A man begin the trip when the hands of the clock are attached between 8 and 9 hours; the journey ends when the child pointer is between 14 and 15 hours and the pointer larger is 180° the other. How long the trip lasted ?
Hello, Apprentice123!
I believe I've translated your problem to English . . .
The minute hand moves at 6 degress/minute.A man begin the trip when the hands of the clock are together between 8:00 and 9:00.
The journey ends when the hour hand is between 14:00 and 15:00 hours
and the minute hand is diametrically opposite. .How long did the trip last?
. . In $\displaystyle t$ minutes, $\displaystyle M$ moves: $\displaystyle 6t$ degrees.
The hour hand moves at ½ degrees/minute.
. . In $\displaystyle t$ minutes, $\displaystyle H$ moves: $\displaystyle \tfrac{1}{2}t$ degrees.
At 8:00, the minute hand is at 0°; the hour hand is at 240°.
. . In $\displaystyle t$ minutes, $\displaystyle M$ is at $\displaystyle 6t$ degrees . . . and $\displaystyle H$ is at $\displaystyle 240 + \tfrac{1}{2}t$ degrees.
The hands are together: .$\displaystyle 6t \:=\:\tfrac{1}{2}t + 240 \quad\Rightarrow\quad t \:=\:\frac{480}{11} \:=\:43\tfrac{7}{11}$
. . The trip started at: .$\displaystyle 8\!:\!43\tfrac{7}{11}$ a.m.
At 2:00, $\displaystyle M$ is at 0°; $\displaystyle H$ is at 60°.
In $\displaystyle t$ minutes, $\displaystyle M$ is at $\displaystyle 6t$ degrees . . . and $\displaystyle H$ is at $\displaystyle 60 + \tfrac{1}{2}t$ degrees.
The hands are 180° apart: .$\displaystyle 6t \:=\:(60+\tfrac{1}{2}t) + 180 \quad\Rightarrow\quad t \:=\:\frac{480}{11} \:=\:43\tfrac{7}{11}$
. . The trip ended at: .$\displaystyle 2\!:\!43\tfrac{7}{11}$ p.m.
Therefore, the trip took exactly 6 hours.