11.2.1 Linear Equations II, PT3 Focus Practice

Posted on May 5, 2018 by user

11.2.1 Linear Equations II, PT3 Focus Practice

Question 1:

It is given that 2x = 6 and 3x + y = 10.

Calculate the value of y.

Solution:

2x = 6

x = 3

Substitute x = 3 into 3x + y = 10

3 (3) + y = 10

y = 10 – 9

y = 1

Question 2:

It is given that y = –1 and x – 3y = –10.

Calculate the value of x.

Solution:

Substitute y = –1 into x – 3y = –10

x – 3 (–1) = –10

x + 3 = –10

x = –10 – 3

x = –13

Question 3:

It is given that 7x – 2y = 19 and x + y = 13.

Calculate the value of y.

Solution:

Using Substitution method

7x – 2y = 19 -------- (1)

x + y = 13 ------- (2)

From equation (2),

x = 13 – y ------- (3)

Substitute equation (3) into equation (1),

7x – 2y = 19

7 (13 – y) – 2y = 19

91 – 7y – 2y = 19

– 9y = 19 – 91

– 9y = –72

y = 8

Question 4:

It is given that 2x – y = 5 and 3x – 2y = 8.

Calculate the value of x.

Solution:

Using Elimination method

2x – y = 5 -------- (1)

3x – 2y = 8 ------- (2)

(1) × 2: 4x – 2y = 10 -------- (3)

3x – 2y = 8 ------- (2)

(3) – (2): x – 0 = 2

x = 2

Question 5:

It is given that x + 2y = 4 and x + 6y = –4.

Calculate the value of x.

Solution:

Using Elimination method

x + 2y = 4 -------- (1)

x + 6y = –4 ------- (2)

(1) × 3: 3x + 6y = 12 -------- (3)

x + 6y = –4 ------- (2)

(3) – (2): 2x = 12 – (–4)

2x = 16

x = 8

6.3.1 The Methods and Substances Used in Food Technology (Structured Questions)

Posted on May 2, 2018 by user

Question 1:
Diagram 1 shows foods that have been processed through three different methods.

(a) Write down the food processing methods, P, Q and R, in Diagram 1. [3 marks]

(b) What happens to the water content in the food that has been processed through method P? [1 mark]

(c) The food is heated at 121 ^oC during the process in method Q.
What is the purpose of the heating? [1 mark]

(d) Fresh milk is heated for 30 minutes in method R.
What is the temperature used? [1 mark]

Answer:
(a)
P: Dehydration
Q: Canning
R: Pasteurisation

(b) Water is removed from the food.

(c) To kill the microorganisms in the food.

(d) 63 ^oC

Question 2:
Diagram 2.1 shows a food label of fruit juice that has been processed through a method of food processing according to the Food Regulations 1985.

Diagram 2.1

(a) Name the preservative added into the fruit juice in Diagram 2.1. [1 mark]

(b) What information is not shown on the food label in Diagram 2.1 as required by the Food Regulations 1985? [1 mark]

(c) Name the food processing method in Diagram 2.1. [1 mark]

(d) Besides fruit juice, state another drink that can be processed by the method in 2(c). [1 mark]

(e) Based on Diagram 2.2, state the temperature and heating duration during the food processing method in 2(c). [1 mark]

Diagram 2.2

Answer:
(a) Benzoic acid.

(b) Net weight / Volume of liquid.

(c) Pasteurization

(d) Milk

(e)
Temperature: 63 ^oC
Heating duration: 30 minutes
OR
Temperature: 72 ^oC
Heating duration: 15 seconds

11.1 Linear Equations II

Posted on May 1, 2018 by user

11.1 Linear Equations II

11.1.1 Linear Equations in Two Variables

1. A linear equation in two variables is an equation which contains only linear terms and involves two variables.

2. If the value of one variable in an equation is known, then the value of the other variable can be determined and vice versa.

Example:

Given that 2x + 3y = 6, find the value of

(a) x when y = 4, (b) y when x = –3

Solution:

(a) Substitute y = 4 into the equation.

2x + 3y = 6

2x + 3 (4) = 6

2x + 12 = 6

2x = 6 – 12

2x = –6

x = –3

(b) Substitute x = –3 into the equation.

