Question

the denominator of a fraction is 1 less than twice the numerator if the numerator and denominator are both are increased by 1 the fraction becomes 3 by 5 find the fraction ​

Answer 1

$\bold {Question:}$

The denominator of a fraction is 1 