2018 free neco mathematics questions and answers expo/runz

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

Mathematics Theory
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

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4i)
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

4ii) perimeter of the sector
Perimeter = titter/360 x 2 pie r + 2r = 17.6 +(2×14) =17.6+28= 45.6cm

4iii) 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

6 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

7a)

A) T3=6 & T7 =30

I)common difference using Tn=a+(n-1)d

In the 3rd term; n =3

=>T3=a+(3-1)d=6

=>a+2d=6 equation (1)

In the 7th term ;n =7

=> T7=a+(7-1)d=30

=>a + 8d=30 equation (1) & (2)

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Simultaneous

A+2d =6 equation (1)

A+ 8d=30 equation (2)

0+(1-6d) =-24

=> -6d= -24

=>d = -24/-6 =4

II) first term put d=4 into equation

11a)
x+y/2 =11
x+y= 11*2
x+y= 22 —eq(i)
x-y= 4 —-eq(ii)
x+y = 22—-(i)

x-y= 4—-(ii)
____________
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

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

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(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
=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

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