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MATHS OBJ:
110: ACBCDDCBAA
1120: CDCABCCCAC
2130: DADBABCDAD
3140: BADADBCACB
4150: ADDDBDADCA
110: ACBCDDCBAA
1120: CDCABCCCAC
2130: DADBABCDAD
3140: BADADBCACB
4150: ADDDBDADCA
2)
Given that y = 2pxˆ² – p² x – 14
AT (3, 10)
10 = 2p(3)² – p² (3) – 14
10 = 18p – 3p² – 14
3p² – 18p + 24 = 0
p² – 6p + 8 = 0
using factor method,
p² – 2p 4p + 8 = 0
p(p2) – 4(p2) = 0
(p4)(p2) = 0
p4 = 0 or p4 = 0
p= 4 or p =2
Given that y = 2pxˆ² – p² x – 14
AT (3, 10)
10 = 2p(3)² – p² (3) – 14
10 = 18p – 3p² – 14
3p² – 18p + 24 = 0
p² – 6p + 8 = 0
using factor method,
p² – 2p 4p + 8 = 0
p(p2) – 4(p2) = 0
(p4)(p2) = 0
p4 = 0 or p4 = 0
p= 4 or p =2
2b)
The lines must be solved simultenously
3y – 2x = 21 —— (1)
4y + 5x = 5 ——(2)
using elimination method,
(4) 3y – 2x = 21
(30 4y + 5x = 5
12y – 8y = 84 ——— (3)
12y + 15x = 15 ——(4)
equ (4) minus equ(3)
23x = 69
x = 69/23
x = 3
Put this into equation (1)
3y 2(3) = 21
3y = 6 = 21
3y = 21 6
3y = 15
y =15/3
y = 5
coordinates of Q is (3, 5)
The lines must be solved simultenously
3y – 2x = 21 —— (1)
4y + 5x = 5 ——(2)
using elimination method,
(4) 3y – 2x = 21
(30 4y + 5x = 5
12y – 8y = 84 ——— (3)
12y + 15x = 15 ——(4)
equ (4) minus equ(3)
23x = 69
x = 69/23
x = 3
Put this into equation (1)
3y 2(3) = 21
3y = 6 = 21
3y = 21 6
3y = 15
y =15/3
y = 5
coordinates of Q is (3, 5)
3a)
The diagonal = 10.2m and 9.3cm
Using Pythagoras theory
Ac² = 10.2² + 93²
Ac² = 104.04 + 86.49
Ac² = 190.53
Ac² = √190.53
Ac² = 13.80
3b)
DRAW THE DIAGRAM
Using Pythagoras theory
5² = 3² + x²
x² = 5² – 3²
X²= 25 – 9
X² = √16
X= 4cm
CosX = adjacent/hyp
= 4/5
Tan X = opp/adj. = 3/4
5cos x – 4tan x
5(4/5) 4(3/4)
20/5 – 12/4
43= 1
DRAW THE DIAGRAM
Using Pythagoras theory
5² = 3² + x²
x² = 5² – 3²
X²= 25 – 9
X² = √16
X= 4cm
CosX = adjacent/hyp
= 4/5
Tan X = opp/adj. = 3/4
5cos x – 4tan x
5(4/5) 4(3/4)
20/5 – 12/4
43= 1
4ai)
sum of angle in a D =180degree
xdegree + 90degree + 180degree – (3x+15)=180degree
xdegree + 90degree + 180degree – 3x+15=180degree
2x=180degree – 255
+2x/2=+75/2
x=37.5
4aii)
<RsQ =180 – (3x+15)
<RsQ =180(3*37.5+15)
=180(112.5 + 15)
=180 – 127.5
<RsQ= 52.5degree
4b)
2N4seven =15Nnine
2*7^2+N*7^1+4*7degree =1*9^2 + 5*9^1+N*9degree
9*49+N*7+4*1=1*81+5*9+N*1
98+7N+4=81+45+N
7N+102=126+N
7NN=126102
6N/6 =24/6
N=4
No6)
6b) Number that passed = 60% × 240 = 144
Number that failed =
240  144 = 96
Therefore; 28+2x+x+14+6+6x+8 = 96
2x + 62 = 96
2x = 96  62
2x = 34
X = 34/2
X = 17
(i) faulty brakes cars = 8+6+x+6x
= 8+6+6
=20
(ii) only one fault = 28+x+2x
=28+3x
=28+3(19)
=28+51
= 79
Number that failed =
240  144 = 96
Therefore; 28+2x+x+14+6+6x+8 = 96
2x + 62 = 96
2x = 96  62
2x = 34
X = 34/2
X = 17
(i) faulty brakes cars = 8+6+x+6x
= 8+6+6
=20
(ii) only one fault = 28+x+2x
=28+3x
=28+3(19)
=28+51
= 79
No7)
No8)
NO9) Using cosine rule,
TQˆ² = 4ˆ² + 6 ˆ² – 2(4)(6) cos30°
TQˆ² = 16 + 36 – 48(0.8660)
TQˆ² = 52 – 41.568
TQˆ² = 10.432
TQ = √10.432
TQ = 3.23CM
From similar triangles;
PT/TQ = PS/SR
4/3.23 = 10/SR
4SR = 32.3
SR = 32.3/4
SR = 8CM (nearest whole number)
=========================
(10a) Using Pythagoras theorem from SPQ
SQ^2 = 12^2 + 5^2
