The speed of a stream is 4 mph. A boat travels 7 miles upstream in the same time it takes to travel 15 miles downstream. What is the speed of the boat still in the water?
2007-12-12 05:17:41 · 9 answers · asked by Caitlin 2 in Science & Mathematics ➔ Mathematics
Let R be the speed of the boat in still water.
The speed upstream will be (R - 4)
The speed downstream will be (R + 4)
So you can write two equations:
7 = (R - 4)T
15 = (R + 4)T
We don't know the time, but we know it is the same time upstream and downstream. Solve each of these for T and then equate them:
T = 7 / (R - 4)
T = 15 / (R + 4)
Equate them:
7 / (R - 4) = 15 (R + 4)
Cross multiply:
15(R - 4) = 7(R + 4)
Distribute 15 and 7 through the parentheses:
15R - 60 = 7R + 28
Subtract 7R from both sides:
8R - 60 = 28
Add 60 to both sides:
8R = 88
Divide both sides by 8:
R = 88/8
R = 11
So the speed in still water is 11 mph.
We can also go from here to solve for T:
7 = (R - 4)T
7 = (11 - 4)T
7 = 7T
7/7 = T
1 = T
It makes sense that you would take 1 hour at 7 mph (11-4) to go 7 miles upstream and 1 hour at 15 mph (11+4) to go 15 miles downstream.
Anyway, the answer is:
The speed of the boat is 11 mph in still water.
2007-12-12 05:23:01 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
A mathematical explanation, not to say the one above me is bad or anything, but teachers probably wouldn't except it on a test.
d = rt
(upstream the 4 mph subtracts from the boats speed)
7 = (r-4) t
(downstream the 4 mph add to the boats speed)
15 = (r+4) t
t = t
7/(r-4) = 15 / (r+4)
solve for r:
7(r+4) = 15(r-4)
7r + 28 = 15r - 60
15r - 7r = 28 + 60
8r=88
r = 11 mph (speed of boat in still water)
2007-12-12 05:20:59 · answer #2 · answered by KEYNARDO 5 · 0⤊ 0⤋
let, the speed of the boat is x
relative upstream speed is x-4
relative downstream speed is x+4.
then,
(x-4)/7 = (x+4)/15
x = 11 mph
2007-12-12 05:43:01 · answer #3 · answered by Subhajit M 1 · 0⤊ 0⤋
you know that d = rt, and t = d/r
let x = rate of boat in still water
going up stream the rate = x - 4
going down stream rate = x + 4
the distance upstream was 7 miles
the distance downstram was 15 miles
the time is the same both ways.....therefore
7 / (x - 4 ) = 15 / ( x + 4 )
cross multiply and you get:
7(x + 4 ) = 15 ( x - 4 )
7x + 28 = 15x - 60
collect like terms and solve for the x
8x = 88
x = 11 miles per hour
2007-12-12 05:28:05 · answer #4 · answered by Ray 5 · 0⤊ 0⤋
if the speed of the boat is v, its relative speed upstream is v-4 and its reltive speed downstream is v+4.
we get that in time t:
(v-4)t = 7 and (v+4)t = 15 therefore
(v-4)/7 = (v+4)/15 or v = 11 mph
2007-12-12 05:22:28 · answer #5 · answered by Christophe G 4 · 0⤊ 0⤋
11 mph
Here's one way of doing it.
downstream the speed equation is
d=vt
upstream the speed equation is
D=Vt
Not the t's are the same.
Solve each for t and set them equal
(d/v) = (D/V)
Cross multiply
dV = Dv
Since v, the speed upstream is equal to the speed of the Boat in still water (B) minus the current (4mph), and V the speed downstream is B+4 substitute the values in:
7(B+4) = 15(B-4)
7B+28 =15B-60
88=8B
B=11 mph
2007-12-12 05:25:38 · answer #6 · answered by Anonymous · 0⤊ 0⤋
11 mph?
add 4 to 7 (because it slows it down)
subtract 4 from 15 (because it adds to the speed of the boat)
gets you 11 with both numbers.
I could be wrong though :)
2007-12-12 05:20:57 · answer #7 · answered by SonnyAngel 2 · 0⤊ 0⤋
distance = speed X time
so time = distance/speed
Just equate the times as follows:
7/(v -4) = 15/(v + 4)
7v + 28 = 15v - 60
8v = 88
v = 11
2007-12-12 05:27:13 · answer #8 · answered by Joe L 5 · 0⤊ 0⤋
you have not any subject, you're working a velocity try based on your connection on your ISP. Your acquire velocity is in accordance with something completely distinctive. it is your Connection velocity minus your isp's connection to the server's isp, minus the servers connection velocity, minus the form of folk downloading of the server, minus any server caps and obstacles, minus the servers hdd velocity. you could probable acquire distinctive issues at 30kps, and that i wager you ought to acquire particular issues at 100kbps besides, if to procure the ultimate connection. additionally you could properly be on a distinctive node then your friends that ought to alter your acquire velocity. yet I labored for a cable enterprise for years and grow to be a excessive velocity cyber web container supervisor and your acquire velocity is universal. you will have firewall settings that are incorrect nonetheless, on your router or utility, yet different than that, i doubt you will get it to alter.
2016-12-31 08:09:32 · answer #9 · answered by sutterfield 3 · 0⤊ 0⤋