VERIFIED WAEC GCE 2015/16 GENERAL MATHEMATICS ANSWERS (ESSAY & OBJ)

                       MATHS OBJ

1-10 CBADBDDBDA
11-20 DCDBACDCCC
21-30 CACBCDCBCA
31-40 ADBABACDB
41-50 DbBBDBCAAD

^=means raise to power

/ means divide

* means times

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

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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
4a. Draw a triangle
3^2 y^2=5^2
9 y^2=25
Y^2=25-9
Y^2=16
Y= sqrt16
Y=4
:. CosX tanX/sinX
4/3 3/4÷3/5
16 15/20 ÷3/5
31/20÷3/5
=31/20*5/3
=31/12
=2 7/12

4b.
From the diagram
200deg 32deg Ydeg=360deg(< at a point) Ydeg 232deg=360deg Ydeg=360deg-232deg Ydeg=128deg Ndeg=128deg(alternate < Xdeg=128deg 180deg Xdeg=308deg

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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 is8) = 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) = 2x (37x) = 75x = 2x X 37x = 75x [Note the small x are Base while big X is multiply] = (2xX) 3Xx 7Xx = 7Xx 5Xx = 2(3x 7) = 7x 5 = 6x 14 = 7x 5 = 6x - 7x = 5 - 14 = - x = - 9 = x = 9

6b) Let the no of boys = x No of girls = 5 x X 5 / X 2 = 5/4 Cross multiply 4(x 5) = 5(x 2) 4x 20 = 5x 10 4x - 5x = 10 - 20 X =10

6bi) Num of girls in class = x 5 = 10 5 = 15

6bii) total num of student = x x 5 = 10 10 5 = 25
6biii) probability of selectin a boy as class

prefect = 10/25. = 2/5

7a )
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

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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
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