solutions to csec maths p2 june 2015 - img1.wsimg.com
Post on 27-Nov-2021
7 Views
Preview:
TRANSCRIPT
Solutions to CSEC Maths P2 June 2015
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 1a part (i)
417
30
Question 1a part (ii)
3.3444
Question 1a part (iii)
7.855
Question 1a part (iv)
1.7
Question 1b part (i)
Using the information provided in the table we know that 3kg of sugar costs
$10.80. Therefore, 1kg of sugar will cost $10.80/3 = $3.60. Hence, π = $3.60.
Question 1b part (ii)
From the information provided we know that 1kg of rice costs $0.80 more than
1kg of flour. From the table, we know that 1kg of flour costs $1.60. Therefore,
1kg of flour will cost $1.60 + $0.80 = $2.40. Hence, π = $2.40.
If 1kg of flour costs $2.40, then 4kg of flour will cost 4 Γ $2.40 = $9.60. Hence,
π = $9.60.
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 1b part (iii)
The tax exclusive sum of Mrs Roweβs bill is $10.80 + $9.60 + $3.20 = $23.60.
Since the tax is 10% of the tax exclusive sum the total tax is 0.10 Γ $23.60 =
$2.36.
The tax inclusive sum of Mrs Roweβs bill is $23.60 + $2.36 = $25.96.
Question 2a part (i)
We know π = 4; π = 2;π = β1. Hence, π β π + π = (4) β (2) + (β1) = 1
Question 2a part (ii)
We know π = 4; π = 2;π = β1.
Hence, 2ππ = 2 Γ (4)2 = 32
Question 2b part (i)
If π millilitres is the amount of juice that is poured out of the bottle then the amount of juice left in the bottle is equal to (500 β π) ππππππππ‘πππ .
Question 2b part (ii)
If π glasses, each containing π millilitres, are poured out then π Γ π millilitres is the total quantity that was poured out of the bottle. The amount of juice left in the bottle is equal
to (500 β (π Γ π)) ππππππππ‘πππ .
Question 2c
From the given information recall: 2π
3+
2 β π
5
The Lowest Common Multiple of the abovementioned fraction is 15, hence:
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
5 Γ (2π) + 3 Γ (2 β π)
15
Simplifying, 10π + 6 β 3π
15
Simplifying into its simplest form: 7π + 6
15
Question 2d part (i)
Let the quantity of mangoes be represented by π₯ and the quantity of pears be
represented by π¦.
Using the information provided can be represented by the pair of simultaneous
equations: 4π₯ + 2π¦ = 24
2π₯ + 3π¦ = 16
Question 2d part (ii)
x represents the quantity of mangoes; y represents the quantity of pears.
Question 2e part (i)
Solution for Part e: From the given information recall:
π3 β 12π
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Factorizing into its simplest form:
π(π2 β 12)
Question 2e part (ii)
From the given information recall:
2π₯2 β 5π₯ + 3
Expanding:
2π₯2 β 2π₯ β 3π₯ + 3
Factorizing:
2π₯(π₯ β 1) β 3(π₯ β 1)
Factorizing into its simplest form: (2π₯ β 3)(π₯β 1)
Question 3a part (i)
From the given information we know that 12 students played neither the guitar
nor the violin.
Question 3a part (ii)
The total number of students in the class, as an expression of x, is stated below:
2π₯ + π₯ + 4 + 12
3π₯ + 16
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 3a part (iii)
The total number of students in the class, as an equation of x, is stated below:
3π₯ + 16 = 40
Question 3a part (iv)
The determine the number of students that play the guitar we must first determine the
unknown variable x,
Transposing the equation, with respect to x, that stated for Part iii yields,
π₯ =40 β 16
3=
24
3= 8
From the information provided in the Venn Diagram, we know that 16 students
played the guitar ONLY and 8 students played BOTH the guitar and the violin.
Therefore, 16 + 8 = 24 students played the guitar.
