Question:A student has read 70 pages of a book, which represents 20% of the total number of pages in the...

1. TRANSLATE the problem information

• Given information:
  • 70 pages have been read
  • These 70 pages represent 20% of the total book
• What this tells us: We have a part (70) and need to find the whole (total pages)

2. INFER the mathematical approach

• Since we know a part and its percentage of the whole, we can find the whole by:
  • Method 1: Divide the part by the percentage (as a decimal)
  • Method 2: Convert percentage to a fraction and multiply by its reciprocal
• Key insight: If \(70\) pages \(= 20\%\) of total, then total \(= 70 \div 0.20\)

3. SIMPLIFY using Method 1 (decimal division)

• Convert \(20\%\) to decimal: \(20\% = 0.20 = 0.2\)
• Set up the division: \(70 \div 0.2 = 350\)

4. Verify using Method 2 (fraction approach)

• Convert \(20\%\) to fraction: \(20\% = \frac{20}{100} = \frac{1}{5}\)
• If \(70\) pages \(= \frac{1}{5}\) of total, then total \(= 70 \times 5 = 350\)

Answer: C. 350

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students confuse the operation needed and multiply instead of divide.

They think: "70 pages is 20% of the book, so the total must be \(70 \times 0.20 = 14\) pages." This reverses the relationship—they're finding 20% of 70 instead of finding what 70 represents 20% of.

This may lead them to select Choice A (14).

Second Most Common Error:

Poor SIMPLIFY execution: Students set up the problem correctly but make arithmetic errors in the division.

They might calculate \(70 \div 0.2\) incorrectly, perhaps getting confused about dividing by decimals or making computational mistakes.

This leads to confusion and guessing among the remaining answer choices.

The Bottom Line:

The key challenge is recognizing that when you have a part and need the whole, you divide the part BY the percentage, not multiply the part BY the percentage. Many students instinctively reach for multiplication when they see percentages, but this problem requires the inverse operation.