We will learn how to solve different types of problems on arranging ratios in ascending order and descending order.
When a ratio is expressed in fraction or in decimal we first need to convert the ratio into whole number to compare the ratios.
The order of a ratio is important to compare two or more ratios. By reversing the antecedent and consequent of a ratio are different ratio is obtained.
Solved problems on comparing and arranging ratios in ascending and descending order:
1. Compare the ratios 1\(\frac{1}{3}\) : 1\(\frac{1}{5}\) and 1.6 : 1.2
Solution:
1\(\frac{1}{3}\) : 1\(\frac{1}{5}\) and 1.6 : 1.2
= \(\frac{4}{3}\) : \(\frac{6}{5}\) and \(\frac{16}{10}\) : \(\frac{12}{10}\)
= \(\frac{4}{3}\) × 15 : \(\frac{6}{5}\) × 15 and \(\frac{16}{10}\) × 10 : \(\frac{12}{10}\) × 10
= 20 : 18 and 16 : 12
= \(\frac{20}{18}\) and \(\frac{16}{12}\)
= \(\frac{10 × 2}{9 × 2}\) and \(\frac{4 × 4}{3 × 4}\)
= \(\frac{10}{9}\) and \(\frac{4}{3}\)
= 10 : 9 and 4 : 3
Now, \(\frac{10}{9}\) and \(\frac{4}{3}\) are to be compared. L.C.M. of 9 and 3 = 9
\(\frac{10}{9}\) = \(\frac{10 × 1}{9 × 1}\) and \(\frac{4}{3}\) = \(\frac{4 × 3}{3 × 3}\)
= \(\frac{10}{9}\) and \(\frac{12}{9}\)
Since, \(\frac{10}{9}\) < \(\frac{12}{9}\)
Therefore, 10 : 9 < 4 : 3
Hence, 1\(\frac{1}{3}\) : 1\(\frac{1}{5}\) < 1.6 : 1.2
2. Compare the ratios 14 : 23, 5 : 12 and 61 : 92 in
ascending order.
Solution:
Given ratios can be written as \(\frac{14}{23}\), \(\frac{5}{12}\) and \(\frac{61}{92}\)
L.C.M. of the denominators 23, 12 and 92 = 276
\(\frac{14}{23}\) = \(\frac{14 × 12}{23 × 12}\) = \(\frac{168}{276}\)
\(\frac{5}{12}\) = \(\frac{5 × 23}{12 × 23}\) = \(\frac{115}{276}\)
and
\(\frac{61}{92}\) = \(\frac{61 × 3}{92 × 3}\) = \(\frac{183}{276}\)
Since, \(\frac{115}{276}\) < \(\frac{168}{276}\) < \(\frac{183}{276}\)
Therefore, \(\frac{5}{12}\) < \(\frac{14}{23}\) < \(\frac{61}{92}\)
Hence, 5 : 12 < 14 : 23 < 61 : 92
3. Arrange the ratios 1 : 3, 5 : 12, 4 : 15 and 2 : 3 in descending order.
Solution:
Given ratios can be written as \(\frac{1}{3}\), \(\frac{5}{12}\), \(\frac{4}{15}\) and \(\frac{2}{3}\)
L.C.M. of the denominators 3, 12, 15 and 3 = 60
\(\frac{1}{3}\) = \(\frac{1 × 20}{3 × 20}\) = \(\frac{20}{60}\)
\(\frac{5}{12}\) = \(\frac{5 × 5}{12 × 5}\) = \(\frac{25}{60}\)
\(\frac{4}{15}\) = \(\frac{4 × 4}{15 × 4}\) = \(\frac{16}{60}\)
and
\(\frac{2}{3}\) = \(\frac{2 × 20}{3 × 20}\) = \(\frac{40}{60}\)
Since, \(\frac{40}{60}\) > \(\frac{25}{60}\) > \(\frac{20}{60}\) > \(\frac{16}{60}\)
Therefore, \(\frac{2}{3}\) > \(\frac{5}{12}\) > \(\frac{1}{3}\) > \(\frac{4}{15}\)
Hence, 2 : 3 > 5 : 12 > 1 : 3 > 4 : 15.
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