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What is the sum of (1)/(2) + 2 (1)/(3) + 4 (1)/(4)?

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Answer

71127\frac{1}{12}

Step-by-step solution

5 steps

  1. Convert the mixed numbers to improper fractions.

    The expression is 12+213+414\frac{1}{2} + 2\frac{1}{3} + 4\frac{1}{4}.
    Convert 2132\frac{1}{3}:

    213=(2×3)+13=6+13=732\frac{1}{3} = \frac{(2 \times 3) + 1}{3} = \frac{6 + 1}{3} = \frac{7}{3}

    Convert 4144\frac{1}{4}:

    414=(4×4)+14=16+14=1744\frac{1}{4} = \frac{(4 \times 4) + 1}{4} = \frac{16 + 1}{4} = \frac{17}{4}

    The expression becomes:

    12+73+174\frac{1}{2} + \frac{7}{3} + \frac{17}{4}

  2. Find the least common multiple (LCM) of the denominators.

    The denominators are 2, 3, and 4.
    Multiples of 2: 2, 4, 6, 8, 10, 12, ...
    Multiples of 3: 3, 6, 9, 12, ...
    Multiples of 4: 4, 8, 12, ...
    The LCM of 2, 3, and 4 is 12.

  3. Convert each fraction to an equivalent fraction with a denominator of 12.

    For 12\frac{1}{2}:

    12=1×62×6=612\frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12}

    For 73\frac{7}{3}:

    73=7×43×4=2812\frac{7}{3} = \frac{7 \times 4}{3 \times 4} = \frac{28}{12}

    For 174\frac{17}{4}:

    174=17×34×3=5112\frac{17}{4} = \frac{17 \times 3}{4 \times 3} = \frac{51}{12}

  4. Add the fractions.

    612+2812+5112=6+28+5112\frac{6}{12} + \frac{28}{12} + \frac{51}{12} = \frac{6 + 28 + 51}{12} =34+5112= \frac{34 + 51}{12} =8512= \frac{85}{12}

  5. Convert the improper fraction back to a mixed number.

    Divide 85 by 12:
    85÷12=785 \div 12 = 7 with a remainder of 85−(12×7)=85−84=185 - (12 \times 7) = 85 - 84 = 1.
    So, 8512=7112\frac{85}{12} = 7\frac{1}{12}.

    The sum is 71127\frac{1}{12}.

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