Solutions to CSEC Maths P2 January 2011
Transcript of Solutions to CSEC Maths P2 January 2011
Solutions to CSEC Maths P2 January 2011
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Question 1a part (i)
Calculate the exact value of
(5.82 + 1.02) ร 2.5
= 34.66 ร 2.5
= 86.65 to 2 decimal places
Question 1a part (ii)
Calculate the exact value of
๐๐๐
๐ ๐๐
โ ๐
๐
=
229
143
โ 3
7
= (22
9 ร
3
14) โ
3
7
= (66
126) โ
3
7
=11
21โ
3
7
=11 โ 3(3)
21
=๐
๐๐
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Question 1b part (i)
Data Given
Employee is paid a basic wage of $9.50 per hour for a 40 hour week and one and a half times
this rate for overtime.
Calculate the basic weekly wage of one employee for a 40-hour work week
๐ต๐๐ ๐๐ ๐ค๐๐๐๐๐ฆ ๐ค๐๐๐ = โ๐๐ข๐๐๐ฆ ๐๐๐ก๐ ร
๐๐ข๐๐๐๐ ๐๐ โ๐๐ข๐๐ ๐๐ ๐ ๐๐๐ ๐๐ ๐ค๐๐๐ ๐ค๐๐๐
= $9.50 ร 40
= $380 ๐๐๐ ๐ค๐๐๐
Question 1b part (ii)
Calculate the overtime wage for an employee who worked 6 hours overtime
๐๐ฃ๐๐๐ก๐๐๐ ๐ ๐๐ก๐ = 1.5 ร $9.50
= $14.25 ๐๐๐ ๐๐ฃ๐๐๐ก๐๐๐ โ๐๐ข๐
๐๐ฃ๐๐๐ก๐๐๐ ๐ค๐๐๐ ๐๐๐ 6 โ๐๐ข๐๐ = $14.25 ร 6
= $85.50
Question 1b part (iii)
Data Given
A total of $12 084.00 in basic and overtime wages were paid to 30 employees by the company in
a certain week.
Calculate the total paid in overtime wages
๐ต๐๐ ๐๐ ๐ค๐๐๐ ๐๐ 30 ๐๐๐๐๐๐ฆ๐๐๐ ๐๐๐ ๐ 40 โ๐๐ข๐ ๐ค๐๐๐ ๐ค๐๐๐
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= ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐๐๐ฆ๐๐๐ ร ๐๐๐ ๐๐ ๐ค๐๐๐๐๐ฆ ๐ค๐๐๐
= 30 ร $380
= $11, 400
๐๐๐ก๐๐ ๐ค๐๐๐๐ ๐๐๐๐ = $12,084 (๐บ๐๐ฃ๐๐ ๐ท๐๐ก๐)
๐๐๐ก๐๐ ๐๐๐๐ ๐๐ ๐๐ฃ๐๐๐ก๐๐๐ ๐ค๐๐๐๐
= ๐๐๐ก๐๐ ๐ค๐๐๐๐ ๐๐๐๐ โ ๐ต๐๐ ๐๐ ๐ค๐๐๐ ๐๐ 30 ๐๐๐๐๐๐ฆ๐๐๐ ๐๐๐ ๐ 40 โ๐๐ข๐ ๐ค๐๐๐ ๐ค๐๐๐
= $12,084 โ $11, 400
= $12,084 โ $11, 400
= $684.00
Question 1b part (iv)
Calculate the total number of overtime hours worked by employees
๐๐๐ก๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐๐ฃ๐๐๐ก๐๐๐ โ๐๐ข๐๐ ๐ค๐๐๐๐๐ ๐๐ฆ ๐๐๐๐๐๐ฆ๐๐๐ = ๐๐๐ก๐๐ ๐๐๐๐ ๐๐ ๐๐ฃ๐๐๐ก๐๐๐ ๐ค๐๐๐๐
๐๐ฃ๐๐๐ก๐๐๐ ๐ ๐๐ก๐
= $684.00
$14.25 ๐๐๐ โ๐๐ข๐
= 48 โ๐๐ข๐๐
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Question 2a
Simplify 2๐ฅ
5โ
๐ฅ
3
2๐ฅ
5โ
๐ฅ
3
= 3(2๐ฅ) โ 5(๐ฅ)
15
= 6๐ฅ โ 5๐ฅ
15
= ๐ฅ
15
Question 2b
Factorise ๐2๐ + 2๐๐
๐2๐ + 2๐๐
= ๐๐(๐ + 2)
Question 2c
Express p as the subject of the formula ๐ =๐2โ๐
๐ก
๐
1 =
๐2โ๐
๐ก
๐๐ก = (๐2 โ ๐) (1)
๐2 = ๐๐ก + ๐
๐ = โ๐๐ก + ๐
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Question 2d part (i)
Data Given
Type of Box Number of Donuts per Box
Small box x
Large Box 2x + 3
In total 8 small boxes and 5 large boxes were sold.
Write an expression in terms of x to represent the Total number of donuts sold.
๐๐๐ก๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐ข๐ก๐ ๐ ๐๐๐ = ๐๐ข๐๐๐๐ ๐๐ ๐ ๐๐๐๐ ๐๐๐ฅ๐๐ ๐ ๐๐๐ + ๐๐ข๐๐๐๐ ๐๐ ๐๐๐ ๐๐๐ฅ๐๐ ๐ ๐๐๐
= 8(๐ฅ) + 5(2๐ฅ + 3)
= 8๐ฅ + 10๐ฅ + 15
= 18๐ฅ + 15
Question 2d part (ii)
Data Given
The total number of donuts sold was 195.