= 144+25
=169
SQ= sqroot of 169
= 13cm
Sin tita= 5/13 = 0.3846
Tita= Sin^1(0.3846)
= 22.6degrees
From PRQ
Sin tita= PR/12
Sin 22.6 = PR/12
Sin 22.6= PR/12
PR= 12xsin 22.6
PR= 12×0.3843
PR= 4.61cm
(10bii)Let the height at which m touches the wall= y
Cos x^degrees= 8/10= 0.8
x^degrees= Cos^1(0.8)
= 36.87degrees
Sin x^degrees = y/12
Sin 36.87= y/12
y= 12xsin36.87
y= 12×0.60000
y= 7.2m
(10a) Using Pythagoras theorem from SPQ
SQ^2 = 12^2 + 5^2
= 144+25
=169
SQ= sqroot of 169
= 13cm
Sin tita= 5/13 = 0.3846
Tita= Sin^1(0.3846)
= 22.6degrees
From PRQ
Sin tita= PR/12
Sin 22.6 = PR/12
Sin 22.6= PR/12
PR= 12xsin 22.6
PR= 12×0.3843
PR= 4.61cm
(10bii)Let the height at which m touches the wall= y
Cos x^degrees= 8/10= 0.8
x^degrees= Cos^1(0.8)
= 36.87degrees
Sin x^degrees = y/12
Sin 36.87= y/12
y= 12xsin36.87
y= 12×0.60000
y= 7.2m
=========================================
13a)
Frequency=16+x+y
16+x+y=30
x+y=3016
x+y=14(eqi)
(900+30x+50y)/30=52
900+30x+50y=52*30
30x+50y=1560900
30x+50y=660
divide through by 10
3x+5y=66(eqii)
From (i)
x+y=14
x=14y(eqiii)
sub for x in eqii
3(14y) +5y=66
423y+5y=66
2y=6642
y=24/2
y=12
feom eqiii
x=1412
x=2
(13b)
TABULATE
Class interval:110,1120,2130,41,50,5160,6170,7180,8190
Freq:1,1,2,5,12,1,4,3,1
Class boundary:0.510.5,10.520.5,20.530.5,30.540.5,40.550.5,50.560.5,60.570.5,70.580.5,80.590.5
(13C)
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13a)
Frequency=16+x+y
16+x+y=30
x+y=3016
x+y=14(eqi)
(900+30x+50y)/30=52
900+30x+50y=52*30
30x+50y=1560900
30x+50y=660
divide through by 10
3x+5y=66(eqii)
From (i)
x+y=14
x=14y(eqiii)
sub for x in eqii
3(14y) +5y=66
423y+5y=66
2y=6642
y=24/2
y=12
feom eqiii
x=1412
x=2
(13b)
TABULATE
Class interval:110,1120,2130,41,50,5160,6170,7180,8190
Freq:1,1,2,5,12,1,4,3,1
Class boundary:0.510.5,10.520.5,20.530.5,30.540.5,40.550.5,50.560.5,60.570.5,70.580.5,80.590.5
(13C)
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Note: That There is nothing like WAEC Mathematics Expo online. Do not be deceived by fraudsters posing with fake Waec answers on the internet.
The General Maths (Core) 2 (Essay) paper will start by 8:30am and will last for 2hrs 30mins while the General Mathematics/ Mathematics (Core) 1 (Objective) exam will commence 2pm and will last for 1hr 30mins.
++++±++++++++++++++++++++++++++++++++++++
Section B [Theory]
Answer any … questions.
Write your answers on the answer booklet provided.
Answer any … questions.
Write your answers on the answer booklet provided.
1. A = {2, 4, 6, 8}, B = {2, 3, 7, 9} and C = {x: 3 < x < 9} are subsets of the universalset
U = {2, 3, 4, 5, 6, 7, 8, 9}. Find
(a) A n(B’nC’);
(b) (AuB) n(BuC).
U = {2, 3, 4, 5, 6, 7, 8, 9}. Find
(a) A n(B’nC’);
(b) (AuB) n(BuC).
2)
In the diagram above, the points A, B, C and D lie on the circle, centre O. TA and TB are tangents touching the circle at A and B respectively.
AÔB = 132°, AĈD = 59° and AOC is a straight line.
AÔB = 132°, AĈD = 59° and AOC is a straight line.
(a) Find ATB.
(b) Find BDA.
(c) Find BDC.
(d) Find OBD.
3.
4. The frequency distribution of the weight of 100 participants in a high jump competition is as
shown below:
shown below:
Weight (kg) 
2029

