Loading...
A college admissions officer predicts that 20 percent of the students who are accepted will not attend the college. According to this prediction, how many students should be accepted to achieve a planned enrollment of \(\mathrm{x}\) students?
Let's start by understanding what's happening in plain English. The college wants exactly \(\mathrm{x}\) students to actually show up and enroll. However, they know from experience that not everyone who gets accepted will actually attend - specifically, \(20\%\) of accepted students won't come.
So the question is asking: if we want \(\mathrm{x}\) students to actually enroll, and we know \(20\%\) of our accepted students won't show up, how many students should we accept in the first place?
Think of it like planning a dinner party. If you want \(10\) people to actually come, but you know \(2\) out of every \(10\) invited guests usually don't show up, you'd need to invite more than \(10\) people to end up with exactly \(10\) attendees.
Process Skill: TRANSLATE - Converting the problem scenario into a clear mathematical relationship
Now let's think about what "\(20\%\) won't attend" really means for our planning.
If \(20\%\) of accepted students don't attend, that means \(80\%\) DO attend. In other words:
This is the key insight: we need to work with the \(80\%\) who will actually show up, not the \(20\%\) who won't.
Let's call the number of students we should accept "\(\mathrm{A}\)" (for Accepted).
In plain English: The number we accept × the percentage who attend = our target enrollment
So: \(\mathrm{A} \times 80\% = \mathrm{x}\) students
Or: \(\mathrm{A} \times 0.8 = \mathrm{x}\)
This equation captures exactly what we need: if we accept \(\mathrm{A}\) students, and \(80\%\) of them attend, we'll end up with exactly \(\mathrm{x}\) enrolled students.
Now we solve for \(\mathrm{A}\) (the number to accept):
Starting with: \(\mathrm{A} \times 0.8 = \mathrm{x}\)
To find \(\mathrm{A}\), we divide both sides by \(0.8\):
\(\mathrm{A} = \mathrm{x} \div 0.8\)
Let's simplify this division:
\(\mathrm{A} = \mathrm{x} \div 0.8 = \mathrm{x} \div \frac{8}{10} = \mathrm{x} \times \frac{10}{8} = \mathrm{x} \times \frac{5}{4} = 1.25\mathrm{x}\)
So we need to accept \(1.25\mathrm{x}\) students.
We should accept \(1.25\mathrm{x}\) students to achieve a planned enrollment of \(\mathrm{x}\) students.
Let's verify this makes sense: If we accept \(1.25\mathrm{x}\) students and \(80\%\) attend, then:
\(1.25\mathrm{x} \times 0.8 = \mathrm{x}\) students will enroll ✓
The answer is D. \(1.25\mathrm{x}\)
Students often focus on the "\(20\%\) won't attend" and try to work directly with this percentage instead of recognizing they need to work with the \(80\%\) who WILL attend. They might think: "If \(20\%\) don't come, I need to accept \(\mathrm{x} + 20\%\) more" leading to incorrect setups like \(\mathrm{A} = \mathrm{x} + 0.2\mathrm{x} = 1.2\mathrm{x}\).
2. Setting up the equation backwardsStudents may confuse what they're solving for and set up the relationship as: "\(\mathrm{x}\) students accepted × attendance rate = number who attend" instead of "\(\mathrm{A}\) students accepted × attendance rate = \(\mathrm{x}\) target enrollment." This leads them to solve for the wrong variable.
3. Misunderstanding the constraintStudents might not clearly identify that \(\mathrm{x}\) represents the final enrollment target (the constraint they must meet exactly), instead treating it as just a reference number. This confusion makes them unsure whether they're solving for total acceptances or something else.
Students often make mistakes when converting \(20\%\) to \(80\%\), or when working with \(0.8\). Common errors include using \(0.2\) instead of \(0.8\) in their calculations, or incorrectly converting between decimal and fraction forms.
2. Division errors with decimalsWhen dividing by \(0.8\), students frequently make computational mistakes. The step \(\mathrm{A} = \mathrm{x} \div 0.8 = \mathrm{x} \times \frac{5}{4} = 1.25\mathrm{x}\) involves multiple conversions that can trip up students, especially the fraction manipulation \(\mathrm{x} \div \frac{8}{10} = \mathrm{x} \times \frac{10}{8}\).
No likely faltering points - once students correctly calculate \(1.25\mathrm{x}\), the answer choice D clearly matches their result.
Step 1: Choose a convenient target enrollment
Let's say we want \(\mathrm{x} = 100\) students to actually enroll at the college. This gives us a concrete number to work with.
Step 2: Understand the attendance pattern
We know that \(20\%\) of accepted students will not attend, which means \(80\%\) of accepted students will attend.
Step 3: Set up the relationship
If we accept some number of students, and \(80\%\) of them attend, we need that \(80\%\) to equal our target of \(100\) students.
Let \(\mathrm{A}\) = number of students we should accept
Then: \(80\%\) of \(\mathrm{A} = 100\)
\(0.8\mathrm{A} = 100\)
Step 4: Solve for acceptances needed
\(\mathrm{A} = 100 \div 0.8 = 125\)
Step 5: Express as a multiple of target enrollment
We need to accept \(125\) students to get \(100\) enrolled students.
\(125 = 1.25 \times 100\)
So we need to accept \(1.25\mathrm{x}\) students to achieve enrollment of \(\mathrm{x}\) students.
Step 6: Verify our answer
If we accept \(125\) students and \(20\%\) don't attend:
Students who don't attend: \(20\% \times 125 = 25\)
Students who do attend: \(125 - 25 = 100\) ✓
The answer is D. \(1.25\mathrm{x}\)