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

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# 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
=====================================

(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)
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.
=====================================

(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
=====================================

(12a)
Tabulate

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

CHECK OUT:  NECO 2020 MATHEMATICS OBJ AND ESSAY QUESTIONS AND ANSWERS

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

(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.
=====================================

(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)
Perimeter of Arc = Φ/360 × 2πr
= 120/360 × 2 × 22/7 × 7
= 1/3 × 44 = 44/3
= 14.67cm
=====================================

(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

CHECK OUT:  NECO 2020 PHYSICS OBJ AND ESSAY QUESTIONS AND ANSWERS

(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
=====================================

(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
=====================================

(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

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

(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.)
=====================================
COMPLETED!

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