Question 3b part (i)
The triangle can be constructed using the information and instructions provided. The
final triangle is given below,
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
The size of angle BAC is 33.69Β°.
Question 3b part (ii)
The point D, such that ABCD is a parallelogram is identified in the diagram below.
Question 4a
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
The function to be used is π¦ = π₯2 β 2π₯ β 3 . The unknown values for π¦, in the table,
can be determined by substituting the corresponding, known values for π₯, into the aforementioned function. Hence,
π¦(β1) = (β1)2 β 2(β1)β 3 = 1 + 2 β 3 = 0
π¦(3) = (3)2 β 3(2) β 3 = 9 β 6 β 3 = 0
The completed Table is provided below,
x -2 -1 0 1 2 3 4
y 5 0 -3 -4 -3 0 5
Question 4b
The graph can be plotted using the (π₯, π¦) coordinates tabulate above and the
instructions provided. The graph for the given function is illustrated below,
Question 4c
The smooth curve drawn through the (π₯,π¦) coordinates is illustrated below,
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 4d part (i)
From the graph, the values for π₯ for which π₯2 β 2π₯ β 3 = 0 are β1 and 3.
Question 4d part (ii)
From the graph, the minimum value of π₯2 β 2π₯ β 3 is β4.
Question 4d part (iii)
The equation of the line of symmetry of the graph of π¦ = π₯2 β 2π₯ β 3 is π₯ = 1.
Question 5a part (i)
If the car is travelling at a constant speed of 54 ππ/β, then after 2.25 βππ’ππ the car
would have travelled (54 ππ/β) Γ (2.25 βππ’ππ ) = 121.5 ππ.
Using the given information, we must first convert the speed that the car travels from ππ/β to π/π . Hence, the speed that the car travels, in π/π is:
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
π ππππ (π/π ) = (5
18) Γ (54) = 15 π/π πΊππ£ππ π‘βππ‘ π‘βπ πππ π‘πππ£πππ
a distance of 315 π, we can determine the time that car took the travel the aforementioned distance using the following equation:
π‘πππ π‘ππππ = (πππ π‘ππππ π‘πππ£πππππ)/π πππππππππ
Question 5a part (ii)
πππ π‘ππππ π‘πππ£πππππ = 315 π and π ππππ = 15π/π , the time taken is: π‘πππ π‘ππππ = (315)/(15) = 21 π ππππππ
Question 5b part (i)
The scale 1 millimetre = 1 meter in the form 1: π₯ is: 1 Γ 10β3 πππ‘πππ βΆ 1 πππ‘ππ
Hence,
1 πππ‘ππ βΆ1
10β3 πππ‘πππ
Which is,
1 πππ‘ππ βΆ 1000 πππ‘πππ
The scale 2 cm = 6 m in the form 1: π₯ is: 2 Γ 10β2 πππ‘πππ βΆ 6 πππ‘πππ
Hence,
1 πππ‘ππ βΆ6
2 Γ 10β2 πππ‘πππ
Which is, 1 πππ‘ππ βΆ 300 πππ‘πππ
Question 5c part (i)
Using the information provided, the distance, in centimetres, from π to π on the map is
6 ππ.
Question 5c part (ii)
As shown the in illustration below there are approximately eighteen (18) 1 ππ2 tiles. All of the whole tiles are coloured in grey. The corresponding pieces of broken 1 ππ2 tiles
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
that forms a full 1 ππ2 tile is appropriately colour coded. Therefore, the area, in square
centimetres of the plane πππ π on the map is 18 ππ2 .
Question 5c part (iii)
From the information provided, we know that 1 ππ βΆ 2,000 ππ. Therefore, the actual distance, in centimetres, between π and π , is 6 ππ Γ 2,000 = 12,000 ππ. The actual
distance, in metres between π and π is, 12,000 Γ 10β2 π = 120 π. The actual distance, in kilometres between π and π is, (120/1000) ππ = 0.12 ππ.
