# 2005 CEMC Gauss (Grade 7) Problems/Problem 23

## 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 $D$.