This question asks for the factor by which the density of seeds increases when soaked in water, given percentage changes in weight and volume.

Understanding Density Calculation

Density ($\rho$) is defined as mass ($m$) per unit volume ($V$). The formula is:

$ \rho = \frac{m}{V} $

Calculating Density Change

Let's assume the initial mass of the seeds is $m_1$ and the initial volume is $V_1$. The initial density is:

$ \rho_1 = \frac{m_1}{V_1} $

When soaked in water:

• The weight (mass) increases by 20%. The new mass, $m_2$, is:

  $ m_2 = m_1 + 0.20 \times m_1 = 1.20 m_1 $

• The volume increases by 10%. The new volume, $V_2$, is:

  $ V_2 = V_1 + 0.10 \times V_1 = 1.10 V_1 $

The new density, $\rho_2$, is:

$ \rho_2 = \frac{m_2}{V_2} = \frac{1.20 m_1}{1.10 V_1} $

Determining the Density Increase Factor

The factor by which the density increases is the ratio of the new density ($\rho_2$) to the initial density ($\rho_1$):

$ \text{Factor} = \frac{\rho_2}{\rho_1} = \frac{\frac{1.20 m_1}{1.10 V_1}}{\frac{m_1}{V_1}} $

Simplify the expression:

$ \text{Factor} = \frac{1.20 m_1}{1.10 V_1} \times \frac{V_1}{m_1} = \frac{1.20}{1.10} $

Now, calculate the numerical value:

$ \text{Factor} = \frac{1.2}{1.1} = \frac{12}{11} \approx 1.0909... $

Rounding to two decimal places, the factor is approximately 1.09.

Conclusion

The density increases by a factor of approximately 1.09.