Trigonometry-3 (CAT-2008)

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 60⁰ to AB. Also C is located between south and southwest of A with AC at 30⁰ 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 = 60⁰;          A = 30⁰ => C = 90⁰ 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)