In Text Questions (Try These Textbook Page No. 9, 10, 11,13) - Chapter 1 - Fractions - Term 3 - 6th Maths Guide Samacheer Kalvi Solutions

You can Download the In Text Questions (Try These Textbook Page No. 9, 10, 11,13) - Chapter 1 - Fractions - Term 3 - 6th Maths Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

(Try These Textbook Page No. 10)
Question $1 .$
Complete the following table. The first one is done for you.

(Try These Textbook Page No. 11)
Question $1 .$
(i) Are $5 \frac{2}{3}$ and $5 \frac{4}{6}$ equal?
Solution:
$5 \frac{2}{3}=\frac{(5 \times 3)+2}{3}=\frac{17}{3}$
$5 \frac{4}{6}=\frac{(5 \times 6)+4}{6}=\frac{34}{6}$
Equivalent fraction of
$\frac{17}{3}=\frac{17 \times 2}{3 \times 2}=\frac{34}{6}$
$\therefore 5 \frac{2}{3}$ and $5 \frac{4}{6}$ are equal

(ii) $\frac{3}{2} \neq 3 \frac{1}{2}$ why?
Solution:
$\begin{aligned}
3 \frac{1}{2} &=\frac{(3 \times 2)+1}{2}=\frac{7}{2} \\
\frac{3}{2} & \neq \frac{7}{2} \\
\therefore \quad \frac{3}{2} & \neq 3 \frac{1}{2} \\
3 \frac{1}{2} \text { means } 3 &+\frac{1}{2}
\end{aligned}$

Question 2.

Convert $3 \frac{1}{3}$ into improper fraction.
Solution:
$\begin{aligned}
&\text { Improper fraction }=\frac{(\text { whole number } \times \text { Denominator })+\text { Numerator }}{\text { Denominator }} \\
&3 \frac{1}{3}=\frac{(3 \times 3)+1}{3}=\frac{9+1}{3}=\frac{10}{3}
\end{aligned}$

Question $3 .$
Convert $\frac{45}{7}$ into mixed fraction.
Solution:

(Try These Textbook Page No. 13)
Question $1 .$
Find the sum of $5 \frac{4}{9}$ and $3 \frac{1}{6}$.
Solution:

\begin{aligned}
5 \frac{4}{9}+3 \frac{1}{6} &=5+\frac{4}{9}+3+\frac{1}{6} \\
&=(5+3)+\frac{4}{9}+\frac{1}{6} \\
&=8+\frac{8}{8}+\frac{3}{8}=8+\frac{8+3}{18}=8+\frac{11}{18}=8 \frac{11}{18} \\
\therefore 5 \frac{4}{9}+3 \frac{1}{6} &=8 \frac{11}{18}
\end{aligned}

Question 2

Subtract $7 \frac{1}{6}$ and $12 \frac{3}{8}$
Solution:
$\begin{aligned}
12 \frac{3}{8}-7 \frac{1}{6} &=(12-7)+\left(\frac{3}{8}-\frac{1}{6}\right) \\
&=5+\frac{(3 \times 3)-(1 \times 4)}{24} \\
&=5+\frac{9-4}{24}=5+\frac{5}{24}=5 \frac{5}{24} \\
12 \frac{3}{8}-7 \frac{1}{6} &=5 \frac{5}{24}
\end{aligned}$

Question 3 .
Subtract the sum of $6 \frac{1}{6}$ and $3 \frac{1}{5}$ from the sum of $9 \frac{2}{3}$ and $2 \frac{1}{2}$.
Solution:

$\begin{aligned}\left(9 \frac{2}{3}+2 \frac{1}{2}\right)-\left(6 \frac{1}{6}+3 \frac{1}{5}\right) &\left.=9+\frac{2}{3}+2+\frac{1}{2}\right)-\left(6+\frac{1}{6}+3+\frac{1}{5}\right) \\ &=9+2)+\frac{(2 \times 2)+(1 \times 3)}{6}-\left[(6+3)+\frac{(1 \times 5)+(1 \times 6)}{30}\right] \\ &=11+\frac{4+3}{6}-\left[9+\frac{5+6}{30}\right] \\ &=11+\frac{7}{6}-\left[9+\frac{11}{30}\right]=[11-9]+\left[\frac{7}{6}-\frac{11}{30}\right] \\ &=2+\left[\frac{(7 \times 5)-11}{30}\right]=2+\left[\frac{35-11}{30}\right]=2+\frac{24}{30}=2 \frac{4}{5} \end{aligned}$