AMC 10 · 2002 · #25
Grade 8 arithmeticWhen 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?
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A list of integers has some mean. Adding $15$ to the list raises the mean by $2$. Adding $1$ to that larger list lowers its mean by $1$. Find how many integers the original list held.
Givens: Appending $15$ to the original list increases the mean by $2$; Appending $1$ to the enlarged list decreases its mean by $1$; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Unknowns: The number of integers in the original list
Understand
Restated: A list of integers has some mean. Adding $15$ to the list raises the mean by $2$. Adding $1$ to that larger list lowers its mean by $1$. Find how many integers the original list held.
Givens: Appending $15$ to the original list increases the mean by $2$; Appending $1$ to the enlarged list decreases its mean by $1$; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra, #6 Guess and Check
The individual integers are unknown and never needed, but the count and the mean are exactly what the questions talk about. Tool #4 (Introduce a Variable) names the count $n$ and the mean $m$, which pins down the sum as $nm$. Tool #13 (Convert to Algebra) then turns each English sentence about the mean into a linear equation, giving a small $2\times 2$ system that solves cleanly. Tool #6 (Guess and Check) is held in reserve for the review: the answer choices are small whole numbers, so the winning count can be replayed on a concrete list to confirm both conditions.
Execute — Answer: A
6.SP.B.5 Step 1 Name the count and the mean
- Let $n$ be how many integers are in the original list and let $m$ be its mean.
- Because the mean is the sum divided by the count, the sum of all the integers is $nm$.
- These two letters are all that the mean questions actually depend on, so the individual integers can stay unknown.
💡 The mean packs the whole list into one number, so count times mean recovers the total.
6.EE.B.6 Step 2 Turn the first change into an equation
- Appending $15$ makes the count $n+1$ and the sum $nm+15$, and the new mean is the old mean plus $2$.
- Writing that as $\dfrac{nm+15}{n+1}=m+2$ and clearing the denominator gives $nm+15=(m+2)(n+1)=nm+m+2n+2$.
- The $nm$ cancels on both sides, leaving $15=m+2n+2$, that is $m+2n=13$.
💡 Multiplying the new mean by the new count must return the new sum, and the shared $nm$ drops out.
6.EE.B.6 Step 3 Turn the second change into an equation
- Now append $1$ to the enlarged list.
- Its count is $n+2$ and its sum is $nm+15+1=nm+16$.
- The new mean is the enlarged mean $m+2$ dropped by $1$, so it equals $m+1$.
- Then $\dfrac{nm+16}{n+2}=m+1$ gives $nm+16=(m+1)(n+2)=nm+2m+n+2$.
- Again $nm$ cancels, leaving $16=2m+n+2$, that is $2m+n=14$.
💡 The second mean is measured from the raised mean, so the drop is one below $m+2$, not below $m$.
8.EE.C.8 Step 4 Solve the two equations together
- The system is $m+2n=13$ and $2m+n=14$.
- From the first, $m=13-2n$.
- Substituting into the second: $2(13-2n)+n=14$, so $26-3n=14$, giving $3n=12$ and $n=4$ (and then $m=5$).
- The original list had $4$ integers, which is choice (A).
💡 Two independent facts about the same $n$ and $m$ lock in exactly one solution.
6.SP.B.5 Let $n$ be how many integers are in the original list and let $m$ be its mean. B 6.EE.B.6 Appending $15$ makes the count $n+1$ and the sum $nm+15$, and the new mean is th 6.EE.B.6 Now append $1$ to the enlarged list. Its count is $n+2$ and its sum is $nm+15+1= 8.EE.C.8 The system is $m+2n=13$ and $2m+n=14$. From the first, $m=13-2n$. Substituting i Review
Reasonableness: Replay the answer on a concrete list of four $5$s, whose mean is $5$ and sum is $20$. Appending $15$ makes the sum $35$ over $5$ numbers, a mean of $7$ — up by $2$, as required. Appending $1$ makes the sum $36$ over $6$ numbers, a mean of $6$ — down by $1$ from $7$, as required. Both conditions hold, so $n=4$ is consistent.
Alternative: Guess and check straight from the choices. Try $n=4$ with all values equal to the mean $m$: the equations $m+2n=13$ and $2m+n=14$ both need $m=5$, and $5+8=13$ and $10+4=14$ both check, so $n=4$ works while the other choices fail the pair of equations.
CCSS standards used (min grade 8)
6.SP.B.5Summarize numerical data sets by reporting number of observations and measures of center (Using mean = sum ÷ count to write the total of the list as $nm$.)6.EE.B.6Use variables to represent numbers and write expressions when solving a problem (Translating each sentence about the changing mean into an equation in $n$ and $m$.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving $m+2n=13$ and $2m+n=14$ together to find $n=4$.)
⭐ When only counts and means are given, name the count and the mean, write the total as count times mean, and each sentence about the mean becomes one clean equation.
⭐ When only counts and means are given, name the count and the mean, write the total as count times mean, and each sentence about the mean becomes one clean equation.
More like this
Same archetype — closest grade level first.