csec math jan2008
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© Omega Education Unit
© Club Omega
CXC MATHEMATICSClub Omega
Solutions – Jan 2008
© All rights reserved. No part of this document may be copied without the written permission of the Author.
Quote: “Luck is the strategic management of risk”
Omega Education UnitMandeville, JamaicaEmail: cxcdirect@live.comwebsite: www.cxcdirect.schools.officelive.com Math Club: http://cxclink.50.forumer.com/
Telephone: 876 469-2775, 876 860-5263
© Omega Education Unit
** Please see the original past paper for the questions.Only the answers will be provided as per copyright obligations.**********************************************************
Q1a. Jan 2008
step 1. simplifying the numerator :
1 17− 3
4= 8
7− 3
4= 32−21
28 = 1128
step 2. Simplifying the denominator:
⇒ 2 12×1
5 = 52× 1
5 = 510 = 1
2
step 3. Divide Numerator by denominator:
⇒1128
÷ 12
⇒1128
× 21 = 11
14
(ii)
2− 0.240.15 = 2− 0.24×100
0.15×100
= 2− 2415 = 30−24
15 = 615 = 2
5**********************************************************
1.b
Cash Price = $319.95
Total hire purchase price = $ 6910×$ 28.50 = $354.05
Difference in Cash & Hire purchase price
= 354.05 – 319.95 = $34.05
Difference as a percentage of cash price
= 34.05319.95
×100 = 9.62%
**********************************************************
Q2.a
if 3 – 2x7
⇒ −2x7 – 4
⇒ −2x3
⇒ 2x−3 multiply both sides by (-1) and reverse sign
⇒ x−3 /2
1. x 2 – xy=x x− y
2. a 2 – 1=a−1 a1
3. 2p – 2q – p2 pq = 2 p−q – p p−q
= 2− p p−q**********************************************************
Money collected for sponge cake = 2 (k +5)Money collected for Chocolate cake = 10kMoney collected for fruit cake = 4×2k =8k
Total money collected = ( 2k + 10 + 10k + 8 k) = 20k + 10If total momey collected = $140.
⇒ 20k + 10 = 140⇒ k = (140 – 10) over 20 = 6.5
© Club Omega - 876 469-2775Email: cxcdirect@live.com, website: www.cxcdirect.schools.officelive.com , Math club : http://cxclink.50.forumer.com/
2
© Omega Education Unit
Q3
1. S∪T = k, p2. S' = n, r, q
*********************************************************
b)
The following diagrams are drawn to demonstrate the theorems:
step1. Find angle BDC
Angles in a triangle = 180
⇒ ∢ DBC = 180−9042 = 48
step2. Find angle ABD
Angle ABD=∢CDB = 48 ( alt. Angles)
Step 3.
∢ ABC = 4842 = 900
Triangle ABD is isoceles so base angles are equal = 480
Hence ∢ ABC = 180−4848 = 840
*********************************************************
© Club Omega - 876 469-2775Email: cxcdirect@live.com, website: www.cxcdirect.schools.officelive.com , Math club : http://cxclink.50.forumer.com/
S T
k, p
U
l, m q
n, r
48D
C
90
42B
4848
B D
alternate angles
A
C
42B 48
C
A
48B D48
84
A
3
© Omega Education Unit
Q4.
Time to travel from A to B
= 14:20 – 7 :30 = 6hrs + 50 min = 6.833 hrs
Avg, Speed = Distancetime = 410 over 6.8333 = 60 km/h
b)
Area of circle of radius r = 3.5 cm = 3.52 = 38.5cm2
Area of square of side 3.5 cm = 3.52 = 12.25cm2
Area of ¼ ( circle) = ¼ (39.5) = 9.625 cm2
Area of shaded portion = area of square - ¼ (area of circle)
= 12.25 – 9.625 = 2.65cm2
Q5. Frequency Table
# Books (x) # Boys = Frequency (y)
0 21 62 173 84 3
1.Total Boys = ∑ f = (2 + 6 + 17 + 8 + 3) = 36
2.Modal # Books read = 2 ( most (17) boys read this amount
of books)
3.Total books read = ∑ f ×x =
2×06×117×28×33×4 = 76 books
4. Mean number of books read
= Total books # boys = 76
36=2.1
P( books >=3) = # boys reading >= 3 books Total Boys
=83
36 = 1136
© Club Omega - 876 469-2775Email: cxcdirect@live.com, website: www.cxcdirect.schools.officelive.com , Math club : http://cxclink.50.forumer.com/
A
410km
7:30 14:20 time6 hrs + 50 min
B
r = 3.5cm
4
© Omega Education Unit
Question 6 – TransformationFrom the diagram shown on the past paper,
BCL is mapped unto FHL by: A clockwise rotation of 1800
about the origin L.
