Examloaded Logo
home
Home
jamb
Jamb
waec
Waec
neco
Neco
nabteb
Nabteb
university-news
News

JAMB 2019 ASSISTANCE WAEC 2019 ASSISTANCE
jamb
Click Here to Get Our Assistance on JAMB 2019 Exams
waec
Click Here to Get Our Assistance On WAEC 2019 Exams

NECO GCE 2018 LIVE QUESTIONS
   Pass All Your subjects     
SUBSCRIBE HERE NOW!!!

Posted: 2018 Neco GCE Timetable for Nov/Dec Exam 
CLICK HERE FOR OUR ANSWER LINK
 



« | »

Neco 2019/2020 ssce Mathematics Syllabus/Brochure Questions & Answers

NECO 2019 MATHEMATICS ANSWERS 

MATHEMATICS OBJ:
1-10 CDAAEABAEC
11-20 AEDDCDCDCC
21-30 CEBDEDCBBC
31-40 CBEEECBDCC
41-50 DBCBCDDBCA
51-60 BCBDCDCCEC

(1a)
Log10(20x-10) – log10(x+3)= log10^5
Log10(20x-10)/x+3) =log10^5
20x-10/x+3 = 5
Cross multiply
20x-10 = 5(x+3)
5(4x-2) = 5(x+3)
4x-x = 3+2
3x= 5
X=3/5

(1b)
Let actual amount be #x
15% of #x = #600
15x/100 =600
x =(100/15) * 600
x = 100 * 40
x = 4,000
Actual amount = 4,000

(2a)
(X^2 Y^-3 Z)^3/4/X^-1 Y^4 Z^5
= (X^2)^3/4/X^-1 * (Y^-3)^3/4/Y^4 * Z^3/4/Z^5
= X^3/2/X^-1 * Y^-9/4/Y^4 * Z^3/4/Z^5
=X^3/2+1 * Y^-9/4-4 * Z^3/4-5
=X^5/2 * Y^-25/4 * Z^-17/4
=X^10/4 * Y^-25/4 * Z^-17/4

=(X^10/Y^25 Z^17)^1/4

(2b)
√2/k + √2 = 1/k – √2
Multiply both sides by (k+√2)(k-√2)
√2(k-√2) = k+√2
√2k-√2 = k+√2
√2k-k = 2+√2
K(√2 -1) = 2+√2
K = 2+√2/√2-1
K = -(2+√2)/1-√2
Rationalizing
K = -(2+√2) * 1+√2/1-√2
K = -(2+√2)(1+√2)/1 – 2
K = (2+√2)(1+√2)
K = 2+2√2 + √2+2
K = 4+3√2

(3)
V = Mg√1 – r²
Square both sides
V² = m²g²(1-r²)
V²/m²g² = 1-r²
r² = 1 – v²/m²g²
r = √1-(v/mg)²
If v = 15, m = 20, and g = 10
r = √1 – (15/20*10)²
r = √1 – (0.075)²
r= √(1.075)(0.925)
r = √0.994375
r = 0.9972

(4ai)
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?

(5a)
Mode = mass with highest frequency = 35kg
Median is the 18th mass
= 40kg.

(5b)
IN A TABULAR FORM

Under Masses(x kg)
30,35,40,45,50,55

Under frequency(f)
5,9,7,6,4,4
Ef = 35

Under X-A
-10, -5, 0, 5, 10, 15

Under F(X-A)
-50, -45, 0, 30, 40, 60
Ef(X – A) = 35

Mean = A + (Ef(X – A)/Ef)
= 40 + 35/35
= 40 + 1
= 41kg

(6)
(a) log2 = 0.3010
Log3 base 10 = 0.4771
(i) Log10 3.6 = Log10 36/10
= log10 36 – log10 base 10
= log10 (9×4) -1
=log10 9+log10 4 – 1
=log10 3² + log10 2² – 1
=2log10 3 + 2log10 2 – 1
= 2(0.4771) +2(0.3010) -1
= 0.9542 + 0.6020 – 1
= 0.5562

Also Read:

  JAMB HISTORY 2019/2020 RECOMMENDED TEXTBOOKS

(6aii)
Log10 0.9
= log10 9/10 = log10 9-log10 10
= 2log10 3 – 1
= 2(0.4771)-1
= -0.0458
= 1.9542

