2025 AMC 8 Problem 14

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To help you master 2025 AMC 8 Problem 14, this page includes:

• The original AMC problem to try, plus the correct answer

• Clear, step-by-step solutions using different methods

• Key ideas, useful formulas, and takeaways you can apply to similar problems

• Related problems from past AMC exams, plus new practice problems that build the same skills

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2025 AMC8 Problem 14

Solution

Solution 1

Algebraic Casework Method

We can express the mean in terms of $N$ and set it equal to twice the median. By solving this equation for $N$, we can test the few possible values of the median based on where $N$ could be placed in the sorted list.

The median of a list with an even number of elements is the average of the two middle elements when sorted from least to greatest.

Step 1: Set up an expression for the new mean

First, find the sum of the original five numbers. Then, write an expression for the mean of the new six-number list, which includes $N$.

$$2+6+7+7+28=50$$

$$\text{New Mean}=\frac{50+N}{6}$$

Step 2: Relate $N$ directly to the median

Use the given condition that the mean is twice the median to solve for $N$ in terms of the median. This simple equation will let us evaluate potential medians quickly without dealing with fractions.

$$\frac{50+N}{6}=2\times \text{Median}$$

$$50+N=12\times \text{Median}$$

$$N=12\times \text{Median}-50$$

Step 3: Test the case where $N$ is 7 or larger

Assume $N\ge 7$. When we insert $N$ into the sorted list $2,6,7,7,28$, it will land after the first $7$. This means the 3rd and 4th numbers in the new list will definitely be $7$ and $7$. We calculate the median, use our equation to find $N$, and verify that the result satisfies our assumption $N\ge 7$.

$$\text{Median}=\frac{7+7}{2}=7$$

$$N=12(7)-50$$

$$N=84-50=34$$

$$34\ge 7\text{(Valid case)}$$

Step 4: Verify the other cases fail (for completeness)

If $N\le 6$, the 3rd and 4th numbers are $6$ and $7$. Substituting the resulting median into our equation yields a value for $N$ that contradicts the assumption. If $N$ is strictly between $6$ and $7$, the 3rd and 4th numbers are $N$ and $7$. Solving for $N$ in this case also yields a contradictory value.

$$\text{For}N\le 6:N=12(6.5)-50=28\text{(False, since}28\text{is not}\le 6)$$

$$\text{For}6<N<7:N=12\left(\frac{N+7}{2}\right)-50$$

$$N=6N-8⟹5N=8⟹N=1.6\text{(False, since}1.6\text{is not between}6\text{and}7)$$

Correct answer: E. $34$
The exact answer is $34$, which corresponds to choice E. To check, substituting $N=34$ into the list gives a sum of $84$, a mean of $84/6=14$, and a median of $7$, satisfying the condition $14=2\times 7$ perfectly.

Solution 2

Multiple-Choice Shortcut

In a multiple-choice test, the answer choices give us extra information. By noticing that all choices are $7$ or larger, we can completely avoid casework and jump straight to the correct median.

Test-Taking Strategy: Allow the bounds of the answer choices to restrict the possible cases you need to check.

Step 1: Analyze the answer choices

Every given answer choice is at least $7$. Since one of them must be the correct value for $N$, we can safely assume $N\ge 7$ without doing any casework algebra.

$$N\in \{7,14,20,28,34\}⟹N\ge 7$$

Step 2: Determine the median directly

Because $N\ge 7$, when we insert $N$ into the sorted list $2,6,7,7,28$, the first two numbers remain $2$ and $6$, and the next two numbers are guaranteed to be $7$ and $7$. Therefore, the 3rd and 4th numbers are both $7$.

$$\text{Sorted list starts with:}2,6,7,7,\dots$$

$$\text{Median}=\frac{7+7}{2}=7$$

Step 3: Calculate $N$

Since the mean is twice the median, the mean must be 14. We can multiply the mean by 6 to find the total sum of the new list, then subtract the sum of the original five numbers to find $N$.

$$\text{Mean}=2\times 7=14$$

$$\text{Total Sum}=14\times 6=84$$

$$N=84-50=34$$

Correct answer: E. $34$
The exact answer is $34$, matching choice E. The multiple-choice constraint safely restricted the median to $7$, and our backward check confirms this choice fits all conditions.

Similar Problems in Real Exams

The following five AMC 8 problems are closely related to AMC 8 2025 Question 14. These real AMC 8 problems test the same key concept—Median, Mode, Range & Order Statistics. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2023 AMC 8 Problem 20

2022 AMC 8 Problem 19

2019 AMC 8 Problem 10

2016 AMC 8 Problem 6

2015 AMC 8 Problem 5

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Mean (Average):

Valid for any finite set of numerical data. Used to express the new average in terms of the original sum and the unknown $N$.

• Median of an Even-Sized List:

The list must be ordered from least to greatest. Because the new list has 6 elements, the median is calculated by averaging the 3rd and 4th elements.

• Verifying Casework Assumptions:

Applies to any piecewise logic or casework in mathematics. In Method 1, assuming $N\le 6$ produced $N=28$, which contradicts the starting assumption and proves that specific case is impossible.

Key Takeaway: Remember This Strategy for Next Time

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