A number \( x \) varies direatly as \( y \) and partly inversly as \( y^{2} \) given that \( x=8 \), when \( y=4 \), and \( x=25 \cdot 16 \), when \( y=5 \), fend \( x \), when \( y=8 \).

To solve the problem, we need to express the relationship between \( x \) and \( y \) based on the given conditions. The statement "a number \( x \) varies directly as \( y \) and partly inversely as \( y^{2} \)" can be expressed mathematically as: \[ x = k \cdot \frac{y}{y^2} = \frac{k}{y} \] where \( k \) is a constant. ### Step 1: Find the constant \( k \) We have two conditions to find \( k \): 1. When \( y = 4 \), \( x = 8 \): \[ 8 = \frac{k}{4} \implies k = 8 \cdot 4 = 32 \] 2. When \( y = 5 \), \( x = 25 \cdot 16 = 400 \): \[ 400 = \frac{k}{5} \implies k = 400 \cdot 5 = 2000 \] ### Step 2: Check for consistency We have two different values for \( k \) from the two conditions. Let's check if they are consistent: 1. From the first condition, \( k = 32 \). 2. From the second condition, \( k = 2000 \). Since these values are not equal, we need to re-evaluate the relationship. ### Step 3: Re-evaluate the relationship The relationship should be: \[ x = k_1 y - \frac{k_2}{y^2} \] where \( k_1 \) and \( k_2 \) are constants. ### Step 4: Set up equations Using the first condition \( (x = 8, y = 4) \): \[ 8 = k_1 \cdot 4 - \frac{k_2}{4^2} \implies 8 = 4k_1 - \frac{k_2}{16} \tag{1} \] Using the second condition \( (x = 400, y = 5) \): \[ 400 = k_1 \cdot 5 - \frac{k_2}{5^2} \implies 400 = 5k_1 - \frac{k_2}{25} \tag{2} \] ### Step 5: Solve the system of equations Now we will solve equations (1) and (2) simultaneously to find \( k_1 \) and \( k_2 \). Let's express \( k_2 \) from equation (1): \[ k_2 = 64 - 64k_1 \tag{3} \] Substituting equation (3) into equation (2): \[ 400 = 5k_1 - \frac{64 - 64k_1}{25} \] Now, let's solve this equation for \( k_1 \) and \( k_2 \). ### Step 6: Solve for \( k_1 \) and \( k_2 \) We will solve the equations now. \[ 400 = 5k_1 - \frac{64 - 64k_1}{25} \] Multiplying through by 25 to eliminate the fraction: \[ 10000 = 125k_1 - (64 - 64k_1) \] This simplifies to: \[ 10000 = 125k_1 - 64 + 64k_1 \] Combining like terms: \[ 10000 + 64 = 189k_1 \] \[ 10064 = 189k_1 \] Now, solving for \( k_1 \): \[ k_1 = \frac{10064}{189} \] Now substituting \( k_1 \) back into equation (3) to find \( k_2 \): \[ k_2 = 64 - 64 \cdot \frac{10064}{189} \] Now, let's calculate these values. ### Step 7: Calculate \( k_1 \) and \( k_2 \) I will calculate \( k_1 \) and \( k_2 \) now. \[ k_1 = \frac{10064}{189} \] Calculating \( k_1 \): \[ k_1 \approx 53.2 \] Now substituting back to find \( k_2 \): \[ k_2 = 64 - 64 \cdot 53.2 \] Calculating \( k_2 \): \[ k_2 \approx -3392 \] ### Step 8: Find \( x \) when \( y = 8 \) Now we can find \( x \) when \( y = 8 \): \[ x = k_1 \cdot 8 - \frac{k_2}{8^2} \] Substituting the values of \( k_1 \) and \( k_2 \): \[ x = 53.2 \cdot 8 - \frac{-3392}{64} \] Calculating \( x \): \[ x = 425.6 + 53.2 = 478.8 \] Thus, the value of \( x \) when \( y = 8 \) is approximately \( 478.8 \).