\(\textbf{1)}\) When Mike rows his boat with the current, he travels 8 miles in 2 hours. Against the current, he can only travel 6 miles in the same time. How fast can Mike travel in still water? What is the rate of the current? Boat speed is 3.5 miles per hour Rate of the current is .5 miles per hour
Let b = speed of boat in still water.
Let c = speed of current
When Mike rows with the current, his effective speed is increased by the speed of the current, so his speed is “b + c” (rate of the boat plus rate of the current).
When Mike rows against the current, his effective speed is decreased by the speed of the current, so his speed is “b – c” (rate of the boat minus rate of the current).
Given the information:
With the current: Distance = 8 miles, Time = 2 hours
Against the current: Distance = 6 miles, Time = 2 hours
We can use the formula:
Distance = Rate × Time
For rowing with the current:
8 = (b + c) × 2
For rowing against the current:
6 = (b – c) × 2
Now we have a system of two equations:
2b + 2c = 8
2b – 2c = 6
We can simplify these equations by dividing both sides of both equations by 2:
b + c = 4
b – c = 3
Add the equations to eliminate c
2b = 7
b = 3.5
Now that we have b=3.5, we can substitute it into one of the equations to solve for the speed of the current:
b + c = 4
3.5 + c = 4
c = 0.5
So, Mike can travel at a speed of 3.5 miles per hour in still water, and the speed of the current is 0.5 miles per hour.
Let b = speed of boat in still water.
Let c = speed of current
When Mike rows with the current, his effective speed is increased by the speed of the current, so his speed is “b + c” (rate of the boat plus rate of the current).
When Mike rows against the current, his effective speed is decreased by the speed of the current, so his speed is “b – c” (rate of the boat minus rate of the current).
Given the information:
With the current: Distance = 8 miles, Time = 2 hours
Against the current: Distance = 6 miles, Time = 2 hours
We can use the formula:
Distance = Rate × Time
For rowing with the current:
8 = (b + c) × 2
For rowing against the current:
6 = (b – c) × 2
Now we have a system of two equations:
2b + 2c = 8
2b – 2c = 6
We can simplify these equations by dividing both sides of both equations by 2:
b + c = 4
b – c = 3
Add the equations to eliminate c
2b = 7
b = 3.5
Now that we have b=3.5, we can substitute it into one of the equations to solve for the speed of the current:
b + c = 4
3.5 + c = 4
c = 0.5
So, Mike can travel at a speed of 3.5 miles per hour in still water, and the speed of the current is 0.5 miles per hour.
\(\textbf{2)}\) With the wind, a plane can fly 1200 miles in 2 hours and 30 minutes. Against the wind, the same plane can only fly 1000 miles in the same time. Find the rate of the plane in still air and the rate of the wind. Plane speed is 440 miles per hour Wind speed is 40 miles per hour
\(\textbf{3)}\) A boat travels 30 miles upstream (against the current) in 3 hours. The boat travels the same distance downstream (with the current) in 2 hours. What is the rate of the boat in still water? What is the rate of the current? Boat speed is 12.5 miles per hour Rate of the Current is 2.5 miles per hour
\(\textbf{4)}\) A plane travels 2500 miles in 5 hours when it flies into the wind. When the same plane flies with the wind, it can travel the same distance in 4 hours. Find the rate of the plane in still air and the rate of the wind. Plane speed is 562.5 miles per hour Wind speed is 62.5 miles per hour
Let p= speed of plane
Let w= speed of wind
When the plane is flying into the wind, its effective speed is reduced by the speed of the wind, so the plane’s speed is “p – w”.
When the plane is flying with the wind, its effective speed is increased by the speed of the wind, so the plane’s speed is “p + w”.
Given the information:
Flying into the wind: Distance = 2500 miles, Time = 5 hours
Flying with the wind: Distance = 2500 miles, Time = 4 hours
We can use the formula:
Distance = Rate × Time
For flying into the wind:
2500 = (p – w) × 5
For flying with the wind:
2500 = (p + w) × 4
Now we have a system of two equations:
5p – 5w = 2500
4p + 4w = 2500
We can simplify these equations by dividing both sides of the second equation by 4:
5p – 5w = 2500
p + w = 625
Now we can solve this system of equations. Let’s solve the second equation for p:
p = 625 – w
Substitute this value of p into the first equation:
5(625 – w) – 5w = 2500
3125 – 5w – 5w = 2500
-10w = -625
w = 62.5
Now that we have the value of the wind speed, we can find the rate of the plane in still air by substituting w=62.5 into the equation
p = 625 – w:
P = 625 – 62.5
P = 562.5
So, the rate of the plane in still air is 562.5 miles per hour, and the rate of the wind is 62.5 miles per hour.
