5.2.2 Ratio, Rates and Proportions (I), PT3 Focus Practice

Posted on July 22, 2017 by user

Question 6:
In a campaign of selling T shirt for Merdeka Day makes an amount of profit, RM x. The profit is divided between Jamal and Rafizi in the ratio of 5 : 3. Jamal receives RM 5400 more than Rafizi. Calculate the value of x.

Solution:
Portion of profit for Jamal = ⅝
Portion of profit for Rafizi = ⅜
Given Jamal receives RM 5400 more than Rafizi,

\begin{array}{l} \frac{2}{8} \to RM 5400 \\ ∴ \frac{1}{8} \to RM 5400 \div 2 = RM 2700 \\ x = 2700 \times 8 \\ = 21600 \end{array}

Question 7:
Karim and Roger share RM300 in the ratio of 2 : 3. Karim gives ⅓ of his share to Mandeep. Then Mandeep received RM60 from Roger.
Find the ratio of Karim’s money to roger’s money to Mandeep’s money.

Solution:

\begin{array}{l} Money received by Karim \\ = \frac{2}{5} \times RM 300 \\ = RM 120 \\ Money received by Roger \\ = \frac{3}{5} \times RM 300 \\ = RM 180 \\ Karim's money after given to Mandeep \\ = RM12 0 \times \frac{2}{3} \\ = RM 80 \\ Roger's money after given to Mandeep \\ = RM18 0 - RM6 0 \\ = RM12 0 \\ Mandeep's money \\ = RM 40 + RM 60 \\ = RM10 0 \\ Ratio = 80 : 120 : 100 \\ = 4 : 6 : 5 \end{array}

Question 8:
Liza received three types of coloured marbles, red, green and yellow in the ratio of 2 : 5 : x. Given that the number of yellow marbles are more than the number of red marbles but less than the number of green marbles.
Calculate the number of yellow marbles that Liza received if the total number of marbles is 120.

Solution:
red : green : yellow = 2 : 5 : x
Given 2 < x < 5, so possible value of x is 3 in order to get a round number from the total of 120 marbles.

\begin{array}{l} \frac{yellow marbles}{total marbles} = \frac{3}{2 + 5 + 3} \\ \frac{yellow marbles}{120} = \frac{3}{10} \\ yellow marbles = \frac{3}{10} \times 120 \\ yellow marbles = 36 \end{array}

Question 9:
John and Mahmud are required to draw a triangle ABC. The ratio of ∠A : ∠B : ∠C of the triangle drawn by John is 4 : 8 : 3 while the triangle drawn by Mahmud is 5 : 5 : 8.
Find the difference between the value of ∠B drawn by John and Mahmud.

Solution:

\begin{array}{l} ∠ B drawn by John \\ = \frac{8}{4 + 8 + 3} \times 180^{o} \\ = 96^{o} \\ ∠ B drawn by Mahmud \\ = \frac{5}{5 + 5 + 8} \times 180^{o} \\ = 50^{o} \end{array}

Question 10:
The number of workers in an office building is 168 and they are placed in level two and level three. The number of workers in level three is 72.
(a) The ratio of workers in level two to level three is x : 3.
Find the value of x.
(b) 162 new workers have just started working in the office building. 3 of them are placed in level two, 93 in level three, while the rest is placed in level four.
Determine the ratio, in the lowest term of the workers of level two to three to four.

Solution:

\begin{array}{l} (a) \\ Ratio of workers in level two to level three = x : 3 \\ x : 3 = (168 - 72) : 72 \\ x : 3 = 96 : 72 \\ \frac{x}{3} = \frac{96}{72} \\ x = \frac{96}{72} \times 3 \\ x = 4 \end{array}

\begin{array}{l} (b) \\ Ratio of workers in level two to three to four \\ = (96 + 3) : (72 + 93) : (162 - 3 - 93) \\ = 99 : 165 : 66 \\ = \frac{99}{33} : \frac{165}{33} : \frac{66}{33} \\ = 3 : 5 : 2 \end{array}

5.2.1 Ratio, Rates and Proportions (I), PT3 Focus Practice

Posted on July 20, 2017 by user

5.2.1 Ratio, Rates and Proportions (I), PT3 Focus Practice

Question 1:

Given x : y = 4 : 5 and x + y = 180. Find the value of x.

