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

Latest revision as of 01:41, 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 $D$ .

Revision as of 01:25, 24 October 2014 view source RTG (talk \| contribs) 195 edits →Solution 3 ← Older edit		Latest revision as of 01:41, 24 October 2014 view source RTG (talk \| contribs) 195 edits →See Also
Line 62:		Line 62:
	== 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=22\|num-a=24}}

2005 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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
CEMC Gauss (Grade 7)

Page

Toolbox

Search