# NECO 2019 GENERAL MATHEMATICS OBJ AND THEORY VERIFIED QUESTIONS AND ANSWER

NECO 2019 GENERAL MATHEMATICS OBJ AND THEORY VERIFIED ANSWER.

MATHS-OBJ!

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(1a)

At the end of year 1

Using; A = P(1 + R/100)N

A = #110,000(1+5/100)

A = #110,000(1.05)

Amount or savings = #115,500.00

At the beginning of year 2,

Principal, p = 115,500 + #50,000 = #165,500.00

At the end of year 2

A = #165,500(1+5/100)¹

A = #165,500 × 1.05

A = #173,775.00

At the beginning of year 3,

Principal, p = #173,775 + #50,000 = #223,775.00

At the end of year 3,

A = #223,775(1+5/100)

A = #223,775 × 1.05

A = 234,963.75

Total savings after 3 years = #234,963.75 + #50,000 = #284,963.75

(1b)

By end of third year

Savings is lesser than #300,000.00 by;

#300,000.00 – 284,963.75

=#15,036.25

= #15,036.25

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(9ai)

x+1⅔x≤ 2⅓x-1¼

x+5x/3≤ 7x/3 – 5/4

Multiply through with (12) 12x + 20x ≤ 28x – 15

32x ≤ 28x -15

32x – 28x ≤ -15

4x ≤ – 15

x ≤ – 15/4

x ≤ – 3¾.

(9aii)

4x-1/3 – 1+2x/5 ≤ 8+2x

Multiply through with (15) 5(4x-1)-3(1+2x)≤ 15(8+2x)

20x-5-3-6x ≤ 120 + 30x

14x-8 ≤ 120 + 8

-16x ≤ 128

x ≥ 128/-16

x ≥ -8

(9b)

Gradient

m=y²-y¹/x²-x¹

m=9-7/6-3

=⅔

Acute angle θ = Tan-¹ (⅔)

θ=Tan-¹ (0.6667)

θ=33.69degree.

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(5a)

x² – 5x – 24 = 0

x² – 5x = 24

x² – 5x + (5/2)² = 24 + (5/2)²

(x-5/2)² = 24 + 25/4

96 + 25/4

(x – 5/2)² = 121/4

x – 5/2 = ±√121/4

x – 5/2 = ± 11/2

x = 5/2 ± 11/2

x = 5/2 + 11/2 or 5/2 or 11/2

x = 16/2 or -6/2

x = 8 or -3

(5b)

S0/2 (3x²-4x+2)dx

= 3x²+¹/3 – 4x¹-¹/2 + 2x/1 + C

= 3x³/3 – 4x²/2 +2x/1 + C

= (x³ – 2x² + 2x + C) dx

= y = x³+¹/4 – 2x²+1/3 + 2x¹+¹/2 + C

= y = x⁴/4 – 2x³/3 + 2x²/2 + C

= y = x⁴/4 – 2x³/3 + 2x + C

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(12a)

Tabulate

Score : 21-30, 31-40, 41-50, 51-60, 61-70, 71-80

Class mark(x) : 25.5, 35.5, 45.5, 55.5, 65.5, 75.5

f : 2,10,12,15,8,3

d(x- x̄): -20, -10, 0, 10, 20, 30

fd : -40, -100, 0, 150, 160, 90

Assumed mean = 45.5

(12b)

Using assumed mean ( x̄) = A.M + Σfd/Σf

x̄ = 45.5+260/50

x̄ = 45.5+5.2 = 50.7

(12c)

Semi inter quartile = Q2-Q7/2

Q3= 3/4 × f

= 3/4 × 50 =150/4 = 37.5

Q1= 1/4 × f

= 1/4 × 50/1 = 12.5

: . Semi inter quartile = 37.5 – 12.5/2 = 25/2

= 12.5

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(2a)

3^2x-y=1

3^2x-y=3^0

2x-y=0————-(1)

16^x/4 = 8^3x-y

2^4x/2^2 = 2^3(3x-y)

2^4x-2 = 2^9x-3y

:. 4x-2 = 9x-3y

4x-9x+3y= 2

-5x+3y=2————(2)

From equation (1):

2x-y=0

y=2x——–(3)

Substitute for y in equation (2)

-5x+3y=2

-5x+3(2x)=2

-5x+6x=2

x=2

Substitute for x in equation (3)

y=2x

y=2(2)=4

:.x=2, y=4

(2b)

x² – 4/3 + x+3/2

2(x² – 4) + 3(x +3)/ 6

2x² – 8 + 3x + 9/6

2x²+3x+1/6

(2x² + 2x)+(x+1)/6

2x(x+1) +1 (x+1)/6.