(a) Calculate the number of donuts in a small box
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๐๐๐ก๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐ข๐ก๐ ๐ ๐๐๐ = 18๐ฅ + 15
195 = 18๐ฅ + 15
18๐ฅ = 195 โ 15
18๐ฅ = 180
๐ฅ = 10
๐ต๐๐ ๐๐ ๐๐ ๐๐๐ฃ๐๐ ๐๐๐ก๐: ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐ข๐ก๐ ๐๐ ๐ ๐๐๐๐ ๐๐๐ฅ = ๐ฅ
โด ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐ข๐ก๐ ๐๐ ๐ ๐๐๐๐ ๐๐๐ฅ = 10
(b) Calculate the number of donuts in a large box
๐ต๐๐ ๐๐ ๐๐ ๐๐๐ฃ๐๐ ๐๐๐ก๐: ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐ข๐ก๐ ๐๐ ๐๐๐๐๐ ๐๐๐ฅ = 2๐ฅ + 3
โด ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐ข๐ก๐ ๐๐ ๐๐๐๐๐ ๐๐๐ฅ = 2(10) + 3
= 23
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Simplify 7 ๐5๐3 ร 2 ๐2๐
= 7 ร 2 ร ๐5+2 ร ๐3+1
= 14 ๐7๐4
Data Given
One carton of milk measures 6 cm by 4 cm by 10 cm.
Calculate, in cm3, the volume of milk in a carton
๐๐๐๐ข๐๐ ๐๐ ๐๐๐๐ ๐๐ ๐ ๐๐๐๐ก๐๐ = ๐ฟ ร ๐ ร ๐ป
= 6 ร 4 ร 10
= 240 ๐๐3
Data Given
For a recipe to make ice-cream 3 litres of milk are required.
Calculate the number of cartons of milk that should be bought to make the ice-cream
1 ๐๐๐ก๐๐ = 1000 ๐๐3
3 ๐๐๐ก๐๐๐ ๐๐ ๐๐๐๐ = 3 ร 1000
Question 3a
Question 3b part (i)
Question 3b part (ii)
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= 3000 ๐๐3 ๐๐ ๐๐๐๐
๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐ก๐๐๐ ๐๐ ๐๐๐๐ ๐๐๐๐ข๐๐๐๐ = ๐๐๐๐ข๐๐ ๐๐ ๐๐๐๐ ๐๐๐๐ข๐๐๐๐
๐๐๐๐ข๐๐ ๐๐ ๐ ๐๐๐๐ก๐๐
= 3000 ๐๐3
240 ๐๐3
= 12.5
Therefore 13 cartons of milk should be purchased make the ice โ cream according to its recipe.
Data Given
One carton of milk is poured into a cylindrical cup of internal diameter 5 cm.
๐ = 3.14
Calculate the height of milk in the cup
๐๐๐๐ข๐๐ ๐๐ ๐๐๐๐ ๐๐ ๐๐๐ ๐๐๐๐ก๐๐ = ๐๐๐๐ข๐๐ ๐๐ ๐๐๐๐ ๐๐ ๐กโ๐ ๐๐ข๐
๐๐๐๐ข๐๐ ๐๐ ๐๐๐๐ ๐๐ ๐กโ๐ ๐๐ข๐ = ๐๐2โ
Where ๐ = ๐๐๐๐๐ข๐ ๐๐ ๐กโ๐ ๐๐ข๐
โ = โ๐๐๐โ๐ก ๐๐ ๐กโ๐ ๐๐๐๐ ๐๐ ๐กโ๐ ๐๐ข๐
๐ = ๐๐๐ก๐๐๐๐๐ ๐๐๐๐๐๐ก๐๐ ๐๐ ๐กโ๐ ๐๐ข๐
2
= 5 ๐๐
2
= 2.5 ๐๐
Question 3b part (iii)
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๐๐๐๐ข๐๐ ๐๐ ๐๐๐๐ ๐๐ ๐กโ๐ ๐๐ข๐ = ๐๐2โ
240 ๐๐3 = (3.14) ร (2.5 ๐๐)2 ร โ
โ =240
(3.14) ร (2.5 )2
โ = 12.23 ๐๐
๐ = ๐๐. ๐ ๐๐ ๐๐ ๐ ๐๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐
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Question 4a part (i)
Data Given
๐ = {๐โ๐๐๐ ๐๐ข๐๐๐๐๐ ๐๐๐๐ 1 ๐ก๐ 12}
๐ป = {๐๐๐ ๐๐ข๐๐๐๐๐ ๐๐๐ก๐ค๐๐๐ 4 ๐๐๐ 12}
๐ฝ = {๐๐๐๐๐ ๐๐ข๐๐๐๐๐ ๐๐๐๐ 1 ๐ก๐ 12}
List the members of ๐ป
๐ป = {๐๐๐ ๐๐ข๐๐๐๐๐ ๐๐๐ก๐ค๐๐๐ 4 ๐๐๐ 12}
= {5, 7, 9, 11}
Question 4a part (ii)
List the members of ๐ฝ
๐ฝ = { ๐๐๐๐๐ ๐๐ข๐๐๐๐๐ ๐๐๐๐ 1 ๐ก๐ 12}
= {2, 3, 5, 7, 11}
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Question 4a part (iii)
Draw a Venn diagram to represent ๐,๐ป and ๐ฝ
Question 4b part (i)
Construct a triangle LMN with angle ๐ฟ๐๐ = 60ยฐ ,๐๐ = 9 ๐๐ ๐๐๐ ๐ฟ๐ = 7 ๐๐
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Question 4b part (ii)
Measure and state the size of ๐๏ฟฝฬ๏ฟฝ๐ฟ
Using a protractor ๐๏ฟฝฬ๏ฟฝ๐ฟ was found to be 48ยฐ.
