CBSE Class 8 Maths Chapter 13 Direct and Inverse Proportions NCERT Solutions

To check if the parking charges are in direct proportion to the parking time, we need to see if the ratio of charges to time is constant for all given intervals.

Let's calculate the charges per hour for each duration:

• For 4 hours: Charge per hour = $\backslash frac\{₹60\}\{4\}\; =\; ₹15$
• For 8 hours: Charge per hour = $\backslash frac\{₹100\}\{8\}\; =\; ₹12.50$
• For 12 hours: Charge per hour = $\backslash frac\{₹140\}\{12\}\; \approx \; ₹11.67$
• For 24 hours: Charge per hour = $\backslash frac\{₹180\}\{24\}\; =\; ₹7.50$

Since the charges per hour ($₹15,\; ₹12.50,\; ₹11.67,\; ₹7.50$) are not the same for all durations, the parking charges are not in direct proportion to the parking time.

This problem involves a direct proportion because if you increase the parts of red pigment, you need to increase the parts of base proportionally to maintain the same mixture ratio.

The given ratio of red pigment to base is 1:8. Let this constant ratio be $k$.

$k\; =\; \backslash frac\{\backslash text\{Parts\; of\; red\; pigment\}\}\{\backslash text\{Parts\; of\; base\}\}\; =\; \backslash frac\{1\}\{8\}$

We need to find the parts of base for different amounts of red pigment:

1. When parts of red pigment = 4:

   $\backslash frac\{4\}\{\backslash text\{Parts\; of\; base\}\}\; =\; \backslash frac\{1\}\{8\}$

   Parts of base = $4\; \backslash times\; 8\; =\; 32$

2. When parts of red pigment = 7:

   $\backslash frac\{7\}\{\backslash text\{Parts\; of\; base\}\}\; =\; \backslash frac\{1\}\{8\}$

   Parts of base = $7\; \backslash times\; 8\; =\; 56$

3. When parts of red pigment = 12:

   $\backslash frac\{12\}\{\backslash text\{Parts\; of\; base\}\}\; =\; \backslash frac\{1\}\{8\}$

   Parts of base = $12\; \backslash times\; 8\; =\; 96$

4. When parts of red pigment = 20:

   $\backslash frac\{20\}\{\backslash text\{Parts\; of\; base\}\}\; =\; \backslash frac\{1\}\{8\}$

   Parts of base = $20\; \backslash times\; 8\; =\; 160$

So, the completed table is:

Parts of red pigment: 1, 4, 7, 12, 20

Parts of base: 8, 32, 56, 96, 160

This is a problem of direct proportion. We are given that 1 part of red pigment requires 75 mL of base. We need to find out how many parts of red pigment are needed for 1800 mL of base.

Let the required parts of red pigment be $x$.

We can set up a proportion:

$\backslash frac\{\backslash text\{Parts\; of\; red\; pigment\}\}\{\backslash text\{Parts\; of\; base\}\}\; =\; \backslash text\{constant\}$

Using the given information:

$\backslash frac\{1\}\{75\}\; =\; \backslash frac\{x\}\{1800\}$

To solve for $x$, we cross-multiply:

$1\; \backslash times\; 1800\; =\; 75\; \backslash times\; x$

$1800\; =\; 75x$

Now, divide both sides by 75:

$x\; =\; \backslash frac\{1800\}\{75\}$

$x\; =\; 24$

Therefore, 24 parts of red pigment should be mixed with 1800 mL of base.