2x + 3y = 6

2 (–3) + 3y = 6

–6 + 3y = 6

3y = 6 + 6

3y = 12

y = 4

3. A linear equation in two variables has many possible solutions.

11.1.2 Simultaneous Linear Equations in Two Variables

1. Two equations are said to be simultaneous linear equations in two variables if

(a) Both are linear equations in two variables, and

(b) Both involve the same variables.

Example: 2x + y = 9, x = 2y + 1

2. The solution of two simultaneous linear equations in two variables is any pair of values (x, y) that satisfied both the equations.

3. Simultaneous linear equations in two variables can be solved by the substitution method or the elimination method.

Example:

Solve the following simultaneous linear equation.

2x + y = 9

3x – y = –4

Solution:

(A) Substitution method

2x + y = 9 -------- (1) ← label the equations as (1) and (2)

3x – y = –4 ------- (2)

From equation (1),

y = 9 – 2x ------- (3) ← expressing y in terms of x.

Substitute equation (3) into equation (2),

3x – (9 – 2x) = –4

3x – 9 + 2x = –4

5x = –4 + 9

5x = 5

x = 1

Substitute x = 1 into equation (1),

2 (1) + y = 9

2 + y = 9

y = 9 – 2

y = 7

The solution is x = 1, y = 7.

(B) Elimination method

2x + y = 9 -------- (1) ← Both equations have the same coefficient of y.

3x – y = –4 ------- (2)

(1) + (2): 2x + 3x = 9 + (–4) ← y + (–y) = 0

5x = 5

x = 1

Substitute x = 1 into equation (1) or (2),

2x + y = 9 -------- (1)

2 (1) + y = 9

y = 9 – 2

y = 7

The solution is x = 1, y = 7.

10.2.2 Transformations II, PT3 Focus Practice

Posted on April 21, 2018 by user

Question 6:
Diagram below shows a ‘Wayang Kulit’ performed by Tok Dalang. The height of the screen use is 1.5 m. The shadow on the screen is ⅔ the height of the screen.

What is the height, in cm, of the puppet used by Tok Dalang?

Solution:

$\begin{array}{l} Height of shadow \\ = \frac{2}{3} of height of screen \\ = \frac{2}{3} \times 1.5 m \\ = 1.0 m \\ \frac{Height of puppet}{Height of shadow} = \frac{0.2}{0.2 + 0.6} \\ \frac{Height of puppet}{1} = \frac{0.2}{0.8} \\ Height of puppet \\ = 0.25 m \\ = 25 cm \end{array}$

Question 7:
Diagram below shows the shadow of a pillar form on the wall from the light of a spotlight.

(a) State the scale factor of the enlargement.
(b) Find the height of the pillar.

Solution:
(a)

$\begin{array}{l} Scale factor = \frac{2 + 6}{2} \\ = \frac{8}{2} \\ = 4 \end{array}$

(b)

$\begin{array}{l} \frac{Height of pillar}{Height of shadow} = \frac{1}{4} \\ \frac{Height of pillar}{1.2} = \frac{1}{4} \\ Height of pillar = \frac{1}{4} \times 1.2 m \\ = 0.3 m \end{array}$

Question 8:
In diagram below, a torch light is used to form a shadow of a candle on the screen.

The height of the candle is 15 cm.
If the height of the shadow of the candle is ¾ of the height of the screen, find the height of the screen.

Solution:

$\begin{array}{l} \frac{Height of candle}{Height of shadow} = \frac{30}{30 + 90} \\ \frac{15 cm}{Height of shadow} = \frac{1}{4} \\ Height of shadow = 15 cm \times 4 \\ = 60 cm \\ \frac{3}{4} of height of screen = 60 cm \\ Height of screen = \frac{4}{3} \times 60 cm \\ = 80 cm \end{array}$

10.2.1 Transformations II, PT3 Focus Practice

Posted on April 19, 2018 by user

10.2.1 Transformations II, PT3 Focus Practice

Question 1:

In the diagram, ∆P’Q’R’ is the image of ∆PQR under an enlargement.

(a) State the scale factor of the enlargement.

(b) If the area of ∆PQR = 7 cm², calculate the area of ∆P’Q’R’.

