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CBSE Class 9 Mathematics Question 10 of 27

Number Systems — Question 3

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Question

Question 3

Express the following in the form pq\dfrac{p}{q}, where p and q are integers and q ≠ 0.

(i) 0.6‾0.\overline{6}

(ii) 0.47‾0.4\overline{7}

(iii) 0.001‾0.\overline{001}

Answer

(i) 0.6‾0.\overline{6}

Let x = 0.6‾0.\overline{6}

x = 0.666666........... (1)

Here one digit 6 is repeated so we multiply both side by 10 in equation (1)

10x = 6.6666.......

we can write

10x = 6.66666.......

10x = 6 + 0.666666...........

From equation (1)

10x = 6 + x

10x - x = 6

9x = 6

x = 69\dfrac{6}{9} = 23\dfrac{2}{3}

Hence, 0.6‾0.\overline{6} = 23\dfrac{2}{3}

(ii) 0.47‾0.4 \overline{7}

Let x = 0.47777777........ (1)

Here one digit 7 is repeated so we multiply both side by 10 in equation (1)

We can write

10x = 4.77777..... (2)

On subtracting equation (2) from equation (1)

10x - x = 4.777777........ - 0.477777.........

9x = 4.300000......

x = 4.39\dfrac{4.3}{9}

Multiplying numerator and denominator by 10 we get,

x = 4.3×109×10\dfrac{4.3 \times 10}{9 \times 10} = 4390\dfrac{43}{90}

Hence, 0.47‾0.4\overline{7} = 4390\dfrac{43}{90}

(iii) 0.001‾0.\overline{001}

Let x = 0.001001001...... (1)

Here three digit 001 are repeating so we multiply both side by 1000 in equation (1)

1000x = 0001.001001........ (2)

On subtracting equation (2) from equation (1)

1000x - x = 0001.001001...... - 0.001001......

999x = 1.0000000.....

x = 1999\dfrac{1}{999}

Hence, 0.001‾0.\overline{001} = 1999\dfrac{1}{999}