Solution:

\begin{array}{l} Given x : y = 4 : 5 and x + y = 180 \\ \frac{x}{x + y} = \frac{4}{4 + 5} \\ \frac{x}{180} = \frac{4}{9} \\ ∴ x = \frac{4}{9} \times 180 \\ = 80 \end{array}

Question 2:

Given x : y = 9 : 5 and x – y = 16. Find the value of x + y.

Solution:

\begin{array}{l} Given x : y = 9 : 5 and x - y = 16 \\ \frac{x + y}{x - y} = \frac{9 + 5}{9 - 5} \\ \frac{x + y}{16} = \frac{14}{4} \\ x + y = \frac{7}{2} \times 16 \\ = 56 \\ x + y = 56 \end{array}

Question 3:

Amy, Rizal and Muthu donated to a charity fund in the ratio of 2 : 1 : 3. The total donation from Amy and Muthu was RM360.

Calculate the amount donated by Rizal.

Solution:

Amy : Rizal : Muthu = 2 : 1 : 3

\begin{array}{l} \frac{Rizal}{Amy + Muthu} = \frac{1}{2 + 3} \\ \frac{Rizal}{R M 360} = \frac{1}{5} \\ Rizal = \frac{1}{5} \times R M 360 \\ = R M 72 \end{array}

Question 4:

In the figure, TU: UV : VW = 5 : 7 : 2.

If TU = 30 cm, what is the length of TW?

Solution:

\begin{array}{l} Given T U : U V : V W = 5 : 7 : 2 \\ T U = 30 c m \\ \frac{T U + U V + V W}{T U} = \frac{5 + 7 + 2}{5} \\ \frac{T W}{30} = \frac{14}{5} \\ T W = \frac{14}{5} \times 30 \\ = 84 c m \end{array}

Question 5:

In diagram below, ABC is a triangle.

The sides of the triangle are in the ratio AB : BC: CA = 4 : 7 : 6.

Find the difference in length, in cm, between AB and CA.

Solution:

\begin{array}{l} A B : B C : C A = 4 : 7 : 6 \\ B C = 35 c m \\ \frac{A B}{B C} = \frac{4}{7} \\ \frac{A B}{35} = \frac{4}{7} \\ A B = \frac{4}{7} \times 35 \\ = 20 c m \\ \frac{C A}{35} = \frac{6}{7} \\ C A = \frac{6}{7} \times 35 \\ = 30 c m \end{array}

Difference in length between AB and CA

= 30 – 20

= 10 cm

4.2.1 Linear Equations I, PT3 Practice 1

Posted on July 5, 2017 by user

4.2.1 Linear Equations I, PT3 Practice 1

Question 1:

Solve the following linear equations.

(a) 4 – 3n = 5n – 4

(b)

\frac{4 m - 2}{10 + m} = \frac{1}{2}

Solution:

(a)

4 – 3n = 5n – 4

–3n –5n = – 4 – 4

–8n = – 8

8n = 8

n = 8/8 = 1

\begin{array}{l} (b) \frac{4 m - 2}{10 + m} = \frac{1}{2} \\ 2 (4 m - 2) = 10 + m \\ 8 m - 4 = 10 + m \\ 8 m - m = 10 + 4 \\ 7 m = 14 \\ m = \frac{14}{7} \\ m = 2 \end{array}

Question 2:

Solve the following linear equations.

\begin{array}{l} (a) \frac{x}{3} = 4 - x \\ (b) \frac{3 (x - 2)}{5} = 9 \end{array}

Solution:

\begin{array}{l} (a) \frac{x}{3} = 4 - x \\ x = 3 (4 - x) \\ x = 12 - 3 x \\ x + 3 x = 12 \\ 4 x = 12 \\ x = \frac{12}{4} \\ x = 3 \end{array}

\begin{array}{l} (b) \frac{3 (x - 2)}{5} = 9 \\ 3 (x - 2) = 9 \times 5 \\ 3 x - 6 = 45 \\ 3 x = 45 + 6 \\ 3 x = 51 \\ x = \frac{51}{3} \\ x = 17 \end{array}

