Solutions to CSEC Maths P2 June 2015 - img1.wsimg.com

Solutions to CSEC Maths P2 June 2015

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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, 𝑋 = $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, 𝑌 = $2.40.

If 1kg of flour costs $2.40, then 4kg of flour will cost 4 × $2.40 = $9.60. Hence,

𝑍 = $9.60.



Question 1b part (iii)

The tax exclusive sum of Mrs Rowe’s 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 × $23.60 =

$2.36.

The tax inclusive sum of Mrs Rowe’s bill is $23.60 + $2.36 = $25.96.


We know 𝑎 = 4; 𝑏 = 2;𝑐 = −1. Hence, 𝑎 − 𝑏 + 𝑐 = (4) − (2) + (−1) = 1


We know 𝑎 = 4; 𝑏 = 2;𝑐 = −1.

Hence, 2𝑎𝑏 = 2 × (4)2 = 32


If 𝑝 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 − 𝑝) 𝑚𝑖𝑙𝑙𝑖𝑙𝑖𝑡𝑟𝑒𝑠.


If 𝑞 glasses, each containing 𝑟 millilitres, are poured out then 𝑞 × 𝑟 millilitres is the total quantity that was poured out of the bottle. The amount of juice left in the bottle is equal

to (500 − (𝑞 × 𝑟)) 𝑚𝑖𝑙𝑙𝑖𝑙𝑖𝑡𝑟𝑒𝑠.

Question 2c

From the given information recall: 2𝑘

3+

2 − 𝑘

5

The Lowest Common Multiple of the abovementioned fraction is 15, hence:



5 × (2𝑘) + 3 × (2 − 𝑘)

15

Simplifying, 10𝑘 + 6 − 3𝑘

15

Simplifying into its simplest form: 7𝑘 + 6

15

Question 2d part (i)

Let the quantity of mangoes be represented by 𝑥 and the quantity of pears be

represented by 𝑦.

Using the information provided can be represented by the pair of simultaneous

equations: 4𝑥 + 2𝑦 = 24

2𝑥 + 3𝑦 = 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:

𝑎3 − 12𝑎



Factorizing into its simplest form:

𝑎(𝑎2 − 12)

Question 2e part (ii)

From the given information recall:

2𝑥2 − 5𝑥 + 3

Expanding:

2𝑥2 − 2𝑥 − 3𝑥 + 3

Factorizing:

2𝑥(𝑥 − 1) − 3(𝑥 − 1)

Factorizing into its simplest form: (2𝑥 − 3)(𝑥− 1)


From the given information we know that 12 students played neither the guitar

nor the violin.


The total number of students in the class, as an expression of x, is stated below:

2𝑥 + 𝑥 + 4 + 12

3𝑥 + 16




The total number of students in the class, as an equation of x, is stated below:

3𝑥 + 16 = 40


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,

𝑥 =40 − 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.


The triangle can be constructed using the information and instructions provided. The

final triangle is given below,



The size of angle BAC is 33.69°.


The point D, such that ABCD is a parallelogram is identified in the diagram below.

Question 4a



The function to be used is 𝑦 = 𝑥2 − 2𝑥 − 3 . The unknown values for 𝑦, in the table,

can be determined by substituting the corresponding, known values for 𝑥, into the aforementioned function. Hence,

𝑦(−1) = (−1)2 − 2(−1)− 3 = 1 + 2 − 3 = 0

𝑦(3) = (3)2 − 3(2) − 3 = 9 − 6 − 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 (𝑥, 𝑦) coordinates tabulate above and the

instructions provided. The graph for the given function is illustrated below,

Question 4c

The smooth curve drawn through the (𝑥,𝑦) coordinates is illustrated below,




From the graph, the values for 𝑥 for which 𝑥2 − 2𝑥 − 3 = 0 are −1 and 3.


From the graph, the minimum value of 𝑥2 − 2𝑥 − 3 is −4.

