Math ?: A boat takes 45 minutes to travel downstream 12 miles, but it takes 3 hours to make the return trip?

Question

determine the rate of the boat in still water. Determine the rate of the current. (hint: switch 45 minutes into hours). I had this math problem on a test in 1998, still don't know how to friggin figure it out :(:(:(.

Gosh DANGET PIKI, you helped me solve this problem that's been in my stack of papers for 10 years almost. And to figure the speed of the boat was just friggin 10 miles per hour, it's humilliating :( garrrrrrrrrrrr

And for anyone who could figure out the type of algebra this is, I mean there are linear inequates, consecutive integer, but what part of algebra is this. 10 points for whoever could figure this because there are soo many who answered.

Answer 1

Downstream------Upstream
12 miles-------------12 miles
45 mins--------------3 hours
(x + y) mph---------(y - x) mph

speed of water = x mph
speed of boat in still water = y mph

x + y = 12 / (3/4)
x + y = 16

y - x = 12 / 3
y - x = 4

y - x = 4
y + x = 16
2y = 20
y = 10

Speed of boat in still water = 10 mph

Answer 2

The boat takes .75 hours to travel downstream 12 miles, so the velocity of the boat downstream is: 12/.75 = 16 mph. Let S be the rate in still water, and C be the rate of the current. Then S + C = 16. Now it takes 3 hours to return, so it was traveling at 12/3 = 4 mph. So we know that S - C = 4. It's a system of two equations:

S + C = 16
S - C = 4

Add the two equations together to get:
2S = 20
S = 10
C = 16 - S = 16 - 10 = 6

The boat travels at 10 mph in still water, and the current is 6 mph.

Answer 3

the downstream was12 over 0.75 of an hour (45 / 60)
12 x 3/4 = 16
so 16 mile/hr (current + boat speed)

3hours (180 mins) is 4 times 45 mins
12 miles in 3 hours is 4 m/hr
so boat speed is 4 m/h greater than downstream current

Boat speed = 10 m/h ... current = 6m/h

If the boat speed was less than half the river speed, it could never get back ... it would be dragged downstream at a slower rate. So boat speed > current

Answer 4

Let r = rate of the current and b=rate of the boat in still water

Then

r+b = 12 miles/(3/4) hours or 16 mph
r-b = 12/3 = 4 mph

So r = 4 + b
then 4+b+b = 16

2b = 12
b = 6 = rate of boat in still water
r = 10 = downstream current

And Meg is a moron.

Answer 5

If the water is flowing at x miles according to hour, and the value of the boat in nonetheless water is y miles according to hour, then the value of the boat downstream is (x + y) miles according to hour. despite the fact that, upstream the water is going interior the different direction from the boat, so the value of the boat upstream is (y - x) miles according to hour. making use of the formulation velocity = distance/time, we get 2 simultaneous equations: (x + y) = 12/0.75 (y - x) = 12/3 i take advantage of 0.75, in view that we are working in hours. If we artwork out the numbers, we get: y + x = sixteen y - x = 4 So if we upload the two equations mutually, we are able to get rid of x: 2y = 20 So dividing the two aspects by potential of two: y = 10 So the value of the boat in nonetheless water is 10 miles according to hour. Substituting y = 10 into the two of the equations will supply x. shall we % equation a million: x + y = sixteen x + 10 = sixteen So getting rid of 10 from the two aspects, we get: x = 6 So the value of the water is 6 miles according to hour.

Answer 6

Let B be the boat speed, and C be the current. We then have: 0.75(B+C) = 12; 3(B-C) = 12. From here, it's just algebra.

Answer 7

let b = boat speed in still water, c = current speed. distance = speed * time. downstream, boat and current speeds add, upstream they subtract, so...
(b + c)(3/4) = 12
(b - c)3 = 12

b + c = 16
b - c = 4
---------------
2b = 20
b = 10 mph

b + c = 16
-(b - c) = -4
----------------
2c = 12
c = 6 mph