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Solve 1 rac34 x ((4)/(9) + (1)/(3)) / (1 rac15 + 1 rac12) using the rule of BODMAS.

Mathematics

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Answer

245486\frac{245}{486}

74 or 1.75\frac{7}{4} \text{ or } 1.75

Step-by-step solution

9 steps

Okay, let's continue with these problems.

.1.a: Solve 134×(49+13)÷(115+112)1\frac{3}{4} \times \left(\frac{4}{9} + \frac{1}{3}\right) \div \left(1\frac{1}{5} + 1\frac{1}{2}\right) using the rule of BODMAS.

  1. Convert all mixed numbers to improper fractions.

    134=1×4+34=741\frac{3}{4} = \frac{1 \times 4 + 3}{4} = \frac{7}{4} 115=1×5+15=651\frac{1}{5} = \frac{1 \times 5 + 1}{5} = \frac{6}{5} 112=1×2+12=321\frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}

    The expression becomes:

    74×(49+13)÷(65+32)\frac{7}{4} \times \left(\frac{4}{9} + \frac{1}{3}\right) \div \left(\frac{6}{5} + \frac{3}{2}\right)
  2. Solve the operations inside the first bracket (Addition).

    (49+13)=(49+1×33×3)=(49+39)=4+39=79\left(\frac{4}{9} + \frac{1}{3}\right) = \left(\frac{4}{9} + \frac{1 \times 3}{3 \times 3}\right) = \left(\frac{4}{9} + \frac{3}{9}\right) = \frac{4+3}{9} = \frac{7}{9}
  3. Solve the operations inside the second bracket (Addition).

    (65+32)\left(\frac{6}{5} + \frac{3}{2}\right)

    The least common multiple of 5 and 2 is 10.

    (6×25×2+3×52×5)=(1210+1510)=12+1510=2710\left(\frac{6 \times 2}{5 \times 2} + \frac{3 \times 5}{2 \times 5}\right) = \left(\frac{12}{10} + \frac{15}{10}\right) = \frac{12+15}{10} = \frac{27}{10}
  4. Substitute the results back into the expression.

    74×79÷2710\frac{7}{4} \times \frac{7}{9} \div \frac{27}{10}
  5. Perform multiplication and division from left to right.

    First, multiply.

    74×79=7×74×9=4936\frac{7}{4} \times \frac{7}{9} = \frac{7 \times 7}{4 \times 9} = \frac{49}{36}

    The expression becomes:

    4936÷2710\frac{49}{36} \div \frac{27}{10}
  6. Perform the division by multiplying by the reciprocal of the divisor.

    4936×1027\frac{49}{36} \times \frac{10}{27}

    Simplify by canceling common factors (10 and 36 are divisible by 2):

    4918×2×5×227=49×518×27\frac{49}{18 \times 2} \times \frac{5 \times 2}{27} = \frac{49 \times 5}{18 \times 27} =245486= \frac{245}{486}

    The final answer is:

    • 245486\boxed{\frac{245}{486}}

    .1.b: Using your knowledge on decimals, solve this: 4.2×1.6×3.40.8×3.2×5.1\frac{4.2 \times 1.6 \times 3.4}{0.8 \times 3.2 \times 5.1}

  7. Rewrite the expression by removing the decimal points.

    Since there are an equal number of decimal places in the numerator and denominator (one in each of the three numbers, totaling three decimal places for both), we can multiply the numerator and denominator by 103=100010^3 = 1000 to clear the decimals without changing the value of the fraction.

    42×16×348×32×51\frac{42 \times 16 \times 34}{8 \times 32 \times 51}
  8. Simplify the expression by canceling common factors.

    42×16×348×32×51\frac{42 \times 16 \times 34}{8 \times 32 \times 51}
    • Cancel 16 with 32: 1632=12\frac{16}{32} = \frac{1}{2}. The expression becomes: 42×1×348×2×51\frac{42 \times 1 \times 34}{8 \times 2 \times 51}
    • Cancel 34 with 51. Both are divisible by 17: 34=17×234 = 17 \times 2 and 51=17×351 = 17 \times 3. So, 3451=23\frac{34}{51} = \frac{2}{3}. The expression becomes: 42×1×28×2×3\frac{42 \times 1 \times 2}{8 \times 2 \times 3}
    • Cancel 2 from the numerator and denominator:
    42×18×3\frac{42 \times 1}{8 \times 3}
    • Cancel 42 with 3: 423=14\frac{42}{3} = 14. The expression becomes: 148\frac{14}{8}
    • Simplify 148\frac{14}{8} by dividing both numerator and denominator by 2:
    14÷28÷2=74\frac{14 \div 2}{8 \div 2} = \frac{7}{4}
  9. Convert the fraction to a decimal (optional, but often expected with decimal problems).

    74=1.75\frac{7}{4} = 1.75

    The final answer is:

    • 74 or 1.75\boxed{\frac{7}{4} \text{ or } 1.75}

    .a: In a class of 70 pupils, 12\frac{1}{2} of them speak English, 25\frac{2}{5} speak... a. What is the percentage of pupils who speak both English...

    This question is incomplete. The information about what the 25\frac{2}{5} of pupils speak is cut off, and the full question for part 'a' is also cut off. Please provide the complete question for me to solve it.

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