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Waec 2016 Mathematics Objective And Theory Answers – May/June Expo

Waec 2016 Mathematics Objective And Theory Answers – May/June Expo

VERIFIED WAEC 2016 MATHEMATICS QUESTIONS AND ANSWERS

Thursday 21st April 2016
MATHEMATICS 2 (ESSAY) 9.30AM – 12.00PM
MATHEMATICS 1 (OBJ) 2.00PM – 3.30PM

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Mathz-obj
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11CCBCABABBA
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31BCDCDCBABA
41ACADBBDADD

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4)

6)

7)
(3b)
h=7cm
T=462cm^2
T=pie sqr + 2 pie rH
462= pie( r ^2 +2*7*r)
462/3.142
=r sqr= =14r
r sqr+ 14r=147
r sqr +14r- 147=0
r= 14+-( (sqr(14))-4*1*147)/2*1
r=-(14 +-(196-4))/2
r=(-14+_12)/2
r=(-14+12)/2 or
r=2/2
=1cm

==================================

(4a)
Pr(3)=x/total
total=25+30+x+28+40+32
=155+x
0.225=x/155+ x
=34.875+0.225x=x
x-0.225x=34.875
=0.775x=34.875 degree
x=34.875/0.775
=45

(4b)
Pr(even)=30+28+40+32
=90
Pr(even)=90/200=9/20
Pr(prime no)=25=30+45+40
=140
Pr(prime no)=140/200=14/20
Pr(even or prime)=9/20 + 14/20

===========================

(5a)
Area PSR=1/2*PR*SM
PR=8+6=14cm
36=1/2 * 14* 6
36=1/2 * 14* 6 *QR
QR=8.6cm
(5b)
diagram
sin45 degree= 10.65/h
h=10.65/sin 45 degree
=10.65/0.7071
h=15.06cm

===============================

(6a)
first term9a)=1.0=1
common diff(d)=3-1=2
last term(L)=101
L=a+(n-1)d
101=1+(n-1)*2
(n-1)*2=100
n-1=100/2
n-1=50
n=501+1
n=51
sum(s)=n/2(2a+(n-1)d)
S=51/2(2*1+(51-1)*2)
s=25.5(2+100)
S=25.5*102=2601

(6b)
UC=95
Bus & Train(B&T)=7
Train & Car( T & C)=3
B&T&C=8
Bus only(x)=B&C only
Bus(b)=47
train(T)=30
(draw the venn diagram)
Bus only= 47-(7+8+x)
x=47-15+x
47=x+x+7+8
47=2x+15
2x=47-15
2x=32
x=32/2
=16
Train only= 30-(7+8+2)
=30-18=12
(6bi)=x=16

(6bii)
number who travelled by at leasttwo means
=7+1+3+8
=19

==================================

(7a)
given;
(x-2)/4=(x+2)/2x
2x*(x-2)=4*(x+2)
2x^2 -4x= 4x+8
2x^2-4x-4x-8=0
2x^2-8x-8=0
x^2-4x-4=0
x^2-4x=4
x^2-(4* 1/2)=4+(2)^2
(x-2)^2=4+4
(x-2)^2=8
(x-2)= sqr root(8)
x-2=+-2.83
x=+-2.83
x=+_2.83+2
x=-2.83+2
x=4.83 or x=-0.83

(7b)
P+Q+S=180 degree(sum of angles in a triangle)
52 degree+ 90 degree+ S= 180 degree
142+ S=180
S=180-142
S=38 degree
(i)52+38+SQT
=180 degree
SQT=180-90
SQT=90 degree
(ii)PQT= 52 degree

==============================

(8a)
diagram
z=110 degree(sum of angles on a straight line)
z=180-110
z=70 dgree
w degree= 70 degree
x+r+w degree= 180 degree(sum of angle in a triangle)
x+2x+70 degree=180 degree
3x=110
x=110/3=36.7 degree

(8b)
B=10, G=12
TB=600.00
TB=TG+600
mean of TB=100+ mean of TG
where TB= total collection of boys
TG=total collection by girls
mean of TB= ET/B
mean of TB= (TG+600)/10
100+ mean of TG= (TG+600)/100
mean of TG= mean of TG/12
100+ mean of G/12= (TG+600)/10
(100+TG)/12=(TG+600)/10
10*(1200+TG)=12*(TG+600)
=12000+10TG=12TG+7200
=12TG-10TG=12000-7200
2TG=4800
TG=N2400
TB=TG+600
TB=600+2400
Tb=N3000

=============================

(11a)
Given (3p+4q)/(3p-4q) = 2
3p+4q=2*(3p-4q)
3p+4q=6p-8q
3p-6p=-4q-8q
=-3p=-12q
p/q=-12/-3
p/q=4/1, P=4, q=1
p:q=4;1

(11b)
(i)
perimeter(p)=PQ+PU+QR
PQ=UT+TS+SR
34=PQ+UR+4+4
PQ=UR
34-8=2PQ
PQ=26/2=13cm
(ii)
13=2+TS+2
TS=13-4=9cm
radius(r)=9/2=4.5cm
Area(A)= pie r sqr+22/7*(4.5)^2
=445.5/7
63.64cm^2

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Posted by on April 19, 2016.

Categories: Uncategorized

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One Response

  1. Where is it

    by mulanie Iceking on Apr 21, 2016 at 8:54 am

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