07. csec maths january 2008 · (ii) required to determine: smallest value of x that satisfies the...
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aths.c
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CSEC JANUARY 2008 MATHEMATICS GENERAL PROFICIENCY (PAPER 2)
Section I
1. a. (i) Required To Calculate:
Calculation: Numerator:
Denominator:
Hence,
(ii) Required To Calculate:
Calculation:
51
212
43
711
´
-
( ) ( )
281128213228
378443
78
43
711
=
-=
-=
-=-
21
51
25
51
212
=
´=´
form)exactin(1411
12
2811212811
51
212
43
711
=
´=
=´
-
15.024.02 -
form)exactin(4.0
6.1215.024.02
=
-=-
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b. Data: Bicycle with prices for cash or hire purchase (H.P.) payments. (i) Required To Calculate: Total hire purchase price for the bicycle. Calculation:
Hire purchase price of the bicycle = Deposit + Total for 10 installments
(ii) Required To Calculate: Difference between hire purchase price and cash
price. Calculation: Difference between the hire purchase price and the cash price
(iii) Required To Express: Difference between 2 prices as a percentage of the
cash price. Solution: The difference in price as a percentage of the cash price
2. a. (i) Required To Solve: . Solution:
(ii) Required To Determine: Smallest value of x that satisfies the above
inequality. Solution:
If , then the smallest x is 0.
( )00.354$
50.28$1000.69$=
+=
05.34$95.319$00.354$
=-=
places)decimal2to(%64.10%642.10
%10095.31905.34
==
÷øö
çèæ ´=
723 <- x
22
421
42372
723
->÷
->-´
<--<-
<-
x
x
xxx
{ },...3,2,1,0=WWxÎ
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b. Required To Factorise: (i) Solution: (i) (ii) This is a difference of 2 squares
And (iii)
c. Data: Table showing types of cakes, cost and the number sold. (i) Required To Find: An expression, in terms of k, for the amount of money
collected for the sale of sponge cakes. Solution: The cost of 2 sponge cakes at each (ii) Required To Find: An expression in terms of k, for the total amount of
collected. Solution: Cost of sponge cakes Cost of 10 chocolate cakes at $k each Cost of 4 fruit cakes at $2k each Total amount of money collected (iii) Data: Total amount of money collected = $140.00
Required To Calculate: k Calculation: Total amount of money collected = $140.00
pqpqpiiiaiixyx +---- 222 22)(,1)(,
yxxxxyx ..2 -=-( )yxx -=
( ) ( )222 11 -=- aa
( )( )1112 +-=- aaa
( ) ( )qppqppqpqp ---=+-- 222 2
( )( )pqp --= 2
( )5$ +k ( )52 +´= k( )102$ += k
( )102$ += k
10´= kk10$=
42 ´= kk8$=
( ) kkk 810102 +++=( )1020$ += k
10140201401020
-==+\
kk
5.62013013020
=
=
=
k
k
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3. a. Data: Given the elements of universal set, U, S and T. (i) Required To Draw: Venn diagram to represent the information given. Solution:
(ii) (a) Required To List: Members of . Solution: as shown.
(b) Required To List: Members of . Solution:
as shown.
b. Data: Quadrilateral ABCD with AB = AD, , and AB parallel to DC.
(i) Required To Calculate: Calculation:
(Co-interior angles are supplementary).
{ }{ }{ }pkTS
qpkTpmlkS
,,,,,,
=Ç==
TS È
{ }qpkmlTS ,,,,=È
S ¢
{ }nqrS ,,=¢
°= 90ˆDCB °= 42ˆCBD
CBA ˆ
°=°-°=
9090180ˆCBA
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(ii) Required To Calculate: Calculation:
(iii) Required To Calculate: Calculation: (data) (Base angles of an isosceles triangle are equal).
(Sum of angles in a triangle = 180°).
4. a. Data: John left Port A at 07:30 hours and travelled to Port B in the same time
zone. He arrives at Port B at 14:20 hours. (i) Required To Find: Duration of the journey. Solution: Departure time from A is 07:30 hours. Arrival time at B is 14:20 hours. Duration of the journey
(ii) Data: John travelled 410 km.
