1. **State the problem:** We have the formula $$N = \frac{6000}{5p + z}$$ where $N$ is the number of radios sold, $p$ is the price per radio, and $z$ is a constant. We know that when $p = 20$, $N = 30$. We want to find how many radios are sold when $p = 40$. 2. **Find the constant $z$:** Substitute $p = 20$ and $N = 30$ into the formula: $$30 = \frac{6000}{5 \times 20 + z}$$ Simplify the denominator: $$30 = \frac{6000}{100 + z}$$ Multiply both sides by $100 + z$: $$30(100 + z) = 6000$$ $$3000 + 30z = 6000$$ Subtract 3000 from both sides: $$30z = 3000$$ Divide both sides by 30: $$z = \cancel{30}z / \cancel{30} = \frac{3000}{30} = 100$$ 3. **Calculate $N$ when $p = 40$:** Substitute $p = 40$ and $z = 100$ into the formula: $$N = \frac{6000}{5 \times 40 + 100} = \frac{6000}{200 + 100} = \frac{6000}{300}$$ Simplify the fraction: $$N = \frac{6000}{300} = 20$$ **Final answer:** When the price is 40, the shop sells **20 radios** per month.