Question

Syed took out a cash advance of \(d\) dollars from a financing company. The company deducts a fee of \(\frac{1}{3}\) of the original advanced amount along with a wire transfer fee of $$\$ 30.00$$. Which of the following represents the final advanced amount that Syed receives after all applied fees, in dollars? A) $$\frac{1}{3} d-30$$ B) $$\frac{1}{3}(d-30)$$ C) $$\frac{2}{3}(d-30)$$ D) $$\frac{2}{3} d-30$$

Short answer

The short answer is: Syed receives a final advanced amount of \(\boxed{\frac{2}{3}d - 30}\) after all applied fees are deducted, which corresponds to Answer Choice (D).

Step-by-step solution

Deduct the financing fee

First, we need to deduct the financing fee, which is \(\frac{1}{3}\) of the original advanced amount \(d\). To do this, we subtract \(\frac{1}{3}d\) from the total advanced amount \(d\): Final amount after financing fee = \(d - \frac{1}{3}d\)

Simplify the expression

Now we have to combine the like terms: Final amount after financing fee = \(\frac{3}{3}d - \frac{1}{3}d\) = \(\frac{2}{3}d\)

Deduct the wire transfer fee

Next, we need to deduct the wire transfer fee of $$30.00$ from the final amount calculated in step 2: Final advanced amount = Final amount after financing fee \( - \$30.00\) = \(\frac{2}{3}d - 30\) Now, we can match the final expression with the answer choices given: The final advanced amount that Syed receives after all applied fees is \(\boxed{\frac{2}{3}d - 30}\), which corresponds to Answer Choice (D).

Key concepts

Cash Advance Problems

Cash advance problems typically involve understanding the terms of a financial agreement and translating those terms into a mathematical expression. In the context of SAT math problems, these types of questions test a student's ability to work with percentages and fees structurally represented in an algebraic way.

For instance, when you're faced with a problem that states someone took out a cash advance with a certain fee deducted from it, like in our example problem involving Syed, it is crucial to identify the different components of the transaction: the initial loan amount, the percentage of the loan taken as a fee, and any flat fees added or subtracted afterwards. The key to solving such problems is systematically breaking down the cash advance into its constituent financial parts and then representing those parts through variables and constants to form an algebraic equation.

Typically, a cash advance will involve a percentage deduction (a fraction of the original amount), and sometimes an additional fixed cost. To solve these problems, you need to:

• Read and understand the terms of the advance carefully.
• Identify the variable representing the advanced amount.
• Determine the fraction or percentage representing the fee.
• Account for any additional costs or fees as a separate term.
• Develop an algebraic expression that encompasses all these elements.
• Solve the expression to find the final amount received.

By following these steps and practicing similar problems, students can become adept at solving cash advance problems efficiently.

Simplifying Expressions

Simplifying expressions is a foundational skill in algebra that involves combining like terms and reducing expressions to their simplest form. This process often requires a student to apply the distributive property, combine like terms, and simplify fractional coefficients.

In SAT math problems, you might encounter expressions with both numerical and variable parts, much like the financing fee example from our Syed problem, where the expression started as \(d - \frac{1}{3}d\). To simplify this expression, you identify the 'like terms,' which in algebra are terms that contain the same variable raised to the same power. The coefficients of these terms can then be added or subtracted as appropriate.

To simplify an expression with fractions, remember to find a common denominator for all terms, which in many cases will be the terms involving the variable. For our provided example, the terms already have a common denominator of 3, so they were easily combined to give \(\frac{2}{3}d\).

Here's a quick guide for simplifying expressions:

• Apply the distributive property if there are parentheses involved.
• Combine like terms by adding or subtracting coefficients.
• Convert to a common denominator before combining fractional terms.
• Simplify complex fractions, where necessary, before combining terms.
• Reduce fractions to their simplest form at the end if possible.

Simplifying expressions requires practice, but mastering this skill is essential for solving more complex algebraic problems.

Word Problems in Algebra

Word problems in algebra require translating a real-world scenario into an algebraic equation that can be solved to find an unknown value. These problems often involve multiple steps and the key to solving them is to understand the story or scenario being presented.

In SAT math sections, word problems test a student's reading comprehension as much as their mathematical prowess. They tell a story and ask the test-taker to solve for a quantity or relationship between quantities that is not immediately obvious from the text alone. To do this, one must:

• Carefully read and understand the narrative of the problem.
• Determine what quantities are unknown and represent them with variables.
• Identify any given information and corresponding algebraic expressions.
• Set up an equation or a system of equations based on the relationships described in the problem.
• Solve the equation(s) for the unknown quantity.

Returning to our Syed example, the 'story' is that he took out a cash advance, from which a fee and a flat rate were deducted. Translating this scenario into algebra meant defining the variable 'd' for the original amount, understanding the fee structure, and constructing the appropriate algebraic expression to find the final amount Syed receives.

Word problems can seem daunting, but with practice, students can develop a systematic approach to decoding and solving these types of questions. It involves meticulous reading, analyzing the situation, and applying algebraic rules to derive a solution.