Waec 2016 Mathematics Obj And Theory Answers – May/June Expo - Gidifans

You Are Here: Home » Exam - Waec » Waec 2016 Mathematics Obj And Theory Answers – May/June Expo - By Mr. Maraj On April 20, 2016 At 4:42 pm

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Mathz-obj
1DCBCABBADA
11CCBCABABBA
21AACCDCDABC
31BCDCDCBABA
41ACADBBDADD

(1a)
(0.09*1.21)/(3.3*0.00025)
=(9*10^-2*1.2*10^1)/(3.3*25*10^-5)
=(1089^10^-4)/(82.5*10^-5)
=1.32*10^(-4+6)
=1.32*10^2

(1b)
p=5600,T=3,A=1200+5600=6800
A=P(1+R/100)^3
6800=5600(1+R/100)^3
(1+R/100)^3=6800/5600
(1+R/100)^3=1.214
1+R/100=cuberoot(1.214)
1+R/100=1.0667
R/100=1.0667-1
R/100=0.0667
R=0.0667*100
R=66.7%
==================

2a)
7(x+4)-2/3(x-6)<_2[x-3(x+5)] 7x+28-2x/3-4<_2[x-3x-15] 24+7x-2x/3<_2x-6x-30 7x-2x/3-2x+6x<_-30-24 11x-2x/3<_-54 33x-2x<_-162 31x<_-162 x<_-162/31 2b) x = tricycles and y = taxi cabs 2x+4y=6b ... (1) x+y=20 ... (2) from equ..(2) y=20-x substitute for y in equ...(1) 2x+4(20-x) = 66 2x+80-4x=66 80-66=4x-2x 14/2 = 2x/2 x=7 number of tricycles = 7 ================ (3a) CLICK HERE

(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 of triangle=1/2bh where h=6
:. 1/2*6/1*b/1=36
6b/2=36/1
6b=72
B=72/6
Base of angle PSR=12cm
Since |TS| //PQ
QR=PR-PQ
QR=12-8
QR=4cm

5b)
Draw d triangle
From angle ABC, to get |BC|
Tan60/1=10cm/|BC|
|BC|=10.65/tan60
|BC|=10.65/1.7321
=6.1486 =6.15m
Hence |BC|=|DE|
From angle AED, to get |AE|
Draw d diagram
|AB|=6.15tan45
6.15*1
=6.15m
There4, th height of the tree =10.65-6.15
=4.50m
=============

9a)
CLICK HERE
x 62 63 64 65 66 67 68
tally 1 iiiii iiiii iiiii ii iiii iiii iiii iiiii iiiii iiiii i ii
freq 1 5 12 14 10 6 2 total = 50
fx 62 315 768 910 660 402 136 tot = 3253
x-x- -3.06 -2.06 -1.06 0.06 0.94 1.94 2.94
(x-x-) 9.3636 4.2436 1.1236 0.0036 0.8836 3.7636 8.6436
f(x-x)2 9.3636 21.218 13.4832 0.0504 8.836 22.5816 17.2872 tot = 92.82

bi) mean = EFX/EF = 3253/50 =65.06

bii) Standard deviation = Square root of EF(x-x)2/Ef

= Square root of 92.82/50
= square root of 1.8564
= 1.36
================

7a)
x-2/4 :x+2/2x
Cross multiply
2x(x-2)=4(x+2)
Opening the bracket
2x^2-4x = 4x+8
Collect like terms together
2x^2-4x-4x-8=0
2x^2-8x-8=0
Solving with completing the square method
2x^2-8x=8
Divide through with the coefficient of x^2
2x^2/2-8x/2=8/2
X^2-4x=4
Half of coefficient of x
(-4 x 1/2)^2 = (-2)^2
X^2-4x+(-2)^2=4+(-2)^2
X^2-4x+(-2)^2=4+4
(x-2)^2=8
X-2:square root 8
X=2 + or – squar root 8
X=2+ square root 8 or x=2- square root 8
X=2+2.828=4.828
X=2-2.828= -0.828
X=4:83 or – 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