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(8ai)

Total surface area

= Total surface of cylinder + Curve surface of hemisphere

= (πr^2+2πrh) + (2πr^2)

= π(r^2 + 2rh) + π(2r^2)

= π[(r^2 + 2rh) + 2r^2]

= π[(7^2 + 2(7)(10) + 2(7)^2]

= π[(49+140) + 98]

= π(287)

= 287πcm^2

Using π=22/7

Total surface area =287×22/7 = 41×22

= 902cm^2

(8aii)

Volume = Volume of cylinder + volume of hemisphere

= πr2h + 2/3πr^3

= π[r^2h + 2/3r^3]

= π[(7^2)(10) + 2/3(7)^3]

= π(490 + 656/3)

= π(2156/3)

= 22/7 × 2156/3

= 22 × 308/3 = 6776/3

= 2258.67cm^3

(8b)

CLICK HERE FOR THE IMAGE

Perimeter of Arc = Φ/360 × 2πr

= 120/360 × 2 × 22/7 × 7

= 1/3 × 44 = 44/3

= 14.67cm

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(3)

Using SOHCAHTOA

|TM| / |MD| = Tan28°

298.5+1.5/|MD| = 0.5317

|MD| = 300/0.5317 = 564.2m

Similarly,

|TM| / |MC| = Tan34°

300/ |MC| = 0.6745

|MC| = 300/0.6745 = 444.8m

Distance between both , ΔCD

= 564.2 – 444.8

= 119.4m

(10ai)

S=t^3 -3t -9t + 1

ds/dt=v

:. 3t^2 – 6t^2 -9

When v=0

3t^2 -6t^2 -9=0

(3t^2 -9t)+(3t-9)

3t(t-3)+3(t-3)=0

(3t+3)(t-3)=0

3t + 3=0

3t= -3

t= -3/3= -1 or t -3=0

t=3seconds

(10aii)

a=dv/dt = 3t^2 -6t -9= 6t -6

a=6t -6

When a=0

6t -6=0

6t=0+6

6t=6

t=6/6

t=1

(10b)

v=3t^2 -6t -9

When t=2seconds

v=3(2)^2 -6(2) -9

= 3*4-9-12-9= -9m/s

v= -9m/s

acceleration, a when t=2seconds

a=6t -6= 6(2) -6= 12-6

a=6m/s^2

(10c)

a=6t -6 = 36-6=30

a=30m/s^2

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(6a)

For X = a=4a , T6=256

ar^5=256

4ar^5/4=256/4

ar^5=64…………(1)

For Y = a=3a, T5=48

ar^4=48

3ar^4/3=48/3

ar^4=16……………(2)

Divide equ (2) by (1):

ar^5/ar^4=64/16

r=4

Substitute for r in equ (2)

ar^4=16

a × 4^4=16

256a/256=16/256

a=1/16

(6ai)

First term of x : a=4a

a=4×1/16=1/4

(6aii)

Sn = a(r^n-1)/r-1

S4 = 1/4(4^4 -1)/4-1

=1/4(256-1)/3

=1/4 × 255/3

=85/4

S4=21.25

(6b)

y(4x+2)^-3

Let u =4x+2, y=u^-3

du/dx=4 , dy/du= -3u^-4

dy/dx=dy/du * du/dx

= -3u^-4 × 4

= -12u^-4

dy/dx= -12(4x+2)^-4

When x =1,

dy/dx= -12(4*1+2)^-4

= -12(4+2)

= -12 * 6^-4

= -12/6^4

= -12/1296

= -1/108

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(7a)

4x^2 – 9y^2 = 19

2x^2 x^2 – 3^2 y^2=19

(2x-3y)(2x+3y)=19

Substitute for 2x+3y=1

2x-3y=19…………(1)

2x+3y=1…………..(2)

Subtract equ (2) from (1)

2x-3y-(2x+3y)=19-1

3x-3y-2x-3y=18

-6y/-6=18/-6

y = -3

Substitute for y in equ (1)

2x-3(-3)=19

2x+9=19

2x/2=10/2

x=5

(7b)

√4.033/0.611 × 0.356

Put No and Log In a tabular form

No | log

4.033 | 0.6056 -> | 0.6056

0.611 | 1.7860+ –

0.356 | 1.5514

0.611×0.356|1.3374->|1.3374

4.033/ | ——–> | 1.2682

0.611×0.356 ÷2

√4.033/0.611 | ——-> | 0.6341

×0.356

Antilog = 4.306

Ans = 4.306

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(4)

Mark: 1-5, 6-10, 11-15, 16-20, 21-25, 26-30

F : 6,4,5,5,6,4

x : 3,8,13,18,23,28

Fx: 18,32,65,90,138,112

x-x: -12.167, -7.167, -2.167, 2.833, 7.833, 12.833

(x-x)^2: 148.0359, 51.3659, 4.6959, 8.0259, 61.3559, 164.6859

f(x-x)^2: 888.2154, 205.4636, 23.4795, 40.1295, 368.1354, 658.7436

Mean(x) = Σfx/Σf = 455/30 =15.167

Variance = Σf(x-x)^2/Σf = 2184.167/30 = 72.8056

= 72.81(approx.)

Standard deviation = √Variance

= √72.84

= 8.533

= 8.53 (approx.)

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COMPLETED!

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