2.2.1 Squares, Square Roots, Cube and Cube Roots, PT3 Practice 1

Posted on June 12, 2017 by user

2.2.1 Squares, Square Roots, Cube and Cube Roots, PT3 Practice 1

Question 1:

Calculate the values of the following:

\begin{array}{l} (a) \sqrt{\frac{50}{98}} \\ (b) \sqrt{1 \frac{17}{64}} \\ (c) \sqrt{81} - \sqrt{0.01} \\ (d) \sqrt{3.24} \end{array}

Solution:

(a) \sqrt{\frac{50}{98}} = \sqrt{\frac{5025}{9849}} = \sqrt{\frac{25}{49}} = \frac{5}{7}

(b) \sqrt{1 \frac{17}{64}} = \sqrt{\frac{81}{64}} = \frac{9}{8} = 1 \frac{1}{8}

\begin{array}{l} (c) \sqrt{81} - \sqrt{0.01} = 9 - \sqrt{\frac{1}{100}} \\ = 9 - \frac{1}{10} \\ = 9 - 0.1 \\ = 8.9 \end{array}

\begin{array}{l} (d) \sqrt{3.24} = \sqrt{3 \frac{24}{100}} = \sqrt{3 \frac{6}{25}} \\ = \sqrt{\frac{81}{25}} \\ = \frac{9}{5} = 1 \frac{4}{5} \end{array}

Question 2:

Calculate the values of the following:

\begin{array}{l} (a) \sqrt[3]{\frac{16}{250}} \\ (b) \sqrt[3]{- \frac{4}{256}} \\ (c) \sqrt[3]{0.008} \\ (d) \sqrt[3]{0.729} \end{array}

Solution:

(a) \sqrt[3]{\frac{16}{250}} = \sqrt[3]{\frac{8}{125}} = \frac{2}{5}

(b) \sqrt[3]{- \frac{4}{256}} = \sqrt[3]{- \frac{1}{64}} = - \frac{1}{4}

\begin{array}{l} (c) \sqrt[3]{0.008} = \sqrt[3]{\frac{8}{1000}} \\ = \frac{2}{10} \\ = 0.2 \end{array}

\begin{array}{l} (d) \sqrt[3]{- 0.729} = \sqrt[3]{- \frac{729}{1000}} \\ = - \frac{9}{10} \\ = - 0.9 \end{array}

Question 3:

Find the value of

\sqrt[3]{3 \frac{3}{8}} + \sqrt{2 \frac{14}{25}} .

Solution:

\begin{array}{l} \sqrt[3]{3 \frac{3}{8}} + \sqrt{2 \frac{14}{25}} = \sqrt[3]{\frac{27}{8}} + \sqrt{\frac{64}{25}} \\ = \frac{3}{2} + \frac{8}{5} \\ = \frac{31}{10} = 3 \frac{1}{10} \end{array}

Question 4:

Find the values of the following:

(a) 1 – (–0.3)³.

(b)

{(2.1 \div \sqrt[3]{27})}^{2}

Solution:

(a)

1 – (–0.3)³= 1 – [(–0.3) × (–0.3) × (–0.3)]

= 1 – (–0.027)

= 1 + 0.027

= 1.027

(b)

\begin{array}{l} {(2.1 \div \sqrt[3]{27})}^{2} = {(2.1 \div 3)}^{2} \\ = {(0.7)}^{2} \\ = 0.49 \end{array}

Question 5:

Find the values of the following:

\begin{array}{l} (a) {(9 + \sqrt[3]{- 8})}^{2} \\ (b) \sqrt{144} \div \sqrt[3]{216} \times {0.3}^{3} \end{array}

Solution:

\begin{array}{l} (a) {(9 + \sqrt[3]{- 8})}^{2} = {[9 + (- 2)]}^{2} \\ = 7^{2} \\ = 49 \end{array}

\begin{array}{l} (b) \sqrt{144} \div \sqrt[3]{216} \times {0.3}^{3} \\ = 144 \div 6 \times (0.3 \times 0.3 \times 0.3) \\ = 24 \times 0.027 \\ = 0.648 \end{array}

2.1 Squares, Square Roots, Cube and Cube Roots

Posted on June 10, 2017 by user

2.1 Squares, Square Roots, Cube and Cube Roots

(A) Squares

The square of a number is the answer you get when you multiply a number by itself.

