Class 9 Math Exercise 1.3 | Question 3

Question 3: Express the following in the form \\frac{p}{q}), where p and q are integers and q ≠ 0:
i) 0.\\overline{6})
ii) 0.4\\overline{7})
iii) 0.\\overline{001})

Explanatory Answer | Exercise 1.3 Question 3

i) 0.\\overline{6})

Let 0.\\overline{6}) = x.
As there is only one digit that is recurring, let us multiply x by 10.
So, 10x = 6.66666666...
We know x = 0.66666666...
Notice that the decimal part of both 10x and x are the same.
We choose the number to multiply x with keeping in mind that the decimal components both parts are same. We can then subtract one from the other and arrive at an integer as the answer for some multiple of x as shown below.

So, 10x can be written as x + 6
i.e., 10x = 6 + x
or 9x = 6
or x = \\frac{6}{9}) = \\frac{2}{3})
So, 0.\\overline{6}) = \\frac{2}{3})

0.4\\overline{7})

0.4\\overline{7}) be x
The recurring part starts from the second decimal place, so the part that is recurring is 7777....
That is the component that we have to subtract to get an integer.
So, what should we multiply x with?
First by 100.

100x = 47.77777....
The decimal component of 100x is 0.777...
So, subtracting x from 100x will not give us an integer because the decimal part of 100x and x are not the same.
So, let us find such a multiple of x which will have the same decimal component as 100x.
The decimal part of 10x will also be the 0.777...
So, let us subtract 10x from 100x.

100x = 47.77777....
10x = 4.77777777....
Therefore, 100x - 10x = 43
90x = 43

Or x = \\frac{43}{90})
The fractional equivalent of 0.4\\overline{7}) is \\frac{43}{90})

0.\\overline{001})

Let 0.\\overline{001}) be x
As there are three digits that are recurring, we must multiply by 1000.
So, 1000x = 1.\\overline{001})
x = 0.001001001001001001001001
Therefore, 1000x = 1 + x
999x = 1

x = \\frac{1}{999})
The fractional equivalent of 0.\\overline{001}) is \\frac{1}{999})