Is there a specific method for these kind of calculations?
The question's from a past paper but we didn't study it.
I can see that you assign the value as x. Multiply by 100 to get a whole number (except it isn't a whole number?).
But how to know that it's got to be subtracted? And how to know what to subtract by?
The idea behind this method is to get rid of the repeating part of the decimal. Let me show you another example to demonstrate.
Let x = 0.333333....
What do we want to multiply this by so we can get rid of the repeating part? Well, we want to shift the decimal over by enough spaces that we can subtract out the repeating part. This will always be the number of decimals repeated. So in this case, there is only one number repeated, so we want to move the decimal over by 1 place: multiply by 10.
10x = 3.3333333.....
Now if we subtract the original number, what happens to the repeating part?
10x = 3.33333333....
...x = 0.33333333.....
10x - x = 9x = 3
Thus x = 3/9 = 1/3.
Another one.
Let x = 0.878787878787....
Here we have two numbers repeating so we want to shift the decimal by two places. So multiply by 100:
100x = 87.8787878787....
......x = .0.8787878787....
99x = 87
x = 87/99 = 29/33
Hopefully that'll give you a good idea as to the method.
-Dan
Hello, GAdams!
Ron is absolutely correct. . [Edit: I meant Dan! . . . *blush*]
I'll take it a step further.
Suppose the repeating cycle does not start at the decimal point.
Example: .N .= .2.7636363 ...
Multiply by 1000: .1000N .= .2763.636363 ...
. .Multiply by 10: . . .10N .= . . 27.636363 ...
. . . . . Subtract: . . 990N .= .2736
. . . . . . . . . . . . . . . . . . . . . 2736 . . . 152
. . . . Therefore: . . . . . N .= .------- .= .----
. . . . . . . . . . . . . . . . . . . . . .990 - . - .55