Example:

(a) 13²= 13 × 13 = 169

(b) (–10)²= (–10) × (–10) = 100

(d) (–0.06)²= (–0.06) × (–0.06) = 0.0036

\begin{array}{l} (e) {(3 \frac{1}{2})}^{2} = {(\frac{7}{2})}^{2} = \frac{7}{2} \times \frac{7}{2} = \frac{49}{4} \\ (f) {(- 1 \frac{2}{7})}^{2} = {(- \frac{9}{7})}^{2} = (- \frac{9}{7}) \times (- \frac{9}{7}) = \frac{81}{49} \end{array}

(B) Perfect Squares

1. Perfect squares are the squares of whole numbers.

2. Perfect squares are formed by multiplying a whole number by itself.

Example:

4 = 2 × 2 9 = 3 × 3 16 = 4 × 4

3. The first twelve perfect squares are:

= 1², 2², 3², 4², 5², 6², 7², 8², 9², 10², 11², 12²

= 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144

(C) Square Roots

1. The square root of a positive number is a number multiplied by itself whose product is equal to the given number.

Example:

\begin{array}{l} (a) \sqrt{169} = \sqrt{13 \times 13} = 13 \\ (b) \sqrt{\frac{25}{64}} = \sqrt{\frac{5 \times 5}{8 \times 8}} = \frac{5}{8} \\ (c) \sqrt{\frac{72}{98}} = \sqrt{\frac{7236}{9849}} = \sqrt{\frac{6 \times 6}{7 \times 7}} = \frac{6}{7} \\ (d) \sqrt{3 \frac{1}{16}} = \sqrt{\frac{49}{16}} = \frac{7}{4} = 1 \frac{3}{4} \\ (e) \sqrt{1.44} = \sqrt{1 \frac{4411}{10025}} = \sqrt{\frac{36}{25}} = \frac{6}{5} = 1 \frac{1}{5} \end{array}

(D) Cubes

1. The cube of a number is obtained when that number is multiplied by itself twice.

Example:

The cube of 3 is written as

3³ = 3 × 3 × 3

= 27

2. The cube of a negative number is negative.

Example:

(–2)³ = (–2) × (–2) × (–2)

= –8

3. The cube of zero is zero. The cube of one is one, 1³ = 1.

(E) Cube Roots

1. The cube root of a number is a number which, when multiplied by itself twice, produces the particular number.

" \sqrt[3]{} "

is the symbol for cube root.

Example:

\begin{array}{l} \sqrt[3]{64} = \sqrt[3]{4 \times 4 \times 4} \\ = 4 \end{array}

\sqrt[3]{64}

is read as ‘cube root of sixty-four’.

2. The cube root of a positive number is positive.

Example:

\begin{array}{l} \sqrt[3]{125} = \sqrt[3]{5 \times 5 \times 5} \\ = 5 \end{array}

3. The cube root of a negative number is negative.

Example:

\begin{array}{l} \sqrt[3]{- 125} = \sqrt[3]{(- 5) \times (- 5) \times (- 5)} \\ = - 5 \end{array}

4. To determine the cube roots of fractions, the fractions should be simplified to numerators and denominators that are cubes of integers.

