recall your d = rt, distance = rate * time.
s = Samantha's speed rate
one thing to bear in mind is that, when Samantha is going upstream, she's not really going "s" mph fast, because she's going against the current, and the current's rate is 4 mph and subtracting speed from her, then she's really going "s - 4" fast.
Likewise, when she's going downstream because she's going with the current, the current is adding speed to her, so she's going "s + 4" fast.
Now, let's say the distance she covered both ways was the same "d" miles.
Let's recall that since there are 60 minutes in 1 hour, 12 minutes is 12/60 = 1/5 of an hour.
Let the the total distance is D and the speed of Samantha's speed in still water is x ( in mph ),
Here, speed of stream = 4 mph,
Thus, the speed in upstream = (x - 4) mph,
Speed in downstream = (x + 4) mph,
We know that,
∴ Time taken in upstream =
And, time taken in downstream =
According to the question,
Equation (1) - 5 equation (2),
0 = x + 4 - 5x + 20
0 = -4x + 24
4x = 24
⇒ x = 6 mph.
Hence, Samantha swims with the speed 6 mph.
We use the equation s = d/t to solve this.
We do not know Samantha's speed, but we know that going downstream, she swims the given distance in 12 minutes. Going downstream, we add the river's speed, 4, to her speed; this gives us
s+4 = d/12
Solving for d, we will multiply both sides by 12:
12(s+4) = (d/12)(12)
12(s+4) = d
Use the distributive property:
12(s)+12(4) = d
12s+48 = d
Going upstream, Samantha swims the given distance in an hour, 60 minutes. Going upstream, we subtract the river's speed, 4, from her speed; this gives us
s-4 = d/60
Solving for d, we will multiply both sides by 60:
60(s-4) = (d/60)(60)
60(s-4) = d
Use the distributive property:
60(s)-60(4) = d
60s-240 = d
Since both equations equal d, we can set them equal to one another:
12s+48 = 60s-240
Subtract 12s from each side:
12s+48-12s = 60s-240-12s
48 = 48s-240
Add 240 to each side:
48+240 = 48s-240+240
288 = 48s
Divide both sides by 48:
288/48 = 48s/48
6 = s
Samantha's speed is 6 mph.
Choose the most logical value for the variable to represent. Let x= Kelli's swimming speed in still water
The option B is correct, i.e. 6 mph.
Suppose Samantha swim in still water at speed of X mph.
If the river flows at 4 mph,
Then Upstream speed = (X-4) mph.
And Downstream speed = (X+4) mph.
Samantha swam upstream for some distance in one hour. She then swam downstream the same river for the same distance in only 12 minutes.
It means Distance upstream = Distance downstream.
Speed upstream x Time upstream = Speed downstream x Time downstream.
(x-4) * 1 = (x+4) * 12/60
x - 4 = 0.2x + 0.8
x - 0.2x = 0.8 + 4
0.8x = 4.8
x = 4.8/0.8 = 6 mph.
Hence, the option B is correct, i.e. 6 mph.
Given :-Samantha swam upstream for some distance in one hour. She then swam downstream the same river for the same distance in only 12 minutes. The river flows at 4 mph.
To find :-How fast can Samantha swim in still water ?
Let us assume that, speed of samantha in still water is x mph.
→ Usual speed of samantha in downstream = speed in still water + river flows. = (x + 4) mph.
→ Usual speed of samantha in upstream = speed in still water - river flows. = (x - 4) mph.
Let us also assume that, distance cover by her is D mile.
we have given that, Samantha swam upstream for some distance in one hour.
→ Distance / usual speed in upstream = 1 hour.
→ D / (x - 4) = 1
→ D = (x - 4) Eqn.(1)
Now, given that, She swam downstream the same river for the same distance in only 12 minutes.
→ Distance / usual speed in downstream = 12 minutes = (12/60) = 1/5 hours.
→ D/(x + 4) = 1/5
→ 5D = (x + 4)
Putting value of D from Eqn.(1) ,
→ 5(x - 4) = (x + 4)
→ 5x - 20 = x + 4
→ 5x - x = 4 + 20
→ 4x = 24
dividing by 4 both sides,
→ x = 6 mph. (Ans.)Hence, samantha can swim at the rate of 6 mph in still water.
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Let s be samantha's speed and d as the distance she travelled.
Find s by equating the two eq. in terms of d since d upstream is just equal to d downstream.