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}

8.1 Solid Geometry III

Posted on March 15, 2018 by user

8.1 Solid Geometry III

8.1.1 Right Prisms and Right Circular Cylinders

8.1.2 Right Pyramids and Right Circular Cones

8.1.3 Spheres

7.2.2 Algebraic Formulae, PT3 Practice

Posted on March 6, 2018 by user

Question 6:

Given 2 a b = 3 a - \frac{b^{2}}{2} . Express a in terms of b .

Solution:

\begin{array}{l} 2 a b = 3 a - \frac{b^{2}}{2} \\ 3 a - 2 a b = \frac{b^{2}}{2} \\ a (3 - 2 b) = \frac{b^{2}}{2} \\ a = \frac{b^{2}}{2 (3 - 2 b)} \end{array}

Question 7:

Given a = \frac{\sqrt{b + 1}}{a} . Express b in terms of a .

Solution:

\begin{array}{l} a = \frac{\sqrt{b + 1}}{a} \\ \sqrt{b + 1} = a^{2} \\ {[{(b + 1)}^{\frac{1}{2}}]}^{2} = {(a^{2})}^{2} \\ b + 1 = a^{4} \\ b = a^{4} - 1 \end{array}

Question 8:

\begin{array}{l} Given x = \frac{5 w}{2 w - y} . \\ (i) Express w in terms of x and y . \\ (ii) Find the value of w when x = 10 and y = 15. \end{array}

Solution:

\begin{array}{l} (i) \\ x = \frac{5 w}{2 w - y} \\ 2 w x - x y = 5 w \\ 2 w x - 5 w = x y \\ w (2 x - 5) = x y \\ w = \frac{x y}{2 x - 5} \\ (ii) \\ w = \frac{(10) (15)}{2 (10) - 5} \\ = \frac{150}{15} \\ = 10 \end{array}

Question 9:
Azmin is h years old. His father is twice his brother’s age. If Azmin is 3 years older than his brother, write a formula for the sum (S) of their age.

Solution:
Azmin’s age = h
Azmin’s brother’s age = h – 3
Azmin’s father’s age = (h – 3) × 2 = 2h – 6

Therefore, the sum (S) of their age
S = h + (h – 3) + (2h – 6)
S = h + h – 3 + 2h – 6
S = 4h – 9

Question 10:
Mei Ling is 12 years older than Ali. In the next four years, Raju will be two times older than Ali. If h represents Ali’s age, write the algebraic expressions that represent the total of their ages, in terms of h, in four years time.

Solution:
Ali’s age in the next four years = h + 4
Mei Ling’s age = (h + 4) + 12 = h + 16
Raju’s age = (h + 4) × 2 = 2h + 8

Therefore, the total (S) of their age
S = (h + 4) + (h + 16) + (2h + 8)
S = 4h + 28

7.2.1 Algebraic Formulae, PT3 Practice

Posted on March 4, 2018 by user

7.2.1 Algebraic Formulae, PT3 Practice

Question 1:

Given m²+ 7 = k, express m in terms of k.

Solution:

\begin{array}{l} m^{2} + 7 = k \\ m^{2} = k - 7 \\ m = \sqrt{k - 7} \end{array}

Question 2:

Given 5km = mn – 2k, express k in terms of m and n.

Solution:

\begin{array}{l} 5 k m = m n - 2 k \\ 5 k m + 2 k = m n \\ k (5 m + 2) = m n \\ k = \frac{m n}{5 m + 2} \end{array}

Question 3:

Given

2 k - \frac{m}{2 n} = m

, express m in terms of k and n.

Solution:

\begin{array}{l} 2 k - \frac{m}{2 n} = m \\ - m - \frac{m}{2 n} = - 2 k \\ m + \frac{m}{2 n} = 2 k \\ \frac{2 n m + m}{2 n} = 2 k \\ \frac{m (2 n + 1)}{2 n} = 2 k \\ m = \frac{4 k n}{2 n + 1} \end{array}

Question 4:

Given

\frac{4 m}{\sqrt{T} - k} = \frac{3 h}{m}

, express T in terms of h and m.

Solution:

\begin{array}{l} \frac{4 m}{\sqrt{T} - k} = \frac{3 h}{m} \\ 4 m^{2} = 3 h \sqrt{T} - 3 h k \\ 3 h \sqrt{T} = 4 m^{2} + 3 h k \\ \sqrt{T} = \frac{4 m^{2} + 3 h k}{3 h} \\ T = {(\frac{4 m^{2} + 3 h k}{3 h})}^{2} \leftarrow Square both sides \end{array}

Question 5:

Given

\sqrt{\frac{8 s - 3 h}{4}} = 2

, express s in terms of h.

Solution:

\begin{array}{l} \sqrt{\frac{8 s - 3 h}{4}} = 2 \\ \frac{8 s - 3 h}{4} = 4 \leftarrow Square both sides \\ 8 s - 3 h = 16 \\ 8 s = 16 + 3 h \\ s = \frac{16 + 3 h}{8} \\ s = 2 + \frac{3 h}{8} \end{array}

7.1 Algebraic Formulae

Posted on March 1, 2018 by user

7.1 Algebraic Formulae

7.1.1 Variables and Constant

1. A variable is a quantity without a fixed value.

2. A constant is a quantity with a fixed value.

7.1.2 Concept of Formulae

1. An algebraic formula is an equation which shows the relationship between several variables and/ or constant.

2. The subject of an algebraic formula is a linear variable expressed in terms of the other variables.

Example 1:

Express x in terms of y, if y = \frac{2 - 3 x}{x} .

Solution:

\begin{array}{l} y = \frac{2 - 3 x}{x} \\ x y = 2 - 3 x \\ x y + 3 x = 2 \\ x (y + 3) = 2 \\ x = \frac{2}{y + 3} \end{array}

Example 2:

Given that s = \frac{p^{2} - q^{2}}{2 u}, make p the subject of the formula .

Solution:

\begin{array}{l} s = \frac{p^{2} - q^{2}}{2 r} \\ 2 r s = p^{2} - q^{2} \\ p^{2} = 2 r s + q^{2} \\ p = \sqrt{2 r s + q^{2}} \end{array}

6.2.3 Algebraic Expressions (III), PT3 Practice

Posted on February 23, 2018 by user

Question 11:
(a)
Expand: x (2 + y)

(b)

\begin{array}{l} Express \frac{5}{6 y} - \frac{3 x - 5}{12 y} as a single \\ fraction in its simplest form . \end{array}

Solution:
(a)
x (2 + y) = 2x + xy

(b)

\begin{array}{l} \frac{5}{6 y} - \frac{3 x - 5}{12 y} \\ = \frac{5 \times 2}{6 y \times 2} - \frac{3 x - 5}{12 y} \\ = \frac{10 - (3 x - 5)}{12 y} \\ = \frac{10 - 3 x + 5}{12 y} \\ = \frac{15 - 3 x}{12 y} \\ = \frac{3 (5 - x)}{12^{4} y} \\ = \frac{5 - x}{4 y} \end{array}

6.2.2 Algebraic Expressions (III), PT3 Practice

Posted on February 22, 2018 by user

6.2.2 Algebraic Expressions (III), PT3 Practice

Question 6:
(a) Simplify each of the following:

\begin{array}{l} (i) \frac{12 m n}{32} \\ (ii) \frac{x^{2} - x y}{x} \end{array}

(b) Express

\frac{1}{2 q} - \frac{2 p - 7}{6 q}

as a single fraction in its simplest form.

Solution:

\begin{array}{l} (a)(i) \frac{12 m n}{32} = \frac{3 m n}{8} \\ (a)(ii) \frac{x^{2} - x y}{x} = \frac{x (x - y)}{x} = x - y \end{array}

(b)

\begin{array}{l} \frac{1}{2 q} - \frac{2 p - 7}{6 q} = \frac{1 \times 3}{2 q \times 3} - \frac{(2 p - 7)}{6 q} \\ = \frac{3 - 2 p + 7}{6 q} \\ = \frac{10 - 2 p}{6 q} \\ = \frac{2 (5 - p)}{36 q} \\ = \frac{5 - p}{3 q} \end{array}

Question 7:

(a) Factorise 2ae + 3af – 6de – 9df

(b) Simplify

\frac{a^{2} - b^{2}}{{(a + b)}^{2}}

Solution:

(a)

2ae + 3af – 6de – 9df = a (2e + 3f ) – 3d (2e + 3f)

= (2e + 3f ) (a – 3d)

(b)

\begin{array}{l} \frac{a^{2} - b^{2}}{{(a + b)}^{2}} = \frac{(a + b) (a - b)}{(a + b) (a + b)} \\ = \frac{a - b}{a + b} \end{array}

Question 8:

(a) Factorise –8c² – 12ac.

(b) Simplify

\frac{a e + a d - 2 b e - 2 b d}{a^{2} - 4 b^{2}} .

Solution:

(a)

–8c²– 12ac

= –4c (2c + 3a)

(b)

\begin{array}{l} \frac{a e + a d - 2 b e - 2 b d}{a^{2} - 4 b^{2}} = \frac{a (e + d) - 2 b (e + d)}{(a + 2 b) (a - 2 b)} \\ = \frac{(e + d) (a - 2 b)}{(a + 2 b) (a - 2 b)} \\ = \frac{e + d}{a + 2 b} \end{array}

Question 9:

(a) Factorise 12x² – 27y²

(b) Simplify

\frac{3 m^{2} - 10 m + 3}{m^{2} - 9} \div \frac{3 m - 1}{m + 3} .

Solution:

(a)

12x²– 27y²= 3 (4x² – 9y²)

= 3(2x – 3y) (2x + 3y)

(b)

\begin{array}{l} \frac{3 m^{2} - 10 m + 3}{m^{2} - 9} \div \frac{3 m - 1}{m + 3} = \frac{(3 m - 1) (m - 3)}{(m + 3) (m - 3)} \times \frac{m + 3}{3 m - 1} \\ = 1 \end{array}

Question 10:

Simplify: \frac{8 m + m n}{3 m} \div \frac{n^{2} - 64}{24}

Solution:

\begin{array}{l} \frac{8 m + m n}{3 m} \div \frac{n^{2} - 64}{24} \\ = \frac{8 m + m n}{3 m} \times \frac{24}{n^{2} - 64} \\ = \frac{m (8 + n)}{3 m} \times \frac{24}{n^{2} - 8^{2}} \\ = \frac{m (8 + n)}{3 m} \times \frac{24_{8}}{(n - 8) (n + 8)} \\ = \frac{8}{n - 8} \end{array}

6.2.1 Algebraic Expressions (III), PT3 Practice

Posted on February 20, 2018 by user

6.2.1 Algebraic Expressions (III), PT3 Practice

Question 1:

(a)(i) Factorise 18a + 3

(a)(ii) Expand –3 (–y + 5)

(b) Express

\frac{5}{6 y} - \frac{3 x - 5}{12 y}

as a single fraction in its simplest form.

Solution:

(a)(i) 18a + 3 = 3(6a + 1)

(a)(ii) –3 (–y + 5) = 3y – 15

(b)

\begin{array}{l} \frac{5}{6 y} - \frac{3 x - 5}{12 y} = \frac{5 \times 2}{6 y \times 2} - \frac{(3 x - 5)}{12 y} \\ = \frac{10 - 3 x + 5}{12 y} \\ = \frac{15 - 3 x}{12 y} \\ = \frac{3 (5 - x)}{412 y} \\ = \frac{5 - x}{4 y} \end{array}

Question 2:

(a) Expand:

(i) 3 (–a + c)

(ii) –5 (a – c)

(b) Factorise 4x + 2

\frac{3 x + 6}{x^{2} - 4} \div \frac{x + 2}{x - 2}

Solution:

(a)(i) 3 (–a + c) = –3a + 3c

(a)(ii) –5 (a – c) = –5a + 5c

(c)

\begin{array}{l} \frac{3 x + 6}{x^{2} - 4} \div \frac{x + 2}{x - 2} = \frac{3 (x + 2)}{(x + 2) (x - 2)} \times \frac{x - 2}{x + 2} \\ = \frac{3}{x + 2} \end{array}

Question 3:

(a) Factorise:

(i) 5m + 25

(ii) 7x + 9xy

(b) Simplify:

\frac{4 x - 12}{4 y} \div \frac{x^{2} - 9}{y z}

Solution:

(a)(i)5m + 25 = 5 (m + 5)

(a)(ii)7x + 9xy = x (7 + 9y)

\begin{array}{l} (b) \frac{4 x - 12}{4 y} \div \frac{x^{2} - 9}{y z} = \frac{4 (x - 3)}{4 y} \times \frac{y z}{(x + 3) (x - 3)} \\ = \frac{z}{x + 3} \end{array}

Question 4:

(a) Factorise completely:

4 – 100n²

(b) Express

\frac{4}{5 x} - \frac{7 - 10 y}{15 x}

as a single fraction in its simplest form.

Solution:

(a)

4 – 100n²= (2 + 10n)(2 – 10n)

(b)

\begin{array}{l} \frac{4}{5 x} - \frac{7 - 10 y}{15 x} = \frac{4 \times 3}{5 x \times 3} - \frac{(7 - 10 y)}{15 x} \\ = \frac{12 - 7 + 10 y}{15 x} \\ = \frac{5 + 10 y}{15 x} \\ = \frac{5 (1 + 2 y)}{315 x} \\ = \frac{1 + 2 y}{3 x} \end{array}

Question 5:

(a) Simplify:

(m – 4n)(m + 4n) – m²

(b) Simplify:

\frac{3 x - 3 y}{x + y} \times \frac{2 x + 2 y}{6 x}

Solution:

(a)

(m – 4n)(m + 4n) – m²

= m²+ 4mn – 4mn – 4n² – m²

= 0

(b)

\begin{array}{l} \frac{3 x - 3 y}{x + y} \times \frac{2 x + 2 y}{6 x} = \frac{3 (x - y)}{x + y} \times \frac{2 (x + y)}{6 x} \\ = \frac{x - y}{x} \end{array}