Example:

\begin{array}{l} \sqrt[3]{\frac{16}{250}} = \sqrt[3]{\frac{168}{250125}} \\ = \sqrt[3]{\frac{8}{125}} \\ = \frac{2}{5} \end{array}

1.2.2 Directed Numbers, PT3 Practice

Posted on June 7, 2017 by user

1.2.1 Directed Numbers, PT3 Practice 2

Question 6:

Calculate the value of

1 - (- 2 + 7 \times 0.15) \div 2 \frac{1}{2}

Solution:

\begin{array}{l} 1 - (- 2 + 7 \times 0.15) \div 2 \frac{1}{2} \\ = 1 - (- 2 + 1.05) \times \frac{2}{5} \\ = 1 - (- 0.95) \times \frac{2}{5} \\ = 1 - (- 0.38) \\ = 1.38 \end{array}

Question 7:

Calculate the value of

1 \frac{1}{8} \times (\frac{4}{5} - \frac{2}{3})

and express the answer as a fraction in its lowest terms.

Solution:

\begin{array}{l} 1 \frac{1}{8} \times (\frac{4}{5} - \frac{2}{3}) \\ = 1 \frac{1}{8} \times (\frac{12}{15} - \frac{10}{15}) \\ = \frac{9^{3}}{8_{4}} \times \frac{2^{1}}{15_{5}} \\ = \frac{3}{20} \end{array}

1.2.1 Directed Numbers, PT3 Practice

Posted on June 3, 2017 by user

1.2.1 Directed Numbers, PT3 Practice 1

Question 1:

Complete the following calculation.

\begin{array}{l} - 4.6 \times (\frac{3}{5} - 1 \frac{3}{4}) = - 4.6 \times (0.6 - x) \\ = - 4.6 \times x \\ = x \end{array}

Solution:

\begin{array}{l} - 4.6 \times (\frac{3}{5} - 1 \frac{3}{4}) = - 4.6 \times (0.6 - 1.75) \\ = - 4.6 \times - 1.15 \\ = 5.29 \end{array}

Question 2:

Complete the following calculation.

\begin{array}{l} - 43 + 35 \div 1 \frac{1}{6} = - 43 + 35 \div \frac{7}{6} \\ = - 43 + 35 \times x \\ = - 43 + x \\ = - 13 \end{array}

Solution:

\begin{array}{l} - 43 + 35 \div 1 \frac{1}{6} = - 43 + 35 \div \frac{7}{6} \\ = - 43 + 35 \times \frac{6}{7} \\ = - 43 + 30 \\ = - 13 \end{array}

Question 3:
Calculate each of the following.

(a) (–9) × 14

(b) 6.7 – 3.2 × (–0.5)

Solution:

(a) (–9) × 14 = –126

(b) 6.7 – 3.2 × (–0.5) = 6.7 – (–1.6)

= 6.7 + 1.6

= 8.3

= 36

Question 4:

Calculate the value of

0.6 + (- 1 \frac{2}{3}) \div (\frac{4}{15}) - 2

Solution:

\begin{array}{l} 0.6 + (- 1 \frac{2}{3}) \div (\frac{4}{15}) - 2 \\ = 0.6 + [- \frac{5}{3} \times \frac{15^{5}}{4}] - 2 \\ = 0.6 + [- \frac{25}{4}] - 2 \\ = 0.6 - 6 \frac{1}{4} - 2 \\ = 0.6 - 6.25 - 2 \\ = - 7.65 \end{array}

Question 5:

Calculate the value of

(- 0.6 \times 1 \frac{1}{4}) - [- 1.6 \times (- 1.5)] \div 0.4

Solution:

\begin{array}{l} (- 0.6 \times 1 \frac{1}{4}) - [- 1.6 \times (- 1.5)] \div 0.4 \\ = ({- 0.6}^{- 0.3} \times \frac{5}{4^{2}}) - [\frac{{- 1.6}^{- 4} \times (- 1.5)}{{0.4}^{1}}] \\ = - \frac{1.5}{2} - 6 \\ = - 0.75 - 6 \\ = - 6.75 \end{array}

1.1 Directed Numbers

Posted on June 1, 2017 by user

1.1 Directed Numbers

1.1.1 Multiplication and Division of Integers
1. Multiplication and division of like signs gives (+)

\begin{array}{l} (+) \times (+) = + (+) \div (+) = + \\ (-) \times (-) = + (-) \div (-) = + \end{array}

