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Solve the following math problems: Q35: 3/8 * 240, Q36: 4 3/7 - 1 1/6

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2 parts
35

90\text{90}

36

311423\frac{11}{42}

Step-by-step solution

6 steps

Here are the solutions to the problems:

Part 35· 2 steps

38×240\frac{3}{8} \times 240
  1. Multiply the numerator by the whole number.

    3×240=7203 \times 240 = 720
  2. Divide the result by the denominator.

    7208=90\frac{720}{8} = 90

    The answer is 90\boxed{\text{90}}.

Part 36

311423\frac{11}{42}

437−1164\frac{3}{7} - 1\frac{1}{6}
  1. Convert the mixed numbers to improper fractions.

    437=(4×7)+37=28+37=3174\frac{3}{7} = \frac{(4 \times 7) + 3}{7} = \frac{28 + 3}{7} = \frac{31}{7} 116=(1×6)+16=6+16=761\frac{1}{6} = \frac{(1 \times 6) + 1}{6} = \frac{6 + 1}{6} = \frac{7}{6}

    The expression becomes:

    317−76\frac{31}{7} - \frac{7}{6}
  2. Find a common denominator for 7 and 6.

    The least common multiple of 7 and 6 is 42.

    317=31×67×6=18642\frac{31}{7} = \frac{31 \times 6}{7 \times 6} = \frac{186}{42} 76=7×76×7=4942\frac{7}{6} = \frac{7 \times 7}{6 \times 7} = \frac{49}{42}
  3. Subtract the fractions.

    18642−4942=186−4942=13742\frac{186}{42} - \frac{49}{42} = \frac{186 - 49}{42} = \frac{137}{42}
  4. Convert the improper fraction back to a mixed number.

    13742=3 with a remainder of 137−(3×42)=137−126=11\frac{137}{42} = 3 \text{ with a remainder of } 137 - (3 \times 42) = 137 - 126 = 11

    So, the mixed number is 311423\frac{11}{42}.
    The answer is 31142\boxed{3\frac{11}{42}}.

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