Question

Juliet is selling photographs as part of a project for her entrepreneurship class. She sells the first 20 photographs for \(\$ 10\) each. Because the first 20 photographs sold so quickly, she raised the price of the photographs to \(\$ 15\) each for the rest of the project. After her expenses, Juliet earns a profit of \(80 \%\) of the revenues from her sales. What is the least number of photographs she must sell for the rest of the project to earn a profit of at least \(\$ 400\) ? A) 18 B) 20 C) 24 D) 32 $$ \frac{p^{\frac{1}{4}} q^{-3}}{p^{-2} q^{\frac{1}{2}}} $$

Short answer

The least number of photographs Juliet must sell at the increased price to achieve a profit of at least $400 is \(20\). So, the correct answer is (B) 20.

Step-by-step solution

Calculate the total profit from the first 20 photographs sold

First, let's find out the total revenue from selling the first 20 photographs. To do this, we multiply the number of photographs (20) with the price per photograph ($10): Revenue from first 20 photographs = 20 * \(10 = \)200 Since she earns a profit of 80% of the revenues, we can find her profit by multiplying the revenue with 0.8: Profit from first 20 photographs = \(200 * 0.8 = \)160

Calculate the minimum revenue needed to achieve a \(400 profit

To find the minimum revenue needed to achieve a \)400 profit, we first need to determine the remaining profit needed. Remaining profit needed = \(400 - \)160 = $240 Now, we will divide the remaining profit needed by the profit percentage (80%) to determine the minimum revenue required: Minimum Revenue needed = \(\frac{\)240}{0.8}\( = \)300.

Calculate the minimum number of photographs needed to be sold at the increased price

We know that the increased price per photograph is $15. To find the least number of photographs that must be sold to reach the minimum revenue, we simply divide the minimum revenue needed by the increased price per photograph: Number of photographs = \(\frac{300}{15}\) = 20

Conclusion

After reviewing the given answer choices, we find that the least number of photographs that Juliet must sell at the increased price to achieve a profit of at least $400 is 20. So, the correct answer is (B) 20.

Key concepts

Algebraic Word Problems

Algebraic word problems are math questions that require one to translate a written narrative into an algebraic expression or equation in order to solve for a variable. These problems test your understanding of algebra along with your ability to parse language and logical reasoning.

For example, in the SAT problem provided, Juliet's photography sales are described in a real-world context. Students must pull out the important information—how many photographs were sold at what price, the percentage of profit, and the total profit desired—then create and solve algebraic expressions to find the answer.

The key to solving these problems is methodical organization of data: identifying the unknown variable, translating words into algebraic equations, and using arithmetic operations correctly. Problems become easier to navigate once you break them down into smaller pieces.

Another tip is to always keep track of units, like dollars or photographs in this case, ensuring that they remain consistent throughout the problem. Mathematics is, after all, a language of its own, and when it comes to algebraic word problems, fluency matters just as much as computational skills.

Profit Calculation

Profit calculation in math is the process of determining the gain from a business transaction or operation after subtracting expenses. It is essential for any business-related math problem, including those that appear on the SAT.

The profit is generally a percentage of the revenue—total money collected from sales. Mathematically, you calculate profit by the formula: \[ \text{Profit} = \text{Revenue} \times \text{Profit\%} \]One of the most common mistakes students make in profit calculation is confusing revenue with profit. Remember, revenue is the total incoming money before expenses, while profit represents what's left after costs are removed.

In our SAT example, Juliet's profit is not simply the price at which she sells her photographs; it is 80% of the revenue she gains from selling them. Understanding this distinction is crucial for pinpointing the correct amount that must be earned to reach a specific profit goal.

Percentage Problems

Percentage problems are common in standardized tests and involve finding parts of a whole as a percent. They require an understanding of fractions, decimals, and, of course, percentages. An easy way to think about percentage is that it represents a number out of 100.

The basic conversions that are useful to remember include:\[ 1\% = 0.01 \]\[ 50\% = 0.5 \]\[ 100\% = 1 \]In a problem where you need to calculate a certain percentage of a number, you multiply the number by the percentage expressed as a decimal. For instance, to get 80% of any amount, you multiply that amount by 0.8.

In Juliet's case, the profit is 80% of her sales revenue, which means every time Juliet calculates her profit, she'll need to multiply her revenue by 0.8. When dealing with percentages, it's also common to reverse the problem and find the original whole, given the percentage and the part—a concept known as 'percent of a number.' This is precisely what's needed in this SAT exercise when determining the revenue required to achieve a given profit.