Problem of the Week
Problem B and Solution
Symbolic Shapes

Problem

In each question, each shape (\(\Box\), \(\bigcirc\), \(\triangle\)) is equal to the same whole number. For example, \(\Box+\Box=8\) implies that \(2 \times \Box=8\), so \(\Box =4\). (The value of each shape can change from question to question.)

Determine the value of each shape in each question.

1. \[\begin{aligned} \Box+\Box&=8 \\ \bigcirc+\Box&=10 \end{aligned}\]

2. \[\begin{aligned} \Box+\Box+\Box&=18 \\ \Box+\bigcirc+\bigcirc&=14\\ \bigcirc+\Box+\triangle&=18 \end{aligned}\]

3. \[\begin{aligned} \bigcirc+\bigcirc+\bigcirc&=24 \\ \Box\times\bigcirc+\Box&=36\\ \triangle\times\Box+\triangle&=25 \end{aligned}\]

4. \[\begin{aligned} 2\times\triangle&=28 \\ \triangle-10&=\bigcirc\\ \bigcirc+\Box+\Box&=10 \end{aligned}\]

5. \[\begin{aligned} \Box+\triangle+\triangle&=21 \\ \triangle+\bigcirc-\Box&=9\\ \Box+\Box&=18 \end{aligned}\]

Extension: Create a similar problem and challenge a classmate to solve it.

Solution

1. Since \(\Box+\Box=8\), we know that \(2\times \Box=8\), and so \(\Box = 8\div 2 = 4\).
   Then, since \(\bigcirc+\Box=10\), we know that \(\bigcirc+4=10\), and so \(\bigcirc=10-4=6\).
   Thus, \(\Box=4\) and \(\bigcirc=6\).

2. Since \(\Box+\Box+\Box=18\), we know that \(3\times \Box=18\), and so \(\Box=18\div 3=6\).
   Then, since \(\Box+\bigcirc+\bigcirc=14\), we know that \(6+\bigcirc+ \bigcirc=14\), and so \(\bigcirc + \bigcirc = 14-6=8\) and \(\bigcirc = 8\div 2 = 4\).
   Then, since \(\bigcirc+\Box +\triangle = 18\), we have \(4+6+\triangle=18\), and so \(\triangle=18-10=8\).
   Thus, \(\Box=6\), \(\bigcirc=4\), and \(\triangle =8\).

3. Since \(\bigcirc+\bigcirc+\bigcirc=24\), we know that \(3\times \bigcirc=24\), and so \(\bigcirc = 24\div 3 = 8\).
   Then, since \(\Box \times \bigcirc + \Box = 36\), we have \(\Box\times 8+\Box = 36\), and so \(9\times \Box=36\). Therefore, \(\Box = 36\div 9 =4.\)
   Now, since \(\triangle \times \Box + \triangle = 25\), we have \(\triangle\times 4+\triangle=25\), and so \(5\times \triangle=25\) and \(\triangle=25\div 5 = 5\).
   Thus, \(\Box=4\), \(\bigcirc=8\), and \(\triangle =5\).

4. Since \(2\times\triangle=28\), we have \(\triangle=28\div 2 = 14\).
   Then, since \(\triangle-10=\bigcirc\), we have \(14-10=\bigcirc\), and so \(\bigcirc=4\).
   Then, since \(\bigcirc+\Box+\Box =10\), we have \(4+\Box + \Box = 10\), and so \(\Box + \Box = 10-4 = 6\). Therefore, \(\Box=6\div 2=3\).
   Thus, \(\Box=3\), \(\bigcirc=4\), and \(\triangle =14\).

5. For this problem, start with the third equation. Since \(\Box + \Box = 18\), we have \(2\times \Box =18\), and so \(\Box = 18\div 2 = 9\).
   Then, since \(\Box+\triangle+\triangle = 21\), we have \(9 +\triangle + \triangle =21\), and so \(\triangle + \triangle = 21-9 = 12\). Thus, \(2\times\triangle =12\), and so \(\triangle = 12\div 2 = 6\).
   Now, since \(\triangle + \bigcirc-\Box = 9\), we have \(6 + \bigcirc -9 = 9\), and so \(\bigcirc = 9-6+9 = 12\).
   Thus, \(\Box=9\), \(\bigcirc=12\), and \(\triangle =6\).