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Nabteb GCE 2016 Mathematics Obj and Theory Answer – Nov/Dec Expo

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Verified Maths obj
1BDDBAACABA
11ADACBDBDDB
21BCDACBACBA
31DBACABCABB
41DCABDCBCDA

WARNING: PLEASE NOTE THAT THE ANSWERS IN PICTURE FORM IS MORE ACCURATE… SCROLL DOWN FOR PICTURES FORMAT ANSWERS

SECTION A
1a)
[2(1/4)-1(3/8)]/[7/16/3/4]
therefore(9/4-11/8)/(7/16*4/3)
[(18-11)/8]/(7/12)
(7/8)/(7/12)
7/8*12/7
=3/2
=1(1/2)

1b)
a^2-b/a^2+3a+2
a^2-2^2/(a^2+a)(2a+2)
=(a-2)(a+2)/(a+2)(a+1)
=a-2/a+1
==============

2a)
without replacement
Pr(non bad)=Pr(1st good and 2nd good)
=8/12*7/11=14/33
Pr(all bad)=Pr(1st bad and 2nd bad)
=4/12*3/11
=12/12*11
=1/11

2b)
(0.054*8.19)/(0.000243)*10^6
=54*8190/243
=2*210
=420
=4.2*10^2
==============

3a)
Arc length=tita/360*2pieR
=85/360*2*22/7*5/1
=18700/2520
=7.421cm
=>7.42cm(2 d.p)
Chord length=2rsin(tita/2)
Chord AB=2*5sin(85/2)
=10*sin42.5
=10*0.6756
=6.756cm
=>6.76(2dp)

3b)
Perimeter of a minor segement=ArcL+2r
Perimeter=tita/360*2pieR+2r
=85/360*2*3.142+(2*5)
=7.367+2*5
=7.367+10
=17.367cm
=>17.37cm(2dp)
==============

4a)
4^x=8sqroot2
2*2x=2^3*2^(1/2)
2^2x=2^(7/2)
2x=7/2
x=7/4

4bi)
log36=log(4*9)
log36=log4+log9
=2log2+2log3
=2(log2+log3)
=2(0.30+0.477)
=2*0.778
=1.556

4bii)
log60=log(6*10)=log(2*3)+10g10
=log2+log3+log10
=0.301+0.477+1
=0.778+1
=1.778
==============

5ai)
Sn=n/2[2a+(n-1)d]
42=3/2(2a+2d)
84=6a+6d———–(1)
T3-T2=8
a+2d-a+d=8
3d=8
d=8/3
d=2(2/3)
from(1)
6a=84-6d
6a=84-6(8/3)
6a=84-16=68
a=68/6
a=11.333
Sum of first 12 terms
S12=12/2(2*34/3)+(12-1)8/3
=6(68/3+88/3)
=6/1*156/3
=312

5b)
log(3a+7)^1/2=1
(3a+7)^1/2(2)=10^1(2)
3a+7=100
3a=100-7
a=93/3
=31
==============

*SECTION B*

6a)
4(2^x)-129(2^x)=-32
let 2^x=A, 2^2x=A^2
Substitute
4A^2-129A+32=0
(4A^2-128A)(1A+32)=0
4A(A-32)-1(A-32)=0
(4A-1)A-32)
4A=1 or A=32
A=1/4 or A=32
A=2^x
32=2^x or 1/4=2^x
2^5=2^x or 2^-2=2^x
x=5 or x=-2

6b)
Make T the subject of x=2sqroot(T^2-x^2)/SY
(sYT)^2=[2sqroot(T^2-x^2)]^2
S^2Y^2T^2=4(T^2-X^2)
4T^2=S^2Y^2T^2+4X^2
T^2=(S^2Y^2T^2+4X^2)/4
T=1/4sqroot(S^2Y^2T^2+4X^2
==============

7a)
3y-2x/2y-3x=1/4 Ealuate y^2-x^2/6xy=-1/4
4(3y-2x)=(2y-3x)
12y-8x=2y-3x
12y-2y=8x-3x
10y=5x
x=2y
y^2-x^2/6xy
=y^2-(2y)^2/6(2y)y
=y^2-4y/12y^2
=-3y^2/12y^2
=-1/4

7bi)
Length of chord AC=2rsin(tita/2)
length of chord=2*14sin30
=28*0.500
=14.00cm

7bii)
Area of shaded portion=Circle Area-Triangle area
=Pie*r^2-1/2absin(tita)
=22/7*14*14-1/2*14*14sin60
=616-84.868
=531.132cm^2
==============

8a)
Sole 6x^2-7x-4=0
a=6 b=-7 c=-4
x=-b+_sqroot(b^2-4ac)/2a
=-(-7)=_sqroot(-7^2-4*6*-4)/2*6
=7+_sqroot(49+96)/12
=7+_sqroot145/12
=7+12.042/12 or 7-12.042/12
=1.5868 or -0.4202
=1.6 or -0.4

8b)
BAD=2*ADC
BAD+BCD=180
BCD=180-100=80
ThereforeBAD=100=2x
ADC=x=100/2=50
ADC+ABC=180
50+ABC=180
ABC+180-50
=130degrees
==============

9ai)
olume=pie*r^2*h
=22/7*11.2/2*11.2/2*18.4
=50778.112/28
=1813.504cm^3

9aii)
l^2=10^2+24^2
l^2=100+576
l=sqroot(676)
l=26

9aiii)
TSA=pie*r^2+pie*r*l
=22/7*24^2+22/7*24*26
=1810.2857+1961.14286
=3771.42856cm^3
=3771.4cm^3(1dp)
==============

10i)
Distance=tita/360*(2*pie*rcosalpha)
tita=30W+20E
tita=50 degrees
=50/360*6400*0.820/360km
=4577.42km
=4600km(2sf)

10ii)
latitude difference=tita=35+32=67degrees
Distance=tita/360*(2*pie*r)
=2692864/360
=7480.18km
=7500(2sf)

10iii)
Aerage speed=total distance/total time
=(4577.42+7480.18)/(9+11)
=12057.60/20
=602.88km/hr
=600km/hr(2sf)
=============

(1, 2 and 3a)
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(3 continuation, 4 and 5a)
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(5 continuation and 6)
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(6 continuation and 7)
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(7 continuation and 8)
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(9 and 10)
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(10 continuation and 11 *)
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(11b)
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GOODLUCK!!!

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