Question 4b part (iii)
Show the point, K, such that KLMN is a parallelogram
Since the opposite sides of a parallelogram are equivalent in length;
๐๐ฟ = ๐๐พ = 7 ๐๐
๐๐ = ๐ฟ๐พ = 9 ๐๐
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Question 5a (i)
Data Given
Equation of a straight line is 3๐ฆ = 2๐ฅ โ 6
Determine the gradient of the line
3๐ฆ = 2๐ฅ โ 6
In the form ๐ฆ = ๐๐ฅ + ๐, where the gradient of the line is given by โmโ.
๐ฆ =2
3๐ฅ โ 2
๐บ๐๐๐๐๐๐๐ก (๐1) = 2
3
Question 5a (ii)
Determine the equation of the line perpendicular to 3๐ฆ = 2๐ฅ โ 6 that passes through the point
(4, 7)
๐บ๐๐๐๐๐๐๐ก ๐๐ ๐ฟ๐๐๐ = ๐2
๐1 ร ๐2 = โ1 ; ๐ ๐๐๐๐ ๐กโ๐ ๐๐๐๐๐ ๐๐๐ ๐๐๐๐๐๐๐๐๐๐ข๐๐๐
๐2 = โ1
๐1
= โ1
23
= โ3
2
๐ธ๐๐ข๐๐ก๐๐๐ ๐๐ ๐ฟ๐๐๐: (๐ฆ โ ๐ฆ1) = ๐ (๐ฅ โ ๐ฅ1)
(๐ฆ โ 7) = โ3
2 (๐ฅ โ 4)
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(๐ฆ โ 7) = โ3
2 ๐ฅ +
12
2
2๐ฆ โ 14 = โ3 ๐ฅ + 12
2๐ฆ = โ3 ๐ฅ + 26
Question 5b part (i)
Data Given: Diagram showing ๐: ๐ฅ โ ๐ฅ2 โ ๐ where ๐ฅ ๐ {3, 4, 5, 6, 7, 8, 9, 10}
Calculate the value of ๐
Using the data given where 4 is mapped onto 11.
๐ฅ2 โ ๐ = 11
๐ = ๐ฅ2 โ 11
๐ = (4)2 โ 11
๐ = 16 โ 11
๐ = 5
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Question 5b part (ii)
Calculate the value of ๐(3)
๐(๐ฅ) = ๐ฅ2 โ ๐
Substituting ๐ฅ = 3 and ๐ = 5 gives
๐(3) = (3)2 โ 5
๐(3) = 9 โ 5
= 4
Question 5b part (iii)
Calculate the value of ๐ฅ when ๐(๐ฅ) = 95
๐(๐ฅ) = ๐ฅ2 โ ๐
๐ฅ2 โ ๐ = 95
๐ฅ2 = 95 + ๐
๐ฅ2 = 95 + 5
๐ฅ2 = 100
๐ฅ = โ100
๐ฅ = ยฑ10
Based on given data the value of x would be 10 when f(x) = 95.
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Question 6 part (i)
Data Given: Line graph showing the monthly sales, in thousands of dollars, at a school cafeteria
for the period January to May 2010.
Copy and complete the table
Month Jan Feb Mar Apr May
Sales in
$Thousands
38 35 27 15 10
Question 6 (ii)
Find the two months in between which there was the greatest decrease in sales
๐ท๐๐๐๐๐๐ ๐ ๐๐๐ก๐ค๐๐๐ ๐ฝ๐๐ ๐๐๐ ๐น๐๐ = $38,000 โ $35,000
= $3,000
๐ท๐๐๐๐๐๐ ๐ ๐๐๐ก๐ค๐๐๐ ๐น๐๐ ๐๐๐ ๐๐๐ = $35,000 โ $27,000
= $8,000
๐ท๐๐๐๐๐๐ ๐ ๐๐๐ก๐ค๐๐๐ ๐๐๐ ๐๐๐ ๐ด๐๐ = $27,000 โ $15,000
= $12,000
๐ท๐๐๐๐๐๐ ๐ ๐๐๐ก๐ค๐๐๐ ๐ด๐๐ ๐๐๐ ๐๐๐ฆ = $15,000 โ $10,000
= $5,000
โด ๐โ๐ ๐๐๐๐๐๐ ๐ก ๐๐๐๐๐๐๐ ๐ ๐๐๐๐ข๐๐๐ ๐๐๐ก๐ค๐๐๐ ๐กโ๐ ๐๐๐๐กโ๐ ๐๐ ๐๐๐๐โ ๐๐๐ ๐ด๐๐๐๐.