Question 3:

Solve the following linear equations.

\begin{array}{l} (a) \frac{3 m}{4} + 15 = 9 \\ (b) 2 m + 8 = 3 (m - 2) \end{array}

Solution:

\begin{array}{l} (a) \frac{3 m}{4} + 15 = 9 \\ \frac{3 m}{4} = 9 - 15 \\ \frac{3 m}{4} = - 6 \\ 3 m = - 6 \times 4 \\ 3 m = - 24 \\ m = \frac{- 24}{3} \\ m = - 8 \end{array}

\begin{array}{l} (b) 2 m + 8 = 3 (m - 2) \\ 2 m + 8 = 3 m - 6 \\ 2 m - 3 m = - 6 - 8 \\ - m = - 14 \\ m = 14 \end{array}

Question 4:

Solve the following linear equations.

\begin{array}{l} (a) 11 + \frac{2 x}{3} = 9 \\ (b) \frac{x - 5}{3} = \frac{x}{6} \end{array}

Solution:

\begin{array}{l} (a) 11 + \frac{2 x}{3} = 9 \\ \frac{2 x}{3} = 9 - 11 \\ \frac{2 x}{3} = - 2 \\ 2 x = - 6 \\ x = \frac{- 6}{2} \\ x = - 3 \end{array}

\begin{array}{l} (b) \frac{x - 5}{3} = \frac{x}{6} \\ 6 (x - 5) = 3 x \\ 6 x - 30 = 3 x \\ 6 x - 3 x = 30 \\ 3 x = 30 \\ x = 10 \end{array}

Question 5:

Solve the following linear equations.

\begin{array}{l} (a) 4 (2 x - 3) = 24 \\ (b) \frac{y}{2} - \frac{y + 4}{3} = 5 \end{array}

Solution:

\begin{array}{l} (a) 4 (2 x - 3) = 24 \\ 8 x - 12 = 24 \\ 8 x = 24 + 12 \\ 8 x = 36 \\ x = \frac{36}{8} \\ x = 4 \frac{1}{2} \end{array}

\begin{array}{l} (b) \frac{y}{2} - \frac{y + 4}{3} = 5 \\ \frac{y \times 3}{2 \times 3} - \frac{2 (y + 4)}{3 \times 2} = 5 \\ \frac{3 y - 2 (y + 4)}{6} = 5 \\ 3 y - 2 y - 8 = 30 \\ y = 30 + 8 \\ y = 38 \end{array}

4.1 Linear Equations I

Posted on July 1, 2017 by user

4.1 Linear Equations I

4.1.1 Equality

1. An equation is a mathematical statement that joins two equal quantities together by an equality sign ‘=’.

Example: 1 km = 1000 m

2. If two quantities are unequal, the symbol ‘≠’ (is not equal) is used.

Example: 9 ÷ 4 ≠ 3

4.1.2 Linear Equations in One Unknown

1. A linear algebraic term is a term with one unknown and the power of unknown is one.

Example: 8x, -7y, 0.5y, 3a, …..

2. A linear algebraic expression contains two or more linear algebraic terms which are joined by a plus or minus sign.

Example:

3x – 4y, 4x + 9, 6x – 2y + 5, ……

3. A linear equation is an equation involving numbers and linear algebraic terms.

Example:

5x – 4 = 11, 4x + 7 = 15, 3y – 2 = 7

4.1.3 Solutions of Linear Equations in One Unknown

1. Solving an equation is a process of finding the values of the unknown in the equation.

2. The number that satisfies the equation is called the solution or root of the equation.

Example 1:

x + 4 = 12

x = 12 – 4 ← (When +4 is moved to the right of the equation, it becomes –4)

x = 8

Example 2:

x – 7 = 11

x = 11 + 7 ← (When –7 is moved to the right of the equation, it becomes +7)

x = 18

Example 3:

\begin{array}{l} 8 x = 16 \\ x = \frac{16}{8} \leftarrow \begin{array}{l} when the multiplier 8 is moved \\ to the right of the equation, it \\ becomes the divisor 8 . \end{array} \\ x = 2 \end{array}

