*2019-NECO-MATHEMATICS-ANSWERS*
*===============================================================*
*MATHS-OBJ*
1-10: DBBCBDEDAC
11-20: DCDAECDEDC
21-30: CBAAECDCAD
31-40: CCACCBDDCC
41-50: CDDDCABDBA
51-60: BCBDBCAADD
*===============================================================*
*KEEP REFRESHING THIS VIP*
*===============================================================*
(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.
*===============================================================*
(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
SEE THE IMAGE(3)
(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(approximation.)
Standard deviation = √Variance
= √72.84
= 8.533
= 8.53 (approximation.)
SEE IMAGE(4)
*===============================================================*
(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)
Typing
SEE IMAGE(7)
*===============================================================*
(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
SEE IMAGE(
*===============================================================*
(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
SEE IMAGE(10)
*===============================================================*
2019 NECO Mathematics Objective and theory Questions and Answers is available at 9Jatechs, get verified Mathematics Objective and theory neco questions and answers midnight before the commencement of the examination.
Monday 24th June
Note:9Jatechs 2019 NECO Mathematics Objective and theory Questions and Answers is N800 MTN Card
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SEE ALSO : 2018 Nov/Dec Neco Gce Physics Practical Questions and Answers | Exam Runs
ALSO SEE: Physics Objective and Essay 2018 Neco Gce Questions and Answers
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*===============================================================*
*MATHS-OBJ*
1-10: DBBCBDEDAC
11-20: DCDAECDEDC
21-30: CBAAECDCAD
31-40: CCACCBDDCC
41-50: CDDDCABDBA
51-60: BCBDBCAADD
*===============================================================*
*KEEP REFRESHING THIS VIP*
*===============================================================*
(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.
*===============================================================*
(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
SEE THE IMAGE(3)
(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(approximation.)
Standard deviation = √Variance
= √72.84
= 8.533
= 8.53 (approximation.)
SEE IMAGE(4)
*===============================================================*
(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)
Typing
SEE IMAGE(7)
*===============================================================*
(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
SEE IMAGE(
*===============================================================*
(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
SEE IMAGE(10)
*===============================================================*
2019 NECO Mathematics Objective and theory Questions and Answers is available at 9Jatechs, get verified Mathematics Objective and theory neco questions and answers midnight before the commencement of the examination.
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Time and Date for Mathematics Objective and theory 2019 NECO Examination.
Monday 24th June
- Paper III: Objective – General Mathematics ‹‹›› 10:00am – 11:45am
- Paper II: Essay – General Mathematics ‹‹›› 12:00noon – 2:30pm
Recommended: See 2019 NECO Official Exam Timetable >> 9Jatechs
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