Difference between revisions of "2009 AMC 10B Problems/Problem 15"

Revision as of 11:49, 17 January 2021

The following problem is from both the 2009 AMC 10B #15 and 2009 AMC 12B #8, so both problems redirect to this page.

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?

$\mathrm{(A)}\ \frac23a + \frac13b\qquad \mathrm{(B)}\ \frac32a - \frac12b\qquad \mathrm{(C)}\ \frac32a + b\qquad \mathrm{(D)}\ \frac32a + 2b\qquad \mathrm{(E)}\ 3a - 2b$

Solution

Solution 1

Let $x$ be the weight of the bucket and let $y$ be the weight of the water in a full bucket. Then we are given that $x + \frac 23y = a$ and $x + \frac 12y = b$ . Hence $\frac 16y = a-b$ , so $y = 6a-6b$ . Thus $x = b - \frac 12 (6a-6b) = -3a + 4b$ . Finally $x + y = \boxed {3a-2b}$ . The answer is $\mathrm{(E)}$ .

Solution 2

Imagine that we take three buckets of the first type, to get rid of the fraction. We will have three buckets and two buckets' worth of water.

On the other hand, if we take two buckets of the second type, we will have two buckets and enough water to fill one bucket.

The difference between these is exactly one bucket full of water, hence the answer is $3a-2b$ .

Solution 3

We are looking for an expression of the form $xa + yb$ .

We must have $x+y=1$ , as the desired result contains exactly one bucket. Also, we must have $\frac 23 x + \frac 12 y = 1$ , as the desired result contains exactly one bucket of water.

At this moment, it is easiest to check that only the options (A), (B), and (E) satisfy $x+y=1$ , and out of these only (E) satisfies the second equation.

Alternatively, we can directly solve the system, getting $x=3$ and $y=-2$ .

Video Solution

https://youtu.be/fXXSk8zUXgo

~savannahsolver

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 33: / Line 33: @@
 Alternatively, we can directly solve the system, getting <math>x=3</math> and <math>y=-2</math>.
+==Video Solution==
+https://youtu.be/fXXSk8zUXgo
+~savannahsolver
 == See also ==

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