Question

Two planes start from a city and fly in opposite directions, one with an average speed of $40$ km/hr greater than that of the other. If they are $4080$ km apart after $6$ hours; find the average speeds of both the planes.

• A. $120$ km/hr and $360$ km/hr
• B. $430$ km/hr and $360$ km/hr
• C. $540$ km/hr and $360$ km/hr
• D. $320$ km/hr and $360$ km/hr

Answer 1

The correct option is D $320$ km/hr and $360$ km/hr

Let speed of one plane be $x$ km/hr , then speed of another plane $x+40$ km/hr
Given, they are $4080$ km apart after $6$ hours
We know that Distance $=$ Speed $\times$ Times
$=>4080=x\left(6\right)+(x+40)6$
$=>4080=6x+6x+240$
$12x=3840$
$=>x=320$
Hence, the speeds of the plane $320km/hr$ and $320+40=360km/hr$