Question

A student claims that if a price is now \(220 \%\) more than it was before, then it is \(320 \%\) of what it was before, and what it was before is \(31.25 \%\) of what it is now. Do you agree? Explain your answer.

Short answer

Yes, the student's statements are correct. The price which is now \(220 \%\) more than it was before is indeed \(320 \%\) of the original price, and the original price is \(31.25 \%\) of the new price.

Step-by-step solution

Analyze the First Statement

The first statement is that the price is now \(220 \%\) more than it was before. When a percentage is stated as more than something, we understand it as the original value plus an additional amount. So, a \(220 \%\) increase implies that the new value is \(220 \%\) of the original plus the original value itself. In percentages, this translates to \(100 \% + 220 \% = 320 \%\). So, the first statement is correct.

Analyze the Second Statement

The second statement is a restatement of the first, that the price is \(320 \%\) of what it was before. As we determined in the first step, this is also correct.

Analyze the Third Statement

The third statement is that what it was before is \(31.25 \%\) of what it is now. To verify this, we need to determine what percentage the original price is of the new price. Given that the new price is \(320 \%\) of the old price, the old price is \(100 \% / 320 \% = 31.25 \%\) of the new price. Therefore, this statement is also correct.

Key concepts

Percentages in Algebra

Understanding percentages in algebra involves recognizing that percentages represent a proportion of a whole. Typically, 100% is considered the full value, and any percentage then represents a part of that whole. For example, saying that a quantity is 50% of another is equivalent to saying it is half of it. Algebra uses variables to represent unknown quantities, and percentages can be tied to variables in equations, which can be solved to find unknown values.

When working with percentages in algebraic contexts, it becomes essential to being comfortable with converting percentages to decimal form to perform mathematical operations easily. For instance, 50% becomes 0.5, because 50 divided by 100 equals 0.5. This conversion can be helpful in solving percentages problems that involve variables. It's also important to note when percentages are added, like in the given exercise, each percent change refers to portions of the original whole, not a compounded figure.

Solving Percentage Problems

Solving percentage problems necessitates a structured approach. First, define what the 'whole' or 'original value' you're comparing to is. In many cases, it might be the original price, initial quantity, or starting value. After the basis is defined, next, identify the 'part' to compare with the 'whole'.

• Convert the percentage to a decimal by dividing by 100.
• Multiply the decimal by the 'whole' to find the 'part'.
• Add or subtract the 'part' from the 'whole' depending on whether it's an increase or decrease.

If you are facing a complex problem that involves more than one percentage change, follow each step carefully and ensure that you take the percentages of the correct values. If you're dealing with percentage increase, always remember that you're adding the calculated 'part' to the original value to find the new total.

Understanding Percentage Increase

A percentage increase indicates how much a value has grown relative to its original value. To determine the percentage increase, calculate the difference between the new value and the original value, then compare this difference to the original value using a percentage.

For instance, if an item costs \(100 and the price increases by \)20, the percentage increase is calculated by dividing the increase (\(20) by the original amount (\)100), resulting in 0.2 or 20%.

In the context of the exercise provided, when it is stated that there is a '220% increase', it means the new value is 220% plus the original 100% of the initial figure, totalling to 320%. This concept is crucial as it prevents misunderstanding when calculating final values after an increase. Mistakes often occur if the increase is simply added as an exact percentage without considering that the original value is inherently an implied 100%.