Required To Calculate: Average speed. Calculation:
b. Data: Diagram showing a circle of radius 3.5 cm, centre O and square OPQR.
DBA ˆ
°=°-°=
484290ˆDBA
DAB ˆ
ADAB =°=\ 48ˆBDA
( )°=
°+°-°=84
4848180ˆDAB
20:14=
hours656
minutes50hours6
50:6
30:7
=
=
-
1kmh60
hours656
km410takentimeTotalcovereddistanceTotalspeedAverage
-=
=
=
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(i) Required To Calculate: Area of the circle. Calculation:
Area of circle
(ii) Required To Calculate: Area of square OPQR.
Calculation: Area of square OPQR
(iii) Required To Calculate: Area of the shaded region. Calculation: Area of the shaded region Area of square OPQR – Area of sector OPR.
5. Data: Bar graph illustrating the number of books read by the boys of a book club. a. Required To Draw: The frequency data representing the information given. Solution: From the bar graph
No. of books read, x No. of boys, f 0 2 1 6 2 17 3 8 4 3
b. Required To Find: The number of boys in the book club.
( )25.3722
=
2cm5.38=
5.35.3 ´=2cm25.12=
=
( )
places)decimal2to(cm63.2cm562.2
5.384125.12
2
2
=
=
-=
å = 36f
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Solution: No. of boys in the book club
c. Required To Find: Modal number of books read.
Solution: Modal number of books read = 2, as corresponds to the maximum frequency, 17.
d. Required To Calculate: Total number of books read.
Calculation: Total number of books read
e. Required To Calculate: The mean number of books read.
Calculation: The mean number of books read, .
This value is a whole number and may be represented by the integer 2.
f. Required To Calculate: Probability a randomly chose boy reads at least books.
Calculation:
6. a. Data: Diagram showing a pattern of congruent right angled triangles as
å= f36=
2=x
( ) ( ) ( ) ( ) ( )43382171602 ´+´+´+´+´=
761224346
=+++=
x
1.23676
=
=
=ååffx
x
( )
3611
3638
boysofno.Totalbooks3readingboysofNo.books3readsboychosenRandomly
=
=+
=
³=³
å f
P
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(i) Required To Describe: Transformation that maps onto . Solution:
L is a common point. Image is re-oriented and congruent. BLF is a straight line.
is mapped onto HFL by a rotation of 180° (clockwise or anti-clockwise) about L.
(ii) Required To Describe: Transformation that maps onto .
Solution:
and have the same ‘orientation’.
is mapped onto by a horizontal shift 4 units to the right and
1 unit vertically downwards. This may be represented by the translation, T
where .
b. (i) Required To Construct: Parallelogram WXYZ, in which, WX = 7.0 cm, WZ = 5.5 cm and . Solution:
BCLD FHLD
HFLBCL D®D
BCL\
BCLD HFGD
BCLD HFGD
BCLD HFGD
T =4-1
æ
èç
ö
ø÷
°= 60ˆZWX
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(ii) Required To Find: The length of WY. Solution: WY = 10.9 cm (by measurement.)
7. Data: Incomplete table for . a. Required To Complete: The table of values for . Solution:
x -1 0 1 2 3 4 5 y 5 0 -3 -4 -3 0 5
When When When
xxy 42 -=xxy 42 -=
0=x ( ) ( )040 2 -=y0=
3=x ( ) ( )343 2 -=y3-=
5=x ( ) ( )545 2 -=y5=
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b. Required To Draw: The graph of . Solution:
c. (ii) Required To Find: Points at which the curve meets the line. Solution: The line and the curve meet at and .
