IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

deep01

THE VALUE OF 0.423 IS??

(a) 419/999

(B)419/990

(C)423/100

(D) NOT

KAB

The answer is (B)

Method:

1)The numerator of the fraction is obtained by subtracting the non-recurring figure from the given figure.

2)The denominator consists of as many 9's as there are recurring figures and as many zeros as there are non recurring figures.

Here the number is 0.4232323......

So 0.423=(423-4)/990=419/990 because 4 is non-recurring and there are two recurring figures(2and 3).

deep01

kab sorry i couldn't understand what is the meaning of the recurring terms which u r using and pls again tell me the method if u can?

vish0001

yaar, there is a method by the use of derivatives as well , a weird method , check this out in an NCERT book in the chapter derivatives !

KAB

The digits which repeats itself infinite number of times in a number are called recurring figures.
Here the number is 0.423232323........
So here 23 is repeated infinite number of times.
So number of recurring figures is 2.

Here 4 is non recurring.

yahiyafirdous · 15/03/2007 14:54 Subject: an interestiong question

Q

15/03/2007 14:54 Subject: an interestiong question

Q-THE VALUE OF 0.423 IS??

Let x=0.0232323_..........

10x=0.232323

1000x=23 + 0.232323=23+10x

x= 23/(1000-10)

Hence 0.4232323 = 0.4 + x=0.4 + 23/(1000-10)=[(400-4) + 23]/(1000-10)

=419/990 Ans

NOTE: This is proof , how other people arrived at this conclusion directly.

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