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Problem of the Week
Problem B and Solution
Maybe Equals, Maybe Not

Problem

  1. For each of the following statements, fill in the \(\bigcirc\) with \(=\), \(<\), or \(>\) to make the statement true.

    1. \(0.5 + 0.24 \bigcirc \frac{3}{4}\)

    2. \(1.10 + \frac{1}{4} \bigcirc \frac{35}{100}\)

    3. \(2 \times 2 \times 5 \bigcirc \frac{100}{5}\)

    4. \(100-24+65 \bigcirc 14 \times 10\)

    5. \(25 \times 40 + 321 \bigcirc 153 \times 10\)

    6. \(2 \times 24 \div 6 \bigcirc \frac{36}{6}\)

    7. \(35 \times 2 \bigcirc 5 \times 14\)

    8. \(2y+6 \bigcirc y+3+y+3\)

  2. In the following statement, determine values of \(y\) that make the statement true when the \(\bigcirc\) is replaced with \(=\), \(<\), and \(>\). \[4y+12 \bigcirc 4 + 2y + 4 + 3y + 4\]

Solution

  1. To determine which symbol we should put in the \(\bigcirc\), we look at the left side and right side of each statement separately.

    1. Left Side Right Side
      \(0.5+24 =0.74\) \(\frac{3}{4} = 0.75\)

      Since \(0.74 < 0.75\), then \(0.5 + 0.24 < \frac{3}{4}\).

    2. Left Side Right Side
      \(1.10 + \frac{1}{4} = 1.10+0.25=1.35\) \(\frac{35}{100} =0.35\)

      Since \(1.35>0.35\), then \(1.10 + \frac{1}{4} > \frac{35}{100}\).

    3. Left Side Right Side
      \(2\times 2\times 5 =4\times 5=20\) \(\frac{100}{5} =20\)

      Since \(20=20\), then \(2 \times 2 \times 5 =\frac{100}{5}\).

    4. Left Side Right Side
      \(100-24+65 =76+65=141\) \(14 \times 10 =140\)

      Since \(141>140\), then \(100-24+65 >14 \times 10\).

    5. Left Side Right Side
      \(25 \times 40 + 321 =1000+321=1321\) \(153 \times 10 =1530\)

      Since \(1321<1530\), then \(25 \times 40 + 321 < 153 \times 10\).

    6. Left Side Right Side
      \(2 \times 24 \div 6 =48\div 6=8\) \(\frac{36}{6} =6\)

      Since \(8>6\), then \(2 \times 24 \div 6 > \frac{36}{6}\).

    7. Left Side Right Side
      \(35 \times 2 =70\) \(5 \times 14 =70\)

      Since \(70=70\), then \(35 \times 2 = 5 \times 14\).

    8. Left Side Right Side
      \(2y+6\) \(y+3+y+3\)
      \(=y+y+3+3=2y+6\)

      Since \(2y+6=2y+6\) for any value of \(y\), then \(2y+6 = y+3+y+3\).

  2. The right side can be rewritten as \(4 + 2y + 4 + 3y + 4=2y+3y+4+4+4=5y+12\). Then our statement becomes \(4y+12 \bigcirc 5y+12\).

    If \(y\) is any value greater than \(0\), such as \(1\), then \(4y+12<5y+12\). If \(y=0\), then \(4y+12=5y+12\). If \(y\) is any value less than \(0\), such as \(-1\), then \(4y+12>5y+12\).