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NECO 2020/2021 Essay/Obj – General Mathematics Expo Answer is out
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NECO 2020/2021 Essay/Obj – General Mathematics Expo Answer is out
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Previous 2018 Neco Mathematics Answers
Monday 4th June 2018
MATHEMATICS OBJ:
110 CDAAEABAEC
1120 AEDDCDCDCC
2130 CEBDEDCBBC
3140 CBEEECBDCC
4150 DBCBCDDBCA
5160 BCBDCDCCEC
MATHEMATICS OBJ:
110 CDAAEABAEC
1120 AEDDCDCDCC
2130 CEBDEDCBBC
3140 CBEEECBDCC
4150 DBCBCDDBCA
5160 BCBDCDCCEC
10a)
Obtuse <BOD + Reflex<BOD = 360degrees (angle at a point)
105 + reflex<BOD = 360degrees
Reflex <BOD= 360  105
=255°
Now 2w = reflex<BOD(angle at centre = twice angle at circumference)
2w =255°
W = 255/2 =127.5°
Also 2x = obtuse<BOD(angle at centre = twice angle at circumference)
2x = 105°
X = 105/2 = 52.5°
Now EDF = y(base angles of an isosceles triangle)
BED=X=52.5°(angles in the same segment)
EFD+EDF=BED (sum of interior angles of a triangle equal exterior angle)
Y+y = 52.5°
2y = 52.5°
Y = 52.5°/2
=26.25°
(10b)
Draw the diagram
Opp/adj = TanR
TB/BR = TanR
100/BR = Tan60°
BR = 100/tan60
BR = 100√3
BR = 100√3 * √3/√3
=100√3/3m OR 57.7m
11a)
x+y/2 =11
x+y= 11*2
x+y= 22 (1)
xy= 4 (11)
x+y = 22(1)

xy= 4(11)
____
2y = 18
y= 18/2
y=9
Substitute y=9 in equ 1
x+9=22
x=229
x=13
x=13, y=9
x+y= 13+9= 22
Sum of the two number
(11b)
(6x + 3) dx
(6x + 3)dx
(6x +3)^6  (6x + 3)^1
(6 x + 3)^5
(7776x^5 + 243)
38,880x/6 + 243
6480 x^6 + 243x
9(720x^6 + 27x)
(11c)
y = x² + 5x  3 (x = 2)
y = 2² + 5(2)  3
y = 4 + 10  3
y = 14  3
y = 11
Gradient of the curve = 11
Correct 1a)
Log 10(2010)log10(+3)=log105
(2010/+3)=log10 =5
2010/+3=5
5(+3)=2010
5+15=2010
15+10=205
25=15*
=25/15
*=5/3=1 2/3
1b)
Discount percent =15%
Discount amount =#600
Actual amount paid on the article =?
Original amount on the article =
15%=#600
15/100 =600
15=600100
15*=60000
*=60000/15
*=#4,000
Therefore actual amount paid on the article
=#4,000#600
=#3,400
Actual amount paid on the article =#3,400
No 4(i)
length of Arc of the sector
Titter= 72?, r = 14cm
L= titter / 360 x 2 pie r
==> L= 72/360 x 2 x 22/7 x 14
=44352/2520 = 17.6cm
(ii) perimeter of the sector
Perimeter = titter/360 x 2 pie r + 2r = 17.6 +(2×14) =17.6+28= 45.6cm
iii) Area of the sector
Area = Titter/360 x pie r? =
72/360 x 22/7 x (14)? = 72 x 22 x 196/2520
Area= 310464/2520 = 123.2cm?
Obtuse <BOD + Reflex<BOD = 360degrees (angle at a point)
105 + reflex<BOD = 360degrees
Reflex <BOD= 360  105
=255°
Now 2w = reflex<BOD(angle at centre = twice angle at circumference)
2w =255°
W = 255/2 =127.5°
Also 2x = obtuse<BOD(angle at centre = twice angle at circumference)
2x = 105°
X = 105/2 = 52.5°
Now EDF = y(base angles of an isosceles triangle)
BED=X=52.5°(angles in the same segment)
EFD+EDF=BED (sum of interior angles of a triangle equal exterior angle)
Y+y = 52.5°
2y = 52.5°
Y = 52.5°/2
=26.25°
(10b)
Draw the diagram
Opp/adj = TanR
TB/BR = TanR
100/BR = Tan60°
BR = 100/tan60
BR = 100√3
BR = 100√3 * √3/√3
=100√3/3m OR 57.7m
11a)
x+y/2 =11
x+y= 11*2
x+y= 22 (1)
xy= 4 (11)
x+y = 22(1)

xy= 4(11)
____
2y = 18
y= 18/2
y=9
Substitute y=9 in equ 1
x+9=22
x=229
x=13
x=13, y=9
x+y= 13+9= 22
Sum of the two number
(11b)
(6x + 3) dx
(6x + 3)dx
(6x +3)^6  (6x + 3)^1
(6 x + 3)^5
(7776x^5 + 243)
38,880x/6 + 243
6480 x^6 + 243x
9(720x^6 + 27x)
(11c)
y = x² + 5x  3 (x = 2)
y = 2² + 5(2)  3
y = 4 + 10  3
y = 14  3
y = 11
Gradient of the curve = 11
Correct 1a)
Log 10(2010)log10(+3)=log105
(2010/+3)=log10 =5
2010/+3=5
5(+3)=2010
5+15=2010
15+10=205
25=15*
=25/15
*=5/3=1 2/3
1b)
Discount percent =15%
Discount amount =#600
Actual amount paid on the article =?
Original amount on the article =
15%=#600
15/100 =600
15=600100
15*=60000
*=60000/15
*=#4,000
Therefore actual amount paid on the article
=#4,000#600
=#3,400
Actual amount paid on the article =#3,400
No 4(i)
length of Arc of the sector
Titter= 72?, r = 14cm
L= titter / 360 x 2 pie r
==> L= 72/360 x 2 x 22/7 x 14
=44352/2520 = 17.6cm
(ii) perimeter of the sector
Perimeter = titter/360 x 2 pie r + 2r = 17.6 +(2×14) =17.6+28= 45.6cm
iii) Area of the sector
Area = Titter/360 x pie r? =
72/360 x 22/7 x (14)? = 72 x 22 x 196/2520
Area= 310464/2520 = 123.2cm?
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