Exercise 1.2 - 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 $1.2$
Miscellaneous Practice Problems
Question $1 .$
Sankari purchased $2 \frac{1}{2} \mathrm{~m}$ cloth to stitch a long skirt and $1 \frac{3}{4} \mathrm{~m}$ cloth to stitch blouse. If the cost is $₹ 120$ per metre then find the cost of cloth purchased by her
Solution:
Cloth to stitch a long skirt $=2 \frac{1}{2} \mathrm{~m}$
Cloth to stitch a blouse $=1 \frac{3}{4} \mathrm{~m}$
Total length of the cloth $=2 \frac{1}{2}+1 \frac{3}{4} m=2+\frac{1}{2}+1+\frac{3}{4} \quad \mathrm{~m}$
$\begin{aligned}
&=(2+1)+\left(\frac{1}{2}+\frac{3}{4}\right) \mathrm{m}=3+\left(\frac{2}{4}+\frac{3}{4}\right) \mathrm{m} \\
&=3+\left(\frac{2+3}{4}\right) \mathrm{m}=3+\frac{5}{4} \mathrm{~m} \\
&=3+1 \frac{1}{4} \mathrm{~m}=3+1+\frac{1}{4} \mathrm{~m}=4+\frac{1}{4} \mathrm{~m}=4 \frac{1}{4} \mathrm{~m}
\end{aligned}$
Cost of cloth per metre $=₹ 120 .$
$\therefore$ Cost of cloth for $4 \frac{1}{4} \mathrm{~m}=4 \frac{1}{4} \times 120=\frac{17}{4} \times 120=17 \times 30=₹ 510$
Cost of cloth purchased by Sankari $=₹ 510$
 

Question 2 .
From his office, a person wants to reach his house on foot which is at a distance of $5 \frac{3}{4} \mathrm{~km}$. If he had walked $2 \frac{1}{2} \mathrm{~km}$, how much distance still he has to walk to reach his house?
Solution:

Distance of the house from office $=5 \frac{3}{4} \mathrm{~km}$
Distance he walked $=2 \frac{1}{2} \mathrm{~km}$
$\therefore$ Remaining distance $=5 \frac{3}{4}-2 \frac{1}{2} \mathrm{~km}$
$\begin{aligned}
&=5+\frac{3}{4}-\left(2+\frac{1}{2}\right)=(5-2)+\left(\frac{3}{4}-\frac{1}{2}\right) \mathrm{km} \\
&=3+\left(\frac{3}{4}-\frac{2}{4}\right) \mathrm{km}=3+\left(\frac{3-2}{4}\right) \mathrm{km} \\
&=3+\frac{1}{4}=3 \frac{1}{4} \mathrm{~km}
\end{aligned}$
Distance still he has to be walked $=3 \frac{1}{4} \mathrm{~km}$
 

Question $3 .$
Which is smaller? The difference between $\frac{1}{2}$ and $3 \frac{2}{3}$ or the sum of $1 \frac{1}{2}$ and $2 \frac{1}{4}$
Solution:
Difference between $2 \frac{1}{2}$ and $3 \frac{2}{3}=3 \frac{2}{3}-2 \frac{1}{2}$
$\begin{aligned}
&=\frac{1 k}{3} \cdot \frac{-6}{2}=\frac{(11 \times 2)-(5 \times 3)}{6} \\
&=\frac{22-15}{6}=\frac{7}{6}=1 \frac{1}{6}
\end{aligned}$
$\frac{6}{6}-1 \frac{6}{3}$

Question $4 .$
Mangai bought $6 \frac{3}{4} \mathrm{~kg}$ of apples. If Kalai bought $1 \frac{1}{2}$ times as Mangai bought, then how many kilograms of apples did Kalai buy?
Solution:
Weight of apples Mangai bought $=6 \frac{3}{4} \mathrm{~kg}$
Weight of apples Kalai bought $=1 \frac{1}{2}$ times of Mangai bought
$=\frac{3 \times 27}{2 \times 4}=\frac{81}{8}=10 \frac{1}{8} \mathrm{~kg}$
$\begin{aligned}
&=1 \frac{1}{2} \times 6 \frac{3}{4}=\frac{3}{2} \times \frac{27}{4} \\
&=\frac{3 \times 27}{2 \times 4}=\frac{81}{8}=10 \frac{1}{8} \mathrm{~kg}
\end{aligned}$
Weight of apples Kalai bought $=10 \frac{1}{8} \mathrm{~kg}$

Question $5 .$
The length of the staircase is $5 \frac{1}{2} \mathrm{~m}$. If one step is set at $\frac{1}{4} \mathrm{~m}$, then how many steps will be there in the staircase?
Solution:
Length of the staircase $=5 \frac{1}{2} \mathrm{~m}$
Distance between each step $=\frac{1}{4} \mathrm{~m}$
$\therefore$ Number of steps in the staircase $=5 \frac{1}{2} \div \frac{1}{4}=\frac{11}{2} \div \frac{1}{4}=\frac{11}{2} \times \frac{4}{1}=22$
There will be 22 steps in the staircase
 

