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Arrange the following ratios in ascending order of magnitude:

    \[\mathbf{2}:\text{ }\mathbf{3},\text{ }\mathbf{17}:\text{ }\mathbf{21},\text{ }\mathbf{11}:\text{ }\mathbf{14}\text{ }\mathbf{and}\text{ }\mathbf{5}:\text{ }\mathbf{7}\]

Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal | Ratio and Proportion
Arrange the following ratios in ascending order of magnitude:

    \[\mathbf{2}:\text{ }\mathbf{3},\text{ }\mathbf{17}:\text{ }\mathbf{21},\text{ }\mathbf{11}:\text{ }\mathbf{14}\text{ }\mathbf{and}\text{ }\mathbf{5}:\text{ }\mathbf{7}\]

It is given that

    \[\mathbf{2}:\text{ }\mathbf{3},\text{ }\mathbf{17}:\text{ }\mathbf{21},\text{ }\mathbf{11}:\text{ }\mathbf{14}\text{ }\mathbf{and}\text{ }\mathbf{5}:\text{ }\mathbf{7}\]

We can write it in fractions as

    \[2/3,\text{ }17/21,\text{ }11/14,\text{ }5/7\]

Here the LCM of

    \[3,\text{ }21,\text{ }14\text{ }and\text{ }7\text{ }is\text{ }42\]

By converting the ratio as equivalent

    \[\begin{array}{*{35}{l}} 2/3\text{ }=\text{ }\left( 2\text{ }\times \text{ }14 \right)/\text{ }\left( 3\text{ }\times \text{ }14 \right)\text{ }=\text{ }28/42 \\ 17/21\text{ }=\text{ }\left( 17\text{ }\times \text{ }2 \right)/\text{ }\left( 21\text{ }\times \text{ }2 \right)\text{ }=\text{ }34/\text{ }42 \\ 11/14\text{ }=\text{ }\left( 11\text{ }\times \text{ }3 \right)/\text{ }\left( 14\text{ }\times \text{ }3 \right)\text{ }=\text{ }33/42 \\ 5/7\text{ }=\text{ }\left( 5\text{ }\times \text{ }6 \right)/\text{ }\left( 7\text{ }\times \text{ }6 \right)\text{ }=\text{ }30/42 \\ \end{array}\]

Now writing it in ascending order

    \[28/42,\text{ }30/42,\text{ }33/42,\text{ }34/42\]

By further simplification

    \[2/3,\text{ }5/7,\text{ }11/14,\text{ }17/21\]

So we get

    \[2:\text{ }3,\text{ }5:\text{ }7,\text{ }11:\text{ }14\text{ }and\text{ }17:\text{ }21\]

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