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2015/2016 Waec GCE MATHEMATICS complete Questions and Answers
10 a )
YX / XZ= XM /MZ
W/10 =8 / 15
15 W= 10 * 8 W= 5. 33 cm
10 bi)
q^ 2 =p ^ 2+ r ^ 2 - 2prcos tita
q^ 2 =20 ^ 2 + 15 - 2 * 20 * 15 c 0s 90
q^ 2 =400+ 225- 0
q^ 2 =625 q= sqroot 625
q= 25 km
10 bii )
p/ sinP= q /sinQ =r /sinR
25 / sin 90 =15 /sinR
sinR= 15 * 1 /25 sinR= 0 . 6
R =sin ^ - 1 (0 . 6 )
R =36 . 86 degrees
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= 278degrees
==================================
9a )
CBA= 180- ( 128- x ) sum of angle on a straight line
CBA= 52 + x
ADC = 180- ( 128- x )
=52 + x
Also BCD = 180- x (angle on a straight line )
DCQ= 180- ( 180- x ) DCQ= 180- 180+x
DCQ= x
x + 52 +x + 76 = 180
2x = 180- 52 - 76
2x / 2= 52 / 2
x = 26 degrees
=====================================
8a )
diagram
(8 + x )^ 2 = x ^ 2+ 32
64 + 16 x + x ^ 2 = x ^ 2 + 1024
16 x = 1024- 64 = 960
therefore 960/16 = 60 x = 960/16
=60
therefore the radius = 60 + 8
=68 cm
=====================================
8b )
diagram
(i ) volume of a pyramid
=1 / 3 AH
2601= 1 / 3 * A * 27
A =7803 /27
=289cm ^ 3
Area of square = 289 t^ 2 = 289
t= sqr rut (289)
l = 17 cm
8bii )
AC ^ 2 = 17 ^ 2 + 17 ^ 2
AC ^ 2 = 289 + 289 Ac ^ 2 = 578
AC =sqr root (578)
AC = 24 . 04 cm
for the triangele COP
CO= 1 / 2 AC
=1 / 2 * 24 . 04 VC ^ 2 = 27 ^ 2 + 12 . 07
VC ^ 2 = 929 + 144. 49
VC = SQR root ( 873. 48 )
=29 . 55 cm
cos tita = ADJ / hyp
cos x = 8. 5 /29 . 55 cos x =0 . 2877
x = cos ^ - 1 0 . 2877
=73 . 66 degree
==================================
7a )
|PQ |^ 2 = |PB | ^ 2 + | BQ| ^ 2
|PQ |^ 2 = (5 - x )^ 2 + X ^ 2
|PQ |^ 2 = 25 - 10 x + x ^ 2 + x ^ 2 [ note ^ means Raise to
power]
|PQ |^ 2 = 2x ^ 2 - 10 x + 25
NB: PQ = QR (side of a square )
Area of PQRS = PQ x QR
= PQ x PQ = PQ^ 2
Area of PQRS = ( 2x ^ 2 - 10 x + 25 ) m square
GIVEN: 2 x ^ 2 - 10 + 25 = 3 /5 of 25
= 2x ^ 2 - 10 x + 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
==================================
(4 a )
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
(4 b )
=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
==================================
(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
(3 ai) Perimeter of garden
= 4x
= 4 x 5
= 20 m
(3 aii )
=Draw The Diagram =
Area Covered By Both Garden And path
= (x + 1) (x + 2 )
= (5 + 1) (5 + 2 )
= 6x 7
= 42 m^ 2
==================================
2a )
1/ x +1 / x + 3 =1 / 2
L. C . M= x ( x +3 )
X +3 + x / x (x +3 )=1 /2
2( 2 x + 3 ) / x ( x + 3 )=1 /2
2( 2 x + 3 )=x ( x +3 )
4x + 6 =x ^ 2+ 3 x
X ^ 2 + 3x = 4x + 6
X ^ 2 + 3x - 4x - 6= 0
X ^ 2 - x - 6= 0
(X ^ 2- 3 x ) +( 2 x - 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 + 2400 y = 39600 - - - > (2 )
From eqtn (I )
X =17 - y
Using elimination method to eliminate Y
2250x + 2250 y = 38250 - - - > (3 )
2250x + 2400 y = 39600 - - - > (4 )
Eqtn 4 - eqtn 3
2400x - 2250x = 39600 - 38250
150x = 1350
X =1350 /150
X =9
D trader bought 9bags of beans
==================================
1a )
1/ 2 log 25 / 4- 2 log 10 4 /5 + log 10 320/ 125
Log(25 /4 )^ 1/ 2 - log (4 / 5)^ 2 + log10 320/ 125
Log10 sqrt 25 /4 - log16 /25 +log 10 320/125
Log10 5 / 2- log 10 16 /25 +log 10 320/125
Log10 5 / 2+ log 10 320/ 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
==================================
6a )
2 X (37 ) x = 75 x
2 X (3 x + 7 ) 10 = (7 x + 5 ) 10
2 (3 x + 7 ) = 7 x + 5
6x + 14 = 7 x + 5
14 - 5 = 7 x - 6 x
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 = 4 x + 20
5x - 4 x = 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
click r3freßh page GOODLUCK
The 2015 waec GCE MATHEMATICS question and answers are now available with Exambomb Help Desk
I PITY THOSE THAT COPY WRONG ANSWERS FROM OTHER
SITES.
