
Find the median of $\dfrac{2}{3},\dfrac{4}{5},\dfrac{1}{2},\dfrac{3}{4},\dfrac{6}{5}$.
Answer
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Hint: The median is the middle number in an ascending or descending list of numbers. Since there are five numbers we have the third number as median. To find it we have to write it in ascending order.
Complete step-by-step answer:
Given the numbers $\dfrac{2}{3},\dfrac{4}{5},\dfrac{1}{2},\dfrac{3}{4},\dfrac{6}{5}$
The median is the middle number in an ascending or descending list of numbers. Since there are five numbers we have the third number as median.
We have to write it in ascending order.
For a fraction, if denominator less than numerator then the number is greater than one and if numerator less than denominator then the number is less than one.
So here $\dfrac{6}{5}$ is greater than one and all others are less than one.
Therefore $\dfrac{6}{5}$ is the greatest of all here.
To compare the remaining numbers we can convert them to a common denominator.
For that let us take the LCM of the denominators.
$LCM(3,5,2,4) = 3 \times 5 \times 4 = 60$
So we can rewrite the fractions using the common denominator.
We can see,
$\dfrac{2}{3} = \dfrac{{2 \times 20}}{{3 \times 20}} = \dfrac{{40}}{{60}}$
$\dfrac{4}{5} = \dfrac{{4 \times 12}}{{5 \times 12}} = \dfrac{{48}}{{60}}$
$\dfrac{1}{2} = \dfrac{{1 \times 30}}{{2 \times 30}} = \dfrac{{30}}{{60}}$
$\dfrac{3}{4} = \dfrac{{3 \times 15}}{{4 \times 15}} = \dfrac{{45}}{{60}}$
Now we can compare the values.
We have,
$\dfrac{{30}}{{60}} < \dfrac{{40}}{{60}} < \dfrac{{45}}{{60}} < \dfrac{{48}}{{60}}$
$ \Rightarrow \dfrac{1}{2} < \dfrac{2}{3} < \dfrac{3}{4} < \dfrac{4}{5}$
And already we have seen that $\dfrac{6}{5}$ is the greatest.
So we have,
$ \Rightarrow \dfrac{1}{2} < \dfrac{2}{3} < \dfrac{3}{4} < \dfrac{4}{5} < \dfrac{6}{5}$
Now we can see $\dfrac{3}{4}$ is the middle number.
Therefore the median is $\dfrac{3}{4}$.
Note: If the numbers were given in ascending or descending order, we can directly write the middle number as the median. If there were even numbers, there will be two middle numbers. In those cases, median is the average of the two middle numbers.
Complete step-by-step answer:
Given the numbers $\dfrac{2}{3},\dfrac{4}{5},\dfrac{1}{2},\dfrac{3}{4},\dfrac{6}{5}$
The median is the middle number in an ascending or descending list of numbers. Since there are five numbers we have the third number as median.
We have to write it in ascending order.
For a fraction, if denominator less than numerator then the number is greater than one and if numerator less than denominator then the number is less than one.
So here $\dfrac{6}{5}$ is greater than one and all others are less than one.
Therefore $\dfrac{6}{5}$ is the greatest of all here.
To compare the remaining numbers we can convert them to a common denominator.
For that let us take the LCM of the denominators.
$LCM(3,5,2,4) = 3 \times 5 \times 4 = 60$
So we can rewrite the fractions using the common denominator.
We can see,
$\dfrac{2}{3} = \dfrac{{2 \times 20}}{{3 \times 20}} = \dfrac{{40}}{{60}}$
$\dfrac{4}{5} = \dfrac{{4 \times 12}}{{5 \times 12}} = \dfrac{{48}}{{60}}$
$\dfrac{1}{2} = \dfrac{{1 \times 30}}{{2 \times 30}} = \dfrac{{30}}{{60}}$
$\dfrac{3}{4} = \dfrac{{3 \times 15}}{{4 \times 15}} = \dfrac{{45}}{{60}}$
Now we can compare the values.
We have,
$\dfrac{{30}}{{60}} < \dfrac{{40}}{{60}} < \dfrac{{45}}{{60}} < \dfrac{{48}}{{60}}$
$ \Rightarrow \dfrac{1}{2} < \dfrac{2}{3} < \dfrac{3}{4} < \dfrac{4}{5}$
And already we have seen that $\dfrac{6}{5}$ is the greatest.
So we have,
$ \Rightarrow \dfrac{1}{2} < \dfrac{2}{3} < \dfrac{3}{4} < \dfrac{4}{5} < \dfrac{6}{5}$
Now we can see $\dfrac{3}{4}$ is the middle number.
Therefore the median is $\dfrac{3}{4}$.
Note: If the numbers were given in ascending or descending order, we can directly write the middle number as the median. If there were even numbers, there will be two middle numbers. In those cases, median is the average of the two middle numbers.
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