AMC 10 · 2009 · #15

Grade 8 arithmetic

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

systems-of-equationslinear-equations-two-varfraction-arithmetic convert-to-algebra ↑ Prerequisites: systems-of-equations

📏 Medium solution 💡 2 insights

Problem

When a bucket is two-thirds full of water, the bucket and water weigh $a$ kilograms. When the bucket is one-half full of water the total weight is $b$ kilograms. In terms of $a$ and $b$ , what is the total weight in kilograms when the bucket is full of water?

Pick an answer.

(A)

$\ \frac23a + \frac13b$

(B)

$\ \frac32a - \frac12b$

(C)

$\ \frac32a + b$

(D)

$\ \frac32a + 2b$

(E)

$\ 3a - 2b$

AMC 10 2009 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.

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Toolkit + CCSS Solution

Understand

Restated: A bucket holds some fixed weight of its own, plus water. When it is two-thirds full the bucket and water together weigh $a$ kilograms; when it is one-half full they weigh $b$ kilograms. Find the total weight, in terms of $a$ and $b$, when the bucket is completely full.

Givens: Bucket two-thirds full of water weighs $a$ kg total.; Bucket one-half full of water weighs $b$ kg total.; The empty bucket's weight is a fixed unknown.; A full bucket's water is a fixed unknown amount.

Unknowns: The total weight when the bucket is completely full of water.

Understand

Plan

Primary tool: #4 Introduce a Variable

Secondary: #13 Convert to Algebra, #7 Identify Subproblems

Two hidden quantities never change: the bucket's own weight and the weight of a full load of water. Name each with a variable, turn the two 'how full' facts into two equations, and the full-bucket weight falls out by combining them.

Execute — Answer: E

#4 Introduce a Variable 6.EE.B.6 Step 1

Name the two hidden weights

Let $B$ be the weight of the empty bucket and let $w$ be the weight of the water when the bucket is completely full.
These two numbers stay the same no matter how full the bucket is.

$$B = \text{empty bucket}, \quad w = \text{full load of water}$$

💡 Give the unchanging parts names so you can track them through every case.

#13 Convert to Algebra 7.EE.B.4 Step 2

Write each fullness as an equation

Two-thirds full means the water present is $\frac23 w$, so the total is $B + \frac23 w = a$.
One-half full means the water present is $\frac12 w$, so the total is $B + \frac12 w = b$.

$$B + \tfrac23 w = a, \qquad B + \tfrac12 w = b$$

💡 A fraction of the full water plus the same bucket gives each measured weight.

#7 Identify Subproblems 8.EE.C.8 Step 3

Subtract to isolate the water

Subtract the second equation from the first.
The bucket weight $B$ cancels, leaving only water: $\left(\frac23 - \frac12\right)w = a - b$, which is $\frac16 w = a - b$.
So $w = 6a - 6b$.

$$\left(\tfrac23 - \tfrac12\right)w = a - b \;\Rightarrow\; \tfrac16 w = a - b \;\Rightarrow\; w = 6a - 6b$$

💡 Lining up the two equations lets the shared bucket weight erase itself.

#13 Convert to Algebra 8.EE.C.7 Step 4

Recover the bucket weight

Put $w = 6a - 6b$ back into $B + \frac12 w = b$.
Then $B = b - \frac12(6a - 6b) = b - 3a + 3b = 4b - 3a$.

$$B = b - \tfrac12(6a - 6b) = 4b - 3a$$

💡 Once the water is known, one equation reveals the bucket.

#4 Introduce a Variable 7.EE.A.1 Step 5

Add bucket plus full water

A full bucket weighs $B + w = (4b - 3a) + (6a - 6b) = 3a - 2b$.
That matches choice (E), so the answer is $\textbf{(E)}\ 3a - 2b$.

$$B + w = (4b - 3a) + (6a - 6b) = 3a - 2b$$

💡 Full weight is just the bucket plus one whole load of water.

[1] #4 6.EE.B.6 Let $B$ be the weight of the empty bucket and let $w$ be the weight of the water

[2] #13 7.EE.B.4 Two-thirds full means the water present is $\frac23 w$, so the total is $B + \fr

[3] #7 8.EE.C.8 Subtract the second equation from the first. The bucket weight $B$ cancels, leav

[4] #13 8.EE.C.7 Put $w = 6a - 6b$ back into $B + \frac12 w = b$. Then $B = b - \frac12(6a - 6b)

[5] #4 7.EE.A.1 A full bucket weighs $B + w = (4b - 3a) + (6a - 6b) = 3a - 2b$. That matches cho

Review

Reasonableness: Test with real numbers: let the empty bucket be $2$ and a full load of water be $6$. Then two-thirds full is $2 + 4 = 6 = a$ and one-half full is $2 + 3 = 5 = b$. A full bucket is $2 + 6 = 8$. The formula gives $3a - 2b = 18 - 10 = 8$, which matches, so (E) is correct.

Alternative: Skip solving for the pieces. Multiply the first equation by $3$ and the second by $2$: $3B + 2w = 3a$ and $2B + w = 2b$. Subtracting gives $B + w = 3a - 2b$ directly — and $B + w$ is exactly the full-bucket weight.

CCSS standards used (min grade 8)

6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Naming the unchanging bucket weight and full-water weight as variables.)
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities (Translating 'two-thirds full' and 'one-half full' into two equations.)
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Subtracting the two equations to eliminate the bucket and solve for the water.)
8.EE.C.7 Solve linear equations in one variable (Back-substituting to find the empty bucket's weight.)
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions (Adding the bucket and full-water expressions to get the final formula.)

⭐ Name the parts that never change, write one equation per clue, then subtract to make the part you don't want disappear.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Name the two hidden weights

Write each fullness as an equation

Subtract to isolate the water

Recover the bucket weight

Add bucket plus full water

Review

CCSS standards used (min grade 8)

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