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Enrollment in City College in 1980 was \(83\frac{1}{3}\) percent of enrollment in 1990. What was the percent increase in the college's enrollment from 1980 to 1990?
Let's break down what the problem is telling us in everyday language:
• We know that enrollment in 1980 was \(83\frac{1}{3}\%\) of enrollment in 1990
• We need to find the percent increase from 1980 to 1990
Think of it this way: if 1980 enrollment was \(83\frac{1}{3}\%\) of 1990 enrollment, that means 1980 had less students than 1990. We want to know by what percentage the enrollment grew from the smaller number (1980) to the larger number (1990).
Process Skill: TRANSLATE - Converting the relationship "1980 was \(83\frac{1}{3}\%\) of 1990" into a clear mathematical understanding
To make our calculations simple, let's choose a convenient number for 1990 enrollment. Since we're dealing with \(83\frac{1}{3}\%\), let's think about what this fraction means:
\(83\frac{1}{3}\% = \frac{83\frac{1}{3}}{100} = \frac{\frac{250}{3}}{100} = \frac{250}{300} = \frac{5}{6}\)
So 1980 enrollment was \(\frac{5}{6}\) of 1990 enrollment.
To make this easy to work with, let's say 1990 enrollment was 6 students. This makes our fraction \(\frac{5}{6}\) very simple to calculate.
Using our convenient numbers:
• 1990 enrollment = 6 students
• 1980 enrollment = \(\frac{5}{6} \times 6 = 5\) students
Let's verify: Is 5 equal to \(83\frac{1}{3}\%\) of 6?
\(5 \div 6 = \frac{5}{6} = 83\frac{1}{3}\%\) ✓
So we went from 5 students in 1980 to 6 students in 1990.
Now we can find the percent increase from 1980 to 1990:
Percent increase = \(\frac{\mathrm{New\ Value} - \mathrm{Old\ Value}}{\mathrm{Old\ Value}} \times 100\%\)
In plain English: We're comparing how much the enrollment grew (from 5 to 6, which is an increase of 1) to the original enrollment (5).
Percent increase = \(\frac{6 - 5}{5} \times 100\%\)
Percent increase = \(\frac{1}{5} \times 100\%\)
Percent increase = \(\frac{1}{5} \times 100\%\)
Percent increase = \(20\%\)
The percent increase in enrollment from 1980 to 1990 was 20%.
This matches answer choice C. 20%.
To double-check our logic: if enrollment grew by 20% from 1980 to 1990, then 1980 enrollment should be about 83% of 1990 enrollment, which matches our given information.
1. Confusing the direction of percentage relationship: Students may misinterpret "1980 was \(83\frac{1}{3}\%\) of 1990" and instead think "1990 was \(83\frac{1}{3}\%\) of 1980." This reversal leads them to believe that 1990 had fewer students than 1980, making them calculate a percent decrease instead of a percent increase.
2. Misunderstanding what baseline to use for percent increase: Students might think they need to find what percent 1990 is of 1980 (which would be 120%) rather than understanding that percent increase specifically compares the change to the original (1980) value.
3. Getting confused between "percent of" and "percent increase": Students may think the answer is simply the difference between 100% and \(83\frac{1}{3}\%\), which gives \(16\frac{2}{3}\%\), not realizing this represents the wrong type of percentage calculation.
1. Arithmetic errors with mixed numbers: Students often struggle converting \(83\frac{1}{3}\%\) to the fraction \(\frac{5}{6}\), or they may incorrectly convert it (like making it \(\frac{83}{300}\) instead of \(\frac{250}{300}\)), leading to wrong numerical values throughout their calculation.
2. Using the wrong values in the percent increase formula: Even with the right setup, students might mistakenly put 1990 enrollment as the denominator instead of 1980 enrollment in the percent increase formula, calculating \(\frac{6-5}{6}\) instead of \(\frac{6-5}{5}\).
3. Calculation errors with fractions: When calculating \(\frac{1}{5} \times 100\%\), students might make basic arithmetic mistakes, especially if they're working with their own chosen numbers instead of the convenient values used in the solution.
1. Selecting the complement percentage: Students who calculated that 1980 was \(16\frac{2}{3}\%\) less than 1990 might incorrectly select answer choice B (\(16\frac{2}{3}\%\)) instead of recognizing they need the percent increase from 1980's perspective.
2. Picking the percentage relationship given in the problem: Students might be tempted to select answer choice E (\(183\frac{1}{3}\%\)) because it relates to the \(83\frac{1}{3}\%\) mentioned in the problem, not realizing this doesn't answer the specific question asked.
This problem is perfect for the smart numbers technique because we need to work with the relationship between two enrollment values.
Step 1: Choose a smart number for 1990 enrollment
Since we need to calculate \(83\frac{1}{3}\%\) of the 1990 enrollment, let's choose a number that makes this calculation clean. \(83\frac{1}{3}\% = \frac{83\frac{1}{3}}{100} = \frac{\frac{250}{3}}{100} = \frac{250}{300} = \frac{5}{6}\).
To make \(\frac{5}{6}\) easy to work with, let's choose 1990 enrollment = 6,000 students (any multiple of 6 works, but 6,000 is convenient).
Step 2: Calculate 1980 enrollment
1980 enrollment = \(83\frac{1}{3}\%\) of 1990 enrollment
1980 enrollment = \(\frac{5}{6} \times 6,000 = 5,000\) students
Step 3: Calculate the percent increase from 1980 to 1990
Increase = 1990 enrollment - 1980 enrollment = 6,000 - 5,000 = 1,000 students
Percent increase = \(\frac{\mathrm{Increase}}{\mathrm{Original}} \times 100\%\)
Percent increase = \(\frac{1,000}{5,000} \times 100\% = \frac{1}{5} \times 100\% = 20\%\)
Verification: Our smart number choice of 6,000 was logical because it's divisible by 6, making the \(\frac{5}{6}\) calculation clean and avoiding messy fractions throughout the problem.