USE OURS IF U WERE DIRECTED HERE
WE CONFIRM BEFORE DISTRIBUTION.
2015/2016 Waec GCE MATHEMATICS complete Questions and Answers
10 a )
YX / XZ= XM /MZ
W/10 =8 / 15
15 W= 10 * 8 W= 5. 33 cm
10 bi)
q^ 2 =p ^ 2+ r ^ 2 - 2prcos tita
q^ 2 =20 ^ 2 + 15 - 2 * 20 * 15 c 0s 90
q^ 2 =400+ 225- 0
q^ 2 =625 q= sqroot 625
q= 25 km
10 bii )
p/ sinP= q /sinQ =r /sinR
25 / sin 90 =15 /sinR
sinR= 15 * 1 /25 sinR= 0 . 6
R =sin ^ - 1 (0 . 6 )
R =36 . 86 degrees
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= 278degrees
==================================
9a )
CBA= 180- ( 128- x ) sum of angle on a straight line
CBA= 52 + x
ADC = 180- ( 128- x )
=52 + x
Also BCD = 180- x (angle on a straight line )
DCQ= 180- ( 180- x ) DCQ= 180- 180+x
DCQ= x
x + 52 +x + 76 = 180
2x = 180- 52 - 76
2x / 2= 52 / 2
x = 26 degrees
=====================================
8a )
diagram
(8 + x )^ 2 = x ^ 2+ 32
64 + 16 x + x ^ 2 = x ^ 2 + 1024
16 x = 1024- 64 = 960
therefore 960/16 = 60 x = 960/16
=60
therefore the radius = 60 + 8
=68 cm
=====================================
8b )
diagram
(i ) volume of a pyramid
=1 / 3 AH
2601= 1 / 3 * A * 27
A =7803 /27
=289cm ^ 3
Area of square = 289 t^ 2 = 289
t= sqr rut (289)
l = 17 cm
8bii )
AC ^ 2 = 17 ^ 2 + 17 ^ 2
AC ^ 2 = 289 + 289 Ac ^ 2 = 578
AC =sqr root (578)
AC = 24 . 04 cm
for the triangele COP
CO= 1 / 2 AC
=1 / 2 * 24 . 04 VC ^ 2 = 27 ^ 2 + 12 . 07
VC ^ 2 = 929 + 144. 49
VC = SQR root ( 873. 48 )
=29 . 55 cm
cos tita = ADJ / hyp
cos x = 8. 5 /29 . 55 cos x =0 . 2877
x = cos ^ - 1 0 . 2877
=73 . 66 degree
==================================
7a )
|PQ |^ 2 = |PB | ^ 2 + | BQ| ^ 2
|PQ |^ 2 = (5 - x )^ 2 + X ^ 2
|PQ |^ 2 = 25 - 10 x + x ^ 2 + x ^ 2 [ note ^ means Raise to
power]
|PQ |^ 2 = 2x ^ 2 - 10 x + 25
NB: PQ = QR (side of a square )
Area of PQRS = PQ x QR
= PQ x PQ = PQ^ 2
Area of PQRS = ( 2x ^ 2 - 10 x + 25 ) m square
GIVEN: 2 x ^ 2 - 10 + 25 = 3 /5 of 25
= 2x ^ 2 - 10 x + 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
==================================
(4 a )
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
(4 b )
=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
==================================
(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
(3 ai) Perimeter of garden
= 4x
= 4 x 5
= 20 m
(3 aii )
=Draw The Diagram =
Area Covered By Both Garden And path
= (x + 1) (x + 2 )
= (5 + 1) (5 + 2 )
= 6x 7
= 42 m^ 2
==================================
2a )
1/ x +1 / x + 3 =1 / 2
L. C . M= x ( x +3 )
X +3 + x / x (x +3 )=1 /2
2( 2 x + 3 ) / x ( x + 3 )=1 /2
2( 2 x + 3 )=x ( x +3 )
4x + 6 =x ^ 2+ 3 x
X ^ 2 + 3x = 4x + 6
X ^ 2 + 3x - 4x - 6= 0
X ^ 2 - x - 6= 0
(X ^ 2- 3 x ) +( 2 x - 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 + 2400 y = 39600 - - - > (2 )
From eqtn (I )
X =17 - y
Using elimination method to eliminate Y
2250x + 2250 y = 38250 - - - > (3 )
2250x + 2400 y = 39600 - - - > (4 )
Eqtn 4 - eqtn 3
2400x - 2250x = 39600 - 38250
150x = 1350
X =1350 /150
X =9
D trader bought 9bags of beans
==================================
1a )
1/ 2 log 25 / 4- 2 log 10 4 /5 + log 10 320/ 125
Log(25 /4 )^ 1/ 2 - log (4 / 5)^ 2 + log10 320/ 125
Log10 sqrt 25 /4 - log16 /25 +log 10 320/125
Log10 5 / 2- log 10 16 /25 +log 10 320/125
Log10 5 / 2+ log 10 320/ 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
==================================
6a )
2 X (37 ) x = 75 x
2 X (3 x + 7 ) 10 = (7 x + 5 ) 10
2 (3 x + 7 ) = 7 x + 5
6x + 14 = 7 x + 5
14 - 5 = 7 x - 6 x
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 = 4 x + 20
5x - 4 x = 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
click r3freßh page GOODLUCK
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