Question

If \(p\) varies directly as \(q,\) and \(p\) is 846 when \(q\) is \(135,\) find \(q\) when \(p\) is \(448 .\)

Short answer

\( q = \frac{448 \times 135}{846} = 70.588 \approx 71 \) (rounded to the nearest whole number)

Step-by-step solution

Understand Direct Variation

Direct variation means that when one variable increases, the other variable increases at a constant rate. The formula for direct variation is given by: \( p = kq \) where \( k \) is the constant of variation.

Find the Constant of Variation

Since \( p \) is 846 when \( q \) is 135, we can substitute these values into the formula to find the constant \( k \). So, \( 846 = k \times 135 \). Dividing both sides by 135, we find that \( k = \frac{846}{135} \).

Calculate the Constant of Variation

Calculate the constant of variation \( k \) by dividing 846 by 135: \( k = \frac{846}{135} \).

Use the Constant to Find the New Value

Now we use the calculated constant \( k \) to find the new value of \( q \) when \( p \) is 448. We have the formula \( 448 = kq \). To find \( q \), divide both sides of the equation by \( k \): \( q = \frac{448}{k} \).

Substitute the Value of \( k \) and Solve for \( q \)

Substituting the value of \( k \) we found earlier, we get \( q = \frac{448}{\frac{846}{135}} \). Now, solve for \( q \) by simplifying the fraction.

Simplify the Fraction and Find \( q \)

Simplify the fraction by multiplying 448 by the reciprocal of \( \frac{846}{135} \), which gives us \( q = \frac{448 \times 135}{846} \). Upon calculating, we determine the value of \( q \).

Key concepts

Constant of Variation

When we talk about the constant of variation in a direct variation relationship, we're referring to the consistent factor that relates two variables to each other. Imagine you're on a road trip, and the distance you travel is directly related to the time you've driven – that fixed rate at which distance changes with time is akin to the constant of variation.

Mathematically, if two quantities, let's call them 'A' and 'B,' are in direct variation, we'd express this as \( A = kB \), where \( k \), is the constant of variation. Think of it as a recipe: for every cup of flour (\text{A}), you need \text{k} cups of sugar (\text{B}) to bake the perfect cake every time. If 'A' changes, 'B' will change at a rate that's always multiplied by this constant 'k'.

Returning to the problem at hand, we calculated the constant of variation (\text{k}) by solving the equation \( 846 = k \times 135 \) from the given condition that when \text{p} is 846, \text{q} is 135. By finding the constant, \( k = 6.3 \), we established the definite relationship between \text{p} and \text{q}: thrice the constancy, always the consistency.

Proportional Relationship

A proportional relationship is the best buddy of direct variation. It's a partnership where one quantity is always a constant multiple of the other - if you have a scenario of 'the more you work, the more you earn', you're looking at direct proportionality. It implies a situation of fairness, where one variable scales up or down strictly in tandem with another.

In terms of algebra, if we have two variables, call them 'X' and 'Y', that are directly proportional, we describe this relationship with \( X = mY \), where \( m \) is the magnifying constant that scales Y up to X. This 'm' is essentially the constant of variation we've talked about before.

This concept is the cornerstone of our exercise. Given that \( p \) and \( q \) are directly proportional, we uncovered their secret connection via the constant of variation. When asked to find one variable given another, it's this proportionality that guided us to the solution: proportional thinking for predictable outcomes!

Algebraic Formulas

Essential Tools for Mathematical Relationships

Algebraic formulas are the Swiss Army knife in your math backpack. They're the general expressions that capture the essence of mathematical relationships and provide a framework for tackling a wide range of problems, much like the one at hand.

Armed with the formula for direct variation, \( p = kq \) (where \text{p} and \text{q} are two variables in direct proportion and \text{k} is our constant of variation), we can dive into any sea of numbers and emerge with the pearl we seek. Solving the given problem involved using this formula twice: first, to find the constant \text{k}, and then to calculate the new value of \text{q} when \text{p} is known.

It is this commanding power of algebraic formulas that enables us to transition from understanding the general law (\text{the more you have of this, the more you get of that}) to applying it to find specific solutions. Embracing algebraic formulas means embracing mathematical mastery.