Solution:

(a)

$Scale factor, k = \frac{P' Q'}{P Q} = \frac{6}{3} = 2$

(b)

Area of image = k²× Area of object

Area of ∆P’Q’R’ = 2²× 7

= 28 cm²

Question 2:

In diagram below, OP’Q’ is the image of OPQ under an enlargement with centre O.

Given PQ = 4cm, calculate the length, in cm, of P’Q’.

Solution:

$\begin{array}{l} \frac{P' Q'}{P Q} = \frac{O P'}{O P} \\ \frac{P' Q'}{4} = \frac{9}{3} \\ P' Q' = \frac{9}{3} \times 4 \\ = 12 cm \end{array}$

Question 3:

In diagram below, PQ’R’S’ is the image of PQRS under an enlargement.

Calculate the length, in cm, of SS’.

Solution:

$\begin{array}{l} \frac{P S}{P S'} = \frac{P Q}{P Q'} \\ \frac{P S}{36} = \frac{4}{16} \\ P S = \frac{4}{16} \times 36 \\ = 9 cm \\ S S' = 36 - 9 \\ = 27 cm \end{array}$

Question 4:

On the Cartesian plane, Q’R’S’ is the image of ∆ QRS under an enlargement of centre T.

State the coordinates of T.

Solution:

Coordinates of T = (4, 4).

Question 5:

In diagram below, quadrilateral AFGH is the image of ABCD under an enlargement.

(a) Find the scale factor of the enlargement.

(b) The area represented by the quadrilateral AFGH is 15cm². Find the area, in cm², represented by the shaded region.

Solution:

(a)

$\begin{array}{l} Scale factor \\ = \frac{3}{6 + 3} = \frac{3}{9} = \frac{1}{3} \end{array}$

(b)

Area of image = k²× Area of object

$\begin{array}{l} 15 = {(\frac{1}{3})}^{2} \times Area of object \\ Area of object = 15 \times 9 = 135 \\ Area of shaded region = 135 - 15 \\ = 120 c m^{2} \end{array}$

10.1 Transformations II

Posted on April 16, 2018 by user

10.1 Transformations II

10.1.1 Similarity

Two shapes are similar if

(a) the corresponding angles are equal and

(b) the corresponding sides are proportional.

Example:

Quadrilateral ABCD is similar to quadrilateral JKLM because

$\begin{array}{l} ∠ A = ∠ J = 90^{o} \\ ∠ B = ∠ K = 50^{o} \\ ∠ C = ∠ L = 130^{o} \\ ∠ D = ∠ M = 90^{o} \end{array}$

(All the corresponding angles are equal.)

$\begin{array}{l} \frac{A B}{J K} = \frac{5}{10} = \frac{1}{2} \\ \frac{B C}{K L} = \frac{4}{8} = \frac{1}{2} \\ \frac{C D}{L M} = \frac{2.5}{5} = \frac{1}{2} \\ \frac{A D}{J M} = \frac{3}{6} = \frac{1}{2} \end{array}$

(All the corresponding sides are proportional.)

10.1.2 Enlargement

1. Enlargement is a type of transformation where all the points of an object move from a fixed point at a constant ratio.

2. The fixed point is known as the centre of enlargement and the constant ratio is known as the scale factor.

$Scale factor = \frac{length of side of image}{length of side of object}$

3. The object and the image are similar.

4. If A’ is the image of A under an enlargement with centre O and scale factor k, then

$\frac{O A'}{O A} = k$

if k > 0, then the image is on the same side of the object.
if k < 0, then the image is on the opposite side of the object.
if –1 < k < 1, then the size of the image is a reduction of the size of the object.

5. Area of image = k² × area of object.

9.2.2 Scale Drawings, PT3 Focus Practice

Posted on April 3, 2018 by user

Question 6:
Diagram below shows an irregular pentagon.

By using the scale 1 : 200, complete the scale drawing in the answer space provided. The grid has equal square with sides of 1 cm.

Answer:

Solution:

Question 7:
Diagram below shows polygon.

(a) If the polygon is redrawn using the scale 1 : 500, calculate the length of side drawn for the side 15 m.
(b) On the square grids in the answer space, redraw the polygon using the scale 1 : 500. The grid has equal squares with sides of 1 cm.

