AMC 10 · 2007 · #10
Grade 7 arithmeticThe Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is 20, the father is 48 years old, and the average age of the mother and children is 16. How many children are in the family?
Pick an answer.
AMC 10 2007 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 family is a mother, a father, and some children. The whole family's ages average $20$. The father is $48$. The mother and the children together average $16$. Find how many children there are.
Givens: The average age of the whole family is $20$; The father is $48$ years old; The average age of the mother together with the children is $16$; The family is a mother, a father, and an unknown number of children; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Unknowns: The number of children in the family
Understand
Restated: A family is a mother, a father, and some children. The whole family's ages average $20$. The father is $48$. The mother and the children together average $16$. Find how many children there are.
Givens: The average age of the whole family is $20$; The father is $48$ years old; The average age of the mother together with the children is $16$; The family is a mother, a father, and an unknown number of children; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra, #6 Guess and Check
The count of children is unknown, so Tool #4 (Introduce a Variable) names it $n$ and lets every other quantity be written in terms of $n$. The trick that makes an average useful is to turn it back into a total: total age $=$ average $\times$ number of people. Once both totals are written with $n$, Tool #13 (Convert to Algebra) turns the sentence 'father plus the rest equals the whole family' into one equation, which solves for $n$. Tool #6 (Guess and Check) then plugs the answer back to confirm the averages come out right.
Execute — Answer: E
6.EE.B.6 Step 1 Name the unknown, count the people
- Let $n$ be the number of children.
- The whole family is the mother, the father, and the $n$ children, so the family has $n + 2$ people.
- The mother-and-children group leaves out the father, so it has $n + 1$ people.
💡 Giving the unknown a name lets every headcount in the problem be written as a short expression in $n$.
6.SP.A.3 Step 2 Turn each average into a total
- An average is the total divided by the count, so the total is the average times the count.
- The whole family averages $20$ over $n + 2$ people, giving a family total of $20(n + 2)$.
- The mother-and-children group averages $16$ over $n + 1$ people, giving a total of $16(n + 1)$.
💡 Multiplying an average back by the number of people recovers the total you started from.
7.EE.B.4 Step 3 Split the family total two ways
- The whole family is just the father plus the mother-and-children group.
- So the family total also equals the father's age plus the mother-and-children total.
- Setting the two ways of writing the family total equal gives one equation in $n$.
💡 One whole total can be built either all at once or by adding its parts, and both builds must match.
6.EE.B.7 Step 4 Solve the equation
- Expand both sides: $20n + 40 = 48 + 16n + 16$, so $20n + 40 = 16n + 64$.
- Subtract $16n$ from both sides to get $4n + 40 = 64$, then subtract $40$ to get $4n = 24$.
- Divide by $4$: $n = 6$.
- There are $6$ children, which is choice (E).
💡 Collect the $n$ terms on one side and the plain numbers on the other, then undo the multiplication.
6.EE.B.6 Let $n$ be the number of children. The whole family is the mother, the father, a 6.SP.A.3 An average is the total divided by the count, so the total is the average times 7.EE.B.4 The whole family is just the father plus the mother-and-children group. So the f 6.EE.B.7 Expand both sides: $20n + 40 = 48 + 16n + 16$, so $20n + 40 = 16n + 64$. Subtrac Review
Reasonableness: Check the numbers with $n = 6$. The family has $8$ people, so the family total should be $20 \times 8 = 160$. The mother and $6$ children make $7$ people averaging $16$, a total of $16 \times 7 = 112$. Adding the father: $112 + 48 = 160$, which matches the family total exactly. The averages are consistent, so $6$ children is correct.
Alternative: Think in terms of how the father pulls the average up. Every member is compared to the family average of $20$. The father sits $48 - 20 = 28$ above it. The mother and children each average $16$, which is $20 - 16 = 4$ below the family average, so the $n + 1$ of them are a combined $4(n+1)$ below. For the average to balance, the father's surplus must cancel their shortfall: $28 = 4(n+1)$, giving $n + 1 = 7$ and $n = 6$.
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Letting $n$ stand for the number of children and writing the family size $n+2$ and the mother-and-children size $n+1$.)6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Turning each average back into a total with total $=$ average $\times$ count to get $20(n+2)$ and $16(n+1)$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Building the equation $20(n+2) = 48 + 16(n+1)$ from the fact that the family is the father plus the mother-and-children group.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Simplifying to $4n = 24$ and dividing to find $n = 6$.)
⭐ Turn every average back into a total by multiplying by the number of people, then match the whole against the sum of its parts to solve for the unknown.
⭐ Turn every average back into a total by multiplying by the number of people, then match the whole against the sum of its parts to solve for the unknown.
More like this
Same archetype — closest grade level first.