Mathematics (Theory and Obj) Answers

MATHEMATICS VERIFIED OBJECTIVE
1-10: CBBACCCBBA
11-20: ADBBAABCDB
21-30: BCADBCCAAB
31-40: CDCACDBACB
41-50: BDABCDDCAD

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Where ever you see
^ it means raise to power
/ division
* multiplication
X normal X
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SECTION A ANS ALL ====================
1a) (y-1)log4^10= ylog16^10 log4^10 (y-1)= log16^y10
4^(y-1)=16y
4^y-1=4^2y
y-1=2y
-1=2y=y
-1=y
y= -y
1b) let the actual time for 5km/hr be t for 4km/hr=30mint t 4km/hr=0.5 t distance = 4(0.5 t) =2*4t for 5km/hr, time= t distance =5t 1 4t=5t t=2hrs actual distance = 5*2=10km
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2a) 2/3(3x-5)-3/5(2x-3)=3/1 L C M =15
10(3x-5)-9(2x-3)=45
30x-50-18x 27=45
30x-18x=45 50-27
12x-23=45
12x=45 23
12x=68
x=68/12
x=34/6
x=17/3
2b) U'aS=180-(n 88) =180-n-88=92-n
also, u'TQ=18m
80degree 92-n 180-m=180degree
80 92 180-n-m=180degree
352-n-m=180degree -n-m=180-352
-n-m=-172
(n m)= 172
m n=172dgree
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3a) Tan 23.6° = h/50 Cross multiply Tan 23.6° x h/50 h = 50 tan 23.6° = 21.844m
3b) Area ofA = 1/2bh 45 = 1/2 x 10 x h 45 = 5h
h = 9cm
Area of < QTUS = 1/2 ( QT US)h
= 1/2 ( 6 16)9 = 99cm^2
4a) T6=37 T6=a (6-1)d T6=a 5d
a 5d=37 -----(eq1) s6=147
sn=n/2(2a (n-1)d) 147=3(2a 5d) 49=2a 5d
2a 5d=49 ----(eq2)
a 5d=37 ---(eq1)
2a 5d=49 ---(eq2)
a=12
4b) S15=15/2(2(12) 14d) S15=15/2(24 14d) from(1)
a 5d=37 12 5d=37 5d=37-12 5d=25 d=5 S15 = 15/2(24 14(15) S15= 15/2(24 70) S15=15/2*94 S15=15*42 S15=630
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SECTION B ANS 5 QUESTIONS ONLY
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10a) Sin x = 5/13
Using pythagoras rule
M^2 = 13^2 - 5^2 (^ means Raise to power)
M^2 = 169 - 25
M ^2 = 144
M = √144
M = 12
Hence:
Cos x - 2sin x / 2tan x
12/13 - 2(5/13) / 2(5/12) = 12/13 - 10/23 / 5/6
FIND LCM
= 12 - 10/13 / 5/6 = 12/65
10bi) Considering < LMB /MB/^2. = 12^2 - 9.6^2
/MB/^2 = 51.84
/MB/ = √51.84
/MB/ = 7.2m
From < AML
/LA/^2 = 2.8^2 9.6^2
/LA/ ^2 = 100
/LA/ = √100
/LA/ = 10m
10bii) Let the angle be. θ
FromTanθ = 9.6/2.8
Tan θ = 3.4288
θ = Tan^-1 ( 3.4288)
= 73.74°
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