1. Problem 1: Calculate the interest $Z$ for Kapital $K=520$, Zinssatz $p=2.5\%$, Zeit $t=77$ Tage. Formula for simple interest: $$Z = K \times \frac{p}{100} \times \frac{t}{360}$$ Note: We use 360 days as the banking year. Calculate: $$Z = 520 \times \frac{2.5}{100} \times \frac{77}{360}$$ $$= 520 \times 0.025 \times 0.2139$$ $$= 520 \times 0.005347$$ $$= 2.78044$$ So, $Z \approx 2.78$. 2. Problem 2: Given $K=1200$, $Z=4.50$, $p=4.5\%$, find time $t$. Use formula: $$Z = K \times \frac{p}{100} \times \frac{t}{360}$$ Rearranged for $t$: $$t = \frac{Z \times 360}{K \times \frac{p}{100}}$$ Calculate: $$t = \frac{4.50 \times 360}{1200 \times 0.045}$$ $$= \frac{1620}{54}$$ $$= 30$$ days. 3. Problem 3: Given $K=540$, $t=6$ Monate (6 months = 180 days), $Z=29.70$, find $p$. Formula: $$Z = K \times \frac{p}{100} \times \frac{t}{360}$$ Rearranged for $p$: $$p = \frac{Z \times 100 \times 360}{K \times t}$$ Calculate: $$p = \frac{29.70 \times 100 \times 360}{540 \times 180}$$ $$= \frac{1069200}{97200}$$ $$= 11$$\%. 4. Problem 4: Given $p=9\%$, $t=1.5$ Jahre (1.5 years = 540 days), $Z=27$, find $K$. Formula: $$Z = K \times \frac{p}{100} \times \frac{t}{360}$$ Rearranged for $K$: $$K = \frac{Z \times 100 \times 360}{p \times t}$$ Calculate: $$K = \frac{27 \times 100 \times 360}{9 \times 540}$$ $$= \frac{972000}{4860}$$ $$= 200$$. Final answers: 1. $Z = 2.78$ 2. $t = 30$ days 3. $p = 11\%$ 4. $K = 200$