Answer:

$\begin{array}{l} (a) \\ Length of side of the drawing \\ = \frac{15}{5} \\ = 3 cm \end{array}$

(b)

9.2.1 Scale Drawings, PT3 Focus Practice

Posted on April 1, 2018 by user

Question 1:
Diagram below shows polygon P drawn on a grid of equal squares with sides of 1 unit.

On the grid in the answer space, redraw polygon P using the scale

$1 : \frac{1}{2} .$
The grid has equal squares with sides of 1 unit.

Answer:

Question 2:
Diagram below shows polygon W drawn on a grid of equal squares with sides of 1 unit.

On the grid in the answer space, redraw polygon W using the scale 1 : 2.
The grid has equal squares with sides of 1 unit.

Answer:

Question 3:
Diagram below shows a polygon PQRSTU.

(a) If the polygon is redrawn using the scale 1 : 400, calculate the length of PQ.
(b) In the answer space, redraw the polygon using the scale 1 : 400. The grid has equal squares with sides of 1 cm.

Answer:
(a)

$\begin{array}{l} Length of P Q = \frac{4800 cm}{400} \\ = 12 cm \end{array}$

(b)

Question 4:
Diagram below shows a polygon.

On the grid in the answer space, redraw the polygon using the scale 1 : 500. The grid has equal squares with sides of 1 cm.

Answer:

Actual length	Drawing
45 m = 4500 cm	4500 ÷ 500 = 9 cm
20 m = 2000 cm	2000 ÷ 500 = 4 cm
35 m = 3500 cm	3500 ÷ 500 = 7 cm
10 m = 1000 cm	1000 ÷ 500 = 2 cm

Question 5:
Diagram below shows a polygon.

(a) If the polygon is redrawn using the scale 1 : 300, calculate the length, in cm, the drawing for the side 24 m.
(b) In the answer space, redraw the polygon using the scale 1 : 300. The grid has equal squares with sides of 1 cm.

Answer:
(a)

$\begin{array}{l} Length of side of the drawing \\ = \frac{2400 cm}{300} \\ = 8 cm \end{array}$

(b)

8.2.2 Solid Geometry (III), PT3 Focus Practice

Posted on March 19, 2018 by user

Question 6:
Puan Rafidah wants to prepare a right circular cone shaped container in order to fill it with candies for her family Hari Raya Celebration. Diagram below shows the net of a right circular cone.

The circumference of the lid is 44 cm and its height is 24 cm.
Calculate the volume, in cm³ , of the container.

$[π = \frac{22}{7}]$

Solution:

$\begin{array}{l} 2 π r = 44 \\ 2 \times \frac{22}{7} \times r = 44 \\ r = \frac{44}{2} \times \frac{7}{22} \\ r = 7 cm \\ Volume of cone = \frac{1}{3} π r^{2} h \\ = \frac{1}{3} \times \frac{22}{7} \times 7^{2} \times 24 \\ = 1232 {cm}^{3} \end{array}$

Question 7:
A barrel contains 36 l of mineral water. ⅔ of the mineral water is mixed with 6 bottles of syrup, with a volume of 1.5 l each. The drink will be poured into a number of bottles with the same volume of 300 ml.
How many number of bottles needed?

Solution:

$\begin{array}{l} Volume of mineral water \\ = \frac{2}{3} \times 36 l \\ = 24 l \\ Volume of syrup \\ = 6 \times 1.5 l = 9 l \\ Total volume = 24 + 9 \\ = 33 l = 33000 m l \\ Number of bottles needed \\ = \frac{33000}{300} = 110 \end{array}$

Question 8:
A spherical balloon expands such that its volume increases from 36π cm³ to 288π cm³ .
Find the increment of the radius, in cm, of the balloon.