Question 5c part (iv)
To calculate the area, in square meters of the plane πππ π we must use the equation for the area of a trapezium. The equation for the area of a trapezium is described below,
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
From the given diagram, we know π = 3 ππ; π = 6 ππ; β = 4 ππ. Substituting yields:
π΄πππ ππ πππ π =1
2(3 + 6) Γ 4 = 4.5 Γ 4 = 18 ππ2
Therefore, the area of the plane πππ π was calculated to be 18 ππ2 .
Question 6a part (i)
To calculate the volume of water in Tank B, ππ΅ , we use,
ππ΅ = πππ΅2βπ΅
Where; ππ΅ is the radius of the base of Tank B and βπ΅ is the height of Tank B.
Substituting the input parameters and taking π = 3.14 yields,
π = (3.14)Γ (4)2 Γ (5) = 251.2 π2
Question 6a part (ii)
From the information provided, we know that the area of the base of Tank A, π΄π΄ , is
314 π2. To calculate the radius of the base of Tank A, ππ΄ , we use,
ππ΄ = βπ΄π΄
π
Which was derived from the area of a circle π΄ = ππ2 .
Substituting the input parameters and taking π = 3.14 yields,
ππ΄ = β314
3.14= ππ΄ = β100 = Β±10 π2
Since the radius can only be a positive number the radius of the base of Tank A
is 10 π.
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 6a part (iii)
From the information provided, we know that the volume of Tank A, ππ΄ , is eight (8) times greater than the volume of Tank B, ππ΅ . Hence,
ππ΄ = 8ππ΅
To calculate the volume of water in Tank A, ππ΄, we use,
ππ΄ = πππ΄2βπ΄ = 8ππ΅
Where; ππ΄ is the radius of the base of Tank A and βπ΄ is the height of Tank A.
Transposing the above mentioned equation with respect to βπ΄ yields,
βπ΄ =8ππ΅
πππ΄2
Substituting the input parameters and taking π = 3.14 yields,
βπ΄ =8 Γ (251.2)
(3.14) Γ (10)2= 6.4 π
Question 6b part (i)
The Scale Factor is β2.
Question 6b part (ii)
The Scale Factor is negative because the image is on the opposite side of the object and
is flipped upside down.
Question 6b part (iii)
The length of ππ is β13 units, therefore, the length of πβ²πβ² is 2β13 units.
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 6b part (iv)
The length of the base, π, of triangle πππ is 2 ππ. The height, β, of triangle πππ is 3 ππ.
To determine the area of the triangle recall,
π΄ =βπ
2
Substituting the input parameters yields,
π΄ =(3 Γ 2)
2= 3 ππ2
i. The length of the base, π, of triangle πβ²πβ²π β² is 4 ππ. The height, β, of triangle πβ²πβ²π β² is 6 ππ. To determine the area of the triangle recall,
π΄ =βπ
2
Substituting the input parameters yields,
π΄ =(6 Γ 4)
2= 12 ππ2
Therefore, the area of triangle πβ²πβ²π β² is 4 times the area of triangle πππ which 12 ππ2 .
Question 7a
The completed table is shown below:
Month July August September October November
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Sales in $
Thousands 13 20 36 25 15
Question 7b part (i)
August to September
Question 7b part (ii)
July to August
Question 7b part (iii)
The steepness of the lines tells you the rate of sale.
Question 7c
The total sales for the period July to November is equal to the sum of the sales for July,
August, September, October and November, which is $109,000.00. The mean monthly sales for the period July to November, which is five (5) months, is $109,000.00/5 =
$21,800.00. Therefore, the mean monthly sales for the period July to November is $21,800.00.
Question 7d part (i)
From the information provided, we know that the total sale for the period July to
December was $130,00.00. In Part c the total sales for the period July to November was
determined to be $109,000.00.
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
The total sales for the month of December is therefore, $130,000.00 β$109,000.00 = $21,000.00.