Question 4d part (iii)

The equation of the line of symmetry of the graph of 𝑦 = 𝑥2 − 2𝑥 − 3 is 𝑥 = 1.


If the car is travelling at a constant speed of 54 𝑘𝑚/ℎ, then after 2.25 ℎ𝑜𝑢𝑟𝑠 the car

would have travelled (54 𝑘𝑚/ℎ) × (2.25 ℎ𝑜𝑢𝑟𝑠) = 121.5 𝑘𝑚.

Using the given information, we must first convert the speed that the car travels from 𝑘𝑚/ℎ to 𝑚/𝑠. Hence, the speed that the car travels, in 𝑚/𝑠 is:



𝑠𝑝𝑒𝑒𝑑 (𝑚/𝑠) = (5

18) × (54) = 15 𝑚/𝑠𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑐𝑎𝑟 𝑡𝑟𝑎𝑣𝑒𝑙𝑠

a distance of 315 𝑚, we can determine the time that car took the travel the aforementioned distance using the following equation:

𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 = (𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑)/𝑠𝑝𝑒𝑒𝑑𝑆𝑖𝑛𝑐𝑒


𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 = 315 𝑚 and 𝑠𝑝𝑒𝑒𝑑 = 15𝑚/𝑠, the time taken is: 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 = (315)/(15) = 21 𝑠𝑒𝑐𝑜𝑛𝑑𝑠


The scale 1 millimetre = 1 meter in the form 1: 𝑥 is: 1 × 10−3 𝑚𝑒𝑡𝑒𝑟𝑠 ∶ 1 𝑚𝑒𝑡𝑒𝑟

Hence,

1 𝑚𝑒𝑡𝑒𝑟 ∶1

10−3 𝑚𝑒𝑡𝑒𝑟𝑠

Which is,

1 𝑚𝑒𝑡𝑒𝑟 ∶ 1000 𝑚𝑒𝑡𝑒𝑟𝑠

The scale 2 cm = 6 m in the form 1: 𝑥 is: 2 × 10−2 𝑚𝑒𝑡𝑒𝑟𝑠 ∶ 6 𝑚𝑒𝑡𝑒𝑟𝑠

Hence,

1 𝑚𝑒𝑡𝑒𝑟 ∶6

2 × 10−2 𝑚𝑒𝑡𝑒𝑟𝑠

Which is, 1 𝑚𝑒𝑡𝑒𝑟 ∶ 300 𝑚𝑒𝑡𝑒𝑟𝑠

Question 5c part (i)

Using the information provided, the distance, in centimetres, from 𝑄 to 𝑅 on the map is

6 𝑐𝑚.

Question 5c part (ii)

As shown the in illustration below there are approximately eighteen (18) 1 𝑐𝑚2 tiles. All of the whole tiles are coloured in grey. The corresponding pieces of broken 1 𝑐𝑚2 tiles



that forms a full 1 𝑐𝑚2 tile is appropriately colour coded. Therefore, the area, in square

centimetres of the plane 𝑃𝑄𝑅𝑆 on the map is 18 𝑐𝑚2 .

Question 5c part (iii)

From the information provided, we know that 1 𝑐𝑚 ∶ 2,000 𝑐𝑚. Therefore, the actual distance, in centimetres, between 𝑄 and 𝑅, is 6 𝑐𝑚 × 2,000 = 12,000 𝑐𝑚. The actual

distance, in metres between 𝑄 and 𝑅 is, 12,000 × 10−2 𝑚 = 120 𝑚. The actual distance, in kilometres between 𝑄 and 𝑅 is, (120/1000) 𝑘𝑚 = 0.12 𝑘𝑚.

Question 5c part (iv)

To calculate the area, in square meters of the plane 𝑃𝑄𝑅𝑆 we must use the equation for the area of a trapezium. The equation for the area of a trapezium is described below,



From the given diagram, we know 𝑎 = 3 𝑐𝑚; 𝑏 = 6 𝑐𝑚; ℎ = 4 𝑐𝑚. Substituting yields:

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑃𝑄𝑅𝑆 =1

2(3 + 6) × 4 = 4.5 × 4 = 18 𝑐𝑚2

Therefore, the area of the plane 𝑃𝑄𝑅𝑆 was calculated to be 18 𝑐𝑚2 .


