Aptitude - Decimal Fraction - Discussion

6.4356565656.

Let's just solve the digits after.
0.4356565656.

According to the rule;
0.4356.

Now subtract 4356 from the first 2 digits which are not repeating;
4356-43=4313.
According to rule, you put 4313 in the nominator and in the denominator put 9900 (numbers which are not repeating are 2, so put two 9's and the remaining 2 are the repeating digits, so put 00 for that and the final digit will be 9900).

4313/9900.

6 + 4313/9900.

Or if you further solve it then 63713/9900 will be the final answer.

Thanks I hope it will help.

LUK EVRY 1 ITS QUITE SIMPLE:

first let us suppose the decimal as x i.e x=0.232323... eq 1

after dat as we can see the digits r repeating after 2 digits

i.e. 23 so we just have to multiply the decimal no. by 100 and

dat will make 100x=23.232323...eq2

now simply subtract eq2 by 1 u will get the required result

Hay, I will explain it;
Let see;

x = 0.232323232323 ----------> 1
x = 23.23232323..../100 right

After cross multiplication we get,
100 x = 23.2323232323 -------> 2.

after subtracting Eqn 1 &2.
100x - x = 23.23232323 - 0.23232323...
99x = 23 right
Finally, we get;
x = 23/99 That's all.

When a number repeating continiously, add number of 9's.

See formula number 6.

Formulas : http://www.indiabix.com/aptitude/decimal-fraction/formulas

Let r = 0.232323 but 23 keep repeating. So let 100r = 23.232323.

Since the decimal move two times then 100r-r = 23.232323 - 0.232323.

99r = 23.

Therefore r = 23/99.

It is simple. All you need to do is just put as many as nines in denominator which are recurring or under the bar. In numerator remove the bar sign.

There are dots (....) after 23 i.e. 23 is repeated infinitely .232323232323232323...

But only two 'digits' - 2&3 are repeating so taken 2 9's.

Write the repeated figures only once in the numerator and take as many nines in the denominator as is the number of repeating figures.

@Eddy.

I still don't get it. Why should we do 100r-r? Can this concept be explained with another equation perhaps?

Thank you.

If only 2 digits are repeating then you hav to multiply by 100.

If 3 digits are repeating the multiply by 1000 and so on.