2009 AMC 10B Problem 15

P11-15~3 minPrint

Linear Equations & Word Problems Systems of Equations Set up the equation/formula and compute; no special trick needed Test small/specific values or special cases to find or verify the answer

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?

Show solution without answering

Key insight

Going from half full to two-thirds full adds one-sixth of the water, and that one-sixth weighs a - b. Add the last one-third (twice that) to the two-thirds weighing.

Solution

The bucket weighs the same in both weighings. Only the amount of water changes, so compare the two weighings.

1. From half full to two-thirds full, the extra water is $\frac23 - \frac12 = \frac46 - \frac36 = \frac16$ of a full bucket.
2. That extra water is the only difference between the weighings, so one-sixth of the water weighs $a - b$ kilograms.
3. To go from two-thirds full to completely full, we need $\frac13$ more of the water. That is $\frac26$ , which is two of those sixths, weighing $2(a - b)$ .
4. So a full bucket weighs
$a + 2(a - b) = a + 2a - 2b = 3a - 2b.$
5. Check with numbers: say the bucket weighs $1$ kg and a full load of water weighs $6$ kg. Then $a = 1 + 4 = 5$ and $b = 1 + 3 = 4$ . The formula gives $3 \cdot 5 - 2 \cdot 4 = 7$ , and indeed $1 + 6 = 7$ .

The answer is $\boxed{\textbf{(E)}\ 3a - 2b}$ .

Why this works

When two weighings share the same fixed part (the bucket), subtracting them cancels that part and leaves only the difference in the changing part (the water). Once you know what a small slice of water weighs, you can build up to any amount. Checking with made-up numbers catches sign mistakes.

Alternative approach

Plug in easy numbers and test the choices. Take an empty bucket weighing $0$ and $6$ kg of water when full. Then $a = 4$ , $b = 3$ , and the full weight is $6$ . The choices give:

(A) $\frac83 + 1 = \frac{11}{3}$
(B) $6 - \frac32 = \frac92$
(C) $6 + 3 = 9$
(D) $6 + 6 = 12$
(E) $12 - 6 = 6$

Only (E) gives $6$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let the empty bucket weigh $B$ and a full load of water weigh $W$ . The two weighings say
$B + \tfrac23 W = a, \qquad B + \tfrac12 W = b.$
Subtracting the second from the first: $\left(\tfrac23 - \tfrac12\right)W = \tfrac16 W = a - b$ , so $W = 6(a - b)$ .

Then $B = b - \tfrac12 W = b - 3(a - b) = 4b - 3a$ .

The full bucket weighs
$B + W = (4b - 3a) + (6a - 6b) = 3a - 2b.$

The answer is $\boxed{\textbf{(E)}\ 3a - 2b}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Introduction to Algebra, Chapter 7: Proportion (§7.4)
Intermediate Algebra, Chapter 1: Basic Techniques for Solving Equations (§1.4)