Problem 76 - Algebra and Trigonometry

Question

After a \(20 \%\) reduction, a digital camera sold for \(\$ 256\) What was the price before the reduction?

Step-by-Step Solution

Answer

The original price of the digital camera before the 20% reduction is \$320.

Step 1: Identify the given variables

From the problem, we identify that the sale price of the digital camera after a 20% reduction is \$256, this means that the sale price is \$256 and the rate of decrease is 20% or 0.20 in decimal form. Thus we have Sale Price = \$256 and Rate of Decrease = 0.20.

Step 2: Substitute the variables into the formula

Next, we substitute these values into the formula. We have \( \text{Original Price} = \frac{\text{Sale Price}}{(1 - \text{Rate of Decrease})} = \frac{\$256}{1 - 0.20} \).

Step 3: Solve for the Original Price

Now we perform the calculations. Subtracting 0.20 from 1 equals 0.8, then dividing \$256 by 0.8 gives us \$320. Therefore, the original price of the digital camera is \$320.

Key Concepts

Understanding Percent Decrease

Making sense of percent decrease can be quite intuitive with the right approach. Think of it this way: anytime a price drops, it means you are paying less than the original amount. To quantify that change, we use a percentage. A percent decrease tells us by what fraction the original price has been reduced.

For instance, let's say you hear that a store applies a 20% discount on a product. This means that for every dollar you used to pay, you now save 20 cents, effectively earning a discount that translates to 20 cents less per dollar. If you bought an item that was originally \(5, with a 20% decrease, you won't pay \)5; you pay 20% less, meaning the sale price would be \(4. To figure this out mathematically, you calculate 20% of \)5 and subtract that from the original price, which is the essence of understanding percent decrease.

Calculating the Sale Price

Once a percent decrease is established, one can determine the sale price of an item. The sale price is the final amount a customer pays after applying a discount to the original price. You can find the sale price by taking the original price and subtracting the value of the discount. On the other end, if you know the sale price and the percent decrease, you can work backwards to find the original price, which is a common exercise in algebraic problem solving.

For example, with a product that ends up being sold for \(256 after a 20% reduction, the calculation would involve identifying the sale price (in this case \)256), understanding that the 20% reduction means you are paying 80% of the original price, and then using that information to calculate what 100% of the price (the original price) would have been before the discount was applied.

Algebraic Problem-Solving

The process of algebraic problem-solving can be quite fascinating. It often involves forming expressions and equations based on a problem's context and then solving these equations to find unknown values. When dealing with sale price and original price, the algebra comes into play when we translate percentages and prices into an equation that we can solve.

With the camera example, by denoting the original price as 'P', we know after a 20% decrease, 80% of 'P' is equal to the sale price of \(256. The equation looks like this: 0.80 \times P = \)256. Algebraically, you'd solve for 'P' to find the original price. This process encapsulates algebraic problem-solving: turning real-world situations into algebraic expressions and manipulating these expressions to find solutions.