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

(New page: {{duplicate|2009 AMC 10B #15 and 2009 AMC 12B #8}} == Problem == When a bucket is two-thirds full of water, the bucket and water weigh ...)
 
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== Solution ==
 
== Solution ==
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=== Solution 1 ===
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Let <math>x</math> be the weight of the bucket and let <math>y</math> be the weight of the water in a full bucket.  Then we are given that <math>x + \frac 23y = a</math> and <math>x + \frac 12y = b</math>.  Hence <math>\frac 16y = a-b</math>, so <math>y = 6a-6b</math>.  Thus <math>x = b - \frac 12 (6a-6b) = -3a + 4b</math>.  Finally <math>x + y = \boxed {3a-2b}</math>. The answer is <math>\mathrm{(E)}</math>.
 
Let <math>x</math> be the weight of the bucket and let <math>y</math> be the weight of the water in a full bucket.  Then we are given that <math>x + \frac 23y = a</math> and <math>x + \frac 12y = b</math>.  Hence <math>\frac 16y = a-b</math>, so <math>y = 6a-6b</math>.  Thus <math>x = b - \frac 12 (6a-6b) = -3a + 4b</math>.  Finally <math>x + y = \boxed {3a-2b}</math>. The answer is <math>\mathrm{(E)}</math>.
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=== Solution 2 ===
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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.
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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.
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The difference between these is exactly one bucket full of water, hence the answer is <math>3a-2b</math>.
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=== Solution 3 ===
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We are looking for an expression of the form <math>xa + yb</math>.
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We must have <math>x+y=1</math>, as the desired result contains exactly one bucket. Also, we must have <math>\frac 23 x + \frac 12 y = 1</math>, as the desired result contains exactly one bucket of water.
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At this moment, it is easiest to check that only the options (A), (B), and (E) satisfy <math>x+y=1</math>, and out of these only (E) satisfies the second equation.
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Alternatively, we can directly solve the system, getting <math>x=3</math> and <math>y=-2</math>.
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=== Solution 4 ===
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Since <math>a</math> is one bucket plus two-thirds of the total amount of water, and <math>b</math> is one bucket plus one-half of the total amount of water, <math>a-b</math> would equal <math>\cancel{\text{bucket}}+\frac23\cdot\text{water}-(\cancel{\text{bucket}}+\frac12\cdot\text{water})=\frac16\cdot\text{water}</math>. Therefore, <math>a-b</math> is one-sixth of the total mass of the water. Starting from <math>a</math>, we add two one-sixths of the total amount of the water to become full. Therefore, the full bucket of water is <math>a+2(a-b)=a+2a-2b=3a-2b\Rightarrow\textbf{(E)}</math>
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-[[User:Sweetmango77|SweetMango77]]
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== Solution 5 (Using answer choices+Very quick bash)==
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WLOG let the bucket hold 6 units of water. Then b=4, a=3. Then test out all the answer choices to arrive at the answer choice (E)
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~PEKKA
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== Video Solution ==
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https://youtu.be/fXXSk8zUXgo
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~savannahsolver
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I restored the original video solution that was deleted by Whitelisted and replaced with a Rick Roll.
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~IceMatrix
  
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2009|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2009|ab=B|num-b=14|num-a=16}}
 
{{AMC12 box|year=2009|ab=B|num-b=7|num-a=9}}
 
{{AMC12 box|year=2009|ab=B|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 12:37, 3 September 2023

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 $y=-2$.

Solution 4

Since $a$ is one bucket plus two-thirds of the total amount of water, and $b$ is one bucket plus one-half of the total amount of water, $a-b$ would equal $\cancel{\text{bucket}}+\frac23\cdot\text{water}-(\cancel{\text{bucket}}+\frac12\cdot\text{water})=\frac16\cdot\text{water}$. Therefore, $a-b$ is one-sixth of the total mass of the water. Starting from $a$, we add two one-sixths of the total amount of the water to become full. Therefore, the full bucket of water is $a+2(a-b)=a+2a-2b=3a-2b\Rightarrow\textbf{(E)}$

-SweetMango77

Solution 5 (Using answer choices+Very quick bash)

WLOG let the bucket hold 6 units of water. Then b=4, a=3. Then test out all the answer choices to arrive at the answer choice (E) ~PEKKA

Video Solution

https://youtu.be/fXXSk8zUXgo

~savannahsolver

I restored the original video solution that was deleted by Whitelisted and replaced with a Rick Roll. ~IceMatrix

See also

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

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