In a certain school, 1/5 of the students own an older computer, made more than three years ago; 1/30 of...

1. Translate the problem requirements: We need to understand that students can own older computers, newer computers, both types, or no computers. The fractions given represent overlapping groups, and we need to find the overlap between older and newer computer owners.
2. Set up the total student equation: Use the principle that all students must add up to 1 (the whole school), accounting for those who own older computers, newer computers, both, or neither.
3. Apply inclusion-exclusion thinking: Recognize that when we add the fraction owning older computers plus the fraction owning newer computers, we double-count those owning both types, so we need to subtract the overlap once.
4. Solve for the overlap algebraically: Set up the equation where (older) + (newer) - (both) + (neither) = 1, then solve for the "both" category.

Execution of Strategic Approach

1. Translate the problem requirements

Let's think about this step by step using everyday language. Imagine we're surveying all the students in a school about their computers.

We have four possible groups of students:

• Students who own ONLY an older computer (more than 3 years old)
• Students who own ONLY a newer computer (made in last 3 years)
• Students who own BOTH an older AND a newer computer
• Students who own NO computer at all

The problem tells us:

• \(\frac{1}{5}\) of students own an older computer (this includes those who might also own a newer one)
• \(\frac{1}{30}\) of students own a newer computer (this includes those who might also own an older one)
• \(\frac{7}{9}\) of students own no computer

We need to find what fraction own BOTH types. This is the overlap between the older computer group and newer computer group.

Process Skill: TRANSLATE - Converting the problem language into clear mathematical categories

2. Set up the total student equation

Here's the key insight: every single student in the school falls into exactly one of our four categories, and together they must account for 100% of the school.

In plain English: (Students with only older) + (Students with only newer) + (Students with both) + (Students with neither) = All students

Let's call the fraction who own both computers 'x'. Now we can express each group:

• Students with only older computers = (Total with older) - (Those with both) = \(\frac{1}{5} - x\)
• Students with only newer computers = (Total with newer) - (Those with both) = \(\frac{1}{30} - x\)
• Students with both computers = x
• Students with no computers = \(\frac{7}{9}\)

3. Apply inclusion-exclusion thinking

Now we use the fact that all students must add up to the whole school (which is 1 when expressed as a fraction):

(Students with only older) + (Students with only newer) + (Students with both) + (Students with neither) = 1

Substituting what we found:

\((\frac{1}{5} - x) + (\frac{1}{30} - x) + x + \frac{7}{9} = 1\)

Simplifying:

\(\frac{1}{5} - x + \frac{1}{30} - x + x + \frac{7}{9} = 1\)

\(\frac{1}{5} + \frac{1}{30} + \frac{7}{9} - x = 1\)

4. Solve for the overlap algebraically

To add these fractions, we need a common denominator. The LCD of 5, 30, and 9 is 90:

• \(\frac{1}{5} = \frac{18}{90}\)
• \(\frac{1}{30} = \frac{3}{90}\)
• \(\frac{7}{9} = \frac{70}{90}\)

So our equation becomes:

\(\frac{18}{90} + \frac{3}{90} + \frac{70}{90} - x = 1\)

\(\frac{91}{90} - x = 1\)

\(\frac{91}{90} - x = \frac{90}{90}\)

\(x = \frac{91}{90} - \frac{90}{90}\)

\(x = \frac{1}{90}\)

Process Skill: MANIPULATE - Working with fractions systematically to solve for the unknown

4. Final Answer

The fraction of students who own both an older and a newer computer is \(\frac{1}{90}\).

Let's verify: \(\frac{18}{90} + \frac{3}{90} + \frac{70}{90} - \frac{1}{90} = \frac{90}{90} = 1\) ✓

This matches answer choice B. \(\frac{1}{90}\)

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misinterpreting overlapping groups
Students often confuse "\(\frac{1}{5}\) of students own an older computer" with "\(\frac{1}{5}\) of students own ONLY an older computer." The given fractions include students who might own both types of computers, but students may treat these as mutually exclusive groups. This leads to setting up the wrong equation where they simply add all fractions without accounting for overlap.

Faltering Point 2: Incorrect categorization of student groups
Students may fail to recognize that there are exactly four distinct categories: (only older), (only newer), (both), and (neither). Instead, they might think there are only three groups or incorrectly assume that owning "both" is impossible. This fundamental misunderstanding prevents them from setting up the inclusion-exclusion principle correctly.

Errors while executing the approach

Faltering Point 1: Common denominator calculation errors
When finding the LCD of 5, 30, and 9, students often make arithmetic mistakes. They might incorrectly identify the LCD as 30 or 45 instead of 90, leading to wrong fraction conversions. For example, converting \(\frac{7}{9}\) incorrectly as \(\frac{70}{30} = \frac{7}{3}\) instead of \(\frac{70}{90}\).

Faltering Point 2: Sign errors in algebraic manipulation
When rearranging the equation "\(\frac{91}{90} - x = 1\)" to solve for x, students frequently make sign errors. They might incorrectly calculate \(x = \frac{91}{90} + \frac{90}{90} = \frac{181}{90}\) instead of \(x = \frac{91}{90} - \frac{90}{90} = \frac{1}{90}\), particularly when moving terms across the equals sign.

Errors while selecting the answer

Faltering Point 1: Selecting unreduced fractions or decimal equivalents
After calculating \(x = \frac{1}{90}\), some students might convert this to a decimal (≈0.011) and then try to match it with answer choices converted to decimals, potentially making rounding errors. Others might express their answer in a different equivalent form that doesn't exactly match the given choices, leading to confusion in selection.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a smart total number of students

We need a number that works well with the fractions \(\frac{1}{5}\), \(\frac{1}{30}\), and \(\frac{7}{9}\). The LCM of denominators 5, 30, and 9 is 90.

Let's say there are 90 students total in the school.

Step 2: Calculate concrete numbers for each group

• Students with older computers: \(\frac{1}{5} \times 90 = 18\) students
• Students with newer computers: \(\frac{1}{30} \times 90 = 3\) students
• Students with no computers: \(\frac{7}{9} \times 90 = 70\) students

Step 3: Apply the total constraint

All students must be accounted for. Using inclusion-exclusion principle:

(Students with older) + (Students with newer) - (Students with both) + (Students with neither) = Total students

18 + 3 - (Students with both) + 70 = 90

Step 4: Solve for students with both types

91 - (Students with both) = 90

Students with both = 91 - 90 = 1 student

Step 5: Convert back to fraction

Fraction with both computers = \(\frac{1}{90}\)

Verification:

• Students with only older computers: 18 - 1 = 17

• Students with only newer computers: 3 - 1 = 2

• Students with both: 1

• Students with neither: 70

• Total: 17 + 2 + 1 + 70 = 90 ✓

The answer is \(\frac{1}{90}\), which matches choice B.