AMC 10 · 2009 · #8

Grade 8 algebra
systems-of-equationslinear-equations-two-varfraction-arithmetic convert-to-algebra ↑ Prerequisites: systems-of-equations
📏 Medium solution 💡 2 insights
Problem
A container plus its contents weighs known amounts at two different fill levels. Find the weight when it is completely full.

Pick an answer.

(A)
$\frac23a + \frac13b$
(B)
$\frac32a - \frac12b$
(C)
$\frac32a + b$
(D)
$\frac32a + 2b$
(E)
3a - 2b
How to solve
Strategy Introduce a Variable

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.

1STEP 1

Name the two hidden weights

Two hidden weights never change.

B = empty bucket, w = full load of water
2STEP 2

Write each fullness as an equation

Each fill level becomes one equation.

B + 2/3 w = a, B + 1/2 w = b
3STEP 3

Subtract to isolate the water

Subtracting isolates the contents.

(2/3 - 1/2)w = a - b → 1/6 w = a - b → w = 6a - 6b
4STEP 4

Recover the bucket weight

Back-substituting recovers the container.

B = b - 1/2(6a - 6b) = 4b - 3a
5STEP 5

Add bucket plus full water

Adding both gives 3a - 2b, choice (D).

B + w = (4b - 3a) + (6a - 6b) = 3a - 2b
Answer
3a - 2b
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.
💡Key takeaway

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

  • 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