Example 4:

\begin{array}{l} \frac{x}{5} = 3 \\ x = 3 \times 5 \leftarrow \begin{array}{l} the divisor 5 becomes the \\ multiplier 5 when moved \\ to the right of the equation . \end{array} \\ x = 15 \end{array}

3.2.2 Algebraic Expressions II, PT3 Focus Practice

Posted on June 25, 2017 by user

Question 6:

Simplify - 4 (5 x - 3) + 7 x - 1.

Solution:

\begin{array}{l} - 4 (5 x - 3) + 7 x - 1 \\ = - 20 x + 12 + 7 x - 1 \\ = - 13 x + 11 \end{array}

Question 7:

(3 x - 2 y) - (x + 4 y)

Solution:

\begin{array}{l} (3 x - 2 y) - (x + 4 y) \\ = 3 x - 2 y - x - 4 y \\ = 2 x - 6 y \end{array}

Question 8:

Simplify - 3 (2 a - 4 b) - \frac{1}{3} (6 a - 15 b)

Solution:

\begin{array}{l} - 3 (2 a - 4 b) - \frac{1}{3} (6 a - 15 b) \\ = - 6 a + 12 b - 2 a + 5 b \\ = - 8 a + 17 b \end{array}

Question 9:

\frac{1}{3} (- 2 x + 6 y - 9 z) - \frac{1}{6} (- 4 x - 18 y + 24 z)

Solution:

\begin{array}{l} \frac{1}{3} (- 2 x + 6 y - 9 z) - \frac{1}{6} (- 4 x - 18 y + 24 z) \\ = - \frac{2}{3} x + 2 y - 3 z + \frac{2}{3} x + 3 y - 4 z \\ = 5 y - 7 z \end{array}

Question 10:

\frac{1}{2} (a + 6 b c) - \frac{1}{5} (3 + 2 b c - 2 a)

Solution:

\begin{array}{l} \frac{1}{2} (a + 6 b c) - \frac{1}{5} (3 + 2 b c - 2 a) \\ = \frac{1}{2} a + 3 b c - \frac{3}{5} - \frac{2}{5} b c + \frac{2}{5} a \\ = \frac{5 a + 4 a}{10} + \frac{15 b c - 2 b c}{5} - \frac{3}{5} \\ = \frac{9 a}{10} + \frac{13 b c}{5} - \frac{3}{5} \end{array}

3.2.1 Algebraic Expressions II, PT3 Focus Practice

Posted on June 23, 2017 by user

3.2.1 Algebraic Expressions II, PT3 Focus Practice

Question 1:

Calculate the product of each of the following pairs of algebraic terms.

(a) 2rs × 4r²s³t

\begin{array}{l} (b) 1 \frac{2}{5} a^{2} b^{2} \times \frac{5}{14} a b^{3} c^{2} \\ (c) 1 \frac{1}{2} x y \times \frac{4}{9} x^{2} z \end{array}

Solution:

(a)

2rs × 4r²s³t = 2 × r × s× 4 × r × r × s × s × s × t = 8 r³s⁴t

\begin{array}{l} (b) 1 \frac{2}{5} a^{2} b^{2} \times \frac{5}{14} a b^{3} c^{2} \\ = \frac{7^{1}}{5^{1}} a a b b \times \frac{5^{1}}{14^{2}} a b b b c c \\ = \frac{1}{2} a^{3} b^{5} c^{2} \end{array}

\begin{array}{l} (c) 1 \frac{1}{2} x y \times \frac{4}{9} x^{2} z \\ = \frac{3^{1}}{2^{1}} x y \times \frac{4^{2}}{9^{3}} x x z \\ = \frac{2}{3} x^{3} y z \end{array}

Question 2:

Find the quotients of each of the following pairs of algebraic terms.