To calculate the volume of water in Tank B, 𝑉𝐵 , we use,

𝑉𝐵 = 𝜋𝑟𝐵2ℎ𝐵

Where; 𝑟𝐵 is the radius of the base of Tank B and ℎ𝐵 is the height of Tank B.

Substituting the input parameters and taking 𝜋 = 3.14 yields,

𝑉 = (3.14)× (4)2 × (5) = 251.2 𝑚2


From the information provided, we know that the area of the base of Tank A, 𝐴𝐴 , is

314 𝑚2. To calculate the radius of the base of Tank A, 𝑟𝐴 , we use,

𝑟𝐴 = √𝐴𝐴

𝜋

Which was derived from the area of a circle 𝐴 = 𝜋𝑟2 .


𝑟𝐴 = √314

3.14= 𝑟𝐴 = √100 = ±10 𝑚2

Since the radius can only be a positive number the radius of the base of Tank A

is 10 𝑚.




From the information provided, we know that the volume of Tank A, 𝑉𝐴 , is eight (8) times greater than the volume of Tank B, 𝑉𝐵 . Hence,

𝑉𝐴 = 8𝑉𝐵

To calculate the volume of water in Tank A, 𝑉𝐴, we use,

𝑉𝐴 = 𝜋𝑟𝐴2ℎ𝐴 = 8𝑉𝐵

Where; 𝑟𝐴 is the radius of the base of Tank A and ℎ𝐴 is the height of Tank A.

Transposing the above mentioned equation with respect to ℎ𝐴 yields,

ℎ𝐴 =8𝑉𝐵

𝜋𝑟𝐴2


ℎ𝐴 =8 × (251.2)

(3.14) × (10)2= 6.4 𝑚


The Scale Factor is −2.


The Scale Factor is negative because the image is on the opposite side of the object and

is flipped upside down.


The length of 𝑃𝑄 is √13 units, therefore, the length of 𝑃′𝑄′ is 2√13 units.



Question 6b part (iv)

The length of the base, 𝑏, of triangle 𝑃𝑄𝑅 is 2 𝑐𝑚. The height, ℎ, of triangle 𝑃𝑄𝑅 is 3 𝑐𝑚.

To determine the area of the triangle recall,

𝐴 =ℎ𝑏

2

Substituting the input parameters yields,

𝐴 =(3 × 2)

2= 3 𝑐𝑚2

i. The length of the base, 𝑏, of triangle 𝑃′𝑄′𝑅′ is 4 𝑐𝑚. The height, ℎ, of triangle 𝑃′𝑄′𝑅′ is 6 𝑐𝑚. To determine the area of the triangle recall,

𝐴 =ℎ𝑏

2


𝐴 =(6 × 4)

2= 12 𝑐𝑚2

Therefore, the area of triangle 𝑃′𝑄′𝑅′ is 4 times the area of triangle 𝑃𝑄𝑅 which 12 𝑐𝑚2 .

Question 7a

The completed table is shown below:

Month July August September October November



Sales in $

Thousands 13 20 36 25 15


August to September


July to August


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.


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.



The total sales for the month of December is therefore, $130,000.00 −$109,000.00 = $21,000.00.


The line graph, completed to show the sales for December is provided below,

Question 8a

Figure 4 of the Sequence is provided below.



Question 8b

The completed table is provided below.

