Question

Ishmael sells yachts and sold \(\$ 90,000\) worth of yachts this month. That is an increase of \(20 \%\) over last month's sales. But last month was \(20 \%\) less than the month before. What were Ishmael's sales two months ago? A. \(\$ 50,000\) B. \(\$ 75,000\) C. \(\$ 90,200\) D. \(\$ 93,750\)

Short answer

\$ 93,750

Step-by-step solution

- Determine last month's sales

First, identify last month's sales. This month Ishmael sold \$ 90,000 worth of yachts, which is an increase of 20% over last month's sales. Let last month's sales be x. Then:\[ 1.20x = 90,000 \] Solve for x: \[ x = \frac{90,000}{1.20} \] \[ x = 75,000 \] So, last month's sales were \$ 75,000.\}

- Determine sales from two months ago

Next, determine the sales from two months ago. Last month's sales were 20% less than the month before (two months ago). Let the sales from two months ago be y. Then: \[ y - 0.20y = 75,000 \] \[ 0.80y = 75,000 \] Solve for y: \[ y = \frac{75,000}{0.80} \] \[ y = 93,750 \] So, the sales two months ago were \$ 93,750.\}

Key concepts

Percentage Increase

Percentage increase is an important concept in many practical situations, such as determining changes in sales figures. To understand it, let's dive into Ishmael's yacht sales. This month Ishmael's sales were \(90,000, representing a 20% increase over last month.
To calculate the previous month's sales, we need to reverse-engineer the percentage increase. If this month’s sales (90,000) were 20% higher than last month's sales, we can express this as:

• Let last month's sales be x
• 1.20x = 90,000

This equation represents the new sales (90,000) as 120% (or 1.20 times) of last month's sales.
Solving for 1x involves dividing both sides by 1.20, giving us:
x = \frac{90,000}{1.20}
x = 75,000. This tells us that last month's sales were \)75,000.
Understanding how to handle percentage increases can help in various real-world applications, whether calculating sales, finances, or personal budgeting.

Algebraic Equations

Algebraic equations are fundamental in solving many types of math problems, including those involving percentages and sales changes. In Ishmael's case, we used an algebraic equation to identify sales figures from different months.
For instance, after finding that last month's sales were \(75,000, we needed to determine the sales from two months ago. Knowing that last month's sales were 20% less than the previous month (two months ago), we set up another algebraic equation:
The sales from two months ago be y. Last month’s sales being 20% less than y can be expressed as:
y - 0.20y = 75,000 ,

• This simplifies to
  0.80y = 75,000

Solving for y involves dividing both sides by 0.80, giving us:
y = \frac{75,000}{0.80}
y = 93,750. This tells us that the sales from two months ago were \)93,750.
Understanding how to manipulate and solve algebraic equations is a crucial skill, especially when dealing with continuous changes and relationships between variables.

Word Problems

Word problems require careful reading and translation of text into mathematical expressions. They're common in tests like the GED and everyday life scenarios. Let's take Ishmael's sales problem as an example.
The problem states that this month’s sales were $90,000, a 20% increase over last month's sales, which were 20% less than the month before. Breaking it down:

• Identify what you need to find. Here, it's the sales from two months ago.
• Translate each part of the problem into mathematical terms.
• First, find last month’s sales with 1.20x = 90,000 . Solve it to get x=75,000.
• Then, determine the sales from two months ago using 0.80y = 75,000 . Solve it to get y=93,750.

By systematically translating words into equations, you can solve complex problems step by step. Similarly, practice and familiarity with various problem types will enhance your problem-solving skills.