In Text Questions (Try These Textbook Page No. 9, 10, 11,13) - Chapter 1 Fractions Term 3 6th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

(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:

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