Using an equal-armed balance, if $\square \square \square \square$ balances $\odot \circ$ and $\circ \circ \bigcirc$ balances $\triangle \triangle$, which of the following would not balance $\triangle O \square$ ?
(A) $\triangle O \square$
(B) $\square \square \square \triangle$
(C) $\square \square \circ \circ$
(D) $\Delta \Delta \square$
(E)
Solution 1

If $4 \square$ balance $2 \circ$, then $1 \square$ would balance the equivalent of $\frac{1}{2} \circ$.

Similarly, $1 \triangle$ would balance the equivalent of $1 \frac{1}{2} \circ$.

If we take each of the answers and convert them to an equivalent number of $\odot$, we would have:

(A): $1 \frac{1}{2}+1+\frac{1}{2}=3 \circ$

(B): $3\left(\frac{1}{2}\right)+1 \frac{1}{2}=30$

(C): $2\left(\frac{1}{2}\right)+2=3 \circ$

(D): $2\left(1 \frac{1}{2}\right)+\frac{1}{2}=3 \frac{1}{2} \circ$

(E): $1+4\left(\frac{1}{2}\right)=3$ ○

Therefore, $2 \Delta$ and $1 \square$ do not balance the required.

## Solution 2

Since $4 \square$ balance $2 \circ$, then $1 \circ$ would balance $2 \square$.

Therefore, $3 \circ$ would balance $6 \square$, so since $3 \circ$ balance $2 \triangle$, then $6 \square$ would balance $2 \Delta$, or $1 \Delta$ would balance $3 \square$.

We can now express every combination in terms of $\square$ only.

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

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

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

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

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

Therefore, since $1 \Delta, 1 \circ$ and $1 \square$ equals $6 \square$, then it is $2 \Delta$ and $1 \square$ which will not balance with this combination.

## Solution 3

We try assigning weights to the different shapes.

Since $3 \circ$ balance $2 \Delta$, assume that each $\circ$ weighs $2 \mathrm{~kg}$ and each $\triangle$ weighs $3 \mathrm{~kg}$.

Therefore, since $4 \square$ balance $2 \circ$, which weigh $4 \mathrm{~kg}$ combined, then each $\square$ weighs $1 \mathrm{~kg}$.

We then look at each of the remaining combinations.

$1 \triangle, 1 \circ$ and $1 \square$ weigh $3+2+1=6 \mathrm{~kg}$.

$3 \square$ and $1 \triangle$ weigh $3+3=6 \mathrm{~kg}$.

$2 \square$ and $2 \circ$ weigh $2+2 \times 2=6 \mathrm{~kg}$.

$2 \triangle$ and $1 \square$ weigh $2 \times 3+1=7 \mathrm{~kg}$.

$1 \circ$ and $4 \square$ weigh $2+4=6 \mathrm{~kg}$.

Therefore, it is the combination of $2 \Delta$ and $1 \square$ which will not balance the other combinations.