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 C 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

(1)6:15 am (2) 6:30 am (3) 6:45 am (4) 7:00 am (5) 7:15 am

^{0}to AB. Also C is located between south and southwest of A with AC at 30^{0}to AB. The latest time by which Rahim must leave A and still catch the train is closest to(1)6:15 am (2) 6:30 am (3) 6:45 am (4) 7:00 am (5) 7:15 am

Solution follows here:

__Solution:__
Plotting the points A,B and C is the
key here:

AB = 500;

B =
60

^{0}; A = 30^{0}=> C = 90^{0}ABC is a right triangle.
cosB = BC/AB => BC = AB cos60 = 500*(1/2)
= 250

sinB = AC/AB => AC = AB sin60 = 500*(√3/2)
= 250√3

speed of the train = 50 kmph

Time of travel from B to C =
distance/speed = BC/70 = 250/50 = 5 hour

As the train starts at 8:00 am, it
reaches C by 13:00.

“Rahim must reach C at least 15
minutes before the arrival of the train”

=> he must reach C at least by
12:45 pm --(1)

speed of Rahim = 70 kmph

Time of travel from A to C =
distance/speed = AC/70 = 250√3/70

= 6.18 hour = 6 hr 11 min --(2)

From (1) and (2), Rahim must start
latest by 12:45 – 6:11 =

**6:34 am**
Nearest answer is

**6:30 am****Answer(2)**
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