Fractions Class 6 Long Question Answer

Fractions Class 6 Long Question Answer

Fractions Class 6 Long Question Answer

Question 1.

From the sum of [latex]\frac{3}{10}[/latex] and [latex]\frac{7}{15}[/latex], subtract the sum of [latex]\frac{2}{5}[/latex] and [latex]\frac{1}{6}[/latex].

Solution:

Sum of $\frac{3}{10}$ and $\frac{7}{15}=\frac{3}{10}+\frac{7}{15}$

[latex]=\frac{9}{30}+\frac{14}{30}=\frac{23}{30} .[/latex]

Sum of $\frac{2}{5}$ and $\frac{1}{6}=\frac{2}{5}+\frac{1}{6}=\frac{12}{30}+\frac{5}{30}=\frac{17}{30}$.

So, the required difference

= [latex]\frac{23}{30}[/latex] - [latex]\frac{17}{30}[/latex] = [latex]\frac{6}{30}[/latex] = [latex]\frac{1}{5}[/latex].

Question 2.

Arrange the fractions [latex]\frac{2}{9}[/latex], [latex]\frac{2}{3}[/latex], [latex]\frac{8}{21}[/latex] in descending order.

Solution:

Multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54,57, 60, 63, 66, ... and multiples of 21 are 21 42, 63, 84, ...

So, their smallest common multiple is 63. Now, we have :

$\frac{2}{9}=\frac{2 \times 7}{9 \times 7}=\frac{14}{63}, \frac{2}{3}=\frac{2 \times 21}{3 \times 21}=\frac{42}{63}$

and $\frac{8}{21}=\frac{8 \times 3}{21 \times 3}=\frac{24}{63}$.

Here, $\frac{42}{63}>\frac{24}{63}>\frac{14}{63}$.

So, required descending order of given fractions is [latex]\frac{2}{3}[/latex], [latex]\frac{8}{21}[/latex], [latex]\frac{2}{9}[/latex].