(iii) Required To Find: The equation whose roots are the coordinates found
above. Solution:
and are the roots of the equation or .
xxy 42 -=
2=y xxy 42 -= 5.0-=x 5.4=x
5.0-=x 5.4=x 242 =- xx0242 =-- xx
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8. a. Data: Table showing the sum of rational numbers. Required To Complete: The table given.
n SERIES SUM FORMULA 1 1 1
2 1 + 2 3
3 1 + 2 + 3 6
4 1 + 2 + 3 + 4 10
5 1 + 2 + 3 + 4 + 5 15
(i) 6 1 + 2 + 3 + 4 + 5 + 6 21
8 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 36
(ii) n 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + …+
n
The formula for the sum is a constant, , example
When Sum is
When Sum
For (ii), the formula is .
b. Data:
1 + 2 + 3 = 6 and = 36
So too, and
( )( )11121
+
( )( )12221
+
( )( )13321
+
( )( )14421
+
( )( )15521
+
( )( )16621
+
! ! ! !
( )( )18821
+
! ! ! !
( )121
+nn
( )121
+nn
5=n ( )( )15521
+= 15=
15Sum =\
6=n ( )( )16621
+= 21=
21Sum =\
( )121
+nn
2333 6321 =++
104321 =+++ 23333 104321 =+++100=
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(i) Required To Find: Solution: From the table
(ii) Required To Find:
Solution:
c. Required To Find:
Solution:
9. a. Data: Volume, V of a gas varies inversely as the pressure, P, with temperature constant. (i) Required To Find: Equation relating V to P. Solution:
(k is constant of proportion or variation) (ii) Required To Find: k when and . Solution: when
33333333 87654321 +++++++
3687654321 =+++++++
12963687654321 233333333
==+++++++
3333 ...321 n++++
( )
( )2
3333 121...321
121...321
þýü
îíì +=++++\
+=++++
nnn
nnn
33333 12...4321 +++++
( )( )
( )6084784321
78
112122112...4321
23333
==+++\
=
+=+++++
PkV
PV
1
1
´=
µ
8.12=V 500=P
8.12=V 500=P
64005008.12
50018.12
=´=
´=
k
k
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(iii) Required To Calculate: V when . Calculation: When
b. Data: Right-angled triangle of sides .
(i) Required To Find: An equation in terms of a to relate the three sides,
using Pythagoras’ Theorem. Solution:
(ii) Required To Calculate: a
Calculation:
If , the side would compute to be a negative value. Hence, only.
(iii) Required To State: The lengths of the 3 sides of the triangle.
Solution: The lengths of the 3 sides are
480=P
3113
48016400
480
=
´=
=
V
P
( ) ( )1and7, +- aaa
( ) ( ) ( )
04816124914
Theorem'Pythagoras17
2
222
222
=+-
++=+-+
+=-+
aaaaaaa
aaa
( )( )12or4
0412048162
==--=+-
aaaaa
4=a ( )7-a12=a
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The sides are 5, 12 and 13 units.
10. a. Data: School buys x balls and y bats. (i) Required To Find: Inequality for the information given.
Solution: Total number of balls and bats is no more than 30. Hence, …(1)
(ii) Data: School allows no more than $360 to be spent on bats and balls. The cost of a ball is $6 and the cost of a bat is $24. Required To Find: Inequality to represent the information given. Solution: The cost of x balls at $6 each and y bats at $24 each is
Budget allows no more than $360. Similarly,
b. Required To Draw: The graphs of the inequalities shown above, shade the
region that satisfies the inequalities and state the vertices of the feasible region. Solution: and identifies the 1st quadrant. Obtaining 2 points on the line When The line passes through the point (0, 30). When The line passes through the point (30, 0).
yx + 30£30£+ yx
( ) ( ) yxyx 246246 +=´+´
)1...(6046
360246
£+÷
£+
yx
yx
0³x 0³y
30=+ yx0=x 300 =+ y
30=y30=+ yx
0=y 300 =+x30=x
30=+ yx
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The region with the smaller angle corresponds to the region. The region which satisfies is
Obtaining 2 points on the line . When
The line passes through the point (0, 15). When The line passes through the point (60, 0).
