Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60° to AB. Also, C is located between south and southwest of A with AC at 30° to AB. The latest time by which Rahim must leave A and still catch the train is closest to
Explanation:
The angles of the triangle formed by A, B and C tell us that ABC is a right-angle triangle, with right-angle at vertex C, 30° at vertex A and 60° at vertex B.
Since AB = 500 km, in 30°-60°-90° triangle ABC, we get,
AC = 2503 km and BC = 250 km
The train, travels at 50 km/hr. It will travel from B to C (i.e. 250 km) in 5 hours. Since it leaves at 8:00 a.m., it will reach C at 1:00 p.m.
Now, Rahim must be at C latest by 12:45 p.m. (15 minutes before the train)
Travelling at 70 km/hr, he will take approximately 6.2 hours to travel from A to C. Therefore, he must leave at least by
12.75 – 6.2 = 6.55 hours after mid-night.
This is a little after 6:30 a.m. If he leaves by 6:45 a.m., he will not make it to point C 15 minutes before the train arrives.
Hence, option (b).
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