2002 AMC 10B Problem 25

P21-25~4 minPrint

Linear Equations & Word Problems Statistics & Data Systems of Equations Set up the equation/formula and compute; no special trick needed Substitute to simplify (u = x+1/x, shifting, scaling)

When $15$ is appended to a list of integers, the mean is increased by $2$ . When $1$ is appended to the enlarged list, the mean of the enlarged list is decreased by $1$ . How many integers were in the original list?

Show solution without answering

Key insight

Call the count n and the mean m, so the total is mn. Each mean change gives one equation: m + 2n = 13 and 2m + n = 14.

Solution

We do not know the numbers, but a mean only depends on two things: how many numbers there are and their total. So call the original count $n$ and the original mean $m$ . Then the total is $m \times n$ .

1. Add $15$ . Now there are $n + 1$ numbers, the total is $mn + 15$ , and the mean is $m + 2$ . Mean times count is the total:
$(m + 2)(n + 1) = mn + 15.$
Multiply out the left side: $mn + m + 2n + 2 = mn + 15$ . The $mn$ cancels, leaving
$m + 2n = 13.$
2. Then add $1$ . Now there are $n + 2$ numbers, the total is $mn + 16$ , and the mean is $1$ less than $m + 2$ , so it is $m + 1$ :
$(m + 1)(n + 2) = mn + 16.$
Multiply out: $mn + 2m + n + 2 = mn + 16$ , so
$2m + n = 14.$
3. Solve. From the first equation, $m = 13 - 2n$ . Put that into the second: $26 - 4n + n = 14$ , so $3n = 12$ and $n = 4$ . Then $m = 5$ .
4. Check with the list $5, 5, 5, 5$ (mean $5$ ). Add $15$ : total $35$ over $5$ numbers, mean $7$ , up by $2$ . Add $1$ : total $36$ over $6$ numbers, mean $6$ , down by $1$ . Both match.

The answer is $\boxed{\textbf{(A)}\ 4}$ .

Why this works

Every fact about a mean is really a fact about a total, because total equals mean times count. Using the count and the mean as the two unknowns made the big product $mn$ cancel from each equation, leaving two easy equations. Be careful which mean each change is measured from.

Alternative approach

Test the answer choices with a list of equal numbers. Try $n = 4$ with each number equal to $m$ . The first condition says $\frac{4m + 15}{5} = m + 2$ , so $4m + 15 = 5m + 10$ and $m = 5$ . Then the second condition: $\frac{20 + 15 + 1}{6} = 6$ , which is $1$ less than $7$ . It works, so the answer is $4$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let the original list have $n$ integers with mean $m$ , so its sum is $mn$ . The individual entries never matter, only this sum.

Append $15$ : the new list has $n + 1$ entries, sum $mn + 15$ , and mean $m + 2$ :
$mn + 15 = (m + 2)(n + 1) = mn + m + 2n + 2 \quad\Longrightarrow\quad m + 2n = 13.$

Append $1$ to that list: $n + 2$ entries, sum $mn + 16$ , and the mean drops by $1$ from $m + 2$ to $m + 1$ :
$mn + 16 = (m + 1)(n + 2) = mn + 2m + n + 2 \quad\Longrightarrow\quad 2m + n = 14.$

From the first equation $m = 13 - 2n$ ; substituting, $26 - 4n + n = 14$ , so $3n = 12$ and $n = 4$ (with $m = 5$ ).

Sanity check with four $5$ s: adding $15$ gives mean $\frac{35}{5} = 7$ (up $2$ ); adding $1$ gives $\frac{36}{6} = 6$ (down $1$ ).

The answer is $\boxed{\textbf{(A)}\ 4}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Prealgebra, Chapter 13: Data and Statistics
Intermediate Algebra, Chapter 20: Some Advanced Strategies (§20.2)