**Waec 2016 Mathematics Obj And Theory Answers – May/June Expo**- By Franky 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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