2. Multiplication and division of unlike signs gives (–)

\begin{array}{l} (+) \times (-) = - (+) \div (-) = - \\ (-) \times (+) = - (-) \div (+) = - \end{array}

Example:

(a) –25 ÷ 5 = –5

(b) 8 × (–5) = –40

3. Multiplication of 3 integers.

\begin{array}{l} (+) \times (+) \times (+) = (+) \\ (+) \times (+) \times (-) = (-) \\ (+) \times (-) \times (-) = (+) \end{array}

4. Division of 3 integers.

\begin{array}{l} (+) \div (+) \div (+) = (+) \\ (+) \div (+) \div (-) = (-) \\ (+) \div (-) \div (-) = (+) \end{array}

5. The product of an integer and zero is always zero.

Example:

–5 × 0 = 0

6. When zero is divided by any integer except zero, the quotient is zero. Any integer divided by zero is undefined.

Example:

(a) 0 ÷ 9 = 0

(b) –6 ÷ 0 is undefined

1.1.2 Combined Operations of Integers

1. BODMAS → (Brackets of Division, Multiplication, Addition and Subtraction)

1. Operations in the brackets should be carried out first.

2. Followed by × or ÷ from left to right.

3. Followed by + or – from left to right.

Example 1:

(a) –52 ÷ 13 – 15 × 4

(b) 63 ÷ (16 – 7) × (–2)

Solution:

(a)

–52 ÷ 13 – 15 × 4

= (–52 ÷ 13) – (15 × 4) ← (calculate from left to right; ÷ and × are done first)

= –4 – 60

= –64

(b)

63 ÷ (16 – 7) × (–2)

= 63 ÷ 9 × (–2) ← (bracket is done first, then work from left to right)

= 7 × (–2)

= –14

(c)

–30 + 9 × 7 – 16

= –30 + (9 × 7) – 16 ← ( multiply first)

= –30 + 63 – 16

= 17

1.1.3 Fractions

1. Positive fractions are fractions with the positive sign (+), and their values are greater than 0.

2. Negative fractions are fractions with the negative sign (–), and their values are less than 0.

1.1.4 Decimals

1. There are positive decimals and negative decimals.

2. Positive decimals are greater than zero and negative decimals are less than zero.

1.1.5 Directed Numbers

1. Integers, fractions and decimals are directed numbers.

2.The computations of directed numbers is the same as that for whole numbers.

Example 2:

(a)

- \frac{1}{2} + (- 0.37) - (- 5)

(b) [(–28) – (–4)] ÷ (–5.147 – 0.853)

Solution:

(a)

\begin{array}{l} - \frac{1}{2} + (- 0.37) - (- 5) \\ = - 0.5 - 0.37 + 5 \\ = - 0.87 + 5 \\ = 4.13 \end{array}

(b)

[(–28) – (–4)] ÷ (–5.147 – 0.853)

= [–28 + 4] ÷ (–6)

= (–24) ÷ (–6)

= 4

11.2.3 Perimeter and Area, PT3 Practice

Posted on May 7, 2017 by user

Question 9:
In diagram below, ADB is a right-angled triangle and DBFE is a square. C is the midpoint of DB and CH = CD.

Calculate the area, in cm², of the coloured region.

Solution:

\begin{array}{l} Area of △ A B C \\ = \frac{1}{2} \times 6 \times 8 \\ = 24 {cm}^{2} \\ Area of trapezium B C H F \\ = \frac{1}{2} \times (12 + 6) \times 6 \\ = 54 {cm}^{2} \\ Area of C D E F H \\ = (12 \times 12) - 54 \\ = 144 - 54 \\ = 90 {cm}^{2} \\ Area of coloured region \\ = 24 + 90 \\ = 114 {cm}^{2} \end{array}

Question 10:
Diagram below shows a rectangle ACDE.

Calculate the area, in cm², of the coloured region.

