# Exercise 3.1 - Chapter 3 Ratio and Proportion Term 1 6th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024]

Ex $3.1$
Question 1 .

Fill in the blanks.
(i) Ratio of ₹ 3 to ₹ $5=$
(ii) Ratio of $3 \mathrm{~m}$ to $200 \mathrm{~cm}=$
(iii) Ratio of $5 \mathrm{~km} 400 \mathrm{~m}$ to $6 \mathrm{~km}=$
(iv) Ratio of 75 paise to $₹ 2=$
Solution:
(i) $3: 5$
(ii) $3: 2$
Hint: $3 \mathrm{~m}=300 \mathrm{~cm}$
(iii) $9: 10$
Hint: $5 \mathrm{~km} 400 \mathrm{~m}=5400 \mathrm{~m}$ and $6 \mathrm{~km}=6000 \mathrm{~m}$
(iv) $3: 8$
Hint: $₹ 2=200$ paise

Question 2 .
Say whether the following statements are True or False.
(i) The ratio of $130 \mathrm{~cm}$ to $1 \mathrm{~m}$ is $13: 10$.
(ii) One of the terms in a ratio cannot be $1 .$
Solution:
(i) True
(ii) False

Question $3 .$
Find the simplified form of the following ratios.
(i) $15: 20$
(ii) $32: 24$
(iii) $7: 15$
(iv) $12: 27$

(v) $75: 100$
Solution:
(i) $15: 20=\frac{15}{20}=\frac{15 \div 5}{20 \div 5}=\frac{3}{4}=3: 4$
(ii) $32: 24=\frac{32}{24}=\frac{32 \div 8}{24 \div 8}=\frac{4}{3}=4: 3$
(iii) $7: 15=\frac{7}{15}=7: 15$
(iv) $12: 27=\frac{12}{27}=\frac{12 \div 3}{27 \div 3}=\frac{4}{9}=4: 9$
(v) $75: 100=\frac{75}{100}=\frac{75 \div 25}{100 \div 25}=\frac{3}{4}=3: 4$

Question $4 .$
Akilan walks $10 \mathrm{~km}$ in an hour while Selvi walks $6 \mathrm{~km}$ in an hour. Find the simplest ratio of the distance covered by Akilan to that of Selvi.
Solution:
Ratio of the distance covered by Akilan to that of Selvi $=10 \mathrm{~km}: 6 \mathrm{~km}$
\begin{aligned} &=\frac{10}{6} \\ &=\frac{5}{3} \\ &=5: 3 \end{aligned}

Question $5 .$
The cost of parking a bicycle is $₹ 5$ and the cost of parking a scooter is $₹ 15$. Find the simplest ratio of the parking cost of a bicycle to that of a scooter.
Solution:
Parking cost of bicycle $=₹ 5$
Parking cost of Scooter $=₹ 15$
$\frac{\text { Parking cost of bicyle }}{\text { Parking cost of scooter }}=\frac{5}{15}=\frac{1}{3}$
Parking cost of bicycle : scooter $=1: 3$

Question 6 .
Out of 50 students in class 30 are boys. Find the ratio of
(i) number of boys to the number of girls
(ii) number of girls to the total number of students
(iii) number of boys to the total number of students
Solution:
The total number of students $=50$.
The number of boys $=30$.
Then number of girls $=50-30=20$.
(i) Ratio of number of boys to no. of girls $=\frac{30}{20}=\frac{30 \div 10}{20 \div 10}=\frac{3}{2}=3: 2$
(ii) $\frac{\text { Number of girls }}{\text { Total number of students }}=\frac{20}{50}=\frac{20 \div 10}{50 \div 10}=\frac{2}{5}=2: 5$
(iii) $\frac{\text { Number of boys }}{\text { Total number of students }}=\frac{30}{50}=\frac{30 \div 10}{50 \div 10}=\frac{3}{5}=3: 5$

Objective Type Questions
Question $7 .$

The ratio of Rs 1 to 20 paise is
(a) $1: 5$
(b) $1: 2$
(c) $2: 1$
(d) $5: 1$
Solution:
(d) $5: 1$

Question $8 .$
The ratio of $1 \mathrm{~m}$ to $50 \mathrm{~cm}$ is
(a) $1: 50$
(b) $50: 1$
(c) $2: 1$
(d) $1: 2$
Solution:
(c) $2: 1$
Hint: $1 \mathrm{~m}=100 \mathrm{~cm}$

Question $9 .$
The length and breadth of a window are in $1 \mathrm{~m}$ and $70 \mathrm{~cm}$ respectively. The ratio of the length to the breadth is
a) $1: 7$
(b) $7: 1$
(c) $7: 10$
(d) $10: 7$
Solution:
(d) $10: 7$

Question 10 .
The ratio of the number of sides of a triangle to the number of sides of a rectangle is
(a) $4: 3$
(b) $3: 4$
(c) $3: 5$
(d) $3: 2$
Solution:
(b) $3: 4$
Triangle has three sides and a rectangle has four sides.

Question 11 .
If Azhagan is 50 years old and his son is 10 years old then the simplest ratio between the age of Azhagan to his son is
(a) $10: 50$
(b) $50: 10$
(c) $5: 1$
(d) $1: 5$
Solution:
(c) $5: 1$