1. **Problem Statement:** We need to find the value of $X$ given a timeline with cash flows at different times and a 5% interest rate. 2. **Understanding the Problem:** The timeline shows cash flows at times 0, 1, 3, 4, and 5. At times 0 and 1, there are outflows of $X$ each (downward arrows). At time 3, there is an outflow of $X$ and inflows of 20 at time 3, 30 at time 4, and 40 at time 5 (upward arrows). 3. **Formula Used:** To find $X$, we use the principle of the time value of money. The sum of the present values (PV) of all cash flows must be zero for equilibrium: $$\sum \text{PV of inflows} = \sum \text{PV of outflows}$$ The present value of a cash flow $C$ at time $t$ with interest rate $i$ is: $$\text{PV} = \frac{C}{(1+i)^t}$$ 4. **Calculate Present Values:** - Outflows: - At time 0: $X$ (no discounting needed) - At time 1: $\frac{X}{(1+0.05)^1} = \frac{X}{1.05}$ - At time 3: $\frac{X}{(1.05)^3}$ - Inflows: - At time 3: $\frac{20}{(1.05)^3}$ - At time 4: $\frac{30}{(1.05)^4}$ - At time 5: $\frac{40}{(1.05)^5}$ 5. **Set up the equation:** $$X + \frac{X}{1.05} + \frac{X}{(1.05)^3} = \frac{20}{(1.05)^3} + \frac{30}{(1.05)^4} + \frac{40}{(1.05)^5}$$ 6. **Calculate the denominators:** - $(1.05)^1 = 1.05$ - $(1.05)^3 = 1.157625$ - $(1.05)^4 = 1.21550625$ - $(1.05)^5 = 1.2762815625$ 7. **Rewrite the equation:** $$X + \frac{X}{1.05} + \frac{X}{1.157625} = \frac{20}{1.157625} + \frac{30}{1.21550625} + \frac{40}{1.2762815625}$$ 8. **Calculate the right side:** - $\frac{20}{1.157625} \approx 17.27$ - $\frac{30}{1.21550625} \approx 24.69$ - $\frac{40}{1.2762815625} \approx 31.34$ Sum: $17.27 + 24.69 + 31.34 = 73.30$ 9. **Calculate the left side coefficients:** - $1 + \frac{1}{1.05} + \frac{1}{1.157625} = 1 + 0.95238 + 0.86384 = 2.81622$ 10. **Solve for $X$:** $$X \times 2.81622 = 73.30$$ $$X = \frac{73.30}{2.81622} \approx 26.02$$ **Final answer:** $$\boxed{X \approx 26.02}$$