The value of 1.999... in the form frac{p}{q}, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $x = 1.999... = 1.\overline{9} \quad \dots(1)$Multiply both sides by $10$:$10x = 19.999... = 19.\overline{9} \quad \dots(2)$Subtract equation

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let $x = 1.999... = 1.\overline{9} \quad \dots(1)$Multiply both sides by $10$:$10x = 19.999... = 19.\overline{9} \quad \dots(2)$Subtract equation $(1)$ from equation $(2)$:$10x - x = 19.999... - 1.999...$$9x = 18$Divide by $9$:$x = \frac{18}{9} = 2$Since $2$ can be written as $\frac{2}{1},$ where $p=2$ and $q=1$ are integers and $q 
eq 0,$ the value is $2.$Hence,option $(b)$ is the correct answer.

The value of $1.999...$ in the form $\frac{p}{q},$ where $p$ and $q$ are integers and $q \neq 0,$ is

Similar Questions

For each question,select the proper option from four options given,to make the statement true: $.............$ is a rational number between $7$ and $8$.

Insert a rational number and an irrational number between the following:
$6.375289$ and $6.375738$

Rationalise the denominator in each of the following:
$\frac{1}{7-4 \sqrt{3}}$

Simplify: $(\frac{1}{27})^{\frac{-2}{3}}$

Insert a rational number and an irrational number between the following:
$0.15$ and $0.16$

Vedclass Test Series

Exam Paper Generator

Online Exam Module