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Two fractions are inserted between 1/4 and 1/2 so that the difference between any two successive fractions is the same....

GMAT Algebra : (ALG) Questions

Source: Official Guide
Algebra
Sequences
MEDIUM
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Two fractions are inserted between \(\frac{1}{4}\) and \(\frac{1}{2}\) so that the difference between any two successive fractions is the same. Find the sum of the four fractions.

A
\(\frac{18}{12}\)
B
\(\frac{16}{12}\)
C
\(\frac{5}{2}\)
D
\(\frac{9}{19}\)
E
\(\frac{24}{32}\)
Solution
  1. Translate the problem requirements: We need to find two fractions that fit between \(\frac{1}{4}\) and \(\frac{1}{2}\) such that all four fractions form an arithmetic sequence (equal differences), then sum all four fractions.
  2. Set up the arithmetic sequence structure: Recognize that we have four fractions in arithmetic progression: \(\frac{1}{4}\), (first inserted fraction), (second inserted fraction), \(\frac{1}{2}\).
  3. Find the common difference: Since the fractions are equally spaced, calculate the common difference by dividing the total gap by the number of intervals.
  4. Calculate the sum using arithmetic sequence properties: Apply the formula for sum of arithmetic sequence or simply add the four fractions directly.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking us to do in plain English.

We start with two fractions: \(\frac{1}{4}\) and \(\frac{1}{2}\). Think of these as points on a number line. We need to place two more fractions between them so that the spacing between any two consecutive fractions is exactly the same.

Imagine you're placing four friends in a line with equal spacing between each pair of neighbors. If the first friend is at position \(\frac{1}{4}\) and the last friend is at position \(\frac{1}{2}\), where do the middle two friends stand?

This creates what we call an arithmetic sequence - a sequence where each term differs from the previous one by the same amount.

Our four fractions will look like: \(\frac{1}{4}\), ?, ?, \(\frac{1}{2}\)

Then we need to find the sum of all four fractions.

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

2. Set up the arithmetic sequence structure

Now let's think about what "equal differences" means with fractions.

If we call our common difference 'd', then our four fractions are:

  • First fraction: \(\frac{1}{4}\)
  • Second fraction: \(\frac{1}{4} + d\)
  • Third fraction: \(\frac{1}{4} + 2d\)
  • Fourth fraction: \(\frac{1}{4} + 3d\)

But we know the fourth fraction must equal \(\frac{1}{2}\), so:
\(\frac{1}{4} + 3d = \frac{1}{2}\)

This gives us a simple equation to find d!

3. Find the common difference

Let's solve for d using our equation: \(\frac{1}{4} + 3d = \frac{1}{2}\)

First, let's isolate 3d:
\(3d = \frac{1}{2} - \frac{1}{4}\)

To subtract these fractions, we need a common denominator. The smallest common denominator for 2 and 4 is 4:
\(\frac{1}{2} = \frac{2}{4}\)

So: \(3d = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}\)

Therefore: \(d = \frac{1}{4} ÷ 3 = \frac{1}{12}\)

This makes perfect sense! The total gap from \(\frac{1}{4}\) to \(\frac{1}{2}\) is \(\frac{1}{4}\), and we're dividing it into 3 equal parts (since we have 3 intervals between 4 fractions).

Now we can find our four fractions:

  • First: \(\frac{1}{4} = \frac{3}{12}\)
  • Second: \(\frac{1}{4} + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12}\)
  • Third: \(\frac{1}{4} + 2(\frac{1}{12}) = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}\)
  • Fourth: \(\frac{1}{4} + 3(\frac{1}{12}) = \frac{3}{12} + \frac{3}{12} = \frac{6}{12} = \frac{1}{2}\)

4. Calculate the sum using arithmetic sequence properties

Now we simply add up our four fractions. Since they all have the same denominator of 12, this is straightforward:

Sum = \(\frac{3}{12} + \frac{4}{12} + \frac{5}{12} + \frac{6}{12}\)
Sum = \(\frac{3 + 4 + 5 + 6}{12}\)
Sum = \(\frac{18}{12}\)

Let's verify this makes sense: we're adding four fractions that are evenly spaced between \(\frac{1}{4}\) and \(\frac{1}{2}\). The average of these four numbers should be somewhere between \(\frac{1}{4}\) and \(\frac{1}{2}\).

