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

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