1. **State the problem:** We need to find the time between 7 and 8 o'clock when the hands of the clock are in the same straight line but not overlapping. 2. **Recall key formulas:** - The angle between the hour and minute hands is given by $$\theta = 30H - \frac{11}{2}M$$ where $H$ is the hour and $M$ is the minutes past the hour. - For the hands to be in a straight line but not together, the angle between them must be 180 degrees. 3. **Set up the equation for $H=7$:** $$|30 \times 7 - \frac{11}{2} M| = 180$$ 4. **Solve the equation:** $$|210 - \frac{11}{2} M| = 180$$ This gives two cases: - Case 1: $$210 - \frac{11}{2} M = 180$$ $$210 - 180 = \frac{11}{2} M$$ $$30 = \frac{11}{2} M$$ $$M = \frac{30 \times 2}{11} = \frac{60}{11} \approx 5.45\,\text{minutes}$$ - Case 2: $$210 - \frac{11}{2} M = -180$$ $$210 + 180 = \frac{11}{2} M$$ $$390 = \frac{11}{2} M$$ $$M = \frac{390 \times 2}{11} = \frac{780}{11} \approx 70.91\,\text{minutes}$$ Since minutes cannot exceed 60, we discard $M \approx 70.91$. 5. **Final answer:** The hands are in the same straight line but not together at about $$7:05.45$$ (5.45 minutes past 7 o'clock).