Problem 73 - Algebra 1

Question

GOLD PRICES IN LONDON At 9 A.M., an ounce of gold sells for \(\$ 287.56\). At noon, gold sells for \(\$ 286.90\) per ounce. At 4 P.M., the final price for the day is \(\$ 287.37\) per ounce. What is the change in the price per ounce of gold from 9 A.M. to noon?

Step-by-Step Solution

Answer

The change in the price per ounce of gold from 9 A.M. to noon is \(\$0.66\).

Step 1: Identify the Given Prices

The price of gold at 9 A.M. is given as \(\$ 287.56\), and the price at noon is \(\$ 286.90\) per ounce.

Step 2: Calculate the change in price

To get the change in price from 9 A.M. to noon, subtract the price at noon \(\$286.90\) from the 9 A.M. price \(\$287.56\).

Step 3: Obtain the result

After subtraction, the difference will be the change in price per ounce of gold from 9 A.M. to noon.

Key Concepts

Subtraction of Decimals

Subtracting decimals is a foundational skill in mathematics that has practical applications in everyday life, such as in calculating price changes. Suppose you're tracking the price of an item in a market, and it changes throughout the day. To find out exactly how much it fluctuates, you need to subtract the decimal prices. For example, to determine the price change in gold from 9 A.M. to noon based on the prices provided ((9 A.M. price) - (noon price) = change), you align the decimal points one above the other and subtract like you would with whole numbers.

Remember to place the decimal point in the answer directly below the other decimal points to ensure an accurate calculation. It's important to stack the numbers vertically and borrow as necessary, treating the decimal point as if it were any other digit. Executing this correctly should provide you with the correct change in price, which in our case is (9 A.M. price (\(287.56\)) - noon price (\(286.90\))) = (\(0.66\)).

Mathematical Operations

Mathematical operations such as addition, subtraction, multiplication, and division are the building blocks for dealing with numbers. Subtraction is one of these essential operations that enables us to find the difference between quantities. It's a comparatively simple operation but requires careful attention to detail when working with decimals, as described earlier.

When we calculate the change in price as in the gold price example, we are using the concept of subtraction to find this difference. It is critical that you're comfortable with these operations since they are used not only in academia but also in various daily scenarios such as calculating change, discounts, interest rates, and more.

Real-world Applications of Algebra

Algebra is used extensively in the real world to solve problems involving quantities with unknown values and to understand relationships between variables. In the context of our example with gold prices, algebra is used to determine the price change over time.

This involves assigning a variable to represent the unknown change and then creating an equation based on the given information. The difference between the two prices (\(287.56\) and (\(286.90\)) is represented as a subtraction of decimals. The concept of algebra comes into play as we manipulate these numerical expressions to find the value of the change. Moreover, understanding algebra helps business owners, economists, scientists, and many others to make informed decisions by analyzing trends and forecasting possible outcomes using equations and inequalities.