The region with the smaller angle corresponds to the region. The region which satisfies is
£30£+ yx
604 =+ yx0=x 6040 =+ y
15460604
=
=
=
y
y
604 =+ yx0=y ( ) 6004 =+x
60=x604 =+ yx
£604 £+ yx
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om The feasible region is the area in which both previously shaded regions overlap.
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The vertices of the feasible region are O (0, 0), A (0, 15), B (20, 10) and C (30, 0).
c. Data: Profit made of each ball is $1 and profit made on each bat is $3.
(i) Required To Find: The profit for reach of the combinations above. Solution:
Testing the point (0, 15), (30, 0) and (20, 10). When and
yxP 3+=
0=x 15=y( )
45$1530
=+=P
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When and
When and
As a point of interest, the only point to be considered should be (20, 10), where and since the question specifically indicates – a school buys x balls AND y balls. (They cannot buy 0 bats or 0 balls, that is x and y .)
(ii) Required To Find: Maximum profit that may be made.
Solution: The maximum profit, , which occurs when and .
11. a. Data: O is the centre of the circle WXY and
(i) Required To Calculate:
Calculation:
(The angle subtended by a chord at the centre of the circle is twice the angle subtended at the circumference, standing on the same arc).
30=x 0=y( )
30$0330
=+=P
20=x 10=y( )
50$10320
=+=P
20=x 10=y
+Î Z
50$max =P 20=x 10=y
°= 50ˆYXW
YOW ˆ
( )°=°=
100502ˆYOW
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(ii) Required To Calculate: Calculation: (radii)
(Sum of the angles in a triangle = 180°).
b. Data: Three buoys A, B and C, their relative distances apart and their positions. (i) Required To Sketch: Diagram showing the information given.
Solution:
(ii) Required To Calculate: Distance AC. Calculation:
(cos law)
YWO ˆ
OYOW =
°=
°-°=
=
402100180
)(ˆˆ equalaretriangleisoscelesanofanglesbasetheWYOYWO
( ) ( ) ( )( )
m1.134m14.134
80cos75125275125
801090ˆ
222
==
°-+=
°=°-°=
ACACAC
CBA
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(iii) Required To Calculate: Bearing of C from A. Calculation: Let
The bearing of C from A
12. a. Data: Diagram illustrating a vertical pole and tower standing on horizontal ground.
(i) Required To Calculate: Horizontal distance AB. Calculation: (alternate angles).
°=
°=
°=
°=
4.33134.175sin80sin
Law)Sine(80sin1.134
sin75
ˆ
a
a
a
aCAB
°+°= 4.3390degreenearesttheto123°=
°= 5ˆABD
tan5° = 2.5AB
AB = 2.5tan5°
= 28.57m= 28.6m (to 1 decimal place)
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(ii) Required To Calculate: Height of the tower BC. Calculation:
(Opposite sides of a rectangle).
Height of tower
b. No solution has been offered for this question as it is based on latitude and longitude (Earth Geometry) which has been removed from the syllabus)
13. a. Data: P and Q are the midpoints of AB and BC of vector triangle ABC. (i) Required To Sketch: Diagram to show the information given. Solution:
57.28=DE
m39.1020tan57.28
57.2820tan
=°=
=°
CE
CE
39.105.2 +=BC
m9.12m98.12
==
( )
y
y
y
QCBQ
yBC
xPBAP
xAB
211
23
321
3
2
=
=
=
=\
=
==\
=
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(ii) (a) Required To Find: Expression in terms of x and y for . Solution:
(b) Required To Find: Expression in terms of x and y for .
Solution:
(iii) Required To Prove:
Proof:
Q.E.D.
b. Data: and
AC
yxBCABAC32 +=+=
PQ
yx
BQPBPQ
211+=
+=
ACPQ21
=
ACPQ
PQ
yx
yxAC
yxPQ
212
2112
32211
=
=
÷øö
çèæ +=
+=
+=
÷÷ø
öççè
æ-=÷÷
ø
öççè
æ=
61
,43OSOR ÷÷
ø
öççè
æ-
=25
OT
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(i) (a) Required To Express: in the form .
