Help with 8th grade math homework please?

Question

a boat travels 60 km upstream (against the current) in 5 hours. The boat travels the same distance downstream in 3 hours. What is the rate of the boat in still water? What is the rate of the current? THANK YOU SO MUCH

sorry for giving you a thumbs down if you said "b back in a bit..." =] myyyy apoligies.

Thank u all sooooooo much 4 helping me!!!! geesh u guys are smarties! love you alll u rock!!!!!!!!!!!!!!!!!! =]=]=]=]=]=]=]=]=]=]=]=]=]=]=]=]

Answer 1

let x=speed of the boat in still wate & y=speed of current.
upstream
60km/(x-y) km/hr=5 hrs multiply both sides by x-y
60=5(x-y) divide both sides by 5
12=x-y
downstream
60km/(x+y) km/hr=3hrs multiply both sides by x+y
60=3(x+y) divide both sides by 3
20=x+y
12=x-y add
32=2x divide by 2
x=16 km/hr boat speed in still water
substitute 16 for x in either equation
20=x+y=16+y
subtract 16 from each side
y=4 speed of current.

check by substituting x=16 & y=4 into the other equation
x-y=12
16-4=12
12=12

Answer 2

The total velocity of the boat through the water is equal the the velocity of the boat in still water + the velocity of the current. In other words if the boat travels 50 mph east in still water and the current in 10 mph east the boat goes 60 mph east. Conversely if the boat will go 50 mph west in still water, it will travel 40 mph west with the current.

Now let us create equations. We have two unknowns, current speed and boat speed, so we'll two equations. Call the boat speed B, and the current speed C. The first statement tells us When the boat and current velocities are in opposite directions (net speed B - C), the boat moves 60 km in 5 hours or at a speed of 12 km/hour. Thus B + C = 12. The second statement tells us when the boat and current are moving in the same direction (net speed B + C) the boat moves 60 km in 3 hours, and deviding 60 by 3, we get a speed of 20 km/hour. So B + C = 20.

Now we have a system of equations.
B - C = 12
B + C = 20

If we add the two equations together we get the speed of a round trip, and the current will cancel itself out.
B - C (journey out) + B + C (journey back) = 12 (speed out) + 20 (speed back), so 2B = 32. Now divide this by 2 to get the speed of the boat in still water. B = 16 km/hour. Now substitute this value of 16 into one of the equations. 16 - C = 12.
Solve for C. 16 = 12 + C. 16 - 12 = C. C = 4 km/hour.

Now let's check our answer at a net speed of (16 - 4) km/hour for 5 hours, the boat will travel 12*5 km = 60 km. And at a speed of (16+4)km/h for three hours, the boat will travel 20*3 km = 60 km.

Answer 3

Guess I should have said "OK, be back in a bit" first...

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Going against the current, the boats effective rate would be the rate of the boat in still water (b) MINUS the rate of the current (c):

d = rt
60 = (b - c)5

And going downstream, it would be the still water rate PLUS the rate of the current:

d = rt
60 = (b + c)3

5b - 5c = 60
3b + 3c = 60

15b - 15c = 180
15b + 15c = 300
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30b = 480
b = 16

So 3(16) + 3c = 60, and 3c = 60 - 48 = 12 and c = 4.

The still water speed of the boat is 16 km/hr and the current is 4 km/hr.

As a check, going upstream, the boat's effective rate would be 16 - 4 = 12km/hr, so in 5 hours the boat would travel 12 (5) = 60 km. Check!

Going downstream, the boat's effective rate would be 16 + 4 = 20km/hr, so in 3 hours, the boat would travel 20(3) = 60 km. Checkmate.

Answer 4

Distance = rate x time
let's say the rate of the boat in still water is r and the rate of the current is c. So you have...
60 = (r-c) x 5 and
60 = (r+c) x 3
We have two equations and two variables, so we can solve this. If you are learning to solve systems, you would then use whatever method you're using (substitution, linear combination aka addition method, or Cramer's Rule) to solve it.
If you havn't yet learned to solve systems, you must convert this to one equation in one variable, in other words, the substitution method. Since both equations are about the same, let's just take the first one, 60 = 5(r-c) and write r in terms of c. (this is slightly easier than writing c in terms of r, and you'll eventually need both)
60 = 5(r-c) so (divide by 5) 12 = r - c so 12 + c = r
Now plug this into the other equation...
60 = 3(r + c) (divide by 3 to make the numbers easier)
20 = r + c substitute 12 + c = r
20 = (12 + c) + c
20 = 12 + 2c subtract 12 from each side
8 = 2c so 4 = c. Since 12 + c = r, r = 16
Now check:
60 = 5(r-c) = 5(16-4) = 5x12 ok
60 = 3(r+c) = 3(16+4) = 3x20 ok
Finally, state your answer in terms of the question, not just in numbers.
The rate of the boat is 16 mph and the rate of the current is 4 mph.
You're welcome!
(Ps -- please disregard the four thumbs down I got for this answer. I got those for saying "Ill be back in a bit", and then coming back and editing, not for the answer itself. People need to realise that thumbs down is for answers that don't help, eg, "do your own work", "I hate math", etc. )

Answer 5

Let x = rate in still water
Let y = rate of current
Then x-y = rate upstream and y+x = rate downstream
distance = rate X time, so
60 km = (x-y)5 , and
60 km = (x+y)3
x-y = 12
x+y = 20
2x = 32
x= 16 km/hour = rate in still water
16-y = 12
y = 4 km/hour = rate of current

Answer 6

Since 19 is a negative you add that to 45! then you get D. LOL 45=-19 + d +19 +19 64= d Get d by it's self so you add or subtract the opposite.

Answer 7

Boat = x
Current = y
5(x - y) = 60
x - y = 12 .......(1)

3(x + y) = 60
x + y = 20 .......(2)

Solving (1) and (2)
x = 16
y = 4

Answer 8

vb=velocity of boat
vc=velocity of current
vt=total velocity

vt=vb+vc

60km/5hr=vb+-vc, 12km/hr=vb+-vc, vb=12km/hr+vc
60km/3hr=vb+vc, 20km/hr=vb+vc, vb=20km/hr-vc

so you have:
vb=12km/hr + vc
vb=20km/hr - vc

20km/hr=12km/hr + vc
8km/hr=vc
12km/hr = vb + -vc
12km/hr = vb + -8
20km/hr = vb

Answer 9

ill try to find out

Answer 10

whats ur problem im the best math person in my high school class.