(a) \frac{- 48 a^{2} b^{3} c^{4}}{- 16 a b^{2} c^{2}}

\begin{array}{l} (b) - 15 e f^{3} g^{2} \div (- 40 e f^{2} g) \\ (c) 5 a^{3} c^{2} \div \frac{1}{2} a c \end{array}

Solution:

\begin{array}{l} (a) \frac{- 48 a^{2} b^{3} c^{4}}{- 16 a b^{2} c^{2}} \\ = \frac{- 48^{3} a \times a \times b \times b \times b \times c \times c \times c \times c}{- 16^{1} a \times b \times b \times c \times c} \\ = 3 a c^{2} \end{array}

\begin{array}{l} (b) - 15 e f^{3} g^{2} \div (- 40 e f^{2} g) \\ = \frac{- 15^{3} e \times f \times f \times f \times g \times g}{- 40^{8} e \times f \times f \times g} \\ = \frac{3}{8} f g \end{array}

\begin{array}{l} (c) 5 a^{3} c^{2} \div \frac{1}{2} a c = 5 a {^{3}}^{2} c^{2} \times \frac{2}{a c} \\ = 10 a^{2} c \end{array}

Question 3:

(– 4a²b) ÷ 3b²c²× 9ac =

Solution:

\begin{array}{l} (- 4 a^{2} b) \div 3 b^{2} c^{2} \times 9 a c \\ = \frac{- 4 a^{2} b}{3 b^{21} c^{21}} \times 9^{3} a c \\ = \frac{- 12 a^{3}}{b c} \end{array}

Question 4:

2a²b × 14b³c ÷ 56ab²c² =

Solution:

\begin{array}{l} 2 a^{2} b \times 14 b^{3} c \div 56 a b^{2} c^{2} \\ = 2 a^{21} b \times \frac{14 b^{31} c}{56^{2} a b^{2} c^{21}} \\ = \frac{a b^{2}}{2 c} \end{array}

Question 5:

- \frac{2}{3} (3 a - 6 b + \frac{3}{4} c) =

Solution:

\begin{array}{l} - \frac{2}{3} (3 a - 6 b + \frac{3}{4} c) \\ = - \frac{2}{3} \times 3 a - \frac{2}{3} (- 6^{2} b) - \frac{2}{3} (\frac{3}{4^{2}} c) \\ = - 2 a + 4 b - \frac{1}{2} c \end{array}

3.1 Algebraic Expressions II

Posted on June 20, 2017 by user

3.1 Algebraic Expressions II

3.1.1 Algebraic terms in Two or More Unknowns

1. An algebraic term in two or more unknowns is the product of the unknowns with a number.

Example 1:

4a³b = 4 × a × a × a× b

2. The coefficient of an unknown in the given algebraic term is the factor of the unknown.

Example 2:

7ab: Coefficient of ab is 7.

3. Like algebraic terms are algebraic terms with the same unknowns.

3.1.2 Multiplication and Division of Two or More Algebraic Terms

The coefficients and the unknowns of algebraic terms can be multiplied or divided altogether.

Example 3:

Calculate the product of each of the following pairs of algebraic terms.

(a) 5ac × 2bc

(b) –6xy × 5yz

(c) 20 x y \times (- \frac{2}{5} x^{2} y)

Solution:

(a)

5ac × 2bc = 5 × a × c × 2 × b × c = 10abc ²

(b)

–6xy × 5yz = –6 × x × y × 5 × y × z = –30 xy²z

(c)

\begin{array}{l} 20 x y \times (- \frac{2}{5} x^{2} y) \\ = 20^{4} x y \times (- \frac{2}{5}) \times x \times x \times y \\ = - 8 x^{3} y^{2} \end{array}

Example 4:

Find the quotients of each of the following pairs of algebraic terms.

\begin{array}{l} (a) \frac{42 x y z}{7 x y} \\ (b) \frac{12 x y^{2}}{18 x y} \end{array}

(c) 35 p^{2} q r^{2} \div 30 p r

Solution:

\begin{array}{l} (a) \frac{42 x y z}{7 x y} \\ = \frac{42^{6} x \times y \times z}{7 x \times y} = 6 z \end{array}

\begin{array}{l} (b) \frac{12 x y^{2}}{18 x y} \\ = \frac{12^{2} x \times y \times y}{18^{3} x \times y} = \frac{2}{3} y \end{array}

\begin{array}{l} (c) 35 p^{2} q r^{2} \div 30 p r \\ = \frac{35^{7} p \times p \times q \times r \times r}{30^{6} p \times r} \\ = \frac{7}{6} p q r \end{array}

3.1.3 Algebraic Expressions

An algebraic expression contains one or more algebraic terms. These terms are separated by a plus or minus sign.

