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MATHS OBJ:
1-10: ACBCDDCBAA
11-20: CDCABCCCAC
21-30: DADBABCDAD
31-40: BADADBCACB
41-50: 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(p-2) – 4(p-2) = 0
(p-4)(p-2) = 0
p-4 = 0 or p-4 = 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)




3a)
The diagonal = 10.2m and 9.3cm
Using Pythagoras theory
Ac² = 10.2² + 9-3²
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
4-3= 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
7N-N=126-102
6N/6 =24/6
N=4


1)
On February 28th 2012, value = (100-30/100) * #900,00.00
= 70/100 * #900,00
= #630,000.00
On february 28th 2013, value = (100-22/1000 * #630,00
= 78/100 8 #630,000
= #491,400
On february 28th 2014, value = 78/100 8 #491,400
=383,292
On february 28th 2015, value = 78/100 * #383,292
= #298,967.76
 No6)

6b) Number that passed = 60% × 240 = 144
Number that failed =
240 - 144 = 96
Therefore; 28+2x+x+14+6+6-x+8 = 96
2x + 62 = 96
2x = 96 - 62
2x = 34
X = 34/2
X = 17
(i) faulty brakes cars = 8+6+x+6-x
= 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,
|TQ|ˆ² = 4ˆ² + 6 ˆ² – 2(4)(6) cos30°
|TQ|ˆ² = 16 + 36 – 48(0.8660)
|TQ|ˆ² = 52 – 41.568
|TQ|ˆ² = 10.432
TQ = √10.432
TQ = 3.23CM
From similar triangles;
|PT|/|TQ| = |PS|/|SR|
4/3.23 = 10/|SR|
4|SR| = 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
=========================================
13a)
Frequency=16+x+y
16+x+y=30
x+y=30-16
x+y=14--(eqi)
(900+30x+50y)/30=52
900+30x+50y=52*30
30x+50y=1560-900
30x+50y=660
divide through by 10
3x+5y=66--(eqii)
From (i)
x+y=14
x=14-y--(eqiii)
sub for x in eqii
3(14-y) +5y=66
42-3y+5y=66
2y=66-42
y=24/2
y=12
feom eqiii
x=14-12
x=2

(13b)
TABULATE
Class interval:1-10,11-20,21-30,41,50,51-60,61-70,71-80,81-90
Freq:1,1,2,5,12,1,4,3,1
Class boundary:0.5-10.5,10.5-20.5,20.5-30.5,30.5-40.5,40.5-50.5,50.5-60.5,60.5-70.5,70.5-80.5,80.5-90.5

(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.
1. A = {2, 4, 6, 8}, B = {2, 3, 7, 9} and C = {x: 3 < x < 9} are subsets of the universal-set
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.
(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:
Weight (kg)
20-29
30 – 39
40 – 49
SO – 59
60 – 69
70-79
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) semi-interquartile 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/.



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|.
(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]
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.
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.
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 ]
11.
Section A (Objective)
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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