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Wednesday, June 12, 2019

2018 FREE NECO SSCE MATHEMATICS RUNZ | MATHS QUESTIONS AND ANSWERS

Place Advert Here..
Monday 4th June 2018
OBJ – General Mathematics – 10:00am – 11:45am
Essay – General Mathematics – 12:00noon –2:30pm
OBJ & Essay – Basic Electricity – 3:15pm – 5:30pm

+++++++++++++++++++++++++++++++++++

ONLY SUBSCRIBERS GET ANSWERS ON TIME

MATHEMATICS OBJ:
1-10: CDAAEABAEC
11-20:ACDDCDCDAC
21-30: CEBDEDCBBC
31-40: CBEECCBDCC
41-50: DDCBCDDBBA

51-60:BCEDCBBCEE

(1a)
Log 10(20*-10)-log10(*+3)=log105
(20*-10/*+3)=log10 =5
20*-10/*+3=5
5(+3)=20-10
5*+15=20*-10
15+10=20*-5*
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*=600*100
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

(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

(4)
Draw the diagram 

(i) Arc length = Tita/360*2πr
= 72/360*2*22/7*14
=1/5*44*2
=88/5
=17.6cm

(ii) Perimeter of Sector = arc length +2r
=17.6+2(14)
=17.6+28
=45.6cm

(iii) Area of sector = Tita/360*πr²
=72/360*22/7*14/1*14/1
=1/5*22*2*14
=616/5
=123.2cm2

(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

(8a)
x=a+by(eq i)
when y=5 and x=19
19=a+5b(eq ii)
when y=10 and x=34
34=a+10b(eq iii)
solving eq ii and eq iii
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

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

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

(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

(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


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