Question

Express the following in p/q from, where p and q are integers and q not equal to zero 1) 0.00323232.... 2) 0. 001001001 3)0.1343434...

Answer 1

The number will be 0.0032(bar)

Now we will assume this number as x

Now we will multiply x with the number of digits under the bar( in this case 32)

So we will multiply x by 100 and so we will multiply 0.0032(bar) too

We will get
100x = 0.32(bar) — This will be equation number "1"

We will again multiply 100x with 100 to get the second equation and to get rid of the bar in the number after subtraction-
10000x = 32.32(bar)—This will be our second equation "2"

Now we will subtract "1" from "2"

10000x-100x =32.32(bar)-0.32(bar)

The bars will get cancelled and we will get 9900x = 32 and the value of x will come as 32 divided by 9900( 0.323232.....)

Answer 2

Answer and Explanation:

To find : Express the following in p/q from, where p and q are integers and q not equal to zero?

Solution :

1) 0.00323232....

Let $x=0.00323232....$ ....(1)

Multiply both side by 100,

$100x=0.323232...$ .....(2)

Subtract (1) and (2),

$100x-x=(0.323232...)-(0.00323232....)$

$99x=0.32$

$x=\frac{0.32}{99}$

$x=\frac{32}{9900}$

$x=\frac{8}{2475}$

2) 0. 001001001

Let $y=0.001001001$

Remove decimal,

$y=\frac{001001001}{1000000000}$

$y=\frac{1001001}{1000000000}$

3) 0.1343434...

Let $z=0.1343434...$ ....(1)

Multiply both side by 100,

$100z=13.43434...$ .....(2)

Subtract (1) and (2),

$100z-z=(13.43434....)-(0.1343434....)$

$99z=13.3$

$z=\frac{13.3}{99}$

$z=\frac{133}{990}$