We are given two equations in the problem:

1. $2x + 5y - 150 = 0$
2. $kx + 5y = 10$

We are asked to determine the value of $b$ when $(8, b)$ is a solution of the second equation.

Step 1: Solve for $y$ in terms of $x$

First, let's solve the first equation for possible solutions of $x$ and $y$ (whole numbers):

$2x + 5y = 150$

Rearrange for $5y$:

$5y = 150 - 2x$

Now, divide both sides by 5:

$y = \frac{150 - 2x}{5}$

For $y$ to be a whole number, the expression $150 - 2x$ must be divisible by 5. Therefore, $2x \equiv 150 \pmod{5}$, which simplifies to:

$2x \equiv 0 \pmod{5}$

This means $x$ must be a multiple of 5. Let $x = 5n$, where $n$ is an integer. Substituting this into the equation for $y$:

$y = \frac{150 - 2(5n)}{5} = \frac{150 - 10n}{5} = 30 - 2n$

Thus, the number of possible solutions for $x$ and $y$ is based on the values $n$ can take. Since $x$ must be non-negative, $0 \leq x \leq 75$, meaning $0 \leq n \leq 15$. So, there are 16 possible solutions.

Step 2: Solve for $b$ when $(8, b)$ is a solution

Now, substitute $x = 8$ into the second equation $kx + 5y = 10$:

$k(8) + 5b = 10$

This simplifies to:

$8k + 5b = 10$

Solve for $b$:

$5b = 10 - 8k$

$b = \frac{10 - 8k}{5}$

For $b$ to be a whole number, $10 - 8k$ must be divisible by 5. This gives us the condition:

$10 - 8k \equiv 0 \pmod{5}$

Simplifying:

$8k \equiv 10 \pmod{5}$

$3k \equiv 0 \pmod{5}$

Since 3 and 5 are relatively prime, $k$ must be a multiple of 5. Let $k = 5m$, where $m$ is an integer. Substituting this into the equation for $b$:

$b = \frac{10 - 8(5m)}{5} = \frac{10 - 40m}{5} = 2 - 8m$

Step 3: Determine $b$ for different values of $k$

For different values of $m$:

• If $m = 0$, $b = 2$
• If $m = 1$, $b = -6$
• If $m = -1$, $b = 10$

Now, from the options given, none of these values directly match. Therefore, it's possible the equation has been transformed or given a fraction option we missed earlier. Upon solving