a) 800 m
b) 1675 m
c) 848 m
d) 1000 m
Answer: d
Explanation: In 20 minutes the difference between man and son,
= 20 × 20
= 400 m.
Distance traveled by dog when he goes towards the child,
= $$400 \times \frac{{60}}{{40}}$$
= 600 m and time required is 10 minutes.
In 10 min remaining distance between man and child,
= 400 – (20 × 10)
= 200 m.
Time taken by dog to meet the man,
= $$\frac{{200}}{{100}}$$
= 2 min
(Relative speed of dog with child is 40 m/min and same with man is 100 m/min.)
Remaining distance in 2 min,
= 200 – (2 × 20),
= 160 m.
Now, the time taken by dog to meet the child again,
= $$\frac{{160}}{{40}}$$
= 4 min.
In 4 min he covers = 4 × 60 = 240 m.
Now, remaining distance in 4 min = 160 – (4 × 20) = 80 m.
Time required by dog to meet the man once again = $$\frac{{80}}{{100}}$$ = 0.8 min.
Now remaining distance = 80 – (0.8 × 20) = 64 m.
Now, time taken by dog to meet the child again = (64/40) x (8/5) min
Distance travelled by dog = (8/5) x 60 = 96 m
Thus, we can observe that every next time dog just go 2/5th of previous distance to meet the child in the direction of child.
So, we can calculate the total distance covered by dog in the direction of child with the help of GP formula.
Here, first term (a) = 600 and common ratio (r) = 2/5
Sum of the infinite GP = a/(1 – r)
= 600/(1 – 2/5) = (600 x 5)/3 = 1000 m
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