Write the following rational numbers in ascending order:
$(i)$. $\frac{-3}{5},\ \frac{-2}{5},\ \frac{-1}{5}$
$(ii)$. $\frac{1}{3},\ \frac{-2}{9},\ \frac{-4}{3}$
$(iii)$. $\frac{-3}{7},\ \frac{-3}{2},\ \frac{-3}{4}$

AcademicMathematicsNCERTClass 7

Given: 

$(i)$. $\frac{-3}{5},\ \frac{-2}{5},\ \frac{-1}{5}$

$(ii)$. $\frac{-1}{3},\ \frac{-2}{9},\ \frac{-4}{3}$

$(iii)$. $\frac{-3}{7},\ \frac{-3}{2},\ \frac{-3}{4}$


To do: To write the given rational numbers in ascending order.


Solution:


$(i)$. $-\frac{3}{5},\ -\frac{2}{5},\ -\frac{1}{5}$

Here, denominators are the same for each rational number. We just need to compare the numerators of the given rational numbers.

On arranging numerators of the given rational numbers in ascending order we have:

$-3$<$-2$ < $-1$

Therefore, the given rational number in ascending order is as below:

$-\frac{3}{5}$ < $-\frac{2}{5}$ < $-\frac{1}{5}$

$(ii).\ -\frac{1}{3},\ -\frac{2}{9},\ -\frac{4}{3}$

To write the given rational numbers in ascending order, first of all, their denominators have to be equal.

Taking L.C.M of the denominators $3,\ 9$ and $3$ is $9$

Therefore,

$-\frac{1}{3} = \frac{(-1\times 3)}{(3\times 3)} = -\frac{3}{9}$

$-\frac{2}{9} = \frac{(-2\times 1)}{(9\times 1)} = -\frac{2}{9}$

$-\frac{4}{3} = \frac{(-4\times 3)}{(3\times 3)} = -\frac{12}{9}$

On arranging the obtained fractions into ascending order we have,

$-\frac{12}{9}$ < $-\frac{3}{9}$ < $-\frac{2}{9}$

Or,  $-\frac{4}{3}$ < $-\frac{1}{3}$ < $-\frac{2}{9}$

$(iii).\ -\frac{3}{7},\ -\frac{3}{2},\ -\frac{3}{4}$

To write the given rational numbers in ascending order, first of all, their denominators have to be equal.

Taking L.C.M of $7,\ 2$ and $4$ is $28$

So,  $-\frac{3}{7} = \frac{(-3\times 4)}{(7\times 4)} = -\frac{12}{28}$

$-\frac{3}{2} = \frac{(-3\times 14)}{(2\times 14)} = -\frac{42}{28}$

$-\frac{3}{4} = \frac{(-3\times 7)}{(4\times 7)} = -\frac{21}{28}$

On arranging the obtained fractions into ascending order we have:

$-\frac{42}{28}$ < $-\frac{21}{28}$ < $-\frac{12}{28}$

Or, $-\frac{3}{2}$ < $-\frac{3}{4}$ < $-\frac{3}{7}$
raja
Updated on 10-Oct-2022 13:35:25

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