Example 5:

7 – 6a² b + c is an algebraic expression with 3 terms.

3.1.4 Computation Involving Algebraic Expressions

Computation Involving Algebraic Expressions:

(a) 2(3a – 4) = 6a – 8

(b) (15a – 9b) ÷ 3 = 5a – 3b

(c) (6a – 2) – (9 + 4a)

= 6a – 4a – 2 – 9

= 2a – 11

(d) (a² b – 5ab²) – (6a² b – 4abc – 6ab²)

= a² b – 6a² b – 5ab²– (– 6ab²) – (– 4abc)

= –5a² b + ab² + 4abc

2.2.4 Squares, Square Roots, Cube and Cube Roots, PT3 Practice

Posted on June 17, 2017 by user

Question 15:
Match each of the following with the correct value.

Solution:

2.2.3 Squares, Square Roots, Cube and Cube Roots, PT3 Practice

Posted on June 16, 2017 by user

Question 10:

\begin{array}{l} (a) Find the value of \sqrt{0.09} \\ (b) Calculate the value of {(\frac{5}{7} \times \sqrt{49})}^{2} \end{array}

Solution:

\begin{array}{l} (a) \\ \sqrt{0.09} = \sqrt{\frac{9}{100}} = \frac{3}{10} = 0.3 \\ (b) \\ {(\frac{5}{7} \times \sqrt{49})}^{2} = {(\frac{5}{7} \times 7)}^{2} \\ = 5^{2} \\ = 25 \end{array}

Question 11:

\begin{array}{l} (a) Find the value of \sqrt[3]{- \frac{1}{27}} \\ (b) Calculate the value of {(7 + \sqrt[3]{- 8})}^{2} \end{array}

Solution:

\begin{array}{l} (a) \\ \sqrt[3]{- \frac{1}{27}} = \sqrt[3]{- \frac{1}{3} \times - \frac{1}{3} \times - \frac{1}{3}} = - \frac{1}{3} \\ (b) \\ {(7 + \sqrt[3]{- 8})}^{2} = {[7 + (- 2)]}^{2} \\ = 5^{2} \\ = 25 \end{array}

Question 12:

\begin{array}{l} (a) Find the value of {(- \frac{1}{4})}^{3} . \\ (b) Calculate the value of {(4.2 \div \sqrt[3]{27})}^{2} . \end{array}

Solution:

\begin{array}{l} (a) \\ {(- \frac{1}{4})}^{3} = - \frac{1}{4} \times - \frac{1}{4} \times - \frac{1}{4} \\ = - \frac{1}{64} \\ (b) \\ {(4.2 \div \sqrt[3]{27})}^{2} = {(4.2 \div 3)}^{2} \\ = {1.4}^{2} \\ = 1.4 \times 1.4 \\ = 1.96 \end{array}

Question 13:

\begin{array}{l} (a) Find the value of \sqrt[3]{0.008} . \\ (b) Calculate the value of 3^{2} \times \sqrt[3]{- \frac{64}{27}} . \end{array}

Solution:

\begin{array}{l} (a) \\ \sqrt[3]{0.008} = \sqrt[3]{\frac{8}{1000}} \\ = \frac{2}{10} \\ = 0.2 \\ (b) \\ 3^{2} \times \sqrt[3]{- \frac{64}{27}} = 9^{3} \times - \frac{4}{3} \\ = 12 \end{array}