Question 7d part (ii)
The line graph, completed to show the sales for December is provided below,
Question 8a
Figure 4 of the Sequence is provided below.
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 8b
The completed table is provided below.
Figure Number of Unit Triangle Number of Unit Sides
1 1 (3 Γ 1)(1+ 1)
2= 3
2 4 (3 Γ 2)(2+ 1)
2= 9
3 9 (3 Γ 3)(3 + 1)
2= 18
4 16 (3 Γ 4)(4 + 1)
2= 30
12 144 (3 Γ 12)(12+ 1)
2= 234
25 625 (3 Γ 25)(25+ 1)
2= 975
n π2 (3π)(π + 1)
2
Question 9a part (i)
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
From the information provided we know that the teacher marked the examination out
of a maximum of 120 marks. A student who scores 60 marks would receive a percentage score of
60
120Γ 100% = 50%
From the information provided we know that the teacher marked the examination out of a maximum of 120 marks. A student who scores 120 marks
would receive a percentage score of 120
120Γ 100% = 100%
Question 9a part (ii)
Expressing the information in Part i in the format (πππ‘π’ππ π ππππ, πππππππ‘πππ π ππππ)
yields: (60,50%) and (120,100%). Plotting these coordinates on the given graph
sheet:
Question 9a part (iii)
As shown in the graph below, if a candidate is awarded 95 marks on the examination
the candidateβs percentage score will be approximately 79.5%.
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 9a part (iv)
From the information provided, for a student to be awarded a Grade A, her percentage
score must be 85% or more. As shown in the graph below, the minimum mark that the candidate needs to score, to receive a Grade A is 102.
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 9b part (i)
From the information provided we can evaluate π(5) as follows:
Recall,
π(π₯) =π₯2 β 1
3
Substituting,
π(5) =(5)2 β 1
3= 8
Question 9b part (ii)
Given π(π₯) to determine πβ1(π₯).
Recall,
π(π₯) = 3π₯ + 2
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Set,
π¦ = 3π₯ + 2
Transposing with respect to x,
π₯ =π¦ β 2
3
Hence,
πβ1(π¦) =π¦ β 2
3
Writing in terms of π₯,
πβ1(π₯) =π₯ β 2
3
i. To write an expression for ππ(π₯), in the form (π₯ β π)(π₯ + π).
Substituting the function π(π₯) into π(π₯), such that π(π(π₯)):
π(3π₯ + 2) =(3π₯ + 2)2 β 1
3
Factorizing,
ππ(π₯) =(3π₯ + 2 β 1)(3π₯ + 2 + 1)
3
Simplifying,
ππ(π₯) =(3π₯ + 1)(3π₯ + 3)
3
ππ(π₯) =3(3π₯ + 1)(π₯ + 1)
3
ππ(π₯) = (3π₯ + 1)(π₯ + 1)
Question 10a part (i)
The labelled diagram is provided below,
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 10a part (ii)
The length of the flagpole, π·ππ π‘ππππ ππ, can be calculated as follows:
π·ππ π‘ππππ ππ = (π·ππ π‘ππππ ππ΅) β (π·ππ π‘ππππ ππ΅)
π·ππ π‘ππππ ππ΅ can be calculated by using the following trigonometric relationship: π·ππ π‘ππππ ππ΅ = (π·ππ π‘ππππ π΅π ) π‘ππ π‘ππ (β π΅π π)
Substituting the input parameters yields,
π·ππ π‘ππππ ππ΅ = (60) π‘ππ π‘ππ (42Β°) = 54.02 π
π·ππ π‘ππππ ππ΅ can be calculated by using the following trigonometric relationship:
π·ππ π‘ππππ ππ΅ = (π·ππ π‘ππππ π΅π ) π‘ππ π‘ππ (β π΅π π)
Substituting the input parameters yields,
π·ππ π‘ππππ ππ΅ = (60) π‘ππ π‘ππ (35Β°) = 42.01 π
Therefore, the length of the flagpole is: π·ππ π‘ππππ ππ = (54.02) β (42.01) = 12.01 π.
