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Add the following mixed numbers: 1 rac12 + 2 rac13 + 4 rac14

Mathematics

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Answer

81128\frac{1}{12}

Step-by-step solution

5 steps

  1. Convert the mixed numbers to improper fractions.

    112=(1×2)+12=321\frac{1}{2} = \frac{(1 \times 2) + 1}{2} = \frac{3}{2} 213=(2×3)+13=732\frac{1}{3} = \frac{(2 \times 3) + 1}{3} = \frac{7}{3} 414=(4×4)+14=1744\frac{1}{4} = \frac{(4 \times 4) + 1}{4} = \frac{17}{4}

  2. Find the least common multiple (LCM) of the denominators $2, 3,$ and 4.

    The LCM of $2, 3,$ and 4 is 12.

  3. Convert each fraction to an equivalent fraction with the common denominator 12.

    32=3×62×6=1812\frac{3}{2} = \frac{3 \times 6}{2 \times 6} = \frac{18}{12} 73=7×43×4=2812\frac{7}{3} = \frac{7 \times 4}{3 \times 4} = \frac{28}{12} 174=17×34×3=5112\frac{17}{4} = \frac{17 \times 3}{4 \times 3} = \frac{51}{12}

  4. Add the equivalent fractions.

    1812+2812+5112=18+28+5112\frac{18}{12} + \frac{28}{12} + \frac{51}{12} = \frac{18 + 28 + 51}{12} 18+28+5112=9712\frac{18 + 28 + 51}{12} = \frac{97}{12}

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

    Divide 97 by 12:

    97÷12=8 with a remainder of 197 \div 12 = 8 \text{ with a remainder of } 1

    So, 9712=8112\frac{97}{12} = 8\frac{1}{12}.

    The sum is 81128\frac{1}{12}.

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