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  • The value of 1.999 in the form pbyq where p and q are integers and q is not equal to 0, is (A)19by10 (B)1999by1000 (C)2 (D)1by9

The value of 1.999 ___in the form \frac{p}{q}  where p and q are integers and q \neq0, is
(A)\frac{19}{10}
(B)\frac{1999}{1000}
(C)2
(D)\frac{1}{9}

 

Answers (2)

Answer.       [C]
Solution.        
Let x = 1.999……
Since, one digit is repeating, we multiply x by 10
we get, 10x = 19.999……..
so, 10x = 18 + 1.999………
   10x = 18 + x
Therefore, 10x – x = 18, i.e., 9x = 18
i.e., x= \frac{18}{9}= \frac{2}{1}= 2

Hence option C is correct answer.

Posted by

infoexpert27

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Answer.          [C]
Solution.        
Let x = 1.999……
Since, one digit is repeating, we multiply x by 10
we get, 10x = 19.999……..
so, 10x = 18 + 1.999………
   10x = 18 + x
Therefore, 10x – x = 18, i.e., 9x = 18

i.e., x= \frac{18}{9}= \frac{2}{1}= 2

Hence option C is correct answer.

Posted by

infoexpert27

View full answer

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