AMC 10 · 2007 · #5

Grade 8 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

systems-of-equationslinear-equations-two-varidentify-subproblems convert-to-algebraidentify-subproblems ↑ Prerequisites: systems-of-equations

📏 Medium solution 💡 2 insights

Problem

At a school store every pencil has the same price and every notebook has the same price. There, 7 pencils and 8 notebooks together cost $4.15, and 5 pencils and 3 notebooks together cost $1.77. Find the total cost of 16 pencils and 10 notebooks.

Pick an answer.

(A)

$\mathdollar 1.76$

(B)

$\mathdollar 5.84$

(C)

$\mathdollar 6.00$

(D)

$\mathdollar 6.16$

(E)

$\mathdollar 6.32$

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

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.

1STEP 1

Name the prices and write equations

Let p be one pencil's price and n one notebook's, both in dollars: 7p+8n=4.15, 5p+3n=1.77, and the goal is 16p+10n.

7p+8n=4.15, 5p+3n=1.77

2STEP 2

Eliminate the notebooks to find the pencil price

Scale to 24n: 21p+24n=12.45 and 40p+24n=14.16. Subtracting erases n, so 19p=1.71 and one pencil is $0.09.

21p+24n=12.45, 40p+24n=14.16 → 19p=1.71 → p=0.09

3STEP 3

Back-substitute to find the notebook price

Put p=0.09 into 5p+3n=1.77: 0.45+3n=1.77, so 3n=1.32 and one notebook is $0.44.

5(0.09)+3n=1.77 → 3n=1.32 → n=0.44

4STEP 4

Compute the cost of 16 pencils and 10 notebooks

So 16p+10n=16(0.09)+10(0.44)=1.44+4.40=$5.84, which is choice (B).

16(0.09)+10(0.44)=1.44+4.40=5.84 → (B)

Answer

$5.84

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 $4.15, ruling out (A). A fast estimate — tripling the $1.77 order gives 15 pencils and 9 notebooks for $5.31, and we need only one more pencil and one more notebook — lands just below $5.84, confirming (B).

💡Key takeaway

Give the two unknown prices letters, use both facts to solve for each price, then plug them into the order you actually want.

Name the prices and write equations
Eliminate the notebooks to find the pencil price
Back-substitute to find the notebook price
Compute the cost of 16 pencils and 10 notebooks

Name the prices and write equations

Eliminate the notebooks to find the pencil price

Back-substitute to find the notebook price

Compute the cost of 16 pencils and 10 notebooks

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