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#Post#: 1999--------------------------------------------------
Maths Answerd for Waec 24 of april 2014
By: mr nad Date: April 24, 2014, 6:48 am
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(5)Pr(x)=0.5, Pr(X^1)=0.5Pr(Y^1)=0.3 , Pr(Y)=0.7(a)Pr(both
contracts)=Pr(x)*Pr(Y)=0.5*0.7=0.35(b)Pr(exactly one of
thecontracts)=Pr(X)*Pr(Y^1)
+Pr(X^1)*Pr(Y)=(0.5*0.3)+(0.5*0.7)=0.15+0.35=0.5(c)Pr(Ne
ither
of the contrasts)= pr(X^1)*Pr(Y^1)=0.5*0.3=0.15
(12a)Given,
5cos(x+8.5)-1=0C0s(x+8.5)^o=1/5cos(x+8.5)^o=0.2x+8.5
^o=Cos^-1(0.2)x+8.5^o=78.46x=78.46-8.5x=69.96^ox=70(degr
ee)
(12bii)QR^2=(PR^2)+(PQ^2)-2PR*PCOS(RPQ)P^2=32^2+24^2-2(3
2)(24)COS45(DEGREE)P^2=1600-1536*0.7071P^2=1600-1086
.1P^2=513.88P=sqrt(513.88)p=22.67kmDistance
QR=22.67km
(12biii)32/sin(tita)=22.67/sin45(deg),sin(tita)=0.99
81tita=sin^-1(0.9981)tita=86(degree)
=====================
PLS NOTE: Tita means titasign.It look like crossed
zero.r=Sqrt(1765.32) means
r=_/1765.32===================
(9ai)16-2x+6+x+19-3x+5x+8x+4x+7x+4=12545+20x=12520x=125-
4520x=80x=80/20x=4
(9aii)n(PUQnR)=16-2x+5x+6+x=22+4x=22+4(4)=38
(9bi)<WYZ=<WXY=50(deg) correspondingangle
(9bii)
<W=50(degree) AlternateangleYZE=50(degree) (<S in
thesamesegment)YEZ+50(degree)+(50degree)=180(degree)Sum
of angles in a
triangleYEZ+100(degree)=180YEZ=180-100YEZ=80(degree)====
========================(10a)X^2-3x-2x+6=12x^2-5x+6=
12x^2-5x+6-12=0x^2-5x-6=0Factorisingx^2-6+x-6=0x(x-6
)+1(x-6)=0x+1=0
or x-6=0x=-1 or x=6
(10b)PQ^2=7^2+7^2-2(7)(7)cos60(degree)PQ^2=49+49-98*0.5P
Q^2=49PQ=sqrt(49)=7cmPQ
at the centre =7/2=3.5cmArea(A)=
(pie)r^2=(22/7)*(3.5^2)=38.5cm^2
(1a) Simplify
-->(6/10*32*4/100)/ (12/10*8/1000*16/100)
-->(6*32*4/10*1000)*(10*1000*100/12*8*16)
-->(6*32*4*100/12*8*16)
-->(6*2*4*100/12* = 6*100/12
-->=50
(1b)
<AEF=<EBH (corresponding angle)
<EBH=3x(degree)
<GHC=<HBE=7x(degree)(same reason)
120(degree)+7x(degree)+3x(degree)=360(degree)
angle at a
point
120+10x=360
10x=360-120
10x=240
x=240/10=24(degree)
===============================
===============
(2a)
3sqrt(25*3)-sqrt(4*3)+sqrt(27*4)
15sqrt(3)-2sqrt(3)+2sqrt(27)
15sqrt(3)-2sqrt(3)+2sqrt(9*3)
15sqrt(3)-2sqrt(3)+6sqrt(3)=19sqrt(3)
(2b)
124n=232five
(1*n^2)+(2*n^1)+(4*n^o)=(2*5^2)+(3*5^1)
+(2*5^o)
n^2+2n+4=50+15+2
n^2+2n+4-67=0
n^2+2n-63=0
n(n-7)+9n-63=0
n(n-7)+9(n-7)=0
(n+9)(n-7)=0
n=-9 or n=7:. n=7
===============================
==
(3a)
Let p=1/x ,q=1/y
p+q=5 ---->equ(1)
q-p=1 ---->equ(2)
p+q=5 ---->equ(1)
p+q=1 ---->equ(2)
Add equ(1) and equ(2)
2q=6 , q=6/2=3
Substitute q=3 into equation(1)
p+q=5
p+3=5
P=5-3
p=2
Recall p=1/x , 2=1/x, x=1/2
q=1/y, 3=1/y , y=1/3
Therefore: x=1/2 , y=1/3
(3b)
Where the road is good
72=x/0.75
x=72*0.75
x=54km
Where the road is bad
48=x/0.75
x=48*0.75
x=36km
Number of kilometers of good surface is 54km
===============================
===============
(4a)
Area(A)= (tita/360*pieR^2) -(1/2R^2Sin(tita))
504=((90/360)*22/7*r^2)-(1/2*r^2*sin90)
504=(0.25*3.142*r^2)-(0.5r2)
504= 0.7855r^2 - 0.5r^2
504= 0.2855r^2
r^2= 504/0.2855
r^2=1765.32
r=Sqrt(1765.32)
r=42cm
===============================
===============
(4b)
Draw the diagram
<RQS=66(degree) {base angles of isoceles)
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