(6b)
(3√5 – 4√5)(3√5-4√5)/(3√5+4√5)(3√5-4√5)
= 45 – 60 + 80 = 60
45-60+60-80
= 5/35 = 1/7

(7ai)
T3=>a+2d=6(eqi)
T7=>a+6d=30(eqii)
Eqii minus eqi gives
6d-2d=30-6
4d=24
d=24/4
d=6
Common difference=6

(7aii)
Putting d=6 into eqi
a+2(6)=6
a+12=6
a=6-12
a=-6
(7aiii)
10th term T10=a+9d
=-6+9(6)
=-6+54
=48

(7bi)
T3=>ar²=9/2(eqi)
T6=>ar^5=243/16(eqii)
Dividing eqii by eqi
ar^5/ar²=243/16 divided by 9/2
r³=243/16*2/9
r³=27/8
r³=3³/2³
r=3/2
Putting this into eqi
a(3/2)²=9/2
a(9/4)=9/2
a=9/2*4/9
a=4/2=2

(7bii)
Common ratio r=3/2 as above

(8)
x=a+by(eqi)
when y=5 and x=19
19=a+5b(eqii)
when y=10 and x=34
34=a+10b(eqiii)
solving eqii and eqiii
a+10b=34
a+5b=19
=>5b=15
b=15/5=3
putting b=3 in eqii
19=a+5(3)
19=a+15
a=19-15
a=4

(8i)
Putting a=4 and b=3 in eqi
x=4+3y
This is the relationship between xand y

(8ii)
When y=7
x=4+3(7)
x=4+21
x=25

(8b)
3x/x+2 – 5x/3x-1 + 1/5
Find the LCM

3(3x-1)(3x)-3(x+2)(5x)+(+2)3x-1/(x+2)(3x-1)(3)
27x² = 9x – 15x²-30x + 3x²-x+6x-2/3(x+2)(3x-1)

Collect the lime terms
15x² – 34 – 2/3(x+2)(3x-1)

(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)
x-y= 4 —-(11)
x+y = 22—-(1)

x-y= 4—-(11)
____________
2y = 18
y= 18/2
y=9
Substitute y=9 in equ 1
x+9=22
x=22-9
x=13
x=13, y=9
x+y= 13+9= 22
Sum of the two number

Also Read:

  JAMB GEOGRAPHY 2019 RECOMMENDED TEXTBOOKS

(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

(12a)
Pr of Abu to pass = 3/7
Pr of Abu to fail = 1 – 3/7 = 7-3/7 = 4/7

Pr of kuranku to pass = 5/9
Pr of kuranku to fail = 1 – 5/9 = 9 – 5/9 = 4/9

Pr of musa to pass = 12/13
Pr of musa to fail = 1 – 12/13 = 13 – 12/13 = 1/13

Pr of only one of them passing is
=(3/7*4/9*1/13)+(5/9*4/7*1/13)+(12/13*4/7*4/9)
=12/819+ 20/819 + 192/819
=12+20+192/819 = 224/819
= 32/117

(12b)
10Red + 8green + 7blue = 25

(i)
pr of different colour is
Prof(RG)+(RB)+(GB)+(BG)+(BR) +(GR)
=(10/25*8/24)+(10/25*7/24)+(8/25*7/24)+(7/25*8/24)+(7/25*10/24)+(8/25*10/24)
=80/100 + 70/600 + 56/600 + 56/600 + 70/600 + 80/600
= 80+70+56+56+70+80/600
= 412/800 = 103/200

(ii)
pr of atleast one must be
=Pr[RB+BR+GB+BG+BB]
= (10/25*7/24)+(7/25*10/24)+(8/25*7/24)+ (7/25*8/24) + (7/25*7/24)
=70/600+70/600+56/600+56/600+49/600
=70+70+56+56+49
/600
=301/600

Download WordPress Themes
Premium WordPress Themes Download
Download Nulled WordPress Themes
Download Nulled WordPress Themes
lynda course free download
download karbonn firmware
Download WordPress Themes Free
download udemy paid course for free

Posted by on July 29, 2018.

Tags: , , , , , ,

Categories: NECO SSCE GCE 2019 Syllabus Questions

There is love in sharing

0 Responses

Leave a Reply


Your Comment

« | »




Theme by: EL Web Services