The length of the flagpole is 12 π.
Question 10b part (i)
The bearing of 040Β° is shown in the diagram below:
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
Question 10b part (ii)
Using the principles of alternating angles, the measure of β πΎπΏπ is 40Β°.
Question 10b part (iii)
can be calculated using the principles of the Laws of Cosine. Applying the Laws of Cosine to the question yields,
(|πΎπ|)= β(πΎπΏ)2 + (ππΏ)2 β 2(πΎπΏ)(ππΏ)πππ (β πΎπΏπ)
Substituting the input parameters yields:
(|πΎπ|) = β(80)2 + (120)2 β 2(80)(120)πππ (40Β°)
(|πΎπ|) = 78.05 ππ
The length of πΎπ, to the nearest kilometre is 78 ππ.
Question 10b part (iv)
The measure of β πΏπΎπ can be calculated using the Laws of Sine.
Applying the Laws of Sine to the equation yields, |πΏπ|
π ππ π ππ (β πΏπΎπ) =
|πΎπ|
π ππ π ππ (β πΎπΏπ)
Substituting the input parameters yields:
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
120
π ππ π ππ (β πΏπΎπ) =
78
π ππ π ππ (40Β°)
Solving for β πΏπΎπ:
β πΏπΎπ = (120 π ππ π ππ (40Β°)
78) = 81.45Β°
The measure of β πΏπΎπ, to the nearest degree is 81Β°.
i. We know, β πΏπΎπ = 81.45Β°. Therefore, the bearing of π from πΎ is 40Β° + 81.45Β° =
121.45Β°. The bearing of π from πΎ is 121.45Β°.
Question 11a part (i)
The matrix product π΄π΅ is calculated below: π΄π΅ = (1 1 2 3 )(1 2 0 1 ) = (1 + 0 2 + 1 2 + 0 4 + 3 ) = (1 3 2 7 )
Question 11a part (ii)
To prove that the matrix product of π΄ and π΅ is not commutative, the matrix product π΅π΄ must first be calculated:
π΅π΄ = (1 2 0 1 )(1 1 2 3 ) = (1 + 4 1 + 6 0 + 2 0 + 3 ) = (5 7 2 3 )
Since π΄π΅ β π΅π΄ the matrix product of π΄ and π΅ is not commutative
Question 11a part (iii)
To determine π΄β1,
The adjoint matrix of π΄ is provided below,
π΄πππ = (7 β 3 β 2 1 )
The determinant of π΄ is |π΄| = (1 Γ 7) β (3 Γ 2) = 1
The inverse of matrix π΄ is therefore,
www.kerwinspringer.com
Whatsapp +1868 784 0619 to register for premium online classes @ The Student Hub
π΄β1 = (7 β 3 β 2 1 )
Question 11a part (iv)
From the information provided we can solve for π₯ as follows:
The determinant of π, |π|, is: |π| = (2π₯ Γ 3) β (2 Γ 9) = 6π₯ β 18
Since, |π| = 0, |π| = 6π₯ β 18 = 0
Solving for π₯ yields,
π₯ =18
6= 3
Question 11b part (i)
The value of |ππ ββββββ | is as calculated,
|ππ ββββββ | = β(β3)2 + (4)2 = 5
Question 11b part (ii)
The vectors π πββββ β and ππββ ββ expressed in the form (π₯ π¦ ) is given below,
π πββββ β = βππ ββ ββ β + ππββββ β
π πββββ β = β(β3 4 )+ (1 1 ) = (4 β 3 )
ππββ ββ = βππββββ β + ππββ ββ β
ππββ ββ = β(1 1 ) + (5 β 2 ) = (4 β 3 )
Question 11b part (iii)
Since π πββββ β = ππββ ββ both vectors are parallel to each other. Since both vectors connect at S they form a straight line π ππ .
top related