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Question 6 part (iii)
Calculate the mean monthly sales for the period January to May 2011
๐๐๐ก๐๐ ๐๐๐๐กโ๐๐ฆ ๐ ๐๐๐๐ ๐๐๐๐ Jan to May = $ (38,000 + 35,000 + 27,000 + 15,000 + 10,000)
= $125,000
๐๐๐๐ ๐๐๐๐กโ๐๐ฆ ๐ ๐๐๐๐ ๐๐๐๐ Jan to May = ๐๐๐ก๐๐ ๐๐๐๐กโ๐๐ฆ ๐ ๐๐๐๐ ๐๐๐๐ Jan to May
๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐กโ๐
= $125,000
5
= $25,000
Question 6 part(iv)
Data Given: Total sales for the period January to June was $150,000.
Calculate the sales for the month of June
๐๐๐๐๐ ๐๐๐ ๐กโ๐ ๐๐๐๐กโ ๐๐ ๐ฝ๐ข๐๐ = ๐๐๐ก๐๐ ๐ ๐๐๐๐ ๐๐๐ ๐ฝ๐๐ ๐ก๐ ๐ฝ๐ข๐๐ โ ๐๐๐ก๐๐ ๐ ๐๐๐๐ ๐๐๐๐ ๐ฝ๐๐ ๐ก๐ ๐๐๐ฆ
= $ (150,000 โ 125,000)
= $25,000
Question 6 part(v)
Comment on the sales in June compared to the sales for the months January to May
The sales showed an increase when compared to the two previous months; April and May.
The sales were equivalent to the mean monthly sales for the period January to May 2010.
The month of June sales was the only month that displayed an increase since January.
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Question 7 (i)
Data Given: Diagram showing the triangle RST and its transformation RโSโTโ after undergoing a
transformation.
Write the coordinates of R and Rโ
๐ถ๐๐๐๐๐๐๐๐ก๐๐ ๐๐ ๐ = (2, 4)
๐ถ๐๐๐๐๐๐๐๐ก๐๐ ๐๐ ๐ โฒ = (2, 0)
Question 7 (ii)
Describe the transformation which maps triangle RST onto triangle R'S'T'
If the line y = 2 is drawn, it can be seen the vertices of RST and RโSโTโ are located the same
distance away from the line on opposite sides.
RST and RโSโTโ are also identical and RโSโTโ is a lateral inversion of RST.
Therefore the triangle RST was reflected along the line y = 2 to give the triangle RโSโTโ.
Question 7 (iii) part (a)
Data Given: RST undergoes an enlargement, centre (0, 4), scale factor, 3.
Draw triangle RโโSโโTโโ, the image of triangle RST under the enlargement
For the enlargement; centre C = (0, 4)
๐ถ๐๐๐๐๐๐๐๐ก๐๐ ๐๐ ๐ โฒโฒ = 3 ร ๐ถ๐
= [(3 ร 2), (3 ร 0 + 4)]
= (6, 4)
๐ถ๐๐๐๐๐๐๐๐ก๐๐ ๐๐ ๐โฒโฒ = 3 ร ๐ถ๐
= [(3 ร 3), (3(7 โ 4) + 4)]
= (9, 13)
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๐ถ๐๐๐๐๐๐๐๐ก๐๐ ๐๐ ๐โฒโฒ = 3 ร ๐ถ๐
= [(3 ร 5), (3(1) + 4)]
= (15, 7)
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Question 7 (iii) part b
Data Given: Area of triangle RST = 4 square units
Calculate the area of triangle RโโSโโTโโ
๐ด๐๐๐ ๐๐ ๐ก๐๐๐๐๐๐๐ ๐ โโ๐โโ๐โโ = (๐ ๐๐๐๐ ๐๐๐๐ก๐๐ )2 ร ๐๐๐๐ ๐๐ ๐ก๐๐๐๐๐๐๐ ๐ ๐๐
= (3 )2 ร 4
= 36 ๐ ๐๐ข๐๐๐ ๐ข๐๐๐ก๐
Question 7 (iii) part c
State two geometrical relationships between triangles RST and RโโSโโTโโ
Triangles RST and RโโSโโTโโ similar where RST is the object and RโโSโโTโโ is an enlargement of the
object.
Regarding the sides of RST and RโโSโโTโโ:
๐ โฒโฒ๐โฒโฒ
๐ ๐=
๐โฒโฒ๐โฒโฒ
๐๐=
๐ โฒโฒ๐โฒโฒ
๐ ๐= 3 which is the scale factor
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Question 8a part (i)
Data Given: An answer sheet showing a rectangle, A, of area 20 square units and perimeter 24
units.
