CBSE Class 8 Maths Chapter 8: Comparing Quantities NCERT Solutions

1. The speed of the cycle is 15 km/hr and the speed of the scooter is 30 km/hr. To find the ratio, we divide the speed of the cycle by the speed of the scooter: $$\backslash frac\{\backslash text\{Speed\; of\; cycle\}\}\{\backslash text\{Speed\; of\; scooter\}\}\; =\; \backslash frac\{15\; \backslash text\{\; km/hr\}\}\{30\; \backslash text\{\; km/hr\}\}\; =\; \backslash frac\{15\}\{30\}$$ Simplifying the fraction, we get $$\backslash frac\{1\}\{2\}$$. Therefore, the ratio is 1:2.
2. We need to find the ratio of 50 m to 10 km. First, we must ensure both quantities are in the same unit. Since 1 km = 1000 m, 10 km is equal to $10\; \backslash times\; 1000\; =\; 10000$ m. Now we can find the ratio: $$\backslash frac\{50\; \backslash text\{\; m\}\}\{10\; \backslash text\{\; km\}\}\; =\; \backslash frac\{50\; \backslash text\{\; m\}\}\{10000\; \backslash text\{\; m\}\}\; =\; \backslash frac\{50\}\{10000\}$$ Simplifying this fraction by dividing both numerator and denominator by 50, we get $$\backslash frac\{1\}\{200\}$$. Thus, the ratio is 1:200.
3. We need to find the ratio of 50 paise to ₹ 5. First, convert ₹ 5 to paise. Since ₹ 1 = 100 paise, ₹ 5 is equal to $5\; \backslash times\; 100\; =\; 500$ paise. Now we can find the ratio: $$\backslash frac\{50\; \backslash text\{\; paise\}\}\{₹\; 5\}\; =\; \backslash frac\{50\; \backslash text\{\; paise\}\}\{500\; \backslash text\{\; paise\}\}\; =\; \backslash frac\{50\}\{500\}$$ Simplifying this fraction by dividing both numerator and denominator by 50, we get $$\backslash frac\{1\}\{10\}$$. Therefore, the ratio is 1:10.

1. To convert the ratio 3:4 to a percentage, we first write it as a fraction $\backslash frac\{3\}\{4\}$. Then, we multiply by 100% to get the percentage: $$\backslash frac\{3\}\{4\}\; \backslash times\; 100\; \backslash \%\; =\; 75\; \backslash \%$$
2. To convert the ratio 2:3 to a percentage, we write it as a fraction $\backslash frac\{2\}\{3\}$. Then, we multiply by 100% to find the percentage: $$\backslash frac\{2\}\{3\}\; \backslash times\; 100\; \backslash \%\; =\; \backslash frac\{200\}\{3\}\; \backslash \%\; =\; 66\; \backslash frac\{2\}\{3\}\; \backslash \%$$

The total number of students is 25.

The percentage of students good in mathematics is 72%. To find the number of students good in mathematics, we calculate 72% of 25:

\text{Number of good students} = 72\% \text{ of } 25 = \frac{72}{100} \times 25

Calculating this, we get \frac{72 \times 25}{100} = \frac{1800}{100} = 18 students.

The number of students not good in mathematics is the total number of students minus the number of students who are good in mathematics:

\text{Number of not good students} = 25 - 18 = 7

Therefore, 7 students are not good in mathematics.

Let the total number of matches played by the football team be x.

According to the problem, the team won 10 matches, and this represents 40% of the total matches played. We can write this as an equation:

40\% \text{ of } x = 10

To solve for x, we convert the percentage to a fraction:

\frac{40}{100} \times x = 10

Now, we isolate x:

x = \frac{10 \times 100}{40}

x = \frac{1000}{40}

x = 25

So, the football team played a total of 25 matches.

Let the total amount of money Chameli had in the beginning be ₹ x.

Chameli spent 75% of her money. This means the remaining percentage of her money is 100\% - 75\% = 25\%.

We are given that she had ₹ 600 left, which is 25% of her original money. We can set up the equation:

25\% \text{ of } x = 600

To find the original amount x, we convert the percentage to a fraction:

\frac{25}{100} \times x = 600

Now, we solve for x:

x = \frac{600 \times 100}{25}

x = 600 \times 4

x = 2400

Therefore, Chameli had ₹ 2400 in the beginning.