13a)
(x1,y1) =(2,-3)
for 2x + y =6
y=-2x +6
compare y = mx +6
m2 = -2
for parallel lines, m1=m2
: y-y1/x-x1 = m
y–3/x-2 =-2/1
y + 3 = -2(x -2)
y + 3 = -2x + 4
y = -2x + 4 -3
y = -2x +1

13b)
T={2,3,5,7}
I. X angle Y= (x+y+xy) mod 8
– |2|3|5|7
2|4|5|7|1
3|5|6|0|2
5|7|0|2|4
7|1|2|4|6

13bii)
I)
2angle(D angle 7)
2 angle 4, since (5 angle 7)=4
2 angle 4=6
II)
2 angle n =5 angle 7
Since 5 angle 7=4
:. 2 angle n=4
And n=2 because
2angle2=4 5 angle 7
==================

11a)
3p+4q/3p-4q* 2/1 find p:q
1(rp+4q)= 2(3p-4q)
3p+4q=6p-8q
Collect like terms
3p-6p=-8q-4q
-3p=-12q
Ther4 p:q = -3p/-3=-12q/-3
P=4q, p=4, q=1
P:q= 4:1

11b)
Ts=pq-(2+2)
Ts=pq-4
Circumference of Ts= 2pie(TS)
11bi)
Ts=pq-(2+2)
Ts=pq-4
Circumference of Ts= 2pie(TS)
=2pie(pq-4)
Perimeter=pq+4+4+2+2+2pie(pq-4)
34=pq+12+2pie(pq-4)
34-12=pq+2piePQ-8pie
22=PQ(1+2pie)-8pie
22=pq(1(2*22/7)-8*22/7
22=pq(1+44/7)-176/7
22=PQ(7+44/7)-176/7
22+176/7 =pq(51/7)
154+176/7=Pq(51/7)
330/7=pq(51/7)
330/51=pq
Pq=6.4706
11bii)
TS=PQ-4=6.4706-4=2.4706m
Area of semi circle TS=pieD
=7.7626m^2
Area of rectangle PqRU=|pu|*|Pq|
4**6.4706=25.8824
Area of the cross section=area of rectangle-area of semi circle
=25.8824-7.7626
=18.1198m^2
=========

6a)
Sn=n/2(2a+(n-1)d)
A=1, d=2, n=n
Sn=n/2(2*1+(n-1)2)
N/2(2+2,-2)
=n/2*2=n^2

6b
CLICK HERE
n(E)=95, n(BnT)=7
N(U)=95, n(BnTnC’)=7n(B’nTnC)=3,n(BnTnC)=8
n(BnT’nC’)=x, n(BnT’nC),n(b)=47, n(T)=30
i)
Draw d venn diagram
ii)
To find x,
X+x+7+8=47
2x=47-15=32
X=32/2
=16
iii)
No of those who travelled by at least two means
7+3+16+8=34
==========

8a)
KGF = 110
2x = r = y
if KGF = 110, MGH = 180 – 110
MGH =70degree
HMG = Y
MHG = 180 – Y – 70
MHG = 110 – Y
LHJ = 180 -(110 -Y)
=180 – 110 + Y = 70 + Y
nOW : X + r + 70 + y = 180
2x = r + – y
if 2x = r
x = r/2 =y
😡 + 2x + 70 + 2x = 180
5x = 180 -70 = 110
x = 110/5 = 22degree

8b)
let A collection of boy be X and that of girl be y
Tot collected by boy
600 + Y = x
Tot collected by girl
x – 600 = y
x- y =600 –(i)
Ave collection for boys
x/10 = 100+y/12(multipled by 120)
x/10 * 120 = 100 * 120 + y/12 *120
12x = 12000 + 10y
12x -10y= 12000 —(ii)/x – y = 600 *12
12x – 10y =12000 * 1
12x -12y = 7200—-(iii)
12x – 10y = 12000
= -2y = -4800
y = 4800/2 = 2400
y = 4800/2 = 2400
subt 2400 for y in eq(i)
600 + y = x
600 + 2400 = x
x = 3000,
boys collection = #3000
girls collection = # 24000
total collected = #54000

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