Draw and label
(a) rectangle B of area 27 square units and perimeter 24 units
๐ด๐๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐๐ = ๐ ร ๐ค = 27 ๐ ๐๐ข๐๐๐ ๐ข๐๐๐ก๐
๐๐๐๐๐๐๐ก๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐๐ = 2๐ + 2๐ค = 24 ๐ข๐๐๐ก๐
๐ ร ๐ค = 27 โฆโฆโฆโฆโฆโฆ. Equation 1
2๐ + 2๐ค = 24 .โฆโฆโฆโฆ.. Equation 2
From Equation 2
๐ = 12 โ ๐คโฆโฆโฆโฆโฆโฆEquation 3
Substituting Equation 3 into Equation 1
(12 โ ๐ค)๐ค = 27
๐ค2 โ 12๐ค + 27 = 0
(๐ค โ 9)(๐ค โ 3) = 0
โด ๐ค = 9 ๐๐ ๐ค = 3
Taking w=3;
๐ = 12 โ ๐ค
๐ = 12 โ 3 = 9
Therefore for Rectangle B; length = 9 units and width = 3 units
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(b) rectangle C of area 32 square units and perimeter 24 units
๐ด๐๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐๐ = ๐ ร ๐ค = 32 ๐ ๐๐ข๐๐๐ ๐ข๐๐๐ก๐
๐๐๐๐๐๐๐ก๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐๐ = 2๐ + 2๐ค = 24 ๐ข๐๐๐ก๐
๐ ร ๐ค = 32 โฆโฆโฆโฆโฆโฆ. Equation 1
2๐ + 2๐ค = 24 .โฆโฆโฆโฆ.. Equation 2
From Equation 2
๐ = 12 โ ๐คโฆโฆโฆโฆโฆโฆEquation 3
Substituting Equation 3 into Equation 1
(12 โ ๐ค)๐ค = 32
๐ค2 โ 12๐ค + 32 = 0
(๐ค โ 8)(๐ค โ 4) = 0
โด ๐ค = 8 ๐๐ ๐ค = 4
Taking w=4;
๐ = 12 โ ๐ค
๐ = 12 โ 4 = 8
Therefore for Rectangle B; length = 8 units and width = 4 units
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Question 8a part (ii)
Complete the table to show the dimensions of Rectangles B and C
Rectangle Length Width Area
(square units)
Perimeter
(units)
A 10 2 20 24
B 9 3 27 24
C 8 4 32 24
D
E
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Data Given
Rectangle D has a perimeter of 24 cm, with length, ๐, and width, ๐ค.
Determine the values of ๐ and ๐ค for the area of the rectangle D to be as large as possible
๐ด๐๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐๐, ๐ด = ๐ ร ๐ค
๐๐๐๐๐๐๐ก๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐๐ = 2๐ + 2๐ค = 24 ๐ข๐๐๐ก๐
๐ด = ๐ ร ๐ค โฆโฆโฆโฆโฆโฆ. Equation 1
2๐ + 2๐ค = 24 .โฆโฆโฆโฆ.. Equation 2
From Equation 2
๐ = 12 โ ๐คโฆโฆโฆโฆโฆโฆEquation 3
Substituting Equation 3 into Equation 1
๐ด = (12 โ ๐ค) ร ๐ค
๐ด = 12๐ค โ ๐ค2
At Maximum
๐๐ด
๐๐ค= 0
12 โ 2๐ค = 0
2๐ค = 12
๐ค = 6
When w = 6;
๐ = 12 โ ๐ค
= 12 โ 6
Question 8b
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= 6
Since the maximum area occurs at l = 6 and w = 6 rectangle D is actually a square.
๐๐๐ฅ๐๐๐ข๐ ๐ด๐๐๐ = ๐ ร ๐
= 6 ร 6
= 36 ๐ ๐๐ข๐๐๐ ๐ข๐๐๐ก๐
Question 8c
Data Given
Rectangle E has a perimeter of 36 cm, with length, ๐, and width, ๐ค.
Determine the values of ๐ and ๐ค for the area of the rectangle E to be as large as possible For rectangle E to have a maximum value then it is also a square. Therefore ๐ = ๐ค = ๐
๐๐๐๐๐๐๐ก๐๐ ๐๐ ๐๐๐๐ก๐๐๐๐๐ = 4๐ค = 36 ๐ข๐๐๐ก๐
๐ค = 9 ๐ข๐๐๐ก๐
๐ = 9 ๐ข๐๐๐ก๐
๐๐๐ฅ๐๐๐ข๐ ๐ด๐๐๐ = ๐ ร ๐
= 9 ร 9
= 81 ๐ ๐๐ข๐๐๐ ๐ข๐๐๐ก๐
