CBSE Class 8 Maths Exemplar NCERT Solutions: Chapter 10 Direct and Inverse Proportions

Since u and v vary directly, their ratio is constant. This means $\backslash frac\{u\}\{v\}\; =\; k$, where k is the constant of variation.

Given that when u = 10, v = 15, we can find the constant k:

$$k\; =\; \backslash frac\{10\}\{15\}\; =\; \backslash frac\{2\}\{3\}$$

Now, we check each option to see which pair does not satisfy this constant ratio:

(a) For u = 2 and v = 3, $\backslash frac\{u\}\{v\}\; =\; \backslash frac\{2\}\{3\}$. This is a possible pair.

(b) For u = 8 and v = 12, $\backslash frac\{u\}\{v\}\; =\; \backslash frac\{8\}\{12\}\; =\; \backslash frac\{2\}\{3\}$. This is a possible pair.

(c) For u = 15 and v = 20, $\backslash frac\{u\}\{v\}\; =\; \backslash frac\{15\}\{20\}\; =\; \backslash frac\{3\}\{4\}$. This ratio is not equal to $\backslash frac\{2\}\{3\}$.

(d) For u = 25 and v = 37.5, $\backslash frac\{u\}\{v\}\; =\; \backslash frac\{25\}\{37.5\}\; =\; \backslash frac\{250\}\{375\}\; =\; \backslash frac\{2\; \backslash times\; 125\}\{3\; \backslash times\; 125\}\; =\; \backslash frac\{2\}\{3\}$. This is a possible pair.

Therefore, the pair (15, 20) is not a possible pair of corresponding values of u and v.

Since x and y vary inversely, their product is constant. This means $x\; \backslash times\; y\; =\; k$, where k is the constant of variation.

Given that when x = 10, y = 6, we can find the constant k:

$$k\; =\; 10\; \backslash times\; 6\; =\; 60$$

Now, we check each option to see which pair does not satisfy this constant product:

(a) For x = 12 and y = 5, $x\; \backslash times\; y\; =\; 12\; \backslash times\; 5\; =\; 60$. This is a possible pair.

(b) For x = 15 and y = 4, $x\; \backslash times\; y\; =\; 15\; \backslash times\; 4\; =\; 60$. This is a possible pair.

(c) For x = 25 and y = 2.4, $x\; \backslash times\; y\; =\; 25\; \backslash times\; 2.4\; =\; 60$. This is a possible pair.

(d) For x = 45 and y = 1.3, $x\; \backslash times\; y\; =\; 45\; \backslash times\; 1.3\; =\; 58.5$. This product is not equal to 60.

Therefore, the pair (45, 1.3) is not a possible pair of corresponding values of x and y.

If the land is uniformly fertile, a larger area of land will produce a proportionally larger yield, and a smaller area will produce a proportionally smaller yield. This means that as the area of land increases, the yield increases, and as the area decreases, the yield decreases, maintaining a constant ratio between them. Therefore, the area of land and the yield on it vary directly with each other.

The relationship between the number of teeth and the age of a person is not a simple direct or inverse proportion. A person develops a certain set of teeth (milk teeth) and then later develops another set (permanent teeth) as they grow. The number of teeth changes over specific age ranges, but it doesn't follow a consistent mathematical ratio or product that remains constant throughout life. For instance, a child has fewer teeth than an adult, but this doesn't mean the number of teeth varies inversely with age. Similarly, it doesn't vary directly. The number of teeth is related to developmental stages rather than a continuous proportional relationship with age.