1. **Problem statement:** A train travels at one-third of its usual speed and arrives 30 minutes late. On the return journey, it travels at usual speed for 5 minutes, stops for 4 minutes, and then must increase its speed to reach on time. Find the percentage increase in speed needed. 2. **Define variables:** Let the usual speed be $v$ and the usual time to destination be $t$ (in minutes). 3. **Using the first condition:** Traveling at one-third speed means speed is $\frac{v}{3}$. Time taken at one-third speed is $\frac{\text{distance}}{\text{speed}} = \frac{d}{v/3} = 3t$. Since usual time is $t$, the delay is $3t - t = 2t$. Given delay is 30 minutes, so: $$2t = 30 \implies t = 15 \text{ minutes}$$ 4. **Distance calculation:** Distance $d = v \times t = v \times 15$. 5. **Return journey details:** - Travels at usual speed $v$ for 5 minutes, covering distance $d_1 = v \times 5$. - Stops for 4 minutes. - Remaining distance $d_2 = d - d_1 = 15v - 5v = 10v$. 6. **Time left to reach on time:** Total scheduled time is $t = 15$ minutes. Time spent so far: $5$ minutes travel + $4$ minutes stop = $9$ minutes. Remaining time to reach on time: $15 - 9 = 6$ minutes. 7. **Speed needed for remaining distance:** Speed $s$ must satisfy: $$s = \frac{d_2}{\text{time}} = \frac{10v}{6} = \frac{5v}{3}$$ 8. **Percentage increase in speed:** Increase = $s - v = \frac{5v}{3} - v = \frac{2v}{3}$. Percentage increase: $$\frac{\frac{2v}{3}}{v} \times 100 = \frac{2}{3} \times 100 = 66.67\%$$ Nearest to 67%. **Final answer:** The train must increase its usual speed by approximately 67% to reach on time.