Solution:

$\begin{array}{l} V_{1} = \frac{4}{3} π r_{1}^{3} \\ \frac{4}{3} π r_{1}^{3} = 36 π \\ r_{1}^{3} = 36 π \times \frac{3}{4 π} \\ r_{1}^{3} = 27 \\ r_{1} = 3 cm \\ V_{2} = \frac{4}{3} π r_{2}^{3} \\ \frac{4}{3} π r_{2}^{3} = 288 π \\ r_{2}^{3} = 288 π \times \frac{3}{4 π} \\ r_{2}^{3} = 216 \\ r_{2} = 6 cm \\ Increment = r_{2} - r_{1} \\ = 6 - 3 \\ = 3 cm \end{array}$

Question 9:
A boy is given three pieces of cards by his teacher. The cards consist of one rectangle card and two circle cards of the same size as shown in the diagram below. The circumference of each circle is 66 cm.

He is required to combine the cards to form a right circular cylinder.
Calculate the volume, in cm³, of the right circular cylinder.

Solution:

$\begin{array}{l} 2 π r = 66 \\ 2 \times \frac{22}{7} \times r = 66 \\ r = \frac{66 \times 7}{2 \times 22} \\ = 10.5 cm \\ Volume = π r^{2} h \\ = \frac{22}{7} \times {10.5}^{2} \times 20 \\ = 6930 {cm}^{3} \end{array}$

Question 10:
Diagram below shows cube P and cube Q with the total surface area of 726 cm² and 1014 cm² respectively.

Find the difference in volume between cube P and cube Q.

Solution:

$\begin{array}{l} Area of one surface area of cube P = \frac{726}{6} = 121 {cm}^{2} \\ x (Length) \times x (Width) = 121 \\ x^{2} = 121 \\ x = 11 \\ Area of one surface area of cube Q = \frac{1014}{6} = 169 {cm}^{2} \\ y (Length) \times y (Width) = 169 \\ y^{2} = 169 \\ y = 13 \\ Difference in volume \\ = 13^{3} - 11^{3} \\ = 2197 - 1331 \\ = 866 {cm}^{3} \end{array}$

8.2.1 Solid Geometry III, PT3 Practice

Posted on March 17, 2018 by user

8.2.1 Solid Geometry III, PT3 Practice

Question 1:

The diagram below shows a cone with diameter 14 cm and height 6 cm.

Find the volume of the cone, in cm³.

Solution:

$\begin{array}{l} Volume of a cone \\ = \frac{1}{3} π r^{2} h \\ = \frac{1}{3} \times \frac{22}{7} \times 7^{2} \times 62 \\ = 308 {cm}^{3} \end{array}$

Question 2:

In the pyramid shown, the base is a rectangle.

If the volume is 20 cm², find the height of the pyramid, in cm.

Solution:

$\begin{array}{l} Volume of a pyramid = \frac{1}{3} \times base area \times h \\ \frac{1}{3} \times base area \times h = 20 \\ \frac{1}{3} \times 5 \times 4 \times h = 20 \\ \frac{20}{3} \times h = 20 \\ h = 20 \times \frac{3}{20} \\ h = 3 cm \end{array}$

Question 3:

Diagram below shows a composite solid consisting of a right circular cone, a right circular cylinder and a hemisphere.

The volume of the cylinder is 1650 cm³. Calculate the height, in cm, of the cone.

$[Use π = \frac{22}{7}]$

Solution:

$\begin{array}{l} Volume of a cylinder = π r^{2} h \\ \frac{22}{7} \times r^{2} \times 21 = 1650 \\ r^{2} = \frac{1650 \times 7}{22 \times 21} \\ = 25 \\ r = 5 c m \\ Thus the height of the cone \\ = 39 - 5 - 21 \\ = 13 c m \end{array}$

Question 4:

The cross section of the prism shown is an isosceles triangle.

The volume of the prism, in cm³, is

Solution:

$\begin{array}{l} Height of the △ = \sqrt{10^{2} - 6^{2}} \\ = \sqrt{64} \\ = 8 c m \\ Volume of prism \\ = Area of cross section \times Length \\ = (\frac{1}{2} \times 12^{6} \times 8) \times 16 \\ = 768 c m^{3} \end{array}$

Question 5:

A right circular cone has a volume of 77 cm³ and a circular base of radius 3.5 cm. Calculate its height.

Solution:

$\begin{array}{l} V = \frac{1}{3} π r^{2} h \\ h = \frac{3 V}{π r^{2}} \\ = \frac{3 \times 77}{\frac{22}{7} \times 3.5 \times 3.5} \\ = 6 cm \end{array}$