So:R L ,−180 :C −2,0 H 2 , 0
R L ,−180 : B −2,1 H 2,−1
2 BCL is mapped unto HFG by a translation T represented
by: T = 4−1
soT : C 0,0 F 4,−1
T : B 0,1 H 4, 0
T : L 2, 0 G 6,−1
6.b
Construction details.
1. Draw straight line WX = 7cm2. Construct 600 ∢ZWX using compass3. Measure WZ = 5.5cm4. Set compass to a separation of 7cm and with centre
Z, construct an arc above X. 5. Set compass to a separation of 5.5 cm and with
centre X, construct a second arc to intersect the first arc. The intersection of the two arcs is the point Y.
6. Measure WY = 10.9cm
************************************************
From the diagram: A 2,1 B 1,−2
now recall that : R0,−90 : x , y y ,− , x
Graph
Coordinates of intersection = (4.4, 2) and ( - 0.4, 2)
Equation = x 2 – 4x=2 or f x= x2 – 4x – 2
************************************************
Question 8
n Series Sum Formula
3 1+2+3 6 123 31
n 1+2+3+..........+ n - 12n n1
n Series sum Formula
3 1+2+3 6 123 31
3 132333 62
123 31
2
n 132333 ..+ n3
12n n1
2
8 132333 ...+ 83 362
128 81
2
12 132333 ..+ 123 782
1212 121
2
© Club Omega - 876 469-2775Email: cxcdirect@live.com, website: www.cxcdirect.schools.officelive.com , Math club : http://cxclink.50.forumer.com/
W X7 cm
5.5 cm
600
Z Y
10.9 cm
1- 1
y = 2
52
-2
-1
1
y
x0
- 3
- 4
x -1 0 1 2 3 4 5y 5 0 -3 -4 -3 0 5
3 4
2
3
4
5
( 4.4, 2 )( - 0.4, 2 )
5
© Omega Education Unit
Section II
Inverse variation so:
V ∝ 1P
V = kP ( k is a constant)
finding k
now: 12.8= k500 ⇒ k = 12.8×500=6,400
so if v = 480 P=6,400480 = 13.3
**************************************************
b)
For a right angle triangle h2=x 2 y2 .. ( Pythgoras theorem)
so if h=a1 , x=a , y=a−7
⇒ a1 2=a 2a−7 2
⇒ a 22a1=a 2a 2−14a49
⇒ 0 = a2 – 16a48
⇒ 0 = a−4 a−12
⇒ a = 4, or a = 12
since y must not be negative, we will choose a = 12
the result is shown in the diagram.