Question 14:

\begin{array}{l} (a) Find the value of {(- \frac{3}{4})}^{3} . \\ (b) Calculate the value of \sqrt[3]{2 \frac{10}{27}} \div (2^{2} - 3^{2}) . \end{array}

Solution:

\begin{array}{l} (a) \\ {(- \frac{3}{4})}^{3} = - \frac{3}{4} \times - \frac{3}{4} \times - \frac{3}{4} \\ = - \frac{27}{64} \\ (b) \\ \sqrt[3]{2 \frac{10}{27}} \div (2^{2} - 3^{2}) = \sqrt[3]{\frac{64}{27}} \div (4 - 9) \\ = \frac{4}{3} \div - 5 \\ = \frac{4}{3} \times - \frac{1}{5} \\ = - \frac{4}{15} \end{array}

2.2.2 Squares, Square Roots, Cube and Cube Roots, PT3 Focus Practice

Posted on June 14, 2017 by user

2.2.2 Squares, Square Roots, Cube and Cube Roots, PT3 Focus Practice

Question 6:

Complete the operation steps below by filling in the boxes using suitable numbers.

\begin{array}{l} \sqrt[3]{4 \frac{17}{27}} \div \sqrt{1 \frac{9}{16}} = \sqrt[3]{\frac{}{27}} \div \sqrt{\frac{25}{16}} \\ = \frac{}{3} \div \frac{5}{4} \\ = \frac{}{3} \times \frac{4}{5} \\ = \end{array}

Solution:

\begin{array}{l} \sqrt[3]{4 \frac{17}{27}} \div \sqrt{1 \frac{9}{16}} = \sqrt[3]{\frac{125}{27}} \div \sqrt{\frac{25}{16}} \\ = \frac{5}{3} \div \frac{5}{4} \\ = \frac{5}{3} \times \frac{4}{5} \\ = \frac{4}{3} \end{array}

Question 7:

Complete the operation steps below by filling in the boxes using suitable numbers.

\begin{array}{l} {(\sqrt{2 \frac{7}{9}} - \sqrt[3]{\frac{27}{64}})}^{2} = {(\sqrt{\frac{}{9}} - \frac{3}{})}^{2} \\ = {(\frac{}{12})}^{2} \\ = \end{array}

Solution:

\begin{array}{l} {(\sqrt{2 \frac{7}{9}} - \sqrt[3]{\frac{27}{64}})}^{2} = {(\sqrt{\frac{25}{9}} - \frac{3}{4})}^{2} \\ = {(\frac{5}{3} - \frac{3}{4})}^{2} \\ = {(\frac{(5 \times 4) - (3 \times 3)}{12})}^{2} \\ = {(\frac{11}{12})}^{2} \\ = \frac{121}{144} \end{array}

Question 8:

Complete the operation steps below by filling in the boxes using suitable numbers.

\begin{array}{l} \sqrt[3]{1 \frac{61}{64}} - {0.3}^{2} = \sqrt[3]{\frac{}{64}} - {0.3}^{2} \\ = \frac{}{4} - \\ = \end{array}

Solution:

\begin{array}{l} \sqrt[3]{1 \frac{61}{64}} - {0.3}^{2} = \sqrt[3]{\frac{125}{64}} - {0.3}^{2} \\ = \frac{5}{4} - 0.09 \\ = 1.25 - 0.09 \\ = 1.16 \end{array}

Question 9:

Find the value of:

\begin{array}{l} (a) \sqrt[3]{\frac{10}{27} - 5} \\ (b) \sqrt{\frac{4}{49}} \times \sqrt[3]{- 0.216} \end{array}

Solution:

\begin{array}{l} (a) \sqrt[3]{\frac{10}{27} - 5} = \sqrt[3]{\frac{10 - 135}{27}} \\ = \sqrt[3]{- \frac{125}{27}} \\ = - \frac{5}{3} \end{array}

\begin{array}{l} (b) \sqrt{\frac{4}{49}} \times \sqrt[3]{- 0.216} = \frac{2}{7} \times \sqrt[3]{\frac{216}{1000}} \\ = \frac{2}{7} \times \frac{6}{510} \\ = \frac{6}{35} \end{array}