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Thursday 27th June Paper I: Objective - General Mathematics 10.00 Am - 11.45 Am Paper II: Essay - General Mathematics 12.00 Pm - 2.30 Pm

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Verified Maths OBJ:

1-10 BCABCADDAA
11-20 DDAACDBACA
21-30 ABAACECADB
31-40 BCAEBEBBCD
41-50 EDEDDECDBA
51-60 CADAEDDEAB



(1)
n(AUBUC)=n(A)+n(B)+n(C)-n(AnB)
-n(BnC)+n(AnBnC)
55=21+24+23-6-8-5+2x+x
55=49+3x
3x=6 x=2

(1i)
21-6-5-x = 21-11-2
physics only=8

(1ii) 6+5+8= 19

(2a) Using almighty formular
x= -b+sqrt(b^2- 4ac)/2a
=5+sqrt(25-4(2)(3))/2(2)
=(5+1)/4 or (5-1)/4
= 1 or (1)1/2

(2b)
Volume of cylinder= pieR^2h
352=22/7*4*4*h
h=7

(3a)
RD^2=(37)^2- (12)^2
= 1369-144
RD^2=1225
RD=sqrt(1225)
=35cm Area= 35*12=420cm^2

(3b)
((2p+q)/(p-3q))^1/2=h
sqrt(h)=(2p+q)/(p-3q)
Psqrt(h)- 3qsqrt(h)=2p+q
Psqrt(h)-2P= q+3qsqrt(h)
P(sqrt(h)-2)= q+3qsqrt(h) P=q+3qsrt(h)/sqrt(h)-2

(4a)
(n-2)180=
xn+x=180
x(n+1)
180n-360=xn+x
180n-360=x(n+1) x=180n-360/n+1

(4b)
(x+5)/3-(2x-1)/5=5/6
30((x+5)/3) - 30((2x-1)/5)=
30(5/6)
10x + 50-12x+6=25
56-2x=25 -2x=-31
x=15.5

(5a)
perimeter of sector= (tita/360)
2pieR+2r
=2(9)+120/360*2*22/7*9
=18+18.86
=36.9cm

(5b)
150/h=tan30
h=150/tan30
h=259.8 approx 260

(6a) Snth= n/2(2a+11d) -192=24a+132d
-96=12a+66d -32=4a+22d -16=2a+11d ----equ
(1) -252=18(2a+17d)/2 -504=18(2a+17d)
-28=2a+17d-------equ(2) (-16=2a+11d) -
(-28=2a+17d) 12=-6d d=-2

(6b) from equ(1) -16=29+11(-2) -16=29-22
2a=6 a=3

(6c) Tnth= a+(n-1)d =3+(15-1)(-2) T15th= -25
T4th=a+(4-1)d =3+3(-2) =-3 Product of 15th
and 4th= -3*-25=75

(7ai)
STB= 180-75-40
(sum of angles on a straight line)
STB= 65
SRP= 180-65(opposite angles are
complementary)
=115degree

(7aii)
SQP= 115(angles substended from the
same circumference are equal)

(7aiii)
SPQ=180-115-12 (sum of angle in a
triangle)
=53degree

(7b)
C=R+KT
2800=R+K(6)
2800= R+6k ---------equ(1)
- 3600= R+10K------equ(2)
-800 = -4k
k=200
From eqn(1)
2800=R+200(6)
2800=1200R
R=2(1/3)
at 4hours,
C= 2(1/3) + 200(4)
= 800 +2(1/3)
=802(1/3)
=#802.3





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