AMC 10 · 2002 · #16
Grade 8 arithmeticLet a+1=b+2=c+3=d+4=a+b+c+d+5. What is a+b+c+d?
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: Four expressions $a+1$, $b+2$, $c+3$, $d+4$ and the sum expression $a+b+c+d+5$ are all equal to one another. Find the value of $a+b+c+d$.
Givens: $a+1=b+2=c+3=d+4=a+b+c+d+5$ — a single chain of equalities; Four separate unknowns $a$, $b$, $c$, $d$; The last expression in the chain is the sum $a+b+c+d$ plus $5$; Answer choices: (A) $-5$, (B) $-10/3$, (C) $-7/3$, (D) $5/3$, (E) $5$
Unknowns: The value of $a+b+c+d$
Understand
Restated: Four expressions $a+1$, $b+2$, $c+3$, $d+4$ and the sum expression $a+b+c+d+5$ are all equal to one another. Find the value of $a+b+c+d$.
Givens: $a+1=b+2=c+3=d+4=a+b+c+d+5$ — a single chain of equalities; Four separate unknowns $a$, $b$, $c$, $d$; The last expression in the chain is the sum $a+b+c+d$ plus $5$; Answer choices: (A) $-5$, (B) $-10/3$, (C) $-7/3$, (D) $5/3$, (E) $5$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #15 Organize Information in More Ways, #13 Convert to Algebra
The whole problem hangs on the phrase 'all equal.' Tool #4 (Introduce a Variable) captures that directly: call the shared value $k$. Every expression in the chain then becomes a short equation involving $k$, which turns four tangled unknowns into one manageable letter. Tool #15 (Organize Information in More Ways) is the key move: the target $a+b+c+d$ can be written two different ways — once by adding the four individual equations, and once straight from the fifth expression $a+b+c+d+5=k$. Setting those two versions equal gives a single equation in $k$. Tool #13 (Convert to Algebra) then finishes it by solving that linear equation.
Execute — Answer: B
6.EE.B.6 Step 1 Name the shared value
- Every expression in the chain equals the same number, so give that number a name: let $k$ be the common value.
- Each equality now becomes its own small equation: $a+1=k$, $b+2=k$, $c+3=k$, $d+4=k$.
- Solving each for its variable, $a=k-1$, $b=k-2$, $c=k-3$, $d=k-4$.
💡 When many things are declared equal, naming that one shared value turns each equality into a usable equation.
7.EE.A.1 Step 2 Add to get the sum in terms of k
- Add the four expressions for the individual variables.
- The four copies of $k$ combine to $4k$, and the constants $-1-2-3-4$ add to $-10$.
- So the sum $a+b+c+d$ equals $4k-10$.
💡 Adding the four rewritten variables stacks four $k$'s together and collects the constants into one number.
6.EE.B.6 Step 3 Read the sum a second way
- The chain also says $a+b+c+d+5=k$.
- This is a second, direct description of the same sum: subtract $5$ from both sides to get $a+b+c+d=k-5$.
- Now the one quantity $a+b+c+d$ has been written two different ways.
💡 The last piece of the chain already talks about the sum directly, giving a rival expression for the very thing we want.
8.EE.C.7 Step 4 Match the two versions and solve for k
- The sum $a+b+c+d$ cannot be two different things at once, so its two expressions must be equal: $4k-10=k-5$.
- Subtract $k$ from both sides to get $3k-10=-5$, then add $10$ to get $3k=5$, so $k=\tfrac{5}{3}$.
💡 One quantity equals only one number, so its two descriptions must agree, which pins down $k$.
7.NS.A.1 Step 5 Compute the sum
- Use the direct expression $a+b+c+d=k-5$ with $k=\tfrac{5}{3}$: $\tfrac{5}{3}-5=\tfrac{5}{3}-\tfrac{15}{3}=-\tfrac{10}{3}$.
- So $a+b+c+d=-\tfrac{10}{3}$, which is choice (B).
💡 Plugging the found $k$ back into the simplest expression for the sum finishes the job in one subtraction.
6.EE.B.6 Every expression in the chain equals the same number, so give that number a name 7.EE.A.1 Add the four expressions for the individual variables. The four copies of $k$ co 6.EE.B.6 The chain also says $a+b+c+d+5=k$. This is a second, direct description of the s 8.EE.C.7 The sum $a+b+c+d$ cannot be two different things at once, so its two expressions 7.NS.A.1 Use the direct expression $a+b+c+d=k-5$ with $k=\tfrac{5}{3}$: $\tfrac{5}{3}-5=\ Review
Reasonableness: Check every part of the chain with $k=\tfrac{5}{3}$: $a=\tfrac{5}{3}-1=\tfrac{2}{3}$, $b=-\tfrac{1}{3}$, $c=-\tfrac{4}{3}$, $d=-\tfrac{7}{3}$. Their sum is $\tfrac{2-1-4-7}{3}=-\tfrac{10}{3}$, matching the answer. Testing the chain: $a+1=\tfrac{2}{3}+1=\tfrac{5}{3}$ and $a+b+c+d+5=-\tfrac{10}{3}+5=\tfrac{5}{3}$, so both ends equal $k$. Everything is consistent, and $-\tfrac{10}{3}$ is exactly choice (B).
Alternative: Compare only the fifth expression with each of the first four. From $a+1=a+b+c+d+5$ we get $b+c+d=-4$; likewise $a+c+d=-3$, $a+b+d=-2$, and $a+b+c=-1$. Adding these four equations, each variable appears three times, giving $3(a+b+c+d)=-4-3-2-1=-10$, so $a+b+c+d=-\tfrac{10}{3}$ — no need to solve for $k$ at all.
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions when solving a problem (Naming the common value $k$ and turning each equality in the chain into an equation, including the direct expression $a+b+c+d=k-5$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Adding $(k-1)+(k-2)+(k-3)+(k-4)$ and combining like terms into $4k-10$.)8.EE.C.7Solve linear equations in one variable (Solving $4k-10=k-5$, which has the variable on both sides, to find $k=\tfrac{5}{3}$.)7.NS.A.1Add and subtract rational numbers (Computing $\tfrac{5}{3}-5=-\tfrac{10}{3}$ to get the final sum.)
⭐ When a problem says a pile of expressions are all equal, name that shared value with one letter — then write the thing you want two different ways and set them equal.
⭐ When a problem says a pile of expressions are all equal, name that shared value with one letter — then write the thing you want two different ways and set them equal.
More like this
Same archetype — closest grade level first.