Figure Number of Unit Triangle Number of Unit Sides

1 1 (3 × 1)(1+ 1)

2= 3

2 4 (3 × 2)(2+ 1)

2= 9

3 9 (3 × 3)(3 + 1)

2= 18

4 16 (3 × 4)(4 + 1)

2= 30

12 144 (3 × 12)(12+ 1)

2= 234

25 625 (3 × 25)(25+ 1)

2= 975

n 𝑛2 (3𝑛)(𝑛 + 1)

2




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

120× 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

120× 100% = 100%


Expressing the information in Part i in the format (𝑎𝑐𝑡𝑢𝑎𝑙 𝑠𝑐𝑜𝑟𝑒, 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑠𝑐𝑜𝑟𝑒)

yields: (60,50%) and (120,100%). Plotting these coordinates on the given graph

sheet:


As shown in the graph below, if a candidate is awarded 95 marks on the examination

the candidate’s percentage score will be approximately 79.5%.




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.




From the information provided we can evaluate 𝑔(5) as follows:

Recall,

𝑔(𝑥) =𝑥2 − 1

3

Substituting,

𝑔(5) =(5)2 − 1

3= 8


Given 𝑓(𝑥) to determine 𝑓−1(𝑥).

Recall,

𝑓(𝑥) = 3𝑥 + 2



Set,

𝑦 = 3𝑥 + 2

Transposing with respect to x,

𝑥 =𝑦 − 2

3

Hence,

𝑓−1(𝑦) =𝑦 − 2

3

Writing in terms of 𝑥,

𝑓−1(𝑥) =𝑥 − 2

3

i. To write an expression for 𝑔𝑓(𝑥), in the form (𝑥 − 𝑎)(𝑥 + 𝑏).

Substituting the function 𝑓(𝑥) into 𝑔(𝑥), such that 𝑔(𝑓(𝑥)):

𝑔(3𝑥 + 2) =(3𝑥 + 2)2 − 1

3

Factorizing,

𝑔𝑓(𝑥) =(3𝑥 + 2 − 1)(3𝑥 + 2 + 1)

3

Simplifying,

𝑔𝑓(𝑥) =(3𝑥 + 1)(3𝑥 + 3)

3

𝑔𝑓(𝑥) =3(3𝑥 + 1)(𝑥 + 1)

3

𝑔𝑓(𝑥) = (3𝑥 + 1)(𝑥 + 1)


The labelled diagram is provided below,




The length of the flagpole, 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑃𝑇, can be calculated as follows:

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑃𝑇 = (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑃𝐵) − (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑇𝐵)

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑃𝐵 can be calculated by using the following trigonometric relationship: 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑃𝐵 = (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐵𝑅) 𝑡𝑎𝑛 𝑡𝑎𝑛 (∠𝐵𝑅𝑃)


𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑃𝐵 = (60) 𝑡𝑎𝑛 𝑡𝑎𝑛 (42°) = 54.02 𝑚

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑇𝐵 can be calculated by using the following trigonometric relationship:

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑇𝐵 = (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐵𝑅) 𝑡𝑎𝑛 𝑡𝑎𝑛 (∠𝐵𝑅𝑇)


𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑇𝐵 = (60) 𝑡𝑎𝑛 𝑡𝑎𝑛 (35°) = 42.01 𝑚

Therefore, the length of the flagpole is: 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑃𝑇 = (54.02) − (42.01) = 12.01 𝑚.

The length of the flagpole is 12 𝑚.


The bearing of 040° is shown in the diagram below:




Using the principles of alternating angles, the measure of ∠𝐾𝐿𝑀 is 40°.


can be calculated using the principles of the Laws of Cosine. Applying the Laws of Cosine to the question yields,

(|𝐾𝑀|)= √(𝐾𝐿)2 + (𝑀𝐿)2 − 2(𝐾𝐿)(𝑀𝐿)𝑐𝑜𝑠(∠𝐾𝐿𝑀)

Substituting the input parameters yields:

(|𝐾𝑀|) = √(80)2 + (120)2 − 2(80)(120)𝑐𝑜𝑠(40°)

(|𝐾𝑀|) = 78.05 𝑘𝑚

The length of 𝐾𝑀, to the nearest kilometre is 78 𝑘𝑚.

Question 10b part (iv)

The measure of ∠𝐿𝐾𝑀 can be calculated using the Laws of Sine.

