NCERT Exemplar Solutions
Class 6 - Mathematics - Unit 4: Fractions & Decimals - Problems and Solutions
Question 98

Question. 98

Subtract \(8\dfrac{1}{3}\) from \(\dfrac{100}{9}\).

Answer:

\(\dfrac{25}{9}=2\dfrac{7}{9}\)

Detailed Answer with Explanation:

We want to do: \(\dfrac{100}{9} - 8\dfrac{1}{3}\).

  1. Change the mixed number to an improper fraction.

    Multiply the whole number by the denominator and add the numerator:

    \[8\times 3 + 1 = 24 + 1 = 25\]

    So, \(8\dfrac{1}{3} = \dfrac{25}{3}\).

  2. Make the denominators the same (9).

    Convert \(\dfrac{25}{3}\) to ninths:

    \[\dfrac{25}{3} = \dfrac{25\times 3}{3\times 3} = \dfrac{75}{9}\]

  3. Subtract the fractions.

    Now both have denominator 9:

    \[\dfrac{100}{9} - \dfrac{75}{9} = \dfrac{100 - 75}{9} = \dfrac{25}{9}\]

  4. Write as a mixed number (optional).

    \(25\div 9 = 2\) with remainder \(7\):

    \[\dfrac{25}{9} = 2\dfrac{7}{9}\]

Final answer: \(\dfrac{25}{9}\) or \(2\dfrac{7}{9}\).

NCERT Exemplar Solutions Class 6 – Mathematics – Unit 4: Fractions & Decimals – Problems and Solutions | Detailed Answers