GSEB Class 7 Maths Solutions Chapter 8 Comparing Quantities Exercise 8.1

Question 1. Find the ratio of:
(a) Rs. 5 to 50 paise
(b) 15 kg to 210 g
(c) 9 m to 27 cm
(d) 30 days to 36 hours
Answer:
(a) Rs. 5 to 50 paise:
To find the ratio, we first change Rs. 5 into paise. We know that Rs. 1 equals 100 paise.
So, \( \text{Rs. } 5 = 5 \times 100 \text{ paise} = 500 \text{ paise} \)
Now we can write the ratio of 500 paise to 50 paise:
\( 500 \text{ paise} : 50 \text{ paise} \)
To simplify this ratio, we find the Highest Common Factor (HCF) of 500 and 50, which is 50.
We divide both sides by 50:
\( \frac { 500 }{ 50 } : \frac { 50 }{ 50 } \)
\( = 10 : 1 \)
(b) 15 kg to 210 g:
First, we convert 15 kg into grams. We know that 1 kg equals 1000 g.
So, \( 15 \text{ kg} = 15 \times 1000 \text{ g} = 15000 \text{ g} \)
Now, we write the ratio of 15000 g to 210 g:
\( 15000 \text{ g} : 210 \text{ g} \)
To simplify, we find the HCF of 15000 and 210, which is 30.
We divide both parts by 30:
\( \frac { 15000 }{ 30 } : \frac { 210 }{ 30 } \)
\( = 500 : 7 \)
(c) 9 m to 27 cm:
First, we change 9 meters into centimeters. We know that 1 m equals 100 cm.
So, \( 9 \text{ m} = 9 \times 100 \text{ cm} = 900 \text{ cm} \)
Now, we write the ratio of 900 cm to 27 cm:
\( 900 \text{ cm} : 27 \text{ cm} \)
To simplify this, we find the HCF of 900 and 27, which is 9.
We divide both parts by 9:
\( \frac { 900 }{ 9 } : \frac { 27 }{ 9 } \)
\( = 100 : 3 \)
(d) 30 days to 36 hours:
First, we convert 30 days into hours. We know that 1 day has 24 hours.
So, \( 30 \text{ days} = 30 \times 24 \text{ hours} = 720 \text{ hours} \)
Now, we write the ratio of 720 hours to 36 hours:
\( 720 \text{ hours} : 36 \text{ hours} \)
To simplify this, we find the HCF of 720 and 36, which is 36.
We divide both parts by 36:
\( \frac { 720 }{ 36 } : \frac { 36 }{ 36 } \)
\( = 20 : 1 \)
In simple words: To compare two amounts, always make sure they are in the same units first. Then, divide both numbers by their biggest common factor to simplify the ratio.

Exam Tip: Remember to always convert both quantities to the smallest common unit before finding the ratio to avoid mistakes. For example, convert kilograms to grams, meters to centimeters, and days to hours.

Question 2. In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?
Answer:
**First method**
We are told that for 6 students, 3 computers are available.
So, for 1 student, the number of computers available is \( \frac{3}{6} \).
Now, to find the number of computers needed for 24 students, we multiply the number of students by the rate per student:
Number of computers for 24 students \( = \frac{3}{6} \times 24 \)
\( = \frac{1}{2} \times 24 \)
\( = 12 \)
Therefore, 12 computers are needed for 24 students.
**Second method**
Let 'x' be the number of computers needed for 24 students.
We can set up a proportion based on the given ratio:
\( 3 : x = 6 : 24 \)
This can also be written as a fraction equation:
\( \frac{3}{x} = \frac{6}{24} \)
To solve for x, we can cross-multiply or simplify the fraction on the right side.
\( \frac{6}{24} \) simplifies to \( \frac{1}{4} \).
So, \( \frac{3}{x} = \frac{1}{4} \)
Cross-multiplying gives us:
\( 3 \times 4 = x \times 1 \)
\( 12 = x \)
So, \( x = 12 \)
Thus, 12 computers are required for 24 students.
In simple words: If you know how many computers you need for a small group, you can figure out how many you need for a bigger group by using ratios or by finding out how many computers each student gets.

Exam Tip: For problems involving proportions, always set up the ratio consistently (e.g., computers to students on both sides) and cross-multiply to solve for the unknown value. You can use either method to solve for the missing quantity, but the ratio method is often faster.

Question 3. Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs. Area of Rajasthan = 3 lakh km² and area of UP = 2 lakh km².
(i) How many people are there per km2 in both these States?
(ii) Which State is less populated?
Answer:
(i) To find the number of people per square kilometer (population density) for each state, we divide the total population by the total area.
**For Rajasthan:**
Population \( = 570 \text{ lakhs} \)
Area \( = 3 \text{ lakh km}^2 \)
Population per km\(^2 = \frac{\text{Population}}{\text{Area}} = \frac{570 \text{ lakhs}}{3 \text{ lakh km}^2} = 190 \text{ lakh per km}^2 \)
So, in Rajasthan, there are 190 lakh people for every square kilometer.
**For Uttar Pradesh (UP):**
Population \( = 1660 \text{ lakhs} \)
Area \( = 2 \text{ lakh km}^2 \)
Population per km\(^2 = \frac{\text{Population}}{\text{Area}} = \frac{1660 \text{ lakhs}}{2 \text{ lakh km}^2} = 830 \text{ lakh per km}^2 \)
So, in UP, there are 830 lakh people for every square kilometer.
(ii) To find which state is less populated, we compare their population densities.
We have: \( 190 \text{ lakhs} < 830 \text{ lakhs} \)
Since Rajasthan has 190 lakh people per km\(^2\) and UP has 830 lakh people per km\(^2\), Rajasthan is the less populated state.
In simple words: To see how crowded an area is, you divide the total people by the total land space. The state with fewer people per square kilometer is less crowded.

Exam Tip: When comparing population densities, ensure you are comparing "people per unit area" for both regions. A larger total population does not always mean higher density if the area is also significantly larger. Always calculate the density before making comparisons.