These are very easy to set up with a "chart"
---------Distance-----Rate---Time (D=RxT)
slow way --6x----------x-------6
fast way - 5(x+7)-----x+7------5
so 6x = 5(x+7)
6x=5x+35+
x=35
The speed on the slower trip was 35 mph, and on the return trip 42 mph
A consultant traveled 6 hours to attend a meeting. The return trip took only 5 hours because the speed was 7 miles per hour faster. What was the consultant's speed each way?
Let X = distance one way
V1 = speed going
V2 = V1 + 7 = speed returning
X/V1 = 6
X/V2 = X/(V1+7) = 5
Now you have two equations in two unknowns (X and V1)
X = 6 V1
X = 5 (V1 + 7)
0 = V1 - 35
V1 = 35 mph
V2 = 42 mph
X = 210 miles
To solve this problem, we can use the formula Distance = Rate * Time (D = R * T).
Let's assign some variables:
- X: distance traveled in one direction (in miles)
- V1: speed going (in mph)
- V2: speed returning (in mph)
According to the problem, the consultant traveled for 6 hours in one direction and 5 hours on the return trip because the speed was 7 mph faster.
Using the formula D = R * T, we can create two equations:
1. X/V1 = 6 (equation 1)
This equation represents the distance traveled in one direction divided by the speed going, which gives us the time taken for the one-way trip.
2. X/V2 = X/(V1+7) = 5 (equation 2)
This equation represents the distance traveled in one direction divided by the speed returning, which gives us the time taken for the return trip. Since the speed on the return trip is 7 mph faster, we use V1+7.
Now, we have two equations in two unknowns (X and V1). We can solve them simultaneously to find the values.
From equation 1, we have X = 6V1.
Substituting this value of X in equation 2, we get:
6V1/(V1+7) = 5.
Simplifying this equation further, we have:
6V1 = 5(V1+7).
Expanding the equation, we have:
6V1 = 5V1 + 35.
Subtracting 5V1 from both sides, we get:
V1 = 35 mph.
So, the speed going (V1) is 35 mph.
Since the speed returning (V2) is V1 + 7, we have:
V2 = 35 + 7 = 42 mph.
Therefore, the speed on the slower trip was 35 mph, and on the return trip, it was 42 mph.
To calculate the distance (X), we can substitute the value of V1 in equation 1:
X = 6 * 35 = 210 miles.
Hence, the consultant traveled a distance of 210 miles.