1.

The value of `1.999`….. In the form of `p/q`, where p and q are integers and `q ne 0`, isA. `(19)/(10)`B. `(1999)/(1000)`C. `2`D. `1/9`

Answer» Correct Answer - C
Let `x = 1.999"…."`
Now, `10x = 19.999"…."`
On subtracting Eq. (i) from Eq. (ii), we get
`10x - x = (19.999…) - (1.9999……)`
`rArr 9x = 18`
`:. x = (18)/(9) = 2`


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