NECO 2018 MATHEMATICS OBJ & THEORY ANSWERS - EXAMWAVES

MATHEMATICS OBJ:
1-10: CDAAEABAEC
11-20 :BCDDCDCDAC
21-30 :CEBDEDCBBC
31-40: CBEECCBDCC
41-50: DDCBCDDBBA
51-60: BCE-CBBCEE

 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
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(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

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 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 +(2x14) =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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(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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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)
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
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(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

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