Mugdho wrote:
A man goes to the fair in Funcity with his son and faithful dog. Unfortunately man misses his son which he realises 20 minutes later. The son comes back towards his home at the speed of 20 m/min and man follows him at 40 m/min. The dog runs to the son (child) and comes back to the man (father) to show him the direction of his son. It keeps moving to and fro at 60 m/min between son and father, till the man meets the son. what is the distance travelled by the dog in the direction of the son?
a)2000m
b)3000m
c)1500m
d)1000m
e)900m
When the dog runs toward the son, the dog and the son are moving in the same direction and thus are COMPETING.
When elements compete, SUBTRACT their rates.
Thus, the CATCH-UP RATE = (dog's rate) - (son's rate) = 60-20 = 40 meters per minute
When the dog runs back toward the father, the dog and father are traveling toward each other and thus are WORKING TOGETHER to cover the distance between them.
When elements work together, ADD their rates.
Thus, the RUN-BACK RATE = (dog's rate) + (father's rate) = 60+40 = 100 meters per minute
After 20 minutes, the distance traveled ahead by the son = (son's rate)(time) = 20*20 = 400 meters
Leg 1: Time for the dog to catch up to the son, who is 400 meters ahead \(= \frac{distance}{catch-up-rate} = \frac{400}{40} = 10\) minutesIn 10 minutes:
distance traveled by the dog toward the son = (dog's rate)(time) = 60*10 = 600 metersson's new distance = (son's previous distance) + (son's rate)(time) = 400 + 20*10 = 600 meters
father's distance = (father's rate)(time) = 40*10 = 400 meters
distance between the son and the father = 600-400 = 200 meters
Leg 2: Time for the dog to run 200 meters back to the father \(= \frac{distance}{run-back-rate} = \frac{200}{100} = 2\) minutesIn 2 minutes:
dog's new distance = father's new distance = (father's previous distance) + (father's rate)(time) = 400 + 40*2 = 480 meters
son's new distance = (son's previous distance)+ (son's rate)(time) = 600 + 20*2 = 640 meters
distance between the son and the father = 640-480 = 160 meters
Leg 3: Time for the dog to catch-up to the son, who is 160 meters ahead \(=\frac{ distance}{catch-up-rate} = \frac{160}{40} = 4\) minutesIn 4 minutes:
distance traveled by the dog toward the son = (dog's rate)(time) = 60*4 = 240 metersdog's new distance = (dog's previous distance) + 240 = 480 +240 = 720 meters
son's new distance = (son's previous distance) + (son's rate)(time) = 640 + 4*20 = 720 meters
father's new distance = (father's previous distance) + (father's rate)(time) = 480 + 40*4 = 640 meters
distance between the son and the father = 720-640 = 80 meters
Leg 4: Time for the dog to run 80 meters back to the father \(= \frac{distance}{run-back-rate} = \frac{80}{100} = 0.8\) minuteIn 0.8 minute:
dog's new distance = father's new distance = (father's previous distance) + (father's rate)(time) = 640 + 40*0.8 = 672 meters
son's new distance = (son's previous distance) + (son's rate)(time) = 720 + 20*0.8 = 736 meters
distance between the son and the father = 736-672 = 64 meters
Leg 5: Time for the dog to catch-up to the son, who is 64 meters ahead \(= \frac{distance}{catch-up-rate} = \frac{64}{40} = 1.6\) minutesIn 1.6 minutes:
distance traveled by the dog toward the son = (dog's rate)(time) = 60*1.6 = 96 metersOver the 5 legs, sum of the blue distances = 600+240+96 = 936 meters
Since the distance toward the son in Leg 5 = 96 meters, the next distance toward the son must be LESS than 96 meters.
Since 936+96 = 1032, the total distance traveled by the dog toward the son must be more than 936 meters but less than 1032 meters.