Rectangle Length Width Area
(square units)
Perimeter
(units)
A 10 2 20 24
B 9 3 27 24
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C 8 4 32 24
D 6 6 36 24
E 9 9 81 36
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Question 9a part (i)
Data Given
๐(๐ฅ) = 2๐ฅ โ 7
๐ฅ ๐๐๐ ๐(๐ฅ) = โ๐ฅ + 3
Evaluate ๐(5)
๐(๐ฅ) = 2๐ฅ โ 7
๐ฅ
๐(5) = 2(5) โ 7
5
= 3
5
Question 9a part (ii)
(a) Write an expression in terms of x for ๐โ1(๐ฅ)
๐ฟ๐๐ก ๐ฆ = ๐(๐ฅ)
๐ฆ =2๐ฅ โ 7
๐ฅ
๐๐ค๐๐ก๐โ ๐ฅ ๐๐๐ ๐ฆ
๐ฅ = 2๐ฆ โ 7
๐ฆ
๐ฅ๐ฆ = 2๐ฆ โ 7
2๐ฆ โ ๐ฅ๐ฆ = 7
๐ฆ(2 โ ๐ฅ) = 7
๐ฆ = 7
2 โ ๐ฅ
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โด ๐โ1(๐ฅ) = 7
2โ๐ฅ, ๐ฅ โ 2
(b) Write an expression in terms of ๐ฅ for ๐๐(๐ฅ)
Replacing ๐ฅ in ๐(๐ฅ ) with ๐(๐ฅ)
๐๐(๐ฅ) = โ2๐ฅ โ 7
๐ฅ+ 3
= โ2๐ฅ โ 7
๐ฅ+
3
๐ฅ
= โ2๐ฅ โ 7 + 3๐ฅ
๐ฅ
= โ5๐ฅ โ 7
๐ฅ
= โ5 โ7
๐ฅ , ๐ฅ โ 0
Question 9b part (i)
Express the quadratic function 1 โ 6๐ฅ โ ๐ฅ2, in the form ๐ โ ๐(๐ฅ + โ)2
1 โ 6๐ฅ โ ๐ฅ2 โก ๐ โ ๐(๐ฅ + โ)2
1 โ 6๐ฅ โ ๐ฅ2 โก ๐ โ [๐ (๐ฅ + โ)(๐ฅ + โ)]
1 โ 6๐ฅ โ ๐ฅ2 โก ๐ โ [๐ (๐ฅ2 + 2โ๐ฅ + โ2)]
1 โ 6๐ฅ โ ๐ฅ2 โก ๐ โ ๐๐ฅ2 โ 2๐โ๐ฅ โ ๐โ2
Equating coefficients
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๐ฅ2; โ๐ = โ1
๐ = 11
๐ฅ; โ2๐โ = โ6
โ2(1)โ = โ6
โ2โ = โ6
โ = 3
๐๐๐๐ ๐ก๐๐๐ก; ๐ โ ๐โ2 = 1
๐ โ (1)(3)2 = 1
๐ โ 9 = 1
๐ = 10
โด 1 โ 6๐ฅ โ ๐ฅ2 โก 10 โ (๐ฅ + 3)2 in the form ๐ โ ๐(๐ฅ + โ)2 where ๐ = 10, ๐ = 1 ๐๐๐ โ = 3.
Question 9b part (ii)
(a) State the maximum value of 1 โ 6๐ฅ โ ๐ฅ2
1 โ 6๐ฅ โ ๐ฅ2 โก 10 โ (๐ฅ + 3)2
๐๐๐ฅ๐๐๐ข๐ ๐ฃ๐๐๐ข๐ ๐๐ 1 โ 6๐ฅ โ ๐ฅ2 = 10 โ 0
= 10
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(b) State the equation of the axis of symmetry
๐ฟ๐๐ก ๐ฆ = 1 โ 6๐ฅ โ ๐ฅ2
Rearranging gives
๐ฆ = โ๐ฅ2 โ 6๐ฅ + 1 ๐คโ๐๐โ ๐๐ ๐๐ ๐กโ๐ ๐๐๐๐ ๐ฆ = ๐๐ฅ2 + ๐๐ฅ + ๐
๐ธ๐๐ข๐๐ก๐๐๐ ๐๐ ๐กโ๐ ๐๐ฅ๐๐ ๐๐ ๐ ๐ฆ๐๐๐๐ก๐๐ฆ; ๐ฅ = โ๐
2๐
= โ(โ6)
2(โ1)
= 6
โ2
= โ3
๐ธ๐๐ข๐๐ก๐๐๐ ๐๐ ๐กโ๐ ๐๐ฅ๐๐ ๐๐ ๐ ๐ฆ๐๐๐๐ก๐๐ฆ ๐๐ ๐ฅ = โ3
Question 9b part (iii)
Determine the roots of 1 โ 6๐ฅ โ ๐ฅ2
1 โ 6๐ฅ โ ๐ฅ2 = 0
๐ ๐๐๐๐๐ 1 โ 6๐ฅ โ ๐ฅ2 โก 10 โ (๐ฅ + 3)2
10 โ (๐ฅ + 3)2 = 0
(๐ฅ + 3)2 = 10
๐ฅ + 3 = ยฑโ10
๐ฅ = ยฑโ10 โ 3
๐ฅ = +โ10 โ 3 ๐๐ ๐ฅ = โโ10 โ 3
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๐ฅ = 3.162 โ 3 ๐๐ ๐ฅ = โ3.162 โ 3
๐ฅ = 0.162 ๐๐ ๐ฅ = โ6.162
๐ฅ = 0.16 ๐๐ โ 6.16 ๐๐ ๐ ๐ ๐๐๐๐๐๐ ๐๐๐๐๐๐
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Question 10a part (i)
Data Given
Calculate < ๐๐บ๐น
Triangle OGF is an isosceles triangle since the radius of a circle is constant;
๐ฟ๐๐๐๐กโ ๐๐ ๐๐บ = ๐ฟ๐๐๐๐กโ ๐๐ ๐๐น = ๐๐๐๐๐ข๐ ๐๐ ๐กโ๐ ๐๐๐๐๐๐
Therefore the base angles would be equivalent;
๐๏ฟฝฬ๏ฟฝ๐น = ๐๏ฟฝฬ๏ฟฝ๐บ
=180ยฐ โ 118ยฐ
2 since the total sum of angles in a triangle = 180ยฐ
=62ยฐ
2
= 31ยฐ
Question 10a part (ii)
Calculate < ๐ท๐ธ๐น
SGH is a tangent to the circle so an angle of 90ยฐ is created where it comes into contact with the
circle.
