The amount of gasoline used by a car varies jointly as the distance traveled and the square root of speed.

**Answer:**

Answer:

38 \frac{2}{5}3852 liters

Step-by-step explanation:

A variation is a relationship between two variables that share a ratio. This ratio is called “coefficient of variation”. It shows how one variable varies as the other.

Some variables are directly related. Direct variation means as one value increase, the other also increases at a rate equal to the coefficient of variation. If 2 variables, x and y are directly related, then their relationship can be stated as

y = kxy=kx

k is the coefficient of variation.

Some variables are inversely related. Inverse variation means as one value increase, the other variable decreases at a rate equal to the coefficient of variation. If 2 variables, x and y, are inversely related, their relationship can be stated as

y =\frac{k}{x}y=xk

There are times when more than 2 variables are related. This is called a joint variation. A joint variation can have direct or indirect variations.

Variations are often solved by getting the value of k.

Let’s go back to the problem. Let us assign variables to each quantities.

Let g be the amount of gasoline.

Let d be the distance traveled of the car.

Let s be the speed of the car.

Their relationship can be stated by the variation:

g = k*d*\sqrt{s}g=k∗d∗s

We are given that the car used 25 liters for 100 km. on a 100 km/hr speed. Those are values for the variables we have. We can substitute it to the variation and solve for k.

\begin{gathered}g = k*d\sqrt{s}\\g = 25\\d =100\\s = 100\\25 = k*(100)*\sqrt{100}\\25 = k*(100)*(10)\\25=k*1000\\k = \frac{25}{1000} \\k = \frac{1}{40}\end{gathered}g=k∗dsg=25d=100s=10025=k∗(100)∗10025=k∗(100)∗(10)25=k∗1000k=100025k=401

Our constant of variation is \frac{1}{40}401 .

This means we can solve for the variation if we are missing one of variables.

The problem is asking how many liters of gasoline are needed for a 192km trip at a speed of 64 km/hr.

We are given two values (speed and distance), and we have the constant of variation from earlier.

\begin{gathered}g = k*d*\sqrt{s} \\k = \frac{1}{40}\\d= 192\\s = 64\\\end{gathered}g=k∗d∗sk=401d=192s=64

\begin{gathered}g = \frac{1}{40}*(192)(\sqrt{64}\\\\g = \frac{1}{40} (192)(8)\\\\g = \frac{192}{5}\\\\g = 38 \frac{2}{5}\end{gathered}g=401∗(192)(64g=401(192)(8)g=5192g=3852

We need 38\frac{2}{5}3852 liters of gasoline for the trip.

For more information about variations, click here

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