Exercise 2.1 Solutions
1. Express the following pairs of number as the ratio of the first and second quantities:
(a) $25$ and $35$
Solution: The ratio of $25$ and $35$ is
$\frac{25}{35}$
$= \frac{25 \div 5}{35 \div 5}$
$= \frac{5}{7}$
So, the ratio is $5 : 7$.
(b) $7 \frac{1}{3}$ and $9 \frac{2}{5}$
Solution:
First, converting the mixed fractions into improper fractions, we get
$7 \frac{1}{3} = \frac{(7 \times 3) + 1}{3} = \frac{22}{3}$
$9 \frac{2}{5} = \frac{(9 \times 5) + 2}{5} = \frac{47}{5}$
Now, the ratio of $\frac{22}{3}$ and $\frac{47}{5}$ is
$\frac{22}{3} : \frac{47}{5}$
$= \frac{22}{3} \times \frac{5}{47}$
$= \frac{110}{141}$
So, the ratio is $110 : 141$.
(c) $1 \text{ year } 2 \text{ months}$ and $7 \text{ months}$
Solution: First, convert all quantities to the same unit (months), we obtain
$1 \text{ year } = 12 \text{ months}$
$1 \text{ year } 2 \text{ months} = 12 + 2 = 14 \text{ months}$
Now, the ratio of $14 \text{ months}$ and $7 \text{ months}$ is
$14 : 7 = \frac{14}{7} = \frac{2}{1}$
So, the ratio is $2 : 1$.
(d) $7 \text{ kg}$ and $2 \text{ kg } 300 \text{ gram}$
Solution: Convert both quantities to grams, we get
$1 \text{ kg} = 1000 \text{ gram}$
$7 \text{ kg} = 7 \times 1000 = 7000 \text{ gram}$
$2 \text{ kg } 300 \text{ gram} = (2 \times 1000) + 300 = 2300 \text{ gram}$
Now, the ratio is
$7000 : 2300 = \frac{7000}{2300} = \frac{70}{23}$
So, the ratio is $70 : 23$.
(e) $\text{Tk. } 2$ and $40 \text{ paisa}$
Solution: Convert Taka to Paisa
$\text{Tk. } 1 = 100 \text{ paisa}$
$\text{Tk. } 2 = 2 \times 100 = 200 \text{ paisa}$
Now, the ratio of $200 \text{ paisa}$ and $40 \text{ paisa}$ is
$200 : 40 = \frac{200}{40} = \frac{5}{1}$
So, the ratio is $5 : 1$.
2. Simplify the following ratios:
(a) $9 : 12$
Solution: $9 : 12 = \frac{9 \div 3}{12 \div 3} = \frac{3}{4}$
So, the simplified ratio is $3 : 4$.
(b) $15 : 21$
Solution: $15 : 21 = \frac{15 \div 3}{21 \div 3} = \frac{5}{7}$
So, the simplified ratio is $5 : 7$.
(c) $45 : 36$
Solution: $45 : 36 = \frac{45 \div 9}{36 \div 9} = \frac{5}{4}$
So, the simplified ratio is $5 : 4$.
(d) $65 : 26$
Solution:
$65 : 26 = \frac{65 \div 13}{26 \div 13} = \frac{5}{2}$
So, the simplified ratio is $5 : 2$.
3. Fill in the gaps of the following equivalent ratios:
(a) $2 : 3 = 8 : \square$
Solution: Given,
$\frac{2}{3} = \frac{8}{\square}$
Multiplying the numerator and denominator of the left ratio by $4$ to get $8$ in the numerator, we obtain
$\frac{2 \times 4}{3 \times 4} = \frac{8}{12}$
So, the missing value is $12$.
Answer: $2 : 3 = 8 : 12$
(b) $5 : 6 = \square : 36$
Solution: Given,
$frac{5}{6} = \frac{\square}{36}$
Since $6 \times 6 = 36$, multiply both numerator and denominator by $6$, we get
$\frac{5 \times 6}{6 \times 6} = \frac{30}{36}$
So, the missing value is $30$.
Answer: $5 : 6 = 30 : 36$
(c) $7 : \square = 42 : 54$
Solution: Given,
$\frac{7}{\square} = \frac{42}{54}$
Dividing both numerator and denominator of the right ratio by $6$, we get
$\frac{42 \div 6}{54 \div 6} = \frac{7}{9}$
So, the missing value is $9$.
Answer: $7 : 9 = 42 : 54$
(d) $\square : 9 = 63 : 81$
Solution: Given,
$\frac{\square}{9} = \frac{63}{81}$
Dividing both numerator and denominator of the right ratio by $9$, we have
$\frac{63 \div 9}{81 \div 9} = \frac{7}{9}$
So, the missing value is $7$.