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Based on the given diagram; ๐๐บ๐ = 90ยฐ
๐๐บ๐ท = ๐๐บ๐ โ 65ยฐ
๐๐บ๐ท = 90ยฐ โ 65ยฐ
= 25ยฐ
๐ท๐บ๐น = ๐๐บ๐ท + ๐๐บ๐น
๐ท๐บ๐น = 25ยฐ + 31ยฐ
= 56ยฐ
๐ท๏ฟฝฬ๏ฟฝ๐น = 180ยฐ โ ๐ท๐บ๐น
๐ท๏ฟฝฬ๏ฟฝ๐น = 180ยฐ โ 56ยฐ
= 124ยฐ
Question 10b part (i)
Data Given
J, K and L are three sea ports.
A ship began its journey at L, sailed to K, then to L and returned to J.
The bearing of K from J is 54ยฐ and Z is due east of K.
๐ฝ๐พ = 122 ๐๐ and ๐พ๐ฟ = 60 ๐๐
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Draw a clearly labelled diagram to represent the data given
Question 10b part (ii)
(a) Calculate the measure of angle JKL
๐ฝ๏ฟฝฬ๏ฟฝ๐ฟ = 90ยฐ + 54ยฐ
= 144ยฐ
(b) Calculate the distance JL
Applying cosine rule ๐2 = ๐2 + ๐2 โ 2๐๐ cos๐ด
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Where
๐ = ๐ฝ๐ฟ, ๐ = ๐ฝ๐พ, ๐ = ๐พ๐ฟ ๐๐๐ ๐ด = ๐ฝ๏ฟฝฬ๏ฟฝ๐ฟ
๐2 = ๐2 + ๐2 โ 2๐๐ cos๐ด
๐ฝ๐ฟ2 = ๐ฝ๐พ2 + ๐พ๐ฟ2 โ 2 (๐ฝ๐พ)(๐พ๐ฟ) cos( ๐ฝ๏ฟฝฬ๏ฟฝ๐ฟ)
๐ฝ๐ฟ2 = (122)2 + (60)2 โ 2 (122)(60) cos( 144ยฐ)
๐ฝ๐ฟ2 = 14884 + 3600 โ (โ11844)
๐ฝ๐ฟ2 = 30328
๐ฝ๐ฟ = โ30328
๐ฝ๐ฟ = 174.149
Distance JL = 174. 15 to 2 decimal places
(c) Calculate the bearing of J from L
Applying sine rule to triangle JKL
๐
sin๐ด=
๐
sin๐ต
Where
๐ = ๐ฝ๐พ, ๐ด = ๐พ๏ฟฝฬ๏ฟฝ๐ฝ, ๐ = ๐ฝ๐ฟ and ๐ต = ๐ฝ๏ฟฝฬ๏ฟฝ๐ฟ
๐
sin๐ด=
๐
sin๐ต
๐ฝ๐พ
sin(๐พ๏ฟฝฬ๏ฟฝ๐ฝ)=
๐ฝ๐ฟ
sin( ๐ฝ๏ฟฝฬ๏ฟฝ๐ฟ )
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122
sin(๐พ๏ฟฝฬ๏ฟฝ๐ฝ)=
174.15
sin( 144ยฐ )
sin(๐พ๏ฟฝฬ๏ฟฝ๐ฝ) =122
174.15sin( 144ยฐ )
sin(๐พ๏ฟฝฬ๏ฟฝ๐ฝ) =122
296.282
๐พ๏ฟฝฬ๏ฟฝ๐ฝ = sinโ1 (122
296.282)
๐พ๏ฟฝฬ๏ฟฝ๐ฝ = 24.31ยฐ
๐ต๐๐๐๐๐๐ ๐๐ ๐ฝ ๐๐๐๐ ๐ฟ = 270ยฐ โ ๐พ๏ฟฝฬ๏ฟฝ๐ฝ
= 270ยฐ โ 24.31ยฐ
= 245.7ยฐ
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Question 11a part (i)
Data Given
Under a matrix transformation ๐ = (๐ ๐๐ ๐
), the points V and W were mapped onto Vโ and Wโ:
๐ (3, 5) โ ๐โฒ(5,โ3)
๐ (7, 2) โ ๐โฒ(2,โ7)
Determine the values of ๐, ๐ , ๐ ๐๐๐ ๐
๐ ร ๐ = ๐โฒ
(๐ ๐๐ ๐
) ร (35) = (
5โ3
)
(3๐ + 5๐3๐ + 5๐
) = ( 5โ3
)
Equating terms gives
3๐ + 5๐ = 5โฆโฆโฆโฆโฆโฆโฆโฆโฆ .๐ธ๐๐ข๐๐ก๐๐๐ 1
3๐ + 5๐ = โ3โฆโฆ .โฆ . . . . โฆโฆ .โฆ๐ธ๐๐ข๐๐ก๐๐๐ 2
๐ ร ๐ = ๐โฒ
(๐ ๐๐ ๐
) ร (72) = (
2โ7
)
(7๐ + 2๐7๐ + 2๐
) = ( 2โ7
)
Equating terms gives
7๐ + 2๐ = 2โฆโฆโฆโฆโฆโฆโฆโฆโฆ .๐ธ๐๐ข๐๐ก๐๐๐ 3
7๐ + 2๐ = โ7โฆโฆ .โฆ . . . . โฆโฆ .โฆ๐ธ๐๐ข๐๐ก๐๐๐ 4
3๐ + 5๐ = 5โฆโฆ .โฆ . . . . โฆโฆ .โฆ๐ธ๐๐ข๐๐ก๐๐๐ 1
7๐ + 2๐ = 2โฆโฆ .โฆ . . . . โฆโฆ .โฆ๐ธ๐๐ข๐๐ก๐๐๐ 3
