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c) (1)/(2) + (2)/(7)

Mathematics

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3 parts
c

1114\frac{11}{14}

d

3124\frac{31}{24}

f

1715\frac{17}{15}

Step-by-step solution

5 steps

Here's the rest of the solution for problem c) and the solutions for d), f), and h):

Part c· 1 step

12+27\frac{1}{2} + \frac{2}{7}

Step 2 (continued): Convert the second fraction to an equivalent fraction with the LCM as the denominator.

27=2×27×2=414\frac{2}{7} = \frac{2 \times 2}{7 \times 2} = \frac{4}{14}
  1. Add the equivalent fractions.

    714+414=7+414=1114\frac{7}{14} + \frac{4}{14} = \frac{7+4}{14} = \frac{11}{14}

    The sum is 1114\frac{11}{14}.

    1114\boxed{\frac{11}{14}}

Part d

3124\frac{31}{24}

23+58\frac{2}{3} + \frac{5}{8}

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27
Multiples of 8: 8, 16, 24, 32
LCM: 24

  1. Convert the fractions to equivalent fractions with the LCM as the denominator.

    23=2×83×8=1624\frac{2}{3} = \frac{2 \times 8}{3 \times 8} = \frac{16}{24} 58=5×38×3=1524\frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24}
  2. Add the equivalent fractions.

    1624+1524=16+1524=3124\frac{16}{24} + \frac{15}{24} = \frac{16+15}{24} = \frac{31}{24}

    The sum is 3124\frac{31}{24}.

    3124\boxed{\frac{31}{24}}

Part f

1715\frac{17}{15}

45+39\frac{4}{5} + \frac{3}{9}

First, simplify 39\frac{3}{9} to 13\frac{1}{3}. The problem becomes 45+13\frac{4}{5} + \frac{1}{3}.

Multiples of 5: 5, 10, 15, 20
Multiples of 3: 3, 6, 9, 12, 15, 18
LCM: 15

  1. Convert the fractions to equivalent fractions with the LCM as the denominator.

    45=4×35×3=1215\frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15} 13=1×53×5=515\frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15}
  2. Add the equivalent fractions.

    1215+515=12+515=1715\frac{12}{15} + \frac{5}{15} = \frac{12+5}{15} = \frac{17}{15}

    The sum is 1715\frac{17}{15}.

    1715\boxed{\frac{17}{15}}

Part h

12+511\frac{1}{2} + \frac{5}{11}

Multiples of 2: 2, 4, 6, 8, 10, 12, 1

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