A ship sails for a distance of 150 nautical miles against a current, taking a total of 10 hours. It returns to the starting point, going with the current, in 6 hours. For both trips it moves the same speed through the water.
A system of two equations can be used to find the speed of the ship and the current. Let x represent the ship's speed and y represent the speed of the current. What is the correct equation for the ship's return going with the current?
(x + y)(6) = 150
6x + 10y = 150
(x - y)(10) = 150
Using the correct system of equations, calculate the speed of the ship.
5 knots
15 knots
20 knots
25 knots
2007-06-09 04:41:44 · 4 answers · asked by queenpravato 1 in Science & Mathematics ➔ Mathematics
(x-y)(10)=150 and (x+y)(6)=150 are the correct equations
10x-10y=150....(eqn 1)
6x+6y= 150 ....(eqn.2)
Dividing eqn 1 by 10 and eqn 2 by 6,we get
x-y=15
x+y=25
(Adding) 2x=40
orx=20 Therefore,the speed of the ship is 20 knots
2007-06-09 04:48:03 · answer #1 · answered by alpha 7 · 0⤊ 1⤋
(x + y)(6) = 150
20 knots
2007-06-09 11:46:42 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋
x+y(6)=150 (Speed with current)
x-y(10)=150 (Speed against current)
x+y=25
x-y=15
When x is constant, y(current) = 5 and x(ship's speed) = 20
2007-06-09 12:02:38 · answer #3 · answered by Don 2 · 0⤊ 0⤋
x=25
y=0
(25+0)(6)=150
6(25)+10(0)=150
150+0=150
2007-06-09 11:52:08 · answer #4 · answered by Anonymous · 0⤊ 1⤋