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Solving simultaneously
From Equation 1
๐ =5โ5๐
3
Substituting into Equation 3
7 (5 โ 5๐
3 ) + 2๐ = 2
35 โ 35๐
3+ 2๐ = 2
Multiplying throughout by 3
35 โ 35๐ + 6๐ = 6
โ29๐ = โ29
๐ = 1
๐ =5 โ 5๐
3=
5 โ 5(1)
3= 0
3๐ + 5๐ = โ3โฆโฆ .โฆ . . . . โฆโฆ .โฆ๐ธ๐๐ข๐๐ก๐๐๐ 2
7๐ + 2๐ = โ7โฆโฆ .โฆ . . . . โฆโฆ .โฆ๐ธ๐๐ข๐๐ก๐๐๐ 4
Solving simultaneously
From Equation 2
๐ =โ3โ5๐
3
Substituting into Equation 4
7 (โ3 โ 5๐
3 ) + 2๐ = โ7
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โ21 โ 35๐
3+ 2๐ = โ7
Multiplying throughout by 3
โ21 โ 35๐ + 6๐ = โ21
โ29๐ = 0
๐ = 0
๐ =โ3 โ 5๐
3=
โ3 โ 5(0)
3= โ1
Therefore, ๐ = (๐ ๐๐ ๐
) = ( 0 1โ1 0
)
Question 11a part (ii)
State the coordinates of Z such that ๐ (๐ฅ, ๐ฆ) โ ๐โฒ(5, 1) under the transformation, M
๐ (๐ฅ, ๐ฆ) โ ๐โฒ(5, 1)
( 0 1โ1 0
) (๐ฅ๐ฆ) = (
51)
( 0๐ฅ + 1๐ฆโ1๐ฅ + 0๐ฆ
) = (51)
( ๐ฆโ๐ฅ
) = (51)
Since the matrices are both 2 ร 1, we can equate terms
๐ฆ = 5
โ๐ฅ = 1
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๐ฅ = โ1
๐ (๐ฅ, ๐ฆ) = (โ1, 5).
Question 11a part (iii)
Describe the geometric transformation, M
The transformation matrix, ๐ = ( 0 1โ1 0
) rotates the matrix it is applied to in a clockwise
manner about the origin O at an angle of 90ยฐ.
Question 11b part (i)
Data Given
๐๐โโโโ โ and ๐๐ โโโโ โ are position vectors with respect to the origin,๐.
Point, P = (2, 7) and ๐๐ โโโโ โโ = ( 4โ3
)
(a) Write ๐๐โโ โโ โโ ๐๐ ๐กโ๐ ๐๐๐๐ (๐๐)
Point, P = (2, 7)
Hence
๐๐โโ โโ โโ = (27)
๐โ๐๐โ ๐๐ ๐๐ ๐กโ๐ ๐๐๐๐ (๐๐) ๐คโ๐๐๐ ๐ = 2 ๐๐๐ ๐ = 7.
(b) Write ๐๐ โโ โโ โโ ๐๐ ๐กโ๐ ๐๐๐๐ (๐๐)
๐๐ โโ โโ โโ = ๐๐โโ โโ โโ + ๐๐ โโโโ โโ
๐๐ โโโโ โโ โโ โโ = (27) + (
4โ3
)
= (64)
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๐โ๐๐โ ๐๐ ๐๐ ๐กโ๐ ๐๐๐๐ (๐๐) ๐คโ๐๐๐ ๐ = 6 ๐๐๐ ๐ = 4.
Question 11b part (ii)
Data Given
Point, ๐ = (14,โ2)
(a) Find ๐ ๐โโโโโโ
๐ = (14,โ2)
๐๐โโโโ โ = ( 14โ2
)
๐ ๐โโโโโโ = ๐ ๐โโโโ โ + ๐๐โโโโโโ
๐๐ โโ โโ โ = โ (64) + (
14โ2
)
= ( 8โ6
)
(b) Show P, R and S are collinear.
๐๐โโ โโ = ๐๐ โโโโ โโ + ๐ ๐โโโโโโ
R is a common point as ๐๐ โโโโ โ ๐๐๐ ๐ ๐โโโโ โ are parallel to each other and will lie on the line PS.
Using ๐๐ โโโโ โ ๐๐๐ ๐ ๐โโโโ โ ; in order for the points to be collinear
๐๐ โโโโ โ = ๐ค (๐ ๐โโโโ โ) ๐คโ๐๐๐ ๐ ๐๐ ๐ ๐ ๐๐๐๐๐ ๐๐ข๐๐ก๐๐๐๐
๐๐ โโโโ โโ = ( 4โ3
)
๐ ๐โโโโโโ = ( 8โ6
) = 2 ( 4โ3
) ๐คโ๐๐๐ ๐ = 2
Therefore P, R and S are collinear.