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## GCE SOLUTIONS MATHEMATICS WAEC GCE QUESTIONS & ANSWERS

Complete past questions on waec gce, simplified objective theory for waec gce 2017/2018/2019 are all available –  GCE SOLUTIONS MATHEMATICS WAEC GCE QUESTIONS & ANSWERS – expo for waec/neco/nabteb gce 2017/2018.

Answer all the Questions.

SECTION A.

1A) Without using Mathematical table or calculator, simplify
1/2log (25/4) – 2log(4/5) + Log (320/125)
B) The population of a village increase by 20% Every year. The district assembly grant the village Ghc 15.00 per head at the beginning of every year. If the population of the village was 3,000 in the year 2003. Cal the assembly’s total grant to the village from 2003 to 2007.

## GCE SOLUTIONS MATHEMATICS WAEC GCE QUESTIONS & ANSWERS

2A) If 1/x + 1/x+3 = 1/2 , Cal the value of x
2b) A trader bought some bags of rice and some bags of beans for sale in his shop. Each bag of rice cost N2,250.00 while each bag of beans cost N2,400.00. If the trader bought 17 bags altogether and paid N39,600.00, how many bags of beans did he buy ?

3) A garden. In th form of a square is surrounded by a path which is 1 m wide on two opposite side and 1/2m wide on the other two opp side, if the area of the path is 17m^2 cal. The.

3ai) Perimeter of garden
3aii) Area of the space covered by the garden and the path
4a) If sin x= 3/5, where 0 degree < equal to X < equal to 90 degree. Without using table or cal
Calculate Cos X + Tan X / Sin X. #### GCE SOLUTIONS MATHEMATICS WAEC GCE QUESTIONS & ANSWERS

4b)
Where you see Q the number there is 200 degree
Where you see g the number there is 32 degree
Where you see R the number there is X
So in the diagrame Find the value of x.

5A) Afair die is thrown TWO times
a) Construct a table of the outcomes
b) Calulate the probability that the
(I) Sum of the outcomes is 8;
(II) Product of outcome is less than 10;
(III) Outcome contain at least a 3;

SECTION B
[60 Marks] Answer Five Questions Only From This Section.

All Questions Carry Equal Marks

(6a)
A number is written as 37 in base X. Twice the number is written as 75 in base X. Find the value of X.

(6b)
There are 5 more girls than boys in a class. If 2 boys join the class. The ratio of girls to boys will be 5:4. Find the:
(i) Number of girls in the class
(ii) Total number of pupils in the class
(iii) Probability of selecting a boy as the class prefect.

### GCE SOLUTIONS MATHEMATICS WAEC GCE QUESTIONS & ANSWERS ——————————- (7a) [DIAGRAM] In The Diagram, ABCD, is a square of side 5m, P, Q, R and S are points on the square such that |AP| = |BQ| = |CR| = |DS| = X. (i) Show That The Area of the square PQRS is (2X^2 – 10X + 25)m^2 (ii) Find, correct to two decimal places, the values of X for which The Area of PQRS is 3/5 of that of ABCD. (7b) If 1+a/n-1 = d and 2s = n(a +1), express in terms of n and d only. —————————— (8a) [DIAGRAM] In The diagram, WXYZ is a segment of a circle such that XZ is The perpendicular bisector of WY. If |XZ| = 8cm and |WY| = 64Cm, calculate The radius of the circle.

(8b)
A silo in the form of a pyramid on a square base, has height 27cm and volume 2601cm^3. Calculate:
(i) Length of a side of the base of the silo, correct to 3 significant figures;
(ii) Angle between a sloping face and the base of the silo, correct to the nearest degree.

(9a)
[DIAGRAM] In the diagram, ABCD is a cyclic quadrilateral. AB and DC are produced to meet at P.
AD and BC are produced to meet at Q. If ZDQC = 76°,
/BPC° = 52° and /BCP = X, calculate the value of X.

(9bi)
Using ruler and a pair of compass only, construct a rhombus MNOQ of length 10cm and /MNO = 45°
(ii)Measure;
I. |MO|
II. |NQ|

(10)
[DIAGRAM].,
In the diagram, /YXM = /XZM, ‘YZ’ is a straight line. |XM| = 8cm, |XZ| = 10cm, |YZ| = 15cm and |XY| = w cm.
Find the value of w.

(10b)
A ship leaves port P and sails on the bearing of N45°E to a port Q, 15km away. Its then sails on a bearing of S45°E to port R, 20km away
(i) Represent the information in a diagram
(ii) Calculate, Correct to the nearest whole number, the;
I. Distance from P to Q
II. Bearing of P from R.

#### (11) The following table shows the marks distributions of candidates selected for an award. Mark(%): |56-48|59-61|62-64|65-67|68-70|71-73|74-76|77-79|80-82|83-85| Frequency: |10|28|40|92|140|90|50|30|15|5 (a) Construct a cumulative frequency distribution table. (b) Draw a cumulative frequency curve for the distribution. (c) Use the curve to estimate: i. Median mark ii. Probability of selecting a candidate who scored not more than 72%.

12). P /—————————— S
/
/
/
/
Q /—————————————– R
In The diagrame bellow PS//QR. PT||SR. QR =24cm SR=18cm, PQ= 20cm and PS= 12cm cal. Correct to 3s.f
A) < PQT
B) the heigth of triangle PQT
C) area of PTRS as a percentage of the area of PQRS.

13). Using a scale of 2cm to 2 unit on both axes, draw on a graph sheet two perpendicular axes Ox and Oy for -10 < equal to x < equal to 10 and -10 < equal to y < equal to 10 13b) Draw on the same graph sheet, indicate clearly all vertices and their coordinate I) Triangle ABC with vertices A(4 , 2) B(0 , 2) and C(0 ,8) Ii) The image A1 B1 C1 of ABC under mapping (X/y) ——> (y+1/2x) Where A—-A1 , B—–B1 and C—– C1.

iii) The img A2 B2 C2 of ABC under a clockwise rotation of 90 degree about the origin where A—- A2 , B—–B2 and C—– C2
Iv) The img A3 B3 C3 of ABC under an enlargement with the scale factor -1 where A—- A3 , B—–B3 and C—– C3

Verified MATHS OBJ:
1-10: BAABBCDBCC
21-30: BDCBDABACA
31-40: CACCDCDDDB

=================
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.

### GCE SOLUTIONS MATHEMATICS WAEC GCE QUESTIONS & ANSWERS 8bi) {st/pt/ q2=27^2 +12^2 q=sqrt27^2 + 12^2 q=sqrt729+144 q=sqrt873 q=29.5cm. ================================== 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.

==================================
(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 ================================== (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)
= 6×7
= 42m^2.

### GCE SOLUTIONS MATHEMATICS WAEC GCE QUESTIONS & ANSWERS

==================================
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.

==================================
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.

### GCE SOLUTIONS MATHEMATICS WAEC GCE QUESTIONS & ANSWERS

==================================
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 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=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=26degrees. ===================================== (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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