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Difference between revisions of "2005 CEMC Gauss (Grade 7) Problems/Problem 23"

(Created page with "== Problem == Using an equal-armed balance, if <math>\square\square\square\square</math> balances <math>\bigcirc \bigcirc</math> and <math>\bigcirc \bigcirc \bigcirc</math> bal...")
 
(→‎Solution 3)
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== Solution 3 ==
 
== Solution 3 ==
  
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Will be coming later.
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We try assigning weights to the different shapes.
 +
Since <math>3\bigcirc</math> balance <math>2\triangle</math> , assume that each <math>\bigcirc</math> weighs <math>2 kg</math> and each <math>\triangle</math> weighs <math>3 kg</math>.
 +
Therefore, since <math>4\square</math> balance <math>2\bigcirc</math>, which weigh <math>4 kg</math> combined, then each <math>\square</math> weighs <math>1 kg</math>.
 +
We then look at each of the remaining combinations.
 +
<math>1\triangle</math>, <math>1\bigcirc</math>, and <math>1\square</math> weigh <math>3 + 2 + 1 = 6 kg</math>.
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<math>3\square</math> and <math>1\triangle</math> weigh <math>3 + 3 = 6 kg</math>.
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<math>2\square</math> and <math>2\bigcirc</math> weigh <math>2 + 2\times 2 = 6 kg</math>.
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<math>2\triangle</math> and <math>1\square</math> weigh <math>2\times 3 + 1 = 7 kg</math>.
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<math>1\bigcirc</math> and <math>4\square</math> weigh <math>2 + 4 = 6 kg</math>.
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Therefore, it is the combination of <math>2\triangle</math> and <math>1\square</math> which will not balance the other combinations.  The answer is <math>D</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{CEMC box|year=2005|competition=Gauss (Grade 7)|num-b=12|num-a=24}}
 
{{CEMC box|year=2005|competition=Gauss (Grade 7)|num-b=12|num-a=24}}

Revision as of 01:24, 24 October 2014

Problem

Using an equal-armed balance, if $\square\square\square\square$ balances $\bigcirc \bigcirc$ and $\bigcirc \bigcirc \bigcirc$ balances $\triangle \triangle$, which of the following would not balance $\triangle \bigcirc \square$?

$\text{(A)}\ \triangle \bigcirc \square \qquad \text{(B)}\ \square \square \square \triangle \qquad \text{(C)}\ \square \square \bigcirc \bigcirc \qquad \text{(D)}\ \triangle \triangle \square \qquad \text{(E)}\ \bigcirc \square \square \square \square$

Solution 1

If $4 \square$ balance $2\bigcirc$ , then $1 \square$ would balance the equivalent of $\frac{1}{2}\bigcirc$. Similarly, $1 \triangle$ would balance the equivalent of $\frac{3}{2}\bigcirc$. If we take each of the answers and convert them to an equivalent number of $\bigcirc$, we would have:

$\text{(A)}\ \frac{3}{2} + 1 + \frac{1}{2} = 3\bigcirc$

$\text{(B)}\ 3\times \frac{1}{2} + \frac{3}{2} = 3\bigcirc$

$\text{(C)}\ 2\times \frac{1}{2} + 2 = 3\bigcirc$

$\text{(D)}\ 2\times \frac{3}{2} + \frac{1}{2} = 3\frac{1}{2}\bigcirc$

$\text{(E)}\ 1 + 4\times \frac{1}{2} = 3\bigcirc$

Therefore, $2\triangle$ and $1\square$ do not balance the required. The answer is $D$.

Solution 2

Since $4\square$ balance $2\bigcirc$ , then $1\bigcirc$ would balance $2\square$. Therefore, $3\bigcirc$ would balance $6\square$ , so since $3\bigcirc$ balance $2\triangle$ , then $6\square$ would balance $2\triangle$ , or $1\triangle$ would balance $3\square$. We can now express every combination in terms of only.

$1\triangle$, $1\bigcirc$, and $1\square$ equals $3 + 2 + 1 = 6\square$.

$3\square$ and $1\triangle$ equals $3 + 3 = 6\square$.

$2\square$ and $2\bigcirc$ equals $2 + 2\times 2 = 6\square$.

$2\triangle$ and $1\square$ equals $2\times 3 + 1 = 7\square$.

$1\bigcirc$ and $4\square$ equals $2 + 4 = 6\square$.

Therefore, since $1\triangle$ , $1\bigcirc$, and $1\square$ equals $6\square$ , then it is $2\triangle$ and $1\square$ which will not balance with this combination. Thus, the answer is $D$.

Solution 3

We try assigning weights to the different shapes. Since $3\bigcirc$ balance $2\triangle$ , assume that each $\bigcirc$ weighs $2 kg$ and each $\triangle$ weighs $3 kg$. Therefore, since $4\square$ balance $2\bigcirc$, which weigh $4 kg$ combined, then each $\square$ weighs $1 kg$. We then look at each of the remaining combinations. $1\triangle$, $1\bigcirc$, and $1\square$ weigh $3 + 2 + 1 = 6 kg$. $3\square$ and $1\triangle$ weigh $3 + 3 = 6 kg$. $2\square$ and $2\bigcirc$ weigh $2 + 2\times 2 = 6 kg$. $2\triangle$ and $1\square$ weigh $2\times 3 + 1 = 7 kg$. $1\bigcirc$ and $4\square$ weigh $2 + 4 = 6 kg$. Therefore, it is the combination of $2\triangle$ and $1\square$ which will not balance the other combinations. The answer is $D$.

See Also

2005 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
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Problem 12 		Followed by
Problem 24
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CEMC Gauss (Grade 7)
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