A ship leaves on a long voyage When it is 28 miles from the shore a seaplane whose speed is

Let the ship travels 'x' miles more,

Then the seaplane needs to travel (28+x) miles,
so we get, 28+x = 10x

9x = 28

x = 28/9 = 3.10 miles

Distance from the shore = 28 + 3.1 = 31.1 miles.
11 years agoHelpfull: Yes(14) No(1)
280/9 miles as suggested by PP singh is correct.
12 years agoHelpfull: Yes(3) No(1)
If ship speed is x mph, then plane speed = 10x mph.
relative speed = 9x mph
time taken by plane to catch ship = 28/9x h
distance travelled by ship in that time = x*28/9x =28/9
so,,the distance from the shore is (28+28/9)=280/9 miles
11 years agoHelpfull: Yes(3) No(2)
280/9...correct ans
11 years agoHelpfull: Yes(0) No(1)
280/11... bcoz they were opposite to each other so 10x+x=11x. so 280/11x
10 years agoHelpfull: Yes(0) No(1)
time taken is same for both the ships & seaplanetime =distance/speed...
.if speed of ship=s
then seaplane speed=10s
let distance of ship be x then seaplane distance =28+x
therefore (x / s = (28 + x ) / 10s)
then we get x = 28/9 therfore total distance = 28+28/9=31.1
10 years agoHelpfull: Yes(0) No(0)

A ship leaves on a long voyage. When it is 28 miles from the shore, a seaplane, whose speed is 10 times that of the ship is sent to deliver mail. How far from the shore does the seaplane catch upto with the ship?