AMC 10 · 2003 · #13
Grade 7 algebraThe sum of three numbers is 20. The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?
Pick an answer.
AMC 10 2003 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: Three numbers add up to $20$. The first equals four times the sum of the other two, and the second equals seven times the third. Find the product of the three numbers.
Givens: The three numbers add up to $20$; First number $=$ four times (second $+$ third); Second number $=$ seven times the third number; Answer choices: (A) $28$, (B) $40$, (C) $100$, (D) $400$, (E) $800$
Unknowns: The value of each of the three numbers; The product (first) $\times$ (second) $\times$ (third)
Understand
Restated: Three numbers add up to $20$. The first equals four times the sum of the other two, and the second equals seven times the third. Find the product of the three numbers.
Givens: The three numbers add up to $20$; First number $=$ four times (second $+$ third); Second number $=$ seven times the third number; Answer choices: (A) $28$, (B) $40$, (C) $100$, (D) $400$, (E) $800$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #13 Convert to Algebra
Each sentence in the problem describes a number in terms of the others, so Tool #4 (Introduce a Variable) lets us name the unknowns and turn each sentence into an equation. The smart move is to not solve for all three at once. Tool #7 (Identify Subproblems) splits the work: first treat 'the other two' as a single chunk to pin down the first number, then break that chunk apart to get the second and third. Tool #13 (Convert to Algebra) carries each English condition into a clean equation we can solve one step at a time.
Execute — Answer: A
6.EE.B.6 Step 1 Name the numbers and write the conditions
- Call the three numbers $a$ (first), $b$ (second), and $c$ (third).
- Translate each sentence directly: 'they add to $20$' is $a+b+c=20$; 'the first is four times the sum of the other two' is $a=4(b+c)$; 'the second is seven times the third' is $b=7c$.
- That is three facts for three unknowns, which is exactly enough to lock the numbers down.
💡 Give each number a letter and every sentence becomes an equation you can work with.
6.EE.B.7 Step 2 Treat the other two as one chunk to find the first
- Let the chunk be $b+c$, the sum of the other two.
- The first fact says $a+(b+c)=20$, and the second says $a=4(b+c)$.
- Substitute $a$: $4(b+c)+(b+c)=20$, which is $5(b+c)=20$, so $b+c=4$.
- Then $a=4\times 4=16$.
- The whole $20$ splits into four equal chunks, and the first number grabs four of them.
💡 If the first is four times the rest, the total is five equal parts, so the rest is one part and the first is four.
7.EE.B.4 Step 3 Break the chunk apart for the second and third
- Now use $b+c=4$ together with $b=7c$.
- Replace $b$: $7c+c=4$, which is $8c=4$, so $c=\tfrac{1}{2}$.
- Then $b=7c=\tfrac{7}{2}$.
- Check the sum: $16+3.5+0.5=20$, and $3.5$ is indeed seven times $0.5$, so all three conditions hold.
💡 Splitting a total of $4$ into a $7$-to-$1$ ratio makes eight equal shares, so one share is $\tfrac12$.
6.NS.B.3 Step 4 Multiply the three numbers
- The product is $a\times b\times c = 16 \times 3.5 \times 0.5$.
- Take it in pieces: $3.5 \times 0.5 = 1.75$, and $16 \times 1.75 = 28$.
- So the product of all three numbers is $28$, which is choice (A).
💡 Pair the two decimals first so the last multiplication is just $16$ times a tidy number.
6.EE.B.6 Call the three numbers $a$ (first), $b$ (second), and $c$ (third). Translate eac 6.EE.B.7 Let the chunk be $b+c$, the sum of the other two. The first fact says $a+(b+c)=2 7.EE.B.4 Now use $b+c=4$ together with $b=7c$. Replace $b$: $7c+c=4$, which is $8c=4$, so 6.NS.B.3 The product is $a\times b\times c = 16 \times 3.5 \times 0.5$. Take it in pieces Review
Reasonableness: The three numbers $16$, $3.5$, $0.5$ pass every condition: they sum to $20$, the first ($16$) is four times the other two combined ($4\times 4$), and the second ($3.5$) is seven times the third ($0.5$). Their product $28$ is the smallest answer choice, which fits because two of the numbers are less than $1$-and-a-half, dragging the product down; the larger choices like $400$ or $800$ would need all three numbers to be big, but only the first one is.
Alternative: Reduce everything to the third number $c$ from the start: $b=7c$ and $a=4(b+c)=4(8c)=32c$, so the sum $32c+7c+c=40c=20$ gives $c=\tfrac12$. Then $b=3.5$ and $a=16$, and the product is again $16\times 3.5\times 0.5 = 28$.
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the three numbers $a$, $b$, $c$ and writing each sentence of the problem as an equation.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Collapsing $4(b+c)+(b+c)$ into $5(b+c)=20$ to find the chunk $b+c=4$ and the first number $a=16$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Combining $b+c=4$ with $b=7c$ to get $8c=4$ and solve for the second and third numbers.)6.NS.B.3Fluently add, subtract, multiply, and divide multi-digit decimals (Multiplying $16 \times 3.5 \times 0.5$ to get the final product $28$.)
⭐ When one number is four times all the rest, the total splits into five equal parts, so peel off the big number first, then split what is left by the given ratio.
⭐ When one number is four times all the rest, the total splits into five equal parts, so peel off the big number first, then split what is left by the given ratio.
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