Last year the price of a blouse in February was 15% less than the original price of $60 and the...

1. Translate the problem requirements: We need to track the blouse price through two changes: first a \(15\%\) decrease from \(\$60\) in February, then a \(6\%\) increase from February's price in March. Finally, compare March's price to the original \(\$60\) to find the percent difference.
2. Calculate February's reduced price: Apply the \(15\%\) decrease to the original \(\$60\) price to find February's actual price.
3. Calculate March's increased price: Apply the \(6\%\) increase to February's price to find March's final price.
4. Compare March price to original price: Determine what percent March's price is relative to the original \(\$60\), then identify if it's greater or less than the original.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening to this blouse price step by step. We start with a blouse that originally costs \(\$60\). In February, the store reduces the price by \(15\%\) - imagine this as a sale where customers save 15 cents on every dollar. Then in March, they raise the price from February's sale price by \(6\%\) - think of this as the store increasing the current price, not going back to the original.

Our job is to figure out: after these two price changes, is the March price higher or lower than the original \(\$60\), and by what percentage?

Process Skill: TRANSLATE - Converting the sequence of percentage changes into a clear mathematical path

2. Calculate February's reduced price

Let's find February's price using everyday reasoning. If the price drops by \(15\%\), then customers pay \(85\%\) of the original price (since \(100\% - 15\% = 85\%\)).

February's price = \(85\%\) of \(\$60\)
February's price = \(0.85 \times \$60 = \$51\)

So in February, the blouse costs \(\$51\). This makes sense - a \(15\%\) discount on a \(\$60\) item should give us a price in the low \(\$50\)s.

3. Calculate March's increased price

Now in March, the store increases February's price of \(\$51\) by \(6\%\). This means customers will pay \(106\%\) of February's price (since \(100\% + 6\% = 106\%\)).

March's price = \(106\%\) of February's price
March's price = \(106\%\) of \(\$51\)
March's price = \(1.06 \times \$51 = \$54.06\)

So in March, the blouse costs \(\$54.06\).

4. Compare March price to original price

Now we need to compare March's price of \(\$54.06\) to the original price of \(\$60\).

The difference = \(\$54.06 - \$60 = -\$5.94\)

Since this is negative, March's price is less than the original price. To find the percentage:

Percentage difference = \(\frac{\text{Difference}}{\text{Original price}} \times 100\%\)
Percentage difference = \(\frac{-\$5.94}{\$60} \times 100\%\)
Percentage difference = \(-0.099 \times 100\% = -9.9\%\)

The negative sign tells us the March price is \(9.9\%\) less than the original price.

Final Answer

The price of the blouse in March was \(9.9\%\) less than the original price.

This matches answer choice (E) \(9.9\%\) less.

We can verify this makes sense: even though there was a \(6\%\) increase in March, it wasn't enough to overcome the significant \(15\%\) decrease that happened in February.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the sequence of percentage changes: Students often confuse which price serves as the base for each calculation. They might think the March increase of \(6\%\) should be applied to the original \(\$60\) price instead of February's reduced price of \(\$51\). This leads to calculating March price as \(\$60 \times 1.06 = \$63.60\) instead of the correct \(\$51 \times 1.06 = \$54.06\).

2. Confusing 'percent greater/less than' with simple percentage calculation: Students may not realize they need to compare the final March price back to the original price. They might stop after calculating March's price or only focus on the individual percentage changes (\(15\%\) and \(6\%\)) without determining the overall effect relative to the starting point.

Errors while executing the approach

1. Arithmetic errors in percentage calculations: Students frequently make calculation mistakes, such as computing \(0.85 \times \$60\) incorrectly or miscalculating \(1.06 \times \$51\). These small errors compound and lead to wrong final percentages.

2. Sign confusion in the final percentage calculation: When calculating \((\$54.06 - \$60) \div \$60\), students may forget that the negative result (\(-\$5.94 \div \$60 = -0.099\)) indicates the March price is LESS than the original, not greater. They might report \(9.9\%\) greater instead of \(9.9\%\) less.

Errors while selecting the answer

1. Selecting the wrong direction (greater vs. less): Even if students calculate the \(9.9\%\) correctly, they may select answer choice (C) '\(9.0\%\) greater' instead of (E) '\(9.9\%\) less' because they lose track of whether the final price is above or below the original price. The presence of both 'greater' and 'less' options in the answer choices makes this directional error particularly dangerous.