Solution:

\begin{array}{l} Using Pythagoras' theorem (Refer Form 2 Chapter 6) \\ F E^{2} = D F^{2} - D E^{2} \\ = 13^{2} - 12^{2} \\ = 169 - 144 \\ = 25 \\ F E = \sqrt{25} = 5 cm \\ A F = 8 - 5 = 3 cm \\ A B = 12 - 8 = 4 cm \\ Area of rectangle A C D E \\ = 8 \times 12 \\ = 96 {cm}^{2} \\ Area of △ A B F \\ = \frac{1}{2} \times 3 \times 4 \\ = 6 {cm}^{2} \\ Area of △ D E F \\ = \frac{1}{2} \times 5 \times 12 \\ = 30 {cm}^{2} \\ Area of coloured region \\ = 96 - 30 - 6 \\ = 60 {cm}^{2} \end{array}

Question 11:
Diagram below shows a sketch of parallelogram shaped garden, PQRS that consists of flower beds and a playground.

Calculate the area, in m², of the flower beds.

Solution:

\begin{array}{l} Area flower bed \\ = (12 \times 14) - (\frac{1}{2} \times 12 \times 7) \\ = 168 - 42 \\ = 126 m^{2} \end{array}

11.2.2 Perimeter and Area, PT3 Practice

Posted on May 5, 2017 by user

Question 5:
In diagram below, PQUV is a square, QRTU is a rectangle and RST is an equilateral triangle.

The perimeter of the whole diagram is 310 cm.
Calculate the length, in cm, of PV.

Solution:

\begin{array}{l} P V = V U = T S = S R = Q P \\ Given perimeter of the whole diagram = 310 cm \\ P V + V U + U T + T S + S R + R Q + Q P = 310 \\ P V + P V + 50 + P V + P V + 50 + P V = 310 \\ 5 P V + 100 = 310 \\ 5 P V = 210 \\ P V = 42 cm \end{array}

Question 6:
In diagram below, ABCD and CGFE are rectangles. M, G, E and N are midpoints of AB, BC, CD and DA respectively.

Calculate the perimeter, in cm, of the coloured region.

Solution:

\begin{array}{l} Using Pythagoras' theorem \\ M G^{2} = M F^{2} + F G^{2} \\ = 5^{2} + 12^{2} \\ = 25 + 144 \\ = 169 \\ M G = 13 cm \\ Perimeter of the coloured region \\ = 13 + 13 + 5 + 12 + 5 + 12 \\ = 60 cm \end{array}

Question 7:
Diagram below shows a trapezium BCDE and a parallelogram ABEF. ABC and FED are straight lines.

The area of ABEF is 72 cm².
Calculate the area, in cm², of trapezium BCDE.

Solution:

\begin{array}{l} Base \times Height = Area of A B F E \\ 9 cm \times Height = 72 {cm}^{2} \\ Height = \frac{72}{9} \\ = 8 cm \\ C D = 8 cm \\ B C = 23 - 9 \\ = 14 cm \\ Area of trapezium B C D E \\ = \frac{1}{2} \times (B C + E D) \times C D \\ = \frac{1}{2} \times (14 + 9) \times 8 \\ = 92 {cm}^{2} \end{array}

Question 8:
In diagram below, ACEF is a trapezium and BCDG is a square.

Calculate the area, in cm², of the coloured region.

Solution:

\begin{array}{l} Area of trapezium A C E F \\ = \frac{1}{2} \times (8 + 15) \times 10 \\ = 115 {cm}^{2} \\ Area of square B C D G \\ = 4 \times 4 \\ = 16 {cm}^{2} \\ Area of trapezium G D E F \\ = \frac{1}{2} \times (4 + 15) \times 6 \\ = 57 {cm}^{2} \\ Area of coloured region \\ = 115 - 16 - 57 \\ = 42 {cm}^{2} \end{array}

11.2.1 Perimeter and Area, PT3 Practice

Posted on May 3, 2017 by user

Question 1:
In the diagram, ABCD is a trapezium and ABEF is a parallelogram.