Average = \(\frac{18}{12} ÷ 4 = \frac{18}{48} = \frac{3}{8}\)

Since \(\frac{1}{4} = \frac{2}{8}\) and \(\frac{1}{2} = \frac{4}{8}\), an average of \(\frac{3}{8}\) is perfectly reasonable.

Final Answer

The sum of the four fractions is \(\frac{18}{12}\).

This matches answer choice A exactly.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding "equal differences" between fractions
Students often struggle to translate "the difference between any two successive fractions is the same" into the arithmetic sequence concept. They might think this means the fractions should be equally spaced as simple ratios (like \(\frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}\)) rather than understanding that we need equal numerical differences between consecutive terms.

2. Incorrectly setting up the sequence boundaries
Some students may misinterpret the problem and think they need to find fractions that come before \(\frac{1}{4}\) and after \(\frac{1}{2}\), or they might include additional fractions beyond the specified four. The problem specifically states "two fractions are inserted between \(\frac{1}{4}\) and \(\frac{1}{2}\)" meaning we have exactly four fractions total: \(\frac{1}{4}\), two unknown fractions, and \(\frac{1}{2}\).

3. Confusing the number of intervals vs. number of fractions
A critical conceptual error occurs when students think there are 4 equal parts instead of 3. Since we have 4 fractions total, there are only 3 intervals between them. This leads to incorrectly calculating the common difference as \(\frac{1}{4} ÷ 4 = \frac{1}{16}\) instead of the correct \(\frac{1}{4} ÷ 3 = \frac{1}{12}\).

Errors while executing the approach

1. Arithmetic errors when finding common denominators
When calculating \(\frac{1}{2} - \frac{1}{4}\), students frequently make errors in finding the common denominator or in the subtraction itself. They might incorrectly compute this as \(\frac{1}{4}\) instead of \(\frac{1}{4}\), or struggle with converting \(\frac{1}{2}\) to \(\frac{2}{4}\) before performing the subtraction.

2. Incorrect fraction arithmetic when building the sequence
Even with the correct common difference \(d = \frac{1}{12}\), students often make computational errors when adding fractions. For example, when calculating \(\frac{1}{4} + \frac{1}{12}\), they might forget to convert \(\frac{1}{4}\) to \(\frac{3}{12}\) first, or they might add denominators instead of keeping the common denominator of 12.

3. Errors in converting between equivalent fraction forms
Students may correctly find the individual fractions but make mistakes when converting them all to the same denominator for addition. They might work with mixed denominators (some in twelfths, some in fourths) or incorrectly convert between equivalent forms like \(\frac{1}{2}\) and \(\frac{6}{12}\).

Errors while selecting the answer

1. Selecting a simplified form when the answer requires a specific form
Students might correctly calculate the sum as \(\frac{18}{12}\) but then automatically simplify it to \(\frac{3}{2}\) or 1.5, not noticing that \(\frac{18}{12}\) appears as option A. They need to recognize that the answer choices are given in specific forms and match their result accordingly.

2. Misreading similar-looking answer choices
The answer choices include fractions that could result from common errors: \(\frac{16}{12}\) (choice B) might result from calculation mistakes, and students might confuse this with the correct \(\frac{18}{12}\). Similarly, they might select \(\frac{5}{2}\) (choice C) thinking it's equivalent to their calculated result without proper verification.

Answer Choices Explained
A
\(\frac{18}{12}\)
B
\(\frac{16}{12}\)
C
\(\frac{5}{2}\)
D
\(\frac{9}{19}\)
E
\(\frac{24}{32}\)
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