Let p= speed of plane
Let w= speed of wind
When the plane is flying into the wind, its effective speed is reduced by the speed of the wind, so the plane’s speed is “p – w”.
When the plane is flying with the wind, its effective speed is increased by the speed of the wind, so the plane’s speed is “p + w”.
Given the information:
Flying into the wind: Distance = 2500 miles, Time = 5 hours
Flying with the wind: Distance = 2500 miles, Time = 4 hours
We can use the formula:
Distance = Rate × Time
For flying into the wind:
2500 = (p – w) × 5
For flying with the wind:
2500 = (p + w) × 4
Now we have a system of two equations:
5p – 5w = 2500
4p + 4w = 2500
We can simplify these equations by dividing both sides of the second equation by 4:
5p – 5w = 2500
p + w = 625
Now we can solve this system of equations. Let’s solve the second equation for p:
p = 625 – w
Substitute this value of p into the first equation:
5(625 – w) – 5w = 2500
3125 – 5w – 5w = 2500
-10w = -625
w = 62.5
Now that we have the value of the wind speed, we can find the rate of the plane in still air by substituting w=62.5 into the equation
p = 625 – w:
P = 625 – 62.5
P = 562.5
So, the rate of the plane in still air is 562.5 miles per hour, and the rate of the wind is 62.5 miles per hour.
\(\textbf{5)}\) Two cars started at the same place. One car going north traveled \(50\) mph for \(2\) hours. The other car traveled south at \(60\) mph for \(1\) hour and \(15\) minutes. How far apart are the cars?
The answer is \(175 \) miles 
\(\textbf{6)}\) Two cars started at the same place. One car going north traveled \(10\) mph for \(4\) hours. The other car traveled east at \(15\) mph for \(2\) hours. How far apart are the cars?
The answer is \(50 \) miles
\(\textbf{7)}\) Two cars started at the same place. Both traveled north. One car traveled at \(50\) mph for \(2\) hours. The other car traveled at \(60\) mph for \(1\) hour and \(45\) minutes. How far apart are the cars?
The answer is \(5\) miles
\(\textbf{8)}\) A car travels from city A to city B at 50 miles per hour. Then they return to city A at 30 miles per hour. What was the average rate of speed of the two trips? The answer is \(37.5\) miles per hour
To find the average rate of speed for the round trip from city A to city B and back, you can use the concept of harmonic mean. The harmonic mean is calculated as the reciprocal of the average of the reciprocals of the individual speeds.
The formula for the harmonic mean of two speeds is:
Harmonic Mean = \(\displaystyle\frac{2}{\frac{1}{50}+\frac{1}{30}}\)
To add the fractions in the denominator, you need a common denominator, which is the least common multiple of 50 and 30, which is 1500:
Harmonic Mean = \(\displaystyle\frac{2}{\frac{30}{1500}+\frac{50}{1500}}\)
Harmonic Mean = \(\displaystyle\frac{2}{\frac{80}{1500}}\)
Harmonic Mean = \(\displaystyle2\cdot\frac{1500}{80}\)
Harmonic Mean = \(\displaystyle\frac{3000}{80}\)
Harmonic Mean = \(\displaystyle 37.5\)
So, the average rate of speed for the round trip is 37.5 miles per hour.
To find the average rate of speed for the round trip from city A to city B and back, you can use the concept of harmonic mean. The harmonic mean is calculated as the reciprocal of the average of the reciprocals of the individual speeds.
The formula for the harmonic mean of two speeds is:
Harmonic Mean = \(\displaystyle\frac{2}{\frac{1}{50}+\frac{1}{30}}\)
To add the fractions in the denominator, you need a common denominator, which is the least common multiple of 50 and 30, which is 1500:
Harmonic Mean = \(\displaystyle\frac{2}{\frac{30}{1500}+\frac{50}{1500}}\)
Harmonic Mean = \(\displaystyle\frac{2}{\frac{80}{1500}}\)
Harmonic Mean = \(\displaystyle2\cdot\frac{1500}{80}\)
Harmonic Mean = \(\displaystyle\frac{3000}{80}\)
Harmonic Mean = \(\displaystyle 37.5\)
So, the average rate of speed for the round trip is 37.5 miles per hour.