Calculate the area, in cm², of the coloured region.

Solution:

\begin{array}{l} Area of trapezium A B C D \\ = \frac{1}{2} \times (8 + 14) \times 10 \\ = 110 {cm}^{2} \\ Area of parallelogram A B E F \\ = 8 \times 6 \\ = 48 {cm}^{2} \\ Area of the shaded region \\ = 110 - 48 \\ = 62 {cm}^{2} \end{array}

Question 2:
Diagram below shows a rectangle ABCD.

Calculate the area, in cm², of the coloured region.

Solution:

\begin{array}{l} The area of the coloured region \\ = Area of rectangle - Area of trapezium \\ = (12 \times 8) - \frac{1}{2} \times (4 + 6) \times 4 \\ = 96 - 20 \\ = 76 {cm}^{2} \end{array}

Question 3:
In diagram below, AEC is a right-angled triangle with an area of 54 cm² and BCDF is a rectangle.

Calculate
(a) the perimeter, in cm, of the coloured region.
(b) the area, in cm², of the coloured region.

Solution:

\begin{array}{l} (a) \\ Given area of △ A C E \\ \frac{1}{2} \times A C \times 9 = 54 \\ A C = 54 \times \frac{2}{9} \\ A C = 12 cm \\ Using Pythagoras' theorem: \\ A E = \sqrt{9^{2} + 12^{2}} \\ = 15 cm \\ Perimeter of coloured region \\ = 6 + 4.5 + 6 + 4.5 + 15 \\ = 36 cm \end{array}

\begin{array}{l} (b) \\ Area of the coloured region \\ = Area of △ A C E - Area of rectangle B C D F \\ = 54 - (6 \times 4.5) \\ = 54 - 27 \\ = 27 {cm}^{2} \end{array}

Question 4:
Diagram below shows a trapezium ABCDE. ABGF is a square with an area of 36 cm².

Calculate
(a) the perimeter, in cm, of the coloured region.
(b) the area, in cm², of the coloured region.

Solution:

\begin{array}{l} (a) \\ Using Pythagoras' theorem: \\ In △ C D H, \\ C D = \sqrt{8^{2} + 6^{2}} \\ = 10 cm \\ A B = B G = G F = F A = \sqrt{36} = 6 cm \\ Perimeter of coloured region \\ = 6 + 10 + 18 + 2 + 6 + 6 \\ = 48 cm \end{array}

\begin{array}{l} (b) \\ Area of the coloured region \\ = Area of trapezium A B C D E - Area of square A B G F \\ = [\frac{1}{2} (12 + 18) \times 8] - 36 \\ = [\frac{1}{2} \times 30 \times 8] - 36 \\ = 120 - 36 \\ = 84 {cm}^{2} \end{array}

8.2.3 Basic Measurements, PT3 Practice

Posted on April 7, 2017 by user

Question 11:
George departed for Kuantan from Kuala Lumpur at 9.30 a.m. He stopped at a rest station for 45 minutes. Then, he continued his journey and arrived at his destination at 3.45 p.m. How long did he travel on the road.

Solution:
Time taken on the road
= (1545 – 0930) – 0045 ← [3.45 p.m. change to 24-hour system = 1545 (1200 + 0345)]
= 0615 – 0045
= 0530
= 5 hours 30 minutes

Question 12:
Rita began doing her homework at 8.25 in the morning. She took a break for 45 minutes to have some exercises and ½ hour to have lunch. If she finished her homework at 4.45 in the evening on the same day, how long did she take to do her homework?

Solution:
Time taken for Rita to finish her homework
= (1645 – 0825) – (0030 + 0045) ← [4.45 p.m. = 1645 (1200 + 0445)]
= 0820 – 0115
= 0705
= 7 hours 5 minutes

Question 13:
Roslan attended a tuition class for 2½ hours. The class started at 7.15 p.m. The class finished 35 minutes earlier due to power failure.
At what time did the class finish?