Applying the Laws of Sine to the equation yields, |𝐿𝑀|

𝑠𝑖𝑛 𝑠𝑖𝑛 (∠𝐿𝐾𝑀) =

|𝐾𝑀|

𝑠𝑖𝑛 𝑠𝑖𝑛 (∠𝐾𝐿𝑀)

Substituting the input parameters yields:



120

𝑠𝑖𝑛 𝑠𝑖𝑛 (∠𝐿𝐾𝑀) =

78

𝑠𝑖𝑛 𝑠𝑖𝑛 (40°)

Solving for ∠𝐿𝐾𝑀:

∠𝐿𝐾𝑀 = (120 𝑠𝑖𝑛 𝑠𝑖𝑛 (40°)

78) = 81.45°

The measure of ∠𝐿𝐾𝑀, to the nearest degree is 81°.

i. We know, ∠𝐿𝐾𝑀 = 81.45°. Therefore, the bearing of 𝑀 from 𝐾 is 40° + 81.45° =

121.45°. The bearing of 𝑀 from 𝐾 is 121.45°.


The matrix product 𝐴𝐵 is calculated below: 𝐴𝐵 = (1 1 2 3 )(1 2 0 1 ) = (1 + 0 2 + 1 2 + 0 4 + 3 ) = (1 3 2 7 )


To prove that the matrix product of 𝐴 and 𝐵 is not commutative, the matrix product 𝐵𝐴 must first be calculated:

𝐵𝐴 = (1 2 0 1 )(1 1 2 3 ) = (1 + 4 1 + 6 0 + 2 0 + 3 ) = (5 7 2 3 )

Since 𝐴𝐵 ≠ 𝐵𝐴 the matrix product of 𝐴 and 𝐵 is not commutative


To determine 𝐴−1,

The adjoint matrix of 𝐴 is provided below,

𝐴𝑎𝑑𝑗 = (7 − 3 − 2 1 )

The determinant of 𝐴 is |𝐴| = (1 × 7) − (3 × 2) = 1

The inverse of matrix 𝐴 is therefore,



𝐴−1 = (7 − 3 − 2 1 )


From the information provided we can solve for 𝑥 as follows:

The determinant of 𝑀, |𝑀|, is: |𝑀| = (2𝑥 × 3) − (2 × 9) = 6𝑥 − 18

Since, |𝑀| = 0, |𝑀| = 6𝑥 − 18 = 0

Solving for 𝑥 yields,

𝑥 =18

6= 3


The value of |𝑂𝑅⃗⃗⃗⃗⃗⃗ | is as calculated,

|𝑂𝑅⃗⃗⃗⃗⃗⃗ | = √(−3)2 + (4)2 = 5


The vectors 𝑅𝑆⃗⃗⃗⃗ ⃗ and 𝑆𝑇⃗⃗ ⃗⃗ expressed in the form (𝑥 𝑦 ) is given below,

𝑅𝑆⃗⃗⃗⃗ ⃗ = −𝑂𝑅⃗⃗ ⃗⃗ ⃗ + 𝑂𝑆⃗⃗⃗⃗ ⃗

𝑅𝑆⃗⃗⃗⃗ ⃗ = −(−3 4 )+ (1 1 ) = (4 − 3 )

𝑆𝑇⃗⃗ ⃗⃗ = −𝑂𝑆⃗⃗⃗⃗ ⃗ + 𝑂𝑇⃗⃗ ⃗⃗ ⃗

𝑆𝑇⃗⃗ ⃗⃗ = −(1 1 ) + (5 − 2 ) = (4 − 3 )


Since 𝑅𝑆⃗⃗⃗⃗ ⃗ = 𝑆𝑇⃗⃗ ⃗⃗ both vectors are parallel to each other. Since both vectors connect at S they form a straight line 𝑅𝑆𝑇 .

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Solutions to CSEC Maths P2 June 2015 - img1.wsimg.com