Nabteb 2016 MATHEMATICS Obj And Theory Answers

MATHEMATICS OBJ:
1DBCBAACBAC
11AADDACDDAC
21CDBCBBDCDC
31ACDADABCBA
41DBCABCABBD
NABTEB MATHEMATICS ANSWERS
1a.
9^11.259^1.75/9^2
= 9^11.25 1.75/9^2
=9^13/9^2
=9^13-2
=9^11
1b
(i) log2 = log10/5
=log10-log5
=1-log5
=1 0.699
=0.30
ii) log15=log3
5
= log3 log5
=0.479 0.699
=1.17
2a)
1/2x + 3/4y = 1 ——– (1)
x-1/2y=-1 ————(2)
multiply equation (1) through by 4
(1/2x + 3/4y) 4 = 14
2x+4y=4 ————-(3)
multiply equation (2) through by 2
(x – 1/2y)2 = -1
2
2x-y=-2 ————–(4)
solve equ (3) and (4) simultaneously
2x+3y=4
2x-y=-2
0+4y =6
4y/4 = 6/4
y=1.5
subtitute y=1.5 into eqa (4) to solve for x
2x-y=-2
2x-(1.5)=-2
2x-1.5=-2
2x=-2+1.5
2x/2 = -0.5/2
x= -0.250
2b)
Volume (v) = 1/3a^2h
given that
V= 180cm^3
h= 15cm
a = ?
180 =1/3a^215
180=15a^2/3
180/5=5a^2/5
a^2= 36
a = root36
a = 6cm
the lenght of the side of the base is 6cm
3a)
Tn= a+(n-1)d
T3= 26
T10= 26+28
T3=a +(3-1)d=26
a+2d=26———-(1)
T10=a+(10-1)d=54
a+ 9d=54———(2)
solve equa(1) and (2) simutaniously
a+2d=26
a +9d=54
-7d/7 = -28/-7
d= 4
subtitute d=4 into equa (1)
to solve for a
a+2d=26
a+24=26
a+8=26
a=26-8
a= 18
i)the first term (a) from above is 18
ii) the common difference (d) from above is 4..
3b)
sum of angle in a polygon
=(n-2)180
where A = 5
=(5-2)180
=3
180 = 540
the sum of polygon is 540
therefore x+x+8+2x-1+x-3
3x =540
8x=540+4
8x/x = 544/8
x= 68
4a.
t=1 x/1-x
(1-x)t=1 x
t-xt=1 x
t-1=(x xt)
t-1=x(t-1)
x=x-1/t 1
t=8.5
x=8.5-1/8.5 1
x=7.5/9.5
x=15/19
4b. 3:4:5
Shortest height = 10.5cm
Let the full length be x
3:4:5
3 4 5=12
3/124=10.5
x=10.5
12/3
x=10.54=42cm
So the longest length=5/12
42cm =35/2
=17.5cm
5a.
(709.18)^2 – (290.82)^2
=(709.18)^2/(290.82)^2
= 502936.27/84576.27
=5.94

b.
Isosceles (trapezium) Area = 1/2(ab)h
=1/2
abh
=1/281212
=1/2
1152
=576cm^2
6a)
2x/3=6-x / 1+x
2x(1+x)=3(6-x)
2x+2x^2=18-3x
2x^2+2x+3x-18=0
2x^2+5x-18=0
2x^2-4x+9x-18=0
(2x^2 -4x)+99x-18)=0
2x(x-2)+9(x-2)=0
(x-2)(2x+9)=0
x-2=0 or 2x+9=0
x-2=0 or 2x+9=0
x=2 or 2x=-9
x=2 or x=-9/2
(6b)
y=2x-1—(1)
y=2x^2 – 3x-4–(ii)
subt for y inj eqn (ii)
2x^2- 3x-4=2x-1
2x^2-3x-2x-4+1=0
2x^2-3x-2x-4+1=0
2x^2-3x-2x-4+1=0
2x^2-5x-3=0
2x^2+x-6x-3=0
(2x^2 + x)- (6x-3)=0
x(2x+1)-3(2x+1)=0
(2x+1)(x-3)=0
(2x+1)(x-3)=0
2x+1=0 or x-3=0
x= -1/2 or x=3
but y=2x-1
when x=3
y=2(3)-1
y=6-1
y=5
when x=-1/2
y=2(-1/2)
y=-1-1
y=-2
(7a)
2/x=5/x-2
multiply both sides by x
x2/x=x5/x – x2
2=5-2x
5-2x=2
-2×2-5
-2x=-3
x=-3/-2
x=3/2
(7c)
vol=1/3
baseheight
=1/3 * 1/2 * 8
5sin30* 14
=20/3 * 1/2 * 14
=46.67
=47cm^3(nearest whole number)
(8a)
U={1,2,3,4…,20}
A={1,2,3,4,….,15}
B={2,4,6,….20}
C={2,3,5,7,11,13,17,19}
(i)
AnBnC
{1,2,3,…15}n{2,4,6,…20)nC
{2,4,6,8,10,12,14}n{2,3,5,7,11,13,17,18}
={2}
(ii)
A^c U (B-C)
{16,17,18,19,20}U{3,4,5,6,7,8,9,10,11,12,13,14,16,17,18,19}
{3,4,5,6,7,8,10,12,13,14,16,17,18,19,20}
(8b)
(i)Pr(T,3)=1/12
(ii)Pr(H and odd no)=3/12
(iii)Pr(H and prime no)=3/12 =1/4
9.
A=1/2bh
45= 1/2
bh
h= 45/1/2b
h= 45/1/2
28
h=45/14
h=3.21m
While b=A/1/2h
b= 45/1/2
3.21
b=45/1.60
b=28.12cm
9b.
Area of a circle is given 68 whole no 4/7cm^2
i. Diameter =A/2=616/2 =308
ii. Circum = Πr^2
=22/7308^2
=3.14
94864
=297872.9
=2.910^6
(14a)
rentable value=N10,000,000
annual rates=75k
10,000,000
=750,000,000
=N7,500,000
(14b)
P=N420,000, R=7 1/2%, n=3yrs
A=P(1+ R/100)^n
=420,000(1+0.075)^3
=420,000(1.075)
=N521,764.69k
C.I=A-P
=521,764.69-420,000
=N101,764.69

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