1. **Stating the problem:** The distance going is 5 km with a speed of 40 km/h. The return time is 10 minutes less than the going time. 2. **Convert 10 minutes to hours:** Since 1 hour = 60 minutes, then $$10 \text{ minutes} = \frac{10}{60} = \frac{1}{6} \text{ hours}$$ 3. **Define variables:** Let the distance from A to B be $x$ km, with $x > 0$. 4. **Write expressions for time:** Time going = $\frac{x}{40}$ hours Time returning = $\frac{x}{v}$ hours, where $v$ is the return speed. 5. **Given condition:** Time returning is 10 minutes (or $\frac{1}{6}$ hours) less than time going: $$\frac{x}{v} = \frac{x}{40} - \frac{1}{6}$$ 6. **Solve for $v$ or $x$:** Multiply both sides by $v \times 40$ to clear denominators: $$40x = vx - \frac{40v}{6}$$ Rearranged: $$40x - vx = -\frac{40v}{6}$$ Factor $x$: $$x(40 - v) = -\frac{40v}{6}$$ 7. **Since $v$ is unknown, but problem asks to find $x$, we use the original problem data:** The going distance is 5 km, so $x = 5$ km. 8. **Check time going:** $$t_{go} = \frac{5}{40} = \frac{1}{8} \text{ hours} = 7.5 \text{ minutes}$$ 9. **Time returning is 10 minutes less, so:** $$t_{return} = t_{go} - \frac{1}{6} = \frac{1}{8} - \frac{1}{6} = -\frac{1}{24} \text{ hours}$$ This is negative, which is impossible, so the problem likely wants to find $x$ such that the return time is 10 minutes less than going time. 10. **Set up equation:** $$\frac{x}{v} = \frac{x}{40} - \frac{1}{6}$$ But $v$ is not given, so assume return speed is different. 11. **Alternatively, if return speed is the same, the problem is inconsistent. So we solve for $x$ assuming return speed is $v$ and $x$ unknown:** 12. **Rewrite:** $$\frac{x}{v} = \frac{x}{40} - \frac{1}{6}$$ Multiply both sides by $v \times 40$: $$40x = vx - \frac{40v}{6}$$ Rearranged: $$40x - vx = -\frac{40v}{6}$$ $$x(40 - v) = -\frac{40v}{6}$$ 13. **Solve for $x$:** $$x = -\frac{40v}{6(40 - v)}$$ Since $x > 0$, numerator and denominator must have opposite signs. 14. **This is a general form; without $v$ given, we cannot find numeric $x$.** 15. **Summary:** a) $10$ minutes = $\frac{1}{6}$ hours. b) The distance $x$ satisfies $$\frac{x}{v} = \frac{x}{40} - \frac{1}{6}$$ where $v$ is return speed. --- **Additional problems:** **Problem 1:** Find integer $x$ such that $$A = \frac{5}{x-3}$$ is an integer. - For $A$ to be integer, $x-3$ must divide 5. - Divisors of 5 are $\pm1, \pm5$. - So $x-3 = \pm1$ or $\pm5$. - Solutions: - $x-3=1 \Rightarrow x=4$ - $x-3=-1 \Rightarrow x=2$ - $x-3=5 \Rightarrow x=8$ - $x-3=-5 \Rightarrow x=-2$ **Problem 2:** Find integer $x$ such that $$B = \frac{-8}{x+1}$$ is integer. - $x+1$ divides $-8$. - Divisors of $-8$ are $\pm1, \pm2, \pm4, \pm8$. - So $x+1 = \pm1, \pm2, \pm4, \pm8$. - Solutions: - $x = 0, -2, 3, -5, 7, -9$ **Problem 3:** Find integer $x$ such that $$C = \frac{7}{2x-1}$$ is integer. - $2x-1$ divides 7. - Divisors of 7 are $\pm1, \pm7$. - So $2x-1 = \pm1$ or $\pm7$. - Solve: - $2x-1=1 \Rightarrow x=1$ - $2x-1=-1 \Rightarrow x=0$ - $2x-1=7 \Rightarrow x=4$ - $2x-1=-7 \Rightarrow x=-3$ **Problem 4:** Find integer $x$ such that $$D = \frac{x+5}{x+2}$$ is integer. - Let $D = k \in \mathbb{Z}$. - Then $x+5 = k(x+2)$ - Rearranged: $$x+5 = kx + 2k$$ $$x - kx = 2k - 5$$ $$x(1-k) = 2k - 5$$ - If $k \neq 1$, then $$x = \frac{2k - 5}{1 - k}$$ - For $x$ integer, denominator divides numerator. - Check integer $k$ values to find integer $x$. **Problem 5:** Find integer $x$ such that $$E = \frac{x-7}{x-3}$$ is integer. - Let $E = m \in \mathbb{Z}$. - Then $x-7 = m(x-3)$ - Rearranged: $$x - 7 = mx - 3m$$ $$x - mx = -3m + 7$$ $$x(1 - m) = -3m + 7$$ - If $m \neq 1$, then $$x = \frac{-3m + 7}{1 - m}$$ - For $x$ integer, denominator divides numerator. - Check integer $m$ values to find integer $x$. **Final answers:** - a) $10$ minutes = $\frac{1}{6}$ hours. - b) $x$ satisfies $$\frac{x}{v} = \frac{x}{40} - \frac{1}{6}$$ with $x > 0$. - Problem 1: $x = -2, 2, 4, 8$ - Problem 2: $x = -9, -5, -2, 0, 3, 7$ - Problem 3: $x = -3, 0, 1, 4$ - Problem 4 and 5 require checking integer values of $k$ and $m$ respectively.