Question

the denominator of a fraction is one more than twice its numerator if the numerator and the denominator are both decreased by 1 then the number obtained is 1/3.find the fraction ​

Answer 1

$\bf{\large{\underline{\underline{Answer:-}}}}$

The required fraction is 3/7.

$\bf{\large{\underline{\underline{Explanation:-}}}}$

Given :- The denominator of a fraction is 1 more than twice its numerator. If numerator and denominator both are decreased by 1 then tje number obtained is 1/3

To find :- Required fraction

Solution :-

Let the numberator of a fraction be 'x'

Denominator of a fraction = 1 more than twice its numerator = 1 more than twice 'x' = 2(x) + 1 = 2x + 1 = 2x + 1

When numerator and denominator both are decreased by 1

Numerator of the fraction When numerator and denominator both are decreased by 1 = x - 1

Denominator of the fraction When numerator and denominator both are decreased by 1 = (2x + 1) - 1 = 2x + 1 - 1 = 2x

Given that fraction turms to 1/3 When numerator and denominator both are decreased by 1

From the information given we can form an equation

Equation formed :-

$\dfrac{x - 1}{2x} = \dfrac{1}{3}$

By cross multiplication :-

$(x - 1)3 = 1(2x)$

$3x - 3 = 2x$

$3x - 2x = 3$

$x = 3$

Numerator of the fraction = x = 3

Denominator of the fraction = 2x + 1 = 2(3) + 1 = 6 + 1 = 7

Therfore the required fraction is 3/7.

$\bf{\large{\underline{\underline{Verification:-}}}}$

Let us check

$\dfrac{3 - 1}{7 - 1} = \dfrac{1}{3}$

$\dfrac{2}{6} = \dfrac{1}{3}$

$\dfrac{2 \div 2}{6 \div 2} = \dfrac{1}{3}$

$\dfrac{1}{3} = \dfrac{1}{3}$

Answer 2

Hey there

refer to attachment