Standardized Test Practice
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How much hay do x horses consume in a single day? What about in d days?
How do you identify different types of variations?
Substitute appropriate values into the equation from Part A.
Example Equation: y = 10xd
Joint variation, see solution.
930 pounds
We know that a horse requires 10 pounds of hay per day in order to remain healthy. We are asked to write an equation that represents the amount of hay necessary to sustain x horses for d days. Let's call the amount of hay y. For each of the x horses, 10 pounds of hay are needed for a single day.
| Number of Horses | Amount of Hay Necessary For Each Day |
|---|---|
| x | 10 x |
| Number of Horses | Number of Days | Amount of Hay Necessary |
|---|---|---|
| x | d | 10 x d |
We can now write an equation representing the amount of hay y needed to sustain x horses for d days. y = 10 x d
Before we decide what type of equation we have obtained in Part A, let's start by recalling the definitions of different types of variations.
| Type of Variation | Definition | Example Equation (k constant) |
|---|---|---|
| Direct Variation | Occurs when one quantity is equal to the product of a constant and another quantity. | y=kx |
| Joint Variation | Occurs when one quantity varies directly as the product of two or more other quantities. | y=kxz |
| Inverse Variation | Occurs when one quantity is equal to the quotient of a constant and another quantity. Also, the product of the quantities is constant. | y=k/x, where x≠0 and y≠0 |
Let's take another look at the equation we have found in Part A. y = 10 x d Remember that y, x, and d are variables, while 10 is a constant. Therefore, we can see that our equation is an example of a joint variation.
We want to find the amount of hay y that 3 horses need for the month of July. We will use the equation we found in Part A.
The amount of hay 3 horses require for the month of July is 930 pounds.