Exercise 1.1 - Chapter 1 Fractions Term 3 6th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $1.1$
Question $1 .$
Fill in the blanks.
(i) $7 \frac{3}{4}+6 \frac{1}{2}=$
(ii) The sum of whole number and a proper fraction is called
(iii) $5 \frac{1}{3}-3 \frac{1}{2}=$
(iv) $8 \div \frac{1}{2}=$
(v) The number which has its own reciprocal is
Solution:
(i) $14 \frac{1}{4}$
(ii) Mixed Fraction
(iii) $1 \frac{5}{6}$
(iv) 16
(v) 1

Question 2.
Say True or False
(i) $3 \frac{1}{2}$ can be written as $3+\frac{1}{2}$.
(ii) The sum of any two proper fractions is always an improper fraction.
(iii) The mixed fraction of $\frac{13}{4}$ is $3 \frac{1}{4}$.
(iv) The reciprocal of an improper fraction is always a proper fraction.
(v) $3 \frac{1}{4} \times 3 \frac{1}{4}=9 \frac{1}{16}$
Solution:
(i) True
(ii) False
(iii) True
(iv) True
(v) False

Question $3 .$
Answer the following :
Solution:
(i) Find the sum of $\frac{1}{7}$ and $\frac{3}{9}$
To find $\frac{1}{7}+\frac{3}{9}$ $\frac{1}{7}=$ $\frac{3}{9}=$
$\begin{aligned}
&\frac{1}{7}=\frac{1 \times 9}{7 \times 9}=\frac{9}{63} \\
&\frac{3}{9}=\frac{3 \times 7}{9 \times 7}=\frac{21}{63}
\end{aligned}$
$\frac{1}{7}+\frac{3}{9}=\frac{9}{63}+\frac{21}{63}=\frac{30}{63}$
Simplifying $\frac{30}{63}=\frac{30 \div 3}{63 \div 3}=\frac{10}{21} \quad \therefore \quad \frac{1}{7}+\frac{3}{9}=\frac{10}{21}$

(ii) What is the total of $3 \frac{1}{3}$ and $4 \frac{1}{6}$.
Total of $3 \frac{1}{3}$ and $4 \frac{1}{6}=3 \frac{1}{3}+4 \frac{1}{6}=3+\frac{1}{3}+4+\frac{1}{6}=(3+4)+\frac{1}{3}+\frac{1}{6}$ $=7+\frac{2}{6}+\frac{1}{6}=7+\frac{3}{6}=7+\frac{1}{2}=7 \frac{1}{2}$ $3 \frac{1}{3}+4 \frac{1}{6}=7 \frac{1}{2}$
(iii) Simplify : $1 \frac{3}{5}+5 \frac{4}{7}$
$\begin{aligned}
1 \frac{3}{5}+5 \frac{4}{7}=1+\frac{3}{5}+5+\frac{4}{7} &=(1+5)+\frac{3}{5} ;+\frac{-4}{7} \\
&=6+\frac{(3 \times 7)+(4 \times 5)}{35}=6+\left(\frac{21+20}{35}\right) \\
&=6+\frac{41}{35}=6+1 \frac{6}{35}=6+1+\frac{6}{35}=7+\frac{6}{35}=7 \frac{6}{35} \\
\therefore 1 \frac{3}{5}+5 \frac{4}{7} &=7 \frac{6}{35}
\end{aligned}$

(iv) Find the difference between $\frac{8}{9}$ and $\frac{2}{7}$
$\begin{aligned}
\frac{8}{9} &=\frac{8 \times 7}{9 \times 7}=\frac{56}{63} \\
\frac{2}{7} &=\frac{2 \times 9}{7 \times 9}=\frac{18}{63} \\
\frac{56}{63} &>\frac{18}{63} \\
\therefore \frac{8}{9} &>\frac{2}{7} \\
\frac{8}{9}-\frac{2}{7} &=\frac{56}{63}-\frac{18}{63}=\frac{56-18}{63}=\frac{38}{63} \\
\frac{8}{9}-\frac{2}{7} &=\frac{38}{63}
\end{aligned}$
(v) Subtract $1 \frac{3}{5}$ and $2 \frac{1}{3}$
$2 \frac{1}{3}-1 \frac{3}{5}=\frac{7}{3}+\frac{8}{5}=\frac{(7 \times 5)-(8 \times 3)}{15}=\frac{35-24}{15}=\frac{11}{15}$
$2 \frac{1}{3}-1 \frac{3}{5}=\frac{11}{15}$
(vi) Simplify: $7 \frac{2}{7}-3 \frac{4}{21}$

Question $4 .$
Convert mixed fraction into improper fractions and vice versa:
(i) $3 \frac{7}{18}$
(ii) $\frac{99}{7}$
(iii) $\frac{47}{6}$
(iv) $12 \frac{1}{9}$
Solution:

(i) $3 \frac{7}{18}$ is the given mixed fraction
$\begin{aligned}
\text { Improper fraction } &=\frac{\text { (whole number } \times \text { Denominator) }+\text { Numerator }}{\text { Denominator }} \\
&=\frac{(3 \times 18)+7}{18}=\frac{54+7}{18}=\frac{61}{18} ; \quad 3 \frac{7}{18}=\frac{61}{18}
\end{aligned}$
(ii) $\frac{99}{7}$ is the given improper fraction.

Question $5 .$
Multiply the following:
(i) $\frac{2}{3} \times 6$
(ii) $8 \times \frac{1}{3} \times 5$
(iii) $\frac{3}{8} \times \frac{4}{5}$
(iv) $3 \frac{5}{7} \times 1 \frac{1}{13}$

Solution:

(i) $\quad \frac{2}{3} \times 6=\frac{12}{3}=4$
(ii) $\quad 8 \frac{1}{3} \times 5=\left(8+\frac{1}{3}\right) \times 5=40+\frac{5}{3}=40+1 \frac{2}{3}$
$\begin{aligned}
&=40+1+\frac{2}{3}=41 \frac{2}{3} \\
&=41 \frac{2}{3}
\end{aligned}$
(iii)
$\begin{aligned}
&\frac{3}{8} \times \frac{4}{5}=\frac{3 \times 4}{8 \times 5}=\frac{12}{40}=\frac{3}{10} \\
&\frac{3}{8} \times \frac{4}{5}=\frac{3}{10}
\end{aligned}$
(iv)
$\begin{aligned}
&3 \frac{5}{7} \times 1 \frac{1}{13}=\frac{26}{7} \times \frac{14}{13} \\
&3 \frac{5}{7} \times 1 \frac{1}{13}=4
\end{aligned}$

Question $6 .$
Divide the following:
(i) $\frac{3}{7} \div 4$
(ii) $\frac{4}{3} \div \frac{5}{9}$
(iii) $4 \frac{1}{5}+3 \frac{3}{4}$
(iv) $9 \frac{2}{3} \div 1 \frac{2}{3}$
Solution:

(i) $\frac{3}{7} \div 4=\frac{3}{7} \times \frac{1}{4}=\frac{3}{28}$
$\frac{3}{7} \div 4=\frac{3}{28}$
(ii) $\begin{aligned} \frac{4}{3} \div \frac{5}{9} &=\frac{4}{3} \times \frac{9}{5} \\ &=\frac{36}{15}=2 \frac{6}{15}=2 \frac{2}{5} \\ \frac{4}{3} \div \frac{5}{9} &=2 \frac{2}{5} \\ \text { (iii) } \begin{aligned} 4 \frac{1}{5} \div 3 \frac{3}{4} &=\frac{21}{5} \div \frac{15}{4}=\frac{21}{5} \times \frac{4}{15} \\ &=\frac{84}{75}=1 \frac{9}{75}=1 \frac{3}{25} \\ 4 \frac{1}{5} \div 3 \frac{3}{4} &=1 \frac{3}{25} \\ 9 \frac{2}{3} \div 1 \frac{2}{3} &=\frac{29}{3} \div \frac{5}{3}=\frac{29}{3} \times \frac{3}{5} \end{aligned} \\ \text { (iv) } \quad \begin{aligned} 9 \frac{2}{3} \div 1 \frac{2}{3} &=5 \frac{29}{5} \\ &=5 \frac{4}{5} \\ &=5 \frac{4}{5} \end{aligned}\end{aligned}$

Question $7 .$
Gowri purchased $3 \frac{1}{2} \mathrm{~kg}$ of tomatoes, $\frac{3}{4} \mathrm{~kg}$ of brinjal and $1 \frac{1}{4} \mathrm{~kg}$ of onion, what is the total weight of vegetables she bought?
Solution:
Weight of tomatoes Gowri purchased $=3 \frac{1}{2} \mathrm{~kg}$
Weight of Brinjal purchased $=\frac{3}{4} \mathrm{~kg}$
$\therefore$ Total weight of the vegetables $=3 \frac{1}{2}+\frac{3}{4}+1 \frac{1}{4} \mathrm{~kg}=3+\frac{1}{2}+\frac{3}{4}+1+\frac{1}{4} \mathrm{~kg}$
$\begin{aligned}
&=(3+1)+\frac{1}{2}+\frac{3}{4}+\frac{1}{4} \mathrm{~kg}=4+\frac{2}{4}+\frac{3}{4}+\frac{1}{4} \mathrm{~kg} \\
&=4+\frac{2+3+1}{4} \mathrm{~kg}=4+\frac{6}{4} \mathrm{~kg} \\
&=4+1+\frac{2}{4} \mathrm{~kg}=5+\frac{1}{2} \mathrm{~kg}=5 \frac{1}{2} \mathrm{~kg}
\end{aligned}$
Total weight of vegetables that Gowri purchased $=5 \frac{1}{2} \mathrm{~kg}$

Question $8 .$
An oil tin contains $3 \frac{3}{4}$ litres of oil of which $2 \frac{1}{2}$ litres of oil is used. How much oil is left over?
Solution:
Total available oil in the tin $=3 \frac{3}{4}$ litres
Quantity of oil used $=2 \frac{1}{2}$ litres.
Quantity of oil left over $=3 \frac{3}{4}-2 \frac{1}{2}=3+\frac{3}{4}-\left(2+\frac{1}{2}\right)=(3-2)+\left(\frac{3}{4}-\frac{1}{2}\right)$
$=1+\left(\frac{3}{4}-\frac{2}{4}\right)=1+\left(\frac{3-2}{4}\right)=1+\frac{1}{4}=1 \frac{1}{4}$
Quantity of oil leftover $=1 \frac{1}{4}$ litres.

Question $9 .$
Nilavan can walk $4 \frac{1}{2} \mathrm{~km}$ in an hour. How much distance will he cover in $3 \frac{1}{2}$ hours?
Solution:
Distance walked by Nilavan in one hour $=4 \frac{1}{2} \mathrm{~km}$.
$\therefore$ Distance walked in $3 \frac{1}{2}$ hours $=4 \frac{1}{2} \times 3 \frac{1}{2} \mathrm{~km}$.
$4 \longdiv { 6 3 }$ $\frac{4}{23}$ $\frac{20}{3}$ $=\frac{9}{2} \times \frac{7}{2} \mathrm{~km}$
$=\frac{63}{4} \mathrm{~km}=15 \frac{3}{4} \mathrm{~km}$
Nilavan walks $15 \frac{3}{4} \mathrm{~km}$ in $3 \frac{1}{2}$ hours

Question $10 .$
Ravi bought a curtain of length $15 \frac{3}{4} \mathrm{~m}$. If he cut the curtain into small pieces each of length $2 \frac{1}{4} \mathrm{~m}$, then how many small curtains will he get?
Solution:
Length of the curtain Ravi had $=15 \frac{3}{4} \mathrm{~m}$
$\frac{63}{A} \times \frac{1}{9}$
Length of the small curtain $=2 \frac{1}{4} \mathrm{~m}$
$\therefore$ Number of pieces $=15 \frac{3}{4} \div 2 \frac{1}{4}=\frac{63}{4} \div \frac{9}{4}=\frac{63}{4} \times \frac{4}{9}=7$
Ravi will get 7 small curtains.
 

Objective Type Questions
Question 11.
Whcih of the following statement is incorrect?
(a) $\frac{1}{2}>\frac{1}{3}$
(b) $\frac{7}{8}>\frac{6}{7}$
(c) $\frac{8}{9}<\frac{9}{10}$
(d) $\frac{10}{11}<\frac{9}{10}$

Solution:
(d) $\frac{10}{11}<\frac{9}{10}$
Hint:
$\frac{10 \times 10}{11 \times 10}<\frac{9 \times 11}{10 \times 11}=\frac{100}{110}<\frac{99}{110}$
 

Question $12 .$
The difference between $\frac{3}{7}$ and is $\frac{2}{7}$
(a) $\frac{13}{63}$
(b) $\frac{1}{9} T$
(c) $\frac{1}{7}$
(d) $\frac{9}{16}$
Solution:
(a) $\frac{13}{63}$
Hint:
$\frac{35}{7}-\frac{2}{9}=\frac{(3 \times 9)-(2 \times 7)}{63}=\frac{27-14}{63}=\frac{13}{63}$

Question $13 .$
The reciprocal of is $\frac{53}{17}$
(a) $\frac{53}{17}$
(b) $5 \frac{3}{17}$
(c) $\frac{17}{53}$
(d) $3 \frac{5}{17}$
Solution:
(c) $\frac{17}{53}$
Hint:
$\frac{\frac{1}{15}}{\frac{15}{17}}=\frac{17}{53}$

Question 14.

If - , then the value of $A$ is
(a) $\frac{6}{7} 42 \frac{A}{49}$
(b) 36
(c) 25
(d) 48
Solution:
(a) 42
 

Question $15 .$
Pugazh has been given four choices for his pocket money by his father. Which of the choices should he take in order to get the maximum money?
$\begin{array}{ll}\text { (a) } \frac{2}{3} \text { of } ₹ 150 & \text { (b) } \frac{3}{5} \text { of } ₹ 150\end{array}$
(c) $\frac{4}{5}$ of $₹ 150$
(d) $\frac{1}{5}$ of ₹ 150
Solution:
(c) $\frac{4}{5}$ of ₹ 150