M=y2-y1/X2-X1

=-1-2/2-3=-3/-1=3


(4bii)

2X+1=X+3

2X-X=3-1

X=2





(6)COMPLETED 

(6)

126y = 86

1 x y² 2 x y¹ + 6 x yº = 86

y² + 2y + 6 = 86

y² + 2y + 6 – 86 = 0

y² + 2y – 80 = 0

y + 10y – 8y – 80 = 0

y (y+10)-8y (y+10) = 0 

y – 8 = 0 or y + 10 = 0

the positive value of y


(6b) Area of triangle = ½ x b x h

Let the Acheal Area = x

Area = ½ x b x h

Base = x – 9x/100 = 93/100

Height = 9x/100 + x = 107x/100

Area = ½ x 93x/100 x 107x/100

=963x/2000

% error = actual Area – wrong/actual area x 100

= x – 963x/2000 /x 100

= 20000x – 963x/20000 x 100

= 19037/20000 x 100

= 95.185%


(6c)

p/100 + 2p + 7 = 11.02 x 100

p + 200p + 700 = 1102

201p = 1102 – 700

201p = 402

P = 402/201

P = N2

P = 200K




(8)

S²â‚ (3x – 1) (x+2) dx

By expansion’

S²â‚ 3x² + 6x – x – 2 dx

S²â‚ 3x² + 5x – 2 dx

By integrating using d formula

Xn+1/n+1

Therefore 3x ²+¹/2+1 + 5x¹+¹/1+1

= 2xº+¹/0+1 + c

Therefore 3x³/3 + 5x²/2 + 2x/1 + c

Therefore x³ + 5x²/2 + 2x

But x = 2 at higher and are at lower by substituting 2 in x than of value of 1

(2)³ + 5/2(2)² + 2(2) – ((1)³ + 3/2(1) + 2 (1)

8 + 5/2 x 4 + 4 – (1 + 5/2 + 2)

8 + 10 + 4 – (3 + 5/2)

22 – 11/2 = = 22-11/2

= 44- 11/2 

= 33/2


(8bi)

T = 2Ï€ square l/g

Dividing both side by 2Ï€

T/2Ï€ = square root of L/g

By squaring both side

(T/2 Ï€)² = (L/g)²

T²/4 Ï€² = l/g

Cross multiplication

gT²/T² = 4Ï€²L/T² 

g = 4Ï€²L/T²


(8bii)

T = (0.4)1/2 = square root of 0.4

L = 0.04   T1 = 3.14

G = 4Ï€²L/T2 

= 4 x (3.14)² x 0.04/(0.4)²

= 4 x 9.8596 x 0.04/0.4 

= 1.578/0.4

g = 3.94




(9a)

2p-q=10.......(1)

3p+q^2=22........(2)

Eq(1) x3 and eq(2) x2

6p-3q=30

6p+2q=44

Subtracting eq(1) from eq(2)

-3q-2q= -14

2q^2+3q= 14

2q^2+3q-14=0

2q^2+7q-4q-14=0

q(2q+7)-2(2q+7)=0

(q-2)(2q+7)=0

q-2=0 or 2q+7=0

q=2 or q= -7/2

Substitute q into eq(1)

2p-q=10

2p-2=10

2p=10+2

2p=12

p=12/2=6

When q= -7/2

2p-q=10

2p-(-7/2)=10

2p+7/2=10

4p+7=10

4p=10-7

4p=3

p=3/4


(9b)

z^2 -25/z^2-9z+20

If Z is undefined 

Z^2-9z+20=0

Using factorization method 

-4z and -5z

Z^2 -4z-5z+20=0

Z(z-4) - 5(z-4)=0

(z-4)(z-5)=0

Z-4=0 or Z-5=0

Z=4 or Z=5

Z=4 or 5

Z is undefined when it is equal to  4 or 5



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