Answer: $7 : 9 = 63 : 81$
4. The ratio of length and breadth of a hall room is $2 : 5$. Putting the possible values of length and breadth in the gaps, complete the following table.
| Breadth of hall room (m.) | 10 | 20 | 40 | 80 | 160 |
| Length of hall room (m.) | $25$ | $50$ | $100$ | $200$ | $400$ |
Solution: The ratio of length to breadth given in the table is
$\frac{\text{Breadth}}{\text{Length}} = \frac{10}{25} = \frac{2}{5}$
2nd column:
$\frac{\text{Breadth}}{50} = \frac{2}{5} \implies \text{Breadth} = \frac{2 \times 50}{5} = 20$
3rd column:
$\frac{40}{\text{Length}} = \frac{2}{5} \implies \text{Length} = \frac{40 \times 5}{2} = 100$
4th column:
$\frac{\text{Breadth}}{200} = \frac{2}{5} \implies \text{Breadth} = \frac{2 \times 200}{5} = 80$
5th column:
$\frac{160}{\text{Length}} = \frac{2}{5} \implies \text{Length} = \frac{160 \times 5}{2} = 400$
5. Identify the equivalent ratios of the followings :
$12 : 18$ ; $6 : 18$ ; $15 : 10$ ; $3 : 2$ ; $6 : 9$ ; $2 : 3$ ; $1 : 3$ ; $2 : 6$ ; $12 : 8$
Solution: Let us simplify each ratio to its lowest form:
$12 : 18 = \frac{12 \div 6}{18 \div 6} = \frac{2}{3} = 2 : 3$
$6 : 18 = \frac{6 \div 6}{18 \div 6} = \frac{1}{3} = 1 : 3$
$15 : 10 = \frac{15 \div 5}{10 \div 5} = \frac{3}{2} = 3 : 2$
$3 : 2$ (already in simplest form)
$6 : 9 = \frac{6 \div 3}{9 \div 3} = \frac{2}{3} = 2 : 3$
$2 : 3$ (already in simplest form)
$1 : 3$ (already in simplest form)
$2 : 6 = \frac{2 \div 2}{6 \div 2} = \frac{1}{3} = 1 : 3$
$12 : 8 = \frac{12 \div 4}{8 \div 4} = \frac{3}{2} = 3 : 2$
Grouping Equivalent Ratios:
Group 1 ($2 : 3$): $12 : 18$, $6 : 9$, and $2 : 3$
Group 2 ($1 : 3$): $6 : 18$, $1 : 3$, and $2 : 6$
Group 3 ($3 : 2$): $15 : 10$, $3 : 2$, and $12 : 8$
6. Express the following simple ratios into the mixed ratio :
(a) $3 : 5$, $5 : 7$, and $7 : 9$
Solution: To find the mixed (compound) ratio, multiply the antecedent (first) terms together and the consequent (second) terms together:
$\text{Antecedents product} = 3 \times 5 \times 7 = 105$
$\text{Consequents product} = 5 \times 7 \times 9 = 315$
The required mixed ratio is
$105 : 315 = \frac{105}{315} = \frac{1}{3}$
Answer: $1 : 3$
(b) $5 : 3$, $7 : 5$, and $9 : 7$
Solution:
$\text{Antecedents product} = 5 \times 7 \times 9 = 315$
$\text{Consequents product} = 3 \times 5 \times 7 = 105$
The required mixed ratio is:
$315 : 105 = \frac{315}{105} = \frac{3}{1}$
Answer: $3 : 1$
7. Express the ratio $9 : 16$ as inverse ratio.
Solution: The inverse ratio of a given ratio $a : b$ is $b : a$. Therefore, the inverse ratio of $9 : 16$ is $16 : 9$.
Answer: $16 : 9$
8. Which one of the following ratios are unit ratios :
(a) $16 : 13$ \quad (b) $13 : 17$ \quad (c) $21 : 21$
Solution: A ratio whose antecedent and consequent are equal (giving a simplified value of $1 : 1$) is called a unit ratio.
(a) $16 : 13 \neq 1 : 1$
(b) $13 : 17 \neq 1 : 1$
(c) $21 : 21 = \frac{21}{21} = 1 : 1$
Answer: (c) $21 : 21$ is a unit ratio.