Solution:
Class finish at
= (1915 + 0230) – 0035 ← [7.15 p.m. = 1915 (1200 + 0715)]
= 2145 – 0035
= 2110 → (9.10 p.m.)

Question 14:
An aeroplane took off from Kuala Lumpur to Singapore at 2.45 p.m. The flight normally takes 55 minutes. The aeroplane was delayed and landed at Singapore at 4.50 p.m.
How long, in minutes, was the aeroplane delayed?

Solution:
Expected arrival time
= 2.45 p.m. + 55 minutes
= 3.40 p.m.

Time delayed
= 4.50 p.m. – 3.40 p.m.
= 1 hour 10 minutes
= 70 minutes

Question 15:
The flight from Kuala Lumpur to Kota Kinabalu takes 1 hour 45 minutes.
Jacky’s flight should leave at 2.15 p.m. but is delayed by 50 minutes.
At what time, in the 24-hour system, he will reach Kota Kinabalu?

Solution:
2.15 p.m. = 1415 hours
Actual departure time
= 1415 hours + 50 minutes
= 1505 hours

Arrival time at Kota Kinabalu
= 1505 + 0145
= 1650

8.2.2 Basic Measurements, PT3 Practice

Posted on April 5, 2017 by user

Question 6:
The mass of a box containing 6 papayas is 21.32 kg. The mass of the box when it is empty is 1.46 kg.
Calculate the average mass, in g, of a papaya.

Solution:
Mass of 6 papayas
= 21.32 – 1.46
= 19.86 kg

Average mass of a papaya
= 19.86 ÷ 6
= 3.31 kg
= 3.31 × 1000 g
= 3310 g

Question 7:
Louis bought 600 g of cookies. Dennis bought twice the mass of cookies that Jackson bought. They bought 1.35 kg of cookies altogether. Calculate the mass, in g, of cookies bought by Jackson.

Solution:
Let Jackson bought w g of cookies.
600 g + 2 × w + w = 1.35 kg
600 g + 3w = 1.35 kg
600 g + 3w = 1350 g
3w = 1350 g – 600 g
w = 750 g ÷ 3
w = 250 g

Question 8:
If the mass of 4 packets of candies is 2.6 kg, what is the mass of 9 packets of the same candies, in kg?

Solution:

\begin{array}{l} Mass of 4 packets of candies = 2 .6 kg \\ Mass of 9 packets of candies = \frac{2.6}{4} \times 9 \\ = 5.85 kg \end{array}

Question 9:
It is given that

\frac{1}{4}

of fruits is supplied to Juice Stall A and

\frac{2}{7}

to Juice Stall B. The remaining 133.25 kg is sold to a fruit stall.
Calculate the mass of fruits that has been supplied to Fruit Stall B.

Solution:

\begin{array}{l} \frac{1}{4} + \frac{2}{7} = \frac{7}{28} + \frac{8}{28} \\ = \frac{15}{28} \\ Remaining fruits sold to fruit stall \\ = 1 - \frac{15}{28} \\ = \frac{13}{28} \\ Total mass of fruits \\ = \frac{28}{13} \times 133.25 \\ = 287 kg \\ Mass of fruits supplied to Juice Stall B \\ = \frac{2}{7} \times 287 \\ = 82 kg \end{array}

Question 10:
The mixture of metal to produce a piece of 50 sen coin are

\frac{2}{3} zinc, \frac{1}{5} nickel

and the rest is copper.
If the mass of copper is 1.6 g, find the total mass, in g, of zinc and nickel.

Solution:

\begin{array}{l} Fraction of copper \\ = 1 - \frac{2}{3} - \frac{1}{5} \\ = \frac{15}{15} - \frac{10}{15} - \frac{3}{15} \\ = \frac{2}{15} \\ Portion of copper = 2 g \\ Portion of zinc and nickel = 13 g \\ 2 \leftrightarrow 1.6 g \\ Total mass of zinc and nickel, \\ 13 \leftrightarrow \frac{13 \times 1.6}{2} = 10.4 g \end{array}