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Two planes move along a circle of circumference 1.2 kms with constant speeds. When they move in different directions, they meet every 15 seconds and when they move in the same direction one plane overtakes the other every 60 seconds. The speed of the slower plane is :
Answer & Solution
Correct Answer:
Option
B
Let their speeds be x m/sec and y m/sec respectively.
Then,
$$\eqalign{ & \frac{{1200}}{{x + y}} = 15 \cr & \Rightarrow x + y = 80.....(i) \cr} $$
And,
$$\eqalign{ & \frac{{1200}}{{x - y}} = 60 \cr & \Rightarrow x - y = 20.....(ii) \cr} $$
Adding (i) and (ii), we get :
2x = 100 or x = 50
Putting x = 50 in (i), we get : y = 30
Hence, speed of slower plane :
= 30 m/sec
= 0.03 km/sec
Then,
$$\eqalign{ & \frac{{1200}}{{x + y}} = 15 \cr & \Rightarrow x + y = 80.....(i) \cr} $$
And,
$$\eqalign{ & \frac{{1200}}{{x - y}} = 60 \cr & \Rightarrow x - y = 20.....(ii) \cr} $$
Adding (i) and (ii), we get :
2x = 100 or x = 50
Putting x = 50 in (i), we get : y = 30
Hence, speed of slower plane :
= 30 m/sec
= 0.03 km/sec
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