Question -10 Linear programming
If x balls and y bats such that :
1. x y ≤ 30 ( no more than 30 items)
2. 6x24y ≤ 360 ( budget is $360
Re-writing equations:
y ≤ 30−x (1)
24y ≤ 360−6x
y ≤ 15−0.25x (2)
see graph below:
Profit is $1 per ball and $3 per bat
So Profit Equation is Profit = x = 3y
The maximun profit occurs at one of the vertices:
Vertex 1 ( 0,15) ⇒ Profit = 1 x 0 + 3 x 15 = $45
Vertex 2 ( 20,10) ⇒ Profit = 1 x 20 + 3 x 10 = $50
Vertex 3 ( 30,0) ⇒ Profit = 1 x 30 + 3 x 0 = $30
so Max Profit = $50 (20 balls and 10 bats)
© Club Omega - 876 469-2775Email: cxcdirect@live.com, website: www.cxcdirect.schools.officelive.com , Math club : http://cxclink.50.forumer.com/
x = 12
h = 13y = 5
10 20 30 40
5
50 60
(20, 10) Max profit = $50
Bats
0 ball
10
15
20
30
25
(30, 0)
(0, 15)
x + y = 30
6x + 24y = 360
y> = 0
x> = 0
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© Omega Education Unit
Q 11. Geometry & Trig
From the drawing given,
1. ∢WOY angle at centre = twice angle at circumference
= ∢WOY =2 50=1000
2. Triangle OWY is isocesles so, base angles are equal
⇒ ∢OWY = ∢OYW = ½ 180−100 = 400
**************************************************
b)
Find AC:
Now AB = c = 125m; BC = a = 75m ; ∢ ABC=800
Use cosine rule since we know 2 sides + included angle
so b2=a2c2−2ac.CosB
⇒ b2=1252752−2 125 75Cos80
⇒ b = 134.1m = AC
Bearing C from A = 900∢ BAC
Now considering Triangle ABC, and using the sine rule:
⇒a
sinA= b
SinB
where ∢ A=∢ BAC
⇒75
sinA= 134.1
sin80
⇒ sin A = 75134.1
×sin80 =0.54914
⇒ A = sin−1 0.54914 = 33.40
therefore Bearing C from A = 9033.4=123.40
© Club Omega - 876 469-2775Email: cxcdirect@live.com, website: www.cxcdirect.schools.officelive.com , Math club : http://cxclink.50.forumer.com/
33.4
b = 134.1m
a=75 m
c=125 mA
C
19080
N
B
N
90 + 33.4 = 123.40
7
© Omega Education Unit
Q12
Now tan 50= 2.5AB
so: AB = 2.5
Tan50 = 28.58m
***************************************************************
Now tan 200= CE28.58
so CE = 28.58 Tan20 = 10.4
and BC = BE + CE = 2.510.4 = 12.9m
© Club Omega - 876 469-2775Email: cxcdirect@live.com, website: www.cxcdirect.schools.officelive.com , Math club : http://cxclink.50.forumer.com/
2.5
5
5D
A B28.58
20
2.5
B
E
C
28.58
10.412.9m
8
© Omega Education Unit
Vectors & matrices
From Triangle laws of vectors
A BB C= A C
⇒ 2x3y=A C
⇒ A C=2 x1.5y
Considering Triangle BPQ
P Q=P BB Q
⇒ P Q=x1.5 y
Note that this is ½ A C
⇒ P Q=½ A C
***************************************************************
13.b
R T =O T – O R = 5−2 - 34 = 2
−6
S R=O R – O S = 34 - −16 = 4
−2
Position Vector OF
now R F=O F – O R
⇒ O F= R FO R = ½ R T O R
⇒ O F = ½ 2−6 + 34 = 41
**************************************************
© Club Omega - 876 469-2775Email: cxcdirect@live.com, website: www.cxcdirect.schools.officelive.com , Math club : http://cxclink.50.forumer.com/
x1.5y
1.5y
x
B
Q
P
A
C
x+1.5y
2(x+1.5y)
9
R ( 3, 4)S ( -1, 6)
T ( 5, -2 )
F ( 4, 1 )
RF = 1/2 ( RT)
o
© Omega Education Unit
14 – Matrices
if AB = C
⇒ 2 1 1 xy −2 = 5 6
⇒ 2 + y = 5and 2x – 2 = 6
⇒ y=3 , and x=4
***************************************************************
Matrix is singular if the determinant = 0
Det of 2 −11 3 = 6 +1 = 7
Therefore matrix is not Singular.***************************************************************
Inverse of matrix is 2 −11 3
−1
= 1/7 3 1−1 2
***************************************************************
matrix times its inverse is:
1/7 3 1−1 2 2 −1
1 3 = 1/7 61 −33−22 16
= 1/7 7 00 7=1 0
0 1 = Identity Matrix (I)
***************************************************************
The Equation is shown below in Matrix form:
2 −11 3 x
y = 07⇒ 2 −1
1 3−1
07 = xy
Inverse was found above as: 1/7 3 1−1 2
⇒ 1/7 3 1−1 2 07 = x
y⇒ x = 1/7 (0 + 7) = 1
⇒ y = 1/7 ( 0 + 14) = 2
⇒ x = 1, and y = 2
© Club Omega - 876 469-2775Email: cxcdirect@live.com, website: www.cxcdirect.schools.officelive.com , Math club : http://cxclink.50.forumer.com/
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