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NSW Y8 Maths

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NSW Y8 Maths Rates and Ratios Dividing Quantity into Given Ratio

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Dividing Quantity into Given Ratio Theory

\begin{multicols}{2} \textbf{Example 1}\\ Divide \(1200 \text{~cm}\) in the ratio \(3: 4: 5\).\\ \textbf{Example 1 solution}\\ \(\begin{aligned} \text { Total number of parts } & =3+4+5 \\ & =12 \end{aligned}\)\\ \(\begin{aligned} \therefore 1 \text { part } & =1200 \div 12 \\ & =100 . \\ 3 \text { parts } & =3 \times 100=300 \\ 4 \text { parts } & =4 \times 100=400 \\ 5 \text { parts } & =54100=500 \end{aligned}\)\\ \(\begin{aligned} \therefore 1\text {st part } & =300 \text{~cm} \\ 2\text {nd part } & =400 \text{~cm} \\ 3\text {rd part } & =500 \text{~cm} \end{aligned}\) \columnbreak \textbf{Example 2}\\ Two numbers are in the ratio \(5: 9\). If the difference between the numbers is 20, find the smaller number.\\ \textbf{Example 2 solution}\\ \(\begin{aligned} \text { Difference in the ratios } & =9-5 \\ & =4 \end{aligned}\)\\ \(\begin{aligned} \therefore 4 \text { parts } & =20 \\ 1 \text { part } & =5 \\ \therefore 5 \text { parts } & =5 \times 5 \\ & =25 \end{aligned}\)\\ \(\therefore\) Smaller number \(=25\). \end{multicols}

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