2. Operations (N)
  3. Lesson 5: Adding Rational Numbers

Answers and Solutions



    1. \(\begin{align*} \dfrac{1}{4} + \dfrac{3}{8} &= \dfrac{2}{8} + \dfrac{3}{8} \\ &= \dfrac{5}{8} \end{align*}\)

    2. \(\begin{align*} \dfrac{5}{14} + \dfrac{1}{2} &= \dfrac{5}{14} + \dfrac{7}{14} \\ &= \dfrac{12}{14} \\ &= \dfrac{6}{7} \end{align*}\)

    3. \(\begin{align*} \dfrac{3}{5} + \dfrac{3}{10} &= \dfrac{6}{10} + \dfrac{3}{10} \\ &= \dfrac{9}{10} \end{align*}\)

    4. \(\begin{align*} \dfrac{7}{32} + \dfrac{3}{4} &= \dfrac{7}{32} + \dfrac{24}{32} \\ &= \dfrac{31}{32} \end{align*}\)

    5. \(\begin{align*} \dfrac{1}{3} + \dfrac{3}{2} &= \dfrac{2}{6} + \dfrac{9}{6} \\ &= \dfrac{11}{6} \\ &= 1\dfrac{5}{6} \end{align*}\)

    1. \(\begin{align*} \dfrac{1}{2} + \dfrac{3}{5} &= \dfrac{5}{10} + \dfrac{6}{10} \\ &= \dfrac{11}{10} \\ &= 1\dfrac{1}{10} \end{align*}\)

    2. \(\begin{align*} \dfrac{4}{7} + \dfrac{3}{10} &= \dfrac{40}{70} + \dfrac{21}{70} \\ &= \dfrac{61}{70} \end{align*}\)

    3. \(\begin{align*} \dfrac{3}{2} + \dfrac{5}{7} &= \dfrac{21}{14} + \dfrac{10}{14} \\ &= \dfrac{31}{14} \\ &= 2\dfrac{3}{14} \end{align*}\)

    4. \(\begin{align*} \dfrac{2}{3} + \dfrac{11}{7} &= \dfrac{14}{21} + \dfrac{33}{21} \\ &= \dfrac{47}{21} \\ &= 2\dfrac{5}{21} \end{align*}\)

    5. \(\begin{align*} \dfrac{4}{3} + \dfrac{13}{8} &= \dfrac{32}{24} + \dfrac{39}{24} \\ &= \dfrac{71}{24} \\ &= 2\dfrac{23}{24} \end{align*}\)
    1. To estimate, round the values so that fractions are "nicer."
    2. Answers may vary.
      We could estimate by rounding the values to the nearest quarter.
      \(\begin{align*} \dfrac{31}{40} + \dfrac{9}{10} + \dfrac{11}{25} + \dfrac{15}{16} &\approx \dfrac{3}{4} + 1 + \dfrac{1}{2} + 1 \\ &= 2 + \dfrac{3}{4} + \dfrac{2}{4} \\ &= 2\dfrac{5}{4} \\ &= 3\dfrac{1}{4} \end{align*}\)
      Remember that this is an approximate answer.

    1. \(\begin{align*} \text{perimeter of rectangle} &= \text{length + length + width + width} \\ &= 4 + 4 + 7 + 7 \\ &= 22 \end{align*}\)
      Therefore, the perimeter of the rectangle is \(22\) cm.

    2. \(\begin{align*} \text{perimeter of rectangle} &= \text{length + length + width + width} \\ &= 12\dfrac{3}{4} + 12\dfrac{3}{4} + 8\dfrac{2}{5} + 8\dfrac{2}{5} \\ &= 24\dfrac{6}{4} + 16\dfrac{4}{5} \\ &= 25\dfrac{2}{4} + 16\dfrac{4}{5} \\ &= 25\dfrac{10}{20} + 16\dfrac{16}{20} \\ &= 41 \dfrac{26}{20} \\ &= 42 \dfrac{6}{20} \\ &= 42 \dfrac{3}{10} \\ \end{align*}\)
      Therefore, the perimeter of the rectangle is \(42\dfrac{3}{10}\) cm.

    1. \(\begin{align*} \dfrac{1}{4} +\dfrac{1}{2} &= \dfrac{1}{4} + \dfrac{2}{4} \\ &= \dfrac{3}{4} \end{align*}\)
    2. Next, we evaluate \(\dfrac{3}{4} + \dfrac{5}{8}\):
      \(\begin{align*} \dfrac{3}{4} + \dfrac{5}{8} &= \dfrac{6}{8} + \dfrac{5}{8} \\ &= \dfrac{11}{8} \\ &= 1 \dfrac{3}{8} \end{align*}\)
    1. The \(\operatorname{LCM}(5,2,3)=30\).  Therefore, the common denominator of all three fractions is \(30\).
    2. \(\dfrac{2}{5}=\dfrac{12}{30} \quad\quad\quad \dfrac{1}{2}=\dfrac{15}{30} \quad\quad\quad \dfrac{4}{3}=\dfrac{40}{30}\)

    3. \(\begin{align*} \dfrac{2}{5} + \dfrac{1}{2} + \dfrac{4}{3} &= \dfrac{12}{30} + \dfrac{15}{30} + \dfrac{40}{30} \\ &= \dfrac{67}{30} \\ &= 2 \dfrac{7}{30} \end{align*}\)
  7. Let's begin by considering the sum of the whole numbers using the pattern.  We want to find a sum that is close to, but less than, \(100\).\[1+4+7+10+13+16+19+22=92\] Now consider the fractional quantities separately.\[\dfrac{2}{3} + \dfrac{5}{6} + \dfrac{8}{9} + \dfrac{11}{12} + \dfrac{14}{15} + \dfrac{17}{18} + \dfrac{20}{21} + \dfrac{23}{24} \lt 8\] This sum is less than \(8\) because we are finding the sum of \(8\) proper fractions.  Overall, it means that the sum of the first eight numbers in this pattern is less than \(100\).  The next number, \(25\dfrac{26}{27}\) will put the sum over \(100\).  As a result, nine numbers are required to get a sum just over \(100\).
    1. Perimeter of 1st Image \(=1+1+1+1=4\)
      Perimeter of 2nd Image \(=1+1+1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2} =5\)
      Perimeter of 3rd Image \(=1+1+1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4} = 5\dfrac{1}{2}\)
    2. Perimeter of 4th Image \(=1+1+1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8} = 5\dfrac{3}{4}\)