Solution:
(b) Required To Express: in the form .
Solution:
(ii) (a) Required To Find: The position vector of F. Solution: If , then F is the midpoint of .
OR
RT ÷÷ø
öççè
æba
÷÷ø
öççè
æ-
=
÷÷ø
öççè
æ-
+÷÷ø
öççè
æ-=
+=
62
25
43OTRORT
SR ÷÷ø
öççè
æba
÷÷ø
öççè
æ-
=
÷÷ø
öççè
æ+÷÷ø
öççè
æ --=
+=
24
43
61ORSOSR
FTRF = RT
÷÷ø
öççè
æ=
÷÷÷÷
ø
ö
çççç
è
æ
-
+
=
14224253
OF
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aths.c
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(b) Required To State: Coordinates of F.
Solution:
If , then F = (4, 1).
14. a. Data: and
. Required To Calculate: x and y. Calculation:
If then
Equating corresponding entries.
÷÷ø
öççè
æ=
÷÷ø
öççè
æ-
+÷÷ø
öççè
æ=
+=
÷÷ø
öççè
æ-
=
÷÷ø
öççè
æ-
=
=
14
31
43
31
62
2121
RFOROF
RTRF
÷÷ø
öççè
æ=14
OF
( ) ÷÷ø
öççè
æ-
==2
1,12
yx
BA ( )65=C
CAB =
( )
( ) ( ) ( ) ( )( )( )222
2121122
112
-+=-´+´´+´=
÷÷ø
öççè
æ-
=
xyxy
yx
AB
CAB =
( ) ( )65222 =-+ xy
352
==+yy
482622
===-
xx
x
3and4 ==\ yx
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b. Data: .
(i) Required To Show: R is a non-singular matrix. Proof: det = 6 + 1 = 7 Since , then R is non-singular.
(ii) Required To Find:
Solution:
(iii) Required To Prove:
Proof:
Q.E.D.
÷÷ø
öççè
æ -=
3112
R
( ) ( )1132 ´--´=R0¹R
1-R
( )( )
÷÷÷÷
ø
ö
çççç
è
æ
-=
÷÷ø
öççè
æ-
--=-
72
71
71
73
2113
711R
IRR =-1
÷÷ø
öççè
æ=
÷÷÷÷
ø
ö
çççç
è
æ
-÷÷ø
öççè
æ -=´ -
2221
1211
1
72
71
71
73
3112
eeee
RR
I
RR
e
e
e
e
=
÷÷ø
öççè
æ=\
=
÷øö
çèæ ´+÷
øö
çèæ ´=
=
÷øö
çèæ -´+÷
øö
çèæ ´=
=
÷øö
çèæ ´-+÷
øö
çèæ ´=
=
÷øö
çèæ -´-+÷
øö
çèæ ´=
-
1001
1723
711
0713
731
0721
712
1711
732
1
22
21
12
11
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aths.c
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(iv) Data: and
Required To Calculate: x and y by matrix method. Calculation:
…(1) …(2)
Equating corresponding entries. and
02 =- yx 73 =+ yx
02 =- yx73 =+ yx
÷÷ø
öççè
æ
÷÷÷÷
ø
ö
çççç
è
æ
-=÷÷
ø
öççè
æ
÷÷ø
öççè
æ=÷÷
ø
öççè
æ´
÷÷ø
öççè
æ=÷÷
ø
öççè
æ´
´
÷÷ø
öççè
æ=÷÷
ø
öççè
æ
÷÷ø
öççè
æ=÷÷
ø
öççè
æ÷÷ø
öççè
æ -
-
--
-
70
72
71
71
73
70
70
70
70
3112
1
11
1
yx
Ryx
I
Ryx
RR
R
yx
R
yx
÷÷ø
öççè
æ=
÷÷÷÷
ø
ö
çççç
è
æ
÷øö
çèæ ´+÷
øö
çèæ ´-
÷øö
çèæ ´+÷
øö
çèæ ´
=
21
7720
71
7710
73
1=x 2=y
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