30 – 39

40 – 49

SO – 59

60 – 69

7079

Number of  
pa rtici pa nts 
10

18

22

25

16

9

(a) Construct the cumulative frequency table.
(b) Draw the cumulative frequency curve.
(c) From the curve, estimate the:
(i) median;
(ii) semiinterquartile range;
(iii) probability that a participant chosen at random weighs at least 60 kg.
(b) Draw the cumulative frequency curve.
(c) From the curve, estimate the:
(i) median;
(ii) semiinterquartile range;
(iii) probability that a participant chosen at random weighs at least 60 kg.
5. Using ruler and a pair of compasses only,
(a) construct a rhombus PQRS of side 7 cm and ÐPQR = 60o;
(b) locate point X such that X lies on the locus of points equidistant from PQ and QR and also equidistant from Q and R;
(c) measure /XR/.
(a) construct a rhombus PQRS of side 7 cm and ÐPQR = 60o;
(b) locate point X such that X lies on the locus of points equidistant from PQ and QR and also equidistant from Q and R;
(c) measure /XR/.
6. (a) The area of trapezium PQRS is 60 cm2. PQ//RS, /PQ/ = 15 cm, /RS/ = 25 cm and ÐPSR = 30o. Calculate the :
i) Perpendicular height of PQRS;
ii) PS.
i) Perpendicular height of PQRS;
ii) PS.
(b) Ade received 3/5 of a sum of money, Nelly 1/3 of the remainder while Austin took the rest. If Austin’s share is greater than Nelly’s share by N3000, how much was did Ade receive?
7. (a) P varies directly as Q and inversely as the square of R. If P = 1 when Q = 8 and R = 2, find the value of Q when P = 3 and R = 5.
(b) An aeroplane flies from town A(20oN, 60oE) to town B(20oN, 20oE).
(i) If the journey takes 6 hours, calculate, correct to 3 significant figures, the average speed of the aeroplane.
(ii) If it then flies due north from town B to town C, 420 km away, calculate, correct to the nearest degree, the latitude of town C.
[Take radius of the earth = 6400 km and π = 3.142]
(b) An aeroplane flies from town A(20oN, 60oE) to town B(20oN, 20oE).
(i) If the journey takes 6 hours, calculate, correct to 3 significant figures, the average speed of the aeroplane.
(ii) If it then flies due north from town B to town C, 420 km away, calculate, correct to the nearest degree, the latitude of town C.
[Take radius of the earth = 6400 km and π = 3.142]
8. Two fair dice are thrown.
M is the event described by “the sum of the scores is 10” and
N is the event described by “the difference between the scores is 3”.
(a) Write out the elements of M and N.
(b) Find the probability of M or N.
(c) Are M and N mutually exclusive? Give reasons.
M is the event described by “the sum of the scores is 10” and
N is the event described by “the difference between the scores is 3”.
(a) Write out the elements of M and N.
(b) Find the probability of M or N.
(c) Are M and N mutually exclusive? Give reasons.
9. A library received a $1,300 grant. It spends 10% of the grant on magazine subscriptions, 35% on new books, 15% to repair damaged books, 30% to buy new furniture and 10% to train library staff.
(a) Represent this information on a pie chart.
(b) Calculate, correct to the nearest whole number, the percentage increase of the amount for buying new books over that of new furniture.
(a) Represent this information on a pie chart.
(b) Calculate, correct to the nearest whole number, the percentage increase of the amount for buying new books over that of new furniture.
10. A sector of a circle with radius 21 cm has an area of 280 cm2.
(a) Calculate, correct to 1 decimal place, the perimeter of the sector.
(b) If the sector is bent such that its straight edges coincide to form a cone, calculate, correct to the nearest degree, the vertical angle of the cone. [Take π = 22/7 ]
(a) Calculate, correct to 1 decimal place, the perimeter of the sector.
(b) If the sector is bent such that its straight edges coincide to form a cone, calculate, correct to the nearest degree, the vertical angle of the cone. [Take π = 22/7 ]
11.
Section A (Objective)
Answer ALL questions in this section.
Shade your answer in the answer sheet provided.
Answer ALL questions in this section.
Shade your answer in the answer sheet provided.
1. Express as a single fraction 5/7 – 2/5.
2. The temperature in a freezer is –18 °C. The outside temperature is 24 °C. Find the difference between the outside temperature and the freezer temperature.
3. The ratio of boys to girls in a class is 4 : 5. What fraction of the class are boys?
4. The ratio of boys to girls in a school is 3 : 4. There are 120 more girls than boys. How many students are in the school?
5.
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