On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow Metro from "Taganskaya" station to "Kievskaya" station, and in the evening - back (see diagram).

![](https://cdn.mathpix.com/cropped/2024_05_08_22585010b2534c7ab9ebg-08.jpg?height=666&width=726&top_left_y=-1&top_left_x=666)

Upon entering the station, the Scientist boards the first arriving train. It is known that trains run in both directions with approximately equal intervals, and that the train traveling on the northern route (via "Belorusskaya") takes 17 minutes to travel from "Kievskaya" to "Taganskaya" or vice versa, while on the southern route (via "Paveletskaya") it takes 11 minutes. Out of habit, the Scientist always calculates everything. Once, he calculated that over many years of observation:
- the train traveling counterclockwise arrives at "Kievskaya" on average 1 minute and 15 seconds after the train traveling clockwise arrives. The same is true for "Taganskaya".

- on average, the Scientist spends 1 minute less on the trip from home to work than on the trip from work to home.

Find the expected interval between trains traveling in the same direction.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then the expected travel time from "Taganskaya" to "Kievskaya" is $11 p + 17(1-p) = 17 - 6p$.

On the return journey from "Kievskaya" to "Taganskaya," the expected travel time is $17 p + 11(1-p) = 11 + 6p$.

According to the condition, $(11 + 6p) - (17 - 6p) = 1$, from which $p = \frac{7}{12}$. Let the interval between trains be $T$. Then $T(1-p) = Y$, where $Y$ is the time between the arrival of a clockwise train and the arrival of a counterclockwise train at the Scientist's favorite stations. Therefore, $\mathrm{ET} = \mathrm{EY}_{/ 1-p} = \frac{5}{4} \cdot \frac{12}{5} = 3$.

## Answer

## 3 minutes.