9. Divide Tk. $550$ in the ratios of $5 : 6$ and $4 : 7$.
Solution:
Part 1: Dividing Tk. $550$ in the ratio $5 : 6$
Sum of ratio terms $= 5 + 6 = 11$
First share $= \text{Tk. } 550 \times \frac{5}{11} = \text{Tk. } 250$
Second share $= \text{Tk. } 550 \times \frac{6}{11} = \text{Tk. } 300$
Part 2: Dividing Tk. $550$ in the ratio $4 : 7$
Sum of ratio terms $= 4 + 7 = 11$
First share $= \text{Tk. } 550 \times \frac{4}{11} = \text{Tk. } 200$
Second share $= \text{Tk. } 550 \times \frac{7}{11} = \text{Tk. } 350$
Answer:
For $5 : 6$ ratio: Tk. $250$ and Tk. $300$
For $4 : 7$ ratio: Tk. $200$ and Tk. $350$
10. The ratio of the age of father and son is $14 : 3$. If the age of father is $56$ years, what is the age of son ?
Solution: Given,
$\frac{\text{Father’s age}}{\text{Son’s age}} = \frac{14}{3}$
Substituting father’s age $= 56$ years, we get
$\frac{56}{\text{Son’s age}} = \frac{14}{3}$
$14 \times \text{Son’s age} = 56 \times 3$
$\text{Son’s age} = \frac{56 \times 3}{14} = 4 \times 3 = 12 \text{ years}$
Answer: The age of the son is $12$ years.
11. The price ratio of two books is $5 : 7$. If the price of the second book is $84$, what is the price of the first book ?
Solution: Given,
$\frac{\text{Price of 1st book}}{\text{Price of 2nd book}} = \frac{5}{7}$
Substituting the price of the second book $= 84$, we obtain
$\frac{\text{Price of 1st book}}{84} = \frac{5}{7}$
$\text{Price of 1st book} = \frac{5 \times 84}{7} = 5 \times 12 = 60$
Answer: The price of the first book is $60$.
12. In the golden ornament of $20 \text{ grams}$ of $18 \text{ carat}$, the ratio of gold and alloy is $3 : 1$. Determine the quantity of gold and alloy in that ornament.
Solution:
Total weight of the ornament = $20 \text{ grams}$
The ratio of gold and alloy = $3 : 1$
Sum of the ratio terms = $3 + 1 = 4$
Quantity of gold = $20 \times \frac{3}{4} = 5 \times 3 = 15 \text{ grams}$
Quantity of alloy = $20 \times \frac{1}{4} = 5 \text{ grams}$
Answer: Quantity of gold is $15 \text{ grams}$ and alloy is $5 \text{ grams}$.
13. The time ratio of going to and coming from school from house of two friends is $2 : 3$. If the distance of school from the house of the first friend is $5 \text{ km}$, what is the distance of school from the house of the second friend?
Solution: Given,
Ratio of distances = $2 : 3$ Distance of school from 1st friend’s house = $5 \text{ km}$
So, we can write: $\frac{\text{Distance of 1st friend’s house}}{\text{Distance of 2nd friend’s house}} = \frac{2}{3}$
$\frac{5}{\text{Distance of 2nd friend’s house}} = \frac{2}{3}$
$2 \times \text{Distance of 2nd friend’s house} = 5 \times 3$
$\text{Distance of 2nd friend’s house} = \frac{15}{2} = 7.5 \text{ km}$
Answer: The distance of school from the house of the second friend is $7.5 \text{ km}$ (or $7 \frac{1}{2} \text{ km}$).
14. The price ratio of two computers is $5 : 6$. If the price of the first computer is Tk. $25,000/-$, what is the price of the second computer? Due to price hike if the price of the first computer is increased by Tk. $5,000/-$, what type of ratio of their prices will be?
Solution:
Part 1: Finding the price of the second computer
Given, $\frac{\text{Price of 1st computer}}{\text{Price of 2nd computer}} = \frac{5}{6}$
$\frac{25000}{\text{Price of 2nd computer}} = \frac{5}{6}$
$5 \times \text{Price of 2nd computer} = 25000 \times 6$
$\text{Price of 2nd computer} = \frac{25000 \times 6}{5} = 5000 \times 6 = 30,000 \text{ Taka}$
Part 2: New ratio after price increase
Increased price of the 1st computer = $25,000 + 5,000 = 30,000 \text{ Taka}$
Price of the 2nd computer = $30,000 \text{ Taka}$
New ratio of their prices = $30000 : 30000 = 1 : 1$
Since both terms of the ratio are equal, it is a Unit Ratio.
Answer:
The price of the second computer is Tk. $30,000/-$
The new price ratio will be a Unit Ratio ($1 : 1$).