Challenge Problems
Question 6 .
By using the following clues, find who am I?
(i) Each of my numerator and denominator is a single digit number.
(ii) The sum of my numerator and denominator is a multiple of $3 .$
(iii) The product of my numerator and denominator is a multiple of 4
Solution:
The numerator may be any one of? $, 2,3,4,5,6,7,8,9$ and the denominator may be any one of $1,2,3,4,5,6$, $7,8,9 .$ Sum of numerator and denominator is a multiple of $3 .$
$\therefore$ Possible proper fractions are $\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{2}{4}, \frac{2}{7}, \frac{3}{6}, \frac{3}{9}, \frac{4}{5}, \frac{4}{8}, \frac{5}{7}, \frac{6}{9}$
Also given the product of numerator and denominator is a multiple of 4 .
$\therefore$ Possible fractions are $\frac{1}{8}, \frac{2}{4}, \frac{4}{5}, \frac{4}{8}$

Question $7 .$
Add the difference between $1 \frac{1}{3}$ and $3 \frac{1}{6}$ and the difference between $4 \frac{1}{6}$ and $2 \frac{1}{3}$
Solution:

We have to find $\left(3 \frac{1}{6}-1 \frac{1}{3}\right)+\left(4 \frac{1}{6}-2 \frac{1}{3}\right)$
$\left[\because 3 \frac{1}{6}>1 \frac{1}{3}\right.$ and $\left.4 \frac{1}{6}>2 \frac{1}{3}\right]$
$=\left(\frac{19}{6}-\frac{4}{3}\right)+\left(\frac{25}{6}-\frac{7}{3}\right)$
$=\left(\frac{19}{6}-\frac{8}{6}\right)+\left(\frac{25}{6}-\frac{14}{6}\right)=\frac{11}{6}+\frac{11}{6}$
$=\frac{22}{6}=3 \frac{4}{6}=3 \frac{2}{3}$

Question $8 .$
What fraction is to be subtracted from $9 \frac{3}{7}$ to get $3 \frac{1}{5}$ ?
Solution:
$9 \frac{3}{7}-($ The number to be subtracted $)=3 \frac{1}{5}$
$\therefore 9 \frac{3}{7}-3 \frac{1}{5}=$ The number to be subtracted
$\therefore$ The number to be subtracted $=\left(9+\frac{3}{7}\right)-\left(3+\frac{1}{5}\right)$
$\begin{aligned}
&=(9-3)+\left(\frac{3}{7} \pi \cdot \frac{71}{5}\right)=6+\left[\frac{(3 \times 5)-(1 \times 7)}{7 \times 5}\right] \\
&=6+\left(\frac{15-7}{35}\right)=6+\frac{8}{35}
\end{aligned}$
$\therefore$ The fraction to be subtracted $=6 \frac{8}{35}$

Question 9 .
The sum of two fractions is $5 \frac{3}{9}$. If one of the fractions is $2 \frac{3}{4}$, find the other fraction.
Solution:

Sum of two fractions $=5 \frac{3}{9}$
One of them $=2 \frac{3}{4}$
The other fraction $=5 \frac{3}{9}-2 \frac{3}{4}$
$\begin{aligned}
&=\frac{48}{9}-\frac{11}{4}=\frac{16}{3} \div \frac{11}{4}=\frac{(16 \times 4)-(11 \times 3)}{3 \times 4} \\
&=\frac{64-33}{12}=\frac{31}{12}=2 \frac{7}{12}
\end{aligned}$

Question 10 .
By what number should $3 \frac{1}{16}$ be multiplied to get $9 \frac{3}{16} ?$
Solution:

Question 11.

Complete the fifth row in the Leibnitz triangle which is based on subtraction

Solution:

Question 12 .
A painted $\frac{3}{8}$ of the wall of which one third is painted in yellow colour. What fraction is the yellow colour of the entire wall.

Solution:
Painted portion of the wall $=\frac{3}{8}$
Of which the yellow colour painted $=\frac{1}{3}$
$\therefore$ Yellow colour in the entire wall $=\frac{1}{3}$ of $\frac{3}{8}$
$=\frac{1}{3} \times \frac{3}{8}=\frac{1}{8}$
$\frac{1}{8}$ of the wall is painted yellow.
 

Question $13 .$
A rabbit has to cover $26 \frac{1}{4} \mathrm{~m}$ to fetch its food. If it covers $1 \frac{3}{4} \mathrm{~m}$ in one jump, thenhow many jumps will it take to fetch its food?

Solution:
Total distance to be covered by the rabbit $=26 \frac{1}{4} \mathrm{~m}$
Distance covered in one jump $=1 \frac{3}{4} \mathrm{~m}$

$\therefore$ The rabbit jumps 15 times to fetch its food.
 

Question $14 .$
Look at the picture and answer the following questions:
(i) What is the distance from School to library via bus stop?
(ii) What is the distance between school and library via Hospital?
(iii) Which is the shortest distance between (i) and (ii)?
(iv) The distance between School and Hospital is ___times the distance between school and Bus stop.

Solution: