2015 WAEC GCE MATHEMATICS ANSWER IS READY.GET IT NOW

2015 WAEC GCE NOV/DEC

EXAM DAY: SATURDAY

PAPER: MATHEMATICS

Math obj
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31ADBABACDB

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1a)
1/2log 25/4- 2log10 4/5+log10 320/125
Log(25/4)^1/2-log(4/5)^2+log10 320/125
Log10 sqrt25/4-log16/25+log10 320/125
Log10 5/2-log10 16/25+log10 320/125
Log10 5/2+log10 320/125-log10 16/25
Log10 [5/2*320/125÷16/25]
Log10 [5/2*320/125*25/16]
Log10 10=1

1b)
% income= 20%
Grant per land =GHC€15.00
Total population from
2003-2007
=1.2*1.2*1.2*1.2*3000=6220.8
Total grant=population*grant per head
=6220.8*15
=ghc€93312
Total grant=GHC€93312


2a)
1/x+1/x+3=1/2
L.C.M=x(x+3)
X+3+x/x(x+3)=1/2
2(2x+3)/x(x+3)=1/2
2(2x+3)=x(x+3)
4x+6=x^2+3x
X^2+3x=4x+6
X^2+3x-4x-6=0
X^2-x-6=0
(X^2-3x)+(2x-6)=0
X(x-3)+2(x-3)=0
(X+2)(x-3)=0
X=-2 or x=3

2b)
Let d bag of rice be X
Let d bag of beans be Y
X+Y=17--->(I)
2250x+2400y=39600--->(2)
From eqtn (I)
X=17-y
Using elimination method to eliminate Y
2250x+2250y=38250--->(3)
2250x+2400y=39600--->(4)
Eqtn 4- eqtn 3
2400x-2250x = 39600-38250
150x=1350
X=1350/150
X=9
D trader bought 9bags of beans


(3)
=Draw The Diagram=

Area of path --> 2[1/2 (x+2)] + 2(x+1) = 17
x + 2 + 2x = 17
3x + 2 = 17
3x = 17 - 2
3x = 15
x = 15/3
x = 5

(3ai) Perimeter of garden
= 4x
= 4 x 5
= 20m

(3aii)
=Draw The Diagram=

Area Covered By Both Garden And path
= (x+1) (x+2)
= (5+1) (5+2)
= 6x7
= 42m^2




(4a)
Given: Sinx = 3/5
i.e
Using pythogoras triple 3,4,5
CosX + TanX/SinX
=4/5 + 3/4 ÷ 3/5
= 16+15/20 ÷ 3/5
= 31/20 ÷ 3/5
= 31/20 x 5/3
= 31/12

(4b)
=Draw The Diagram=

Reflex P'Q'T + R'Q'T + P'Q'R = 360 (sum at a point)
200 + 32 + P'Q'R = 360
232 + P'Q'R = 360
P'Q'R = 360 - 232
P'Q'R = 128°

From the Diagram
Q'R'U = P'Q'R = 128° (alternate angles)
X = Q'R'U + S'R'U
X = 128° + 180
X = 308°


5a)
1. | 2. | 3. | 4. | 5. | 6. |
------------------------------------------
1|1,1|1,2 |1,3 |1,4 |1,5 |1,6 |
-------------------------------------------
2|2,1|2,2 |2,3 |2,4 |2,5|2,6 |
--------------------------------------------
3|3,1|3,2 |3,3 |3,4 |3,5|3,6 |
---------------------------------------------
4|4,1|4,2 |4,3 |4,4 |4,5|4,6 |
---------------------------------------------
5|5,1|5,2 |5,3 |5,4 |5,5|5,6 |
-----------------------------------------------
6|6,1|6,2 |6,3 |6,4 |6,5|6,6 |
------------------------------------------------

B) pr(sum of outcome is 8)
= 4/36 = 1/9

Bii) pr(product of outcom <10)
= 30/36 = 5/6

Biii) pr(outcom contain atleast a 3)
= 24/36 = 4/6
= 2/3


6a)
2 X (37)x = 75x
2 X (3x + 7)10 = (7x + 5)10
2 (3x + 7) = 7x + 5
6x + 14 = 7x + 5
14 - 5 = 7x - 6x
9 = x
:- x = 9

6b)
Let The Number Of boys be X
Number Of girls in class = x + 5
GIVEN: x+5/+2 = 5/4
5(x+2) = 4(x+5)
5x + 10 = 4x + 20
5x - 4x = 20 - 10
x = 10

i)Numbers of girsl in class
= x + 5
= 10 + 5
= 15

ii)Total number of students
= x + x + 5
= 10 + 10 + 5
= 25

iii)Probability of selecting a boy as class prefect
= 10/24
= 2/5
7a)
|PQ|^2 = |PB|^2 + |BQ|^2
|PQ|^2 = (5 - x )^2 + X^2
|PQ|^2 = 25 - 10x + x^2 + x^2 [note ^ means Raise to power]
|PQ|^2 = 2x^2 - 10x + 25

NB: PQ = QR (side of a square)
Area of PQRS = PQ x QR
= PQ x PQ = PQ^2
Area of PQRS = (2x^2 - 10x + 25)m square

GIVEN: 2x^2 - 10 + 25 = 3/5 of 25
= 2x^2 - 10x + 10 = 0
= x^2 - 5 + 5 = 0
Using General Formular mthod
X = -(-5) ± Square Root (-5)^2 - 4(1)(5)/ (2 x 1)
X = 5 ± Square Root 25 - 20/ 2
X = 5 ± Square Root 5 / 2
X = 5 + Square Root 5 / 2 OR 5 - Square Root 5 / 2
X = 3.62 or 1.68

7b )
L + a / n - 1 = d ------------ (I)
2s = n (a + 1) ---------------- (2)
From eq (1)
L + a = d (n - 1) -------------- (3)
Put eq (3 into 2)
2s = nd (n-1)
2s = d (n^2 - n)
S = d ( n^2 - n ) / 2 or
S = dn(n-1)/2





8a)
By using pythagoras
r^2 = (r - 8)^2 + 32^2
r^2 = r^2 - 16r + 64 + 1024
16r = 64 + 1024
r = 1088/16
r = 68
r=68cm

8bi)
{st/pt/
q2=27^2 +12^2
q=sqrt27^2 + 12^2
q=sqrt729+144
q=sqrt873
q=29.5cm

10a)
YX/XZ=XM/MZ
W/10=8/15
15W=10*8 W=5.33cm

10bi)
q^2=p^2+r^2-2prcos tita
q^2=20^2+15-2*20*15 c0s 90
q^2=400+225-0
q^2=625 q=sqroot625
q=25km

10bii)
p/sinP=q/sinQ=r/sinR
25/sin90=15/sinR
sinR=15*1/25 sinR=0.6
R=sin^-1(0.6)
R=36.86degrees
The bearing of p from R
=90+90+90+alpha
alpha=45-36.86 =90+90+90+8.14
=278.14
=278degrees
The bearing of p from R=278degree



(12a)
(i)
<SPQ= 90(degree) ?(angle in
a semi circle?)
<SPR = 21(degree) ?(angle in
the alternate segment
(ii)
<QPR= 90 – 21 = 69 degree
<QSR= 69 degree = QPR?
(angle in the same segment?)
(12bi)
Given
T= sqrt(U/ (1/f+1/g)
square both sides
T(power)2 = (U/ (1/f+1/g)
f + g/fg= U/T(power)2
f + g = U/T(power)2(fg)
Ufg/T(power)2 – g = f
g(Uf/T(power)2 -1) =f
g= f/(uf/T((power)2 )-1)
(12bii)
T=3, f=4, u=5
g = 4/( 5*4/3(power)2) -1
g= 4*9/11 = 36/11
g= 3 (3/11)

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