AMC 10 · 2007 · #5
Grade 8 arithmeticThe school store sells 7 pencils and 8 notebooks for \mathdollar4.15. It also sells 5 pencils and 3 notebooks for \mathdollar1.77. How much do 16 pencils and 10 notebooks cost?
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: At a school store, $7$ pencils and $8$ notebooks together cost $\mathdollar 4.15$, and $5$ pencils and $3$ notebooks together cost $\mathdollar 1.77$. Find the total cost of $16$ pencils and $10$ notebooks.
Givens: $7$ pencils and $8$ notebooks cost $\mathdollar 4.15$; $5$ pencils and $3$ notebooks cost $\mathdollar 1.77$; Every pencil costs the same, and every notebook costs the same; Answer choices: (A) $\mathdollar 1.76$, (B) $\mathdollar 5.84$, (C) $\mathdollar 6.00$, (D) $\mathdollar 6.16$, (E) $\mathdollar 6.32$
Unknowns: The total cost, in dollars, of $16$ pencils and $10$ notebooks
Understand
Restated: At a school store, $7$ pencils and $8$ notebooks together cost $\mathdollar 4.15$, and $5$ pencils and $3$ notebooks together cost $\mathdollar 1.77$. Find the total cost of $16$ pencils and $10$ notebooks.
Givens: $7$ pencils and $8$ notebooks cost $\mathdollar 4.15$; $5$ pencils and $3$ notebooks cost $\mathdollar 1.77$; Every pencil costs the same, and every notebook costs the same; Answer choices: (A) $\mathdollar 1.76$, (B) $\mathdollar 5.84$, (C) $\mathdollar 6.00$, (D) $\mathdollar 6.16$, (E) $\mathdollar 6.32$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra, #11 Work Backwards
The pencil price and the notebook price are the two hidden numbers everything depends on, so Tool #4 (Introduce a Variable) names them $p$ and $n$. Each sentence then becomes an equation, and Tool #13 (Convert to Algebra) turns the two facts into a system that can be solved by lining the equations up and eliminating one variable. Once one price is known, Tool #11 (Work Backwards) substitutes it into a given equation to recover the other price, and the two prices give the final total.
Execute — Answer: B
6.EE.B.6 Step 1 Name the prices and write equations
- Let $p$ be the cost of one pencil and $n$ the cost of one notebook, both in dollars.
- The first fact — $7$ pencils and $8$ notebooks for $\mathdollar 4.15$ — becomes $7p+8n=4.15$.
- The second fact — $5$ pencils and $3$ notebooks for $\mathdollar 1.77$ — becomes $5p+3n=1.77$.
- The goal, the cost of $16$ pencils and $10$ notebooks, is $16p+10n$.
💡 Giving the unknown prices letters lets each sentence turn into an equation you can work with.
8.EE.C.8 Step 2 Eliminate the notebooks to find the pencil price
- Make the notebook terms match so they cancel.
- Multiply the first equation by $3$ and the second by $8$, so both have $24n$: $3(7p+8n)=3(4.15)$ gives $21p+24n=12.45$, and $8(5p+3n)=8(1.77)$ gives $40p+24n=14.16$.
- Subtract the first of these from the second: the $24n$ terms cancel, leaving $19p=1.71$, so $p=1.71\div19=0.09$.
- One pencil costs $\mathdollar 0.09$.
💡 Scaling both equations to the same notebook count lets subtraction erase the notebooks and expose the pencil price.
8.EE.C.8 Step 3 Back-substitute to find the notebook price
- Put $p=0.09$ into the simpler second equation $5p+3n=1.77$.
- Then $5(0.09)=0.45$, so $0.45+3n=1.77$, which gives $3n=1.77-0.45=1.32$ and $n=1.32\div3=0.44$.
- One notebook costs $\mathdollar 0.44$.
💡 Once one price is known, an original equation has just one unknown left, so it unwinds directly.
6.NS.B.3 Step 4 Compute the cost of 16 pencils and 10 notebooks
- Now use the two prices in the target expression: $16p+10n=16(0.09)+10(0.44)$.
- That is $1.44+4.40=5.84$.
- So $16$ pencils and $10$ notebooks cost $\mathdollar 5.84$, which is choice (B).
- The far-too-small (A) $\mathdollar 1.76$ ignores that this order is much larger than the given ones, while (C) $\mathdollar 6.00$, (D) $\mathdollar 6.16$, and (E) $\mathdollar 6.32$ come from small pricing slips.
💡 With both prices in hand, the total is just each price times its count, added together.
6.EE.B.6 Let $p$ be the cost of one pencil and $n$ the cost of one notebook, both in doll 8.EE.C.8 Make the notebook terms match so they cancel. Multiply the first equation by $3$ 8.EE.C.8 Put $p=0.09$ into the simpler second equation $5p+3n=1.77$. Then $5(0.09)=0.45$, 6.NS.B.3 Now use the two prices in the target expression: $16p+10n=16(0.09)+10(0.44)$. Th Review
Reasonableness: Check both prices against the givens: $7(0.09)+8(0.44)=0.63+3.52=4.15$ and $5(0.09)+3(0.44)=0.45+1.32=1.77$ — both match exactly. The size is sensible too: $16$ pencils and $10$ notebooks is a bigger order than either given one, so the answer must exceed $\mathdollar 4.15$, ruling out (A). A fast estimate — tripling the $\mathdollar 1.77$ order gives $15$ pencils and $9$ notebooks for $\mathdollar 5.31$, and we need only one more pencil and one more notebook — lands just below $\mathdollar 5.84$, confirming (B).
Alternative: You can skip solving for the prices entirely by using the answer choices (Tool #3, Eliminate Possibilities). Three copies of the second purchase, $3\times(5p+3n)=15p+9n$, cost $3\times1.77=\mathdollar 5.31$. The target $16p+10n$ is just one pencil and one notebook more than that, so it must be a little above $\mathdollar 5.31$. Only (B) $\mathdollar 5.84$ sits just above $\mathdollar 5.31$; every other choice is either far below or well above, so (B) is forced without ever finding $p$ and $n$.
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the pencil price $p$ and notebook price $n$ and turning each purchase sentence into an equation.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Eliminating $n$ to get $19p=1.71$ (so $p=0.09$) and back-substituting to get $n=0.44$.)6.NS.B.3Fluently add, subtract, multiply, and divide multi-digit decimals (Computing the final total $16(0.09)+10(0.44)=1.44+4.40=5.84$ and the dollar arithmetic throughout.)
⭐ Give the two unknown prices letters, use both facts to solve for each price, then plug them into the order you actually want.
⭐ Give the two unknown prices letters, use both facts to solve for each price, then plug them into the order you actually want.
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