Question

A rational number such that it's deniminator is greater than its numerator by a constant k.If the denominator is doubled is the number obtained is again equal to original number. Find the rational number​

Answer 1

Answer:

Let the numerator be x .

Denominator = x+K.

Now, as per the question,

x/2(x+K) = x/(x+K)

i.e. x/2 = x =>x = 2x

which means x = 0.

So, the rational number will 0/(0+K) i.e. 0/K,where K is any constant.

Answer 2

Basic Concept Used :-

Writing System of Linear Equation from Word Problems

1. Understand the problem.

• Understand all the words used in stating the problem.

• Understand what you are asked to find.

2. Translate the problem to an equation.

• Assign a variable (or variables) to represent the unknown.

• Clearly state what the variable represents.

3. Carry out the plan and solve the problem.

Let's solve the problem now!!

$\begin{gathered}\begin{gathered}\bf \:Let-\begin{cases} &\sf{numerator = x} \\ &\sf{denominator = x + k} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: So, \: rational \: number \: is \: - \begin{cases} &\sf{\dfrac{x}{x + k} }\end{cases}\end{gathered}\end{gathered}$

↝According to statement,

$\begin{gathered}\begin{gathered}\bf \:Now-\begin{cases} &\sf{numerator = x} \\ &\sf{denominator = 2(x + k)} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: So, \: rational \: number \: is \: - \begin{cases} &\sf{\dfrac{x}{2(x + k)} }\end{cases}\end{gathered}\end{gathered}$

↝Now, it is given that if the denominator is doubled, the rational number obtained is again equal to original number.

$\rm :\implies\:\dfrac{x}{ \cancel{x + k}} = \dfrac{x}{2 \: \cancel{(x + k})}$

$\rm :\implies\:2x = x$

$\rm :\longmapsto\:2x - x = 0$

$\rm :\longmapsto\:x = 0$

$\begin{gathered}\begin{gathered}\bf \: So, \sf\:rational \: number \: is \: - \begin{cases} &\sf{\dfrac{x}{x + k} = \dfrac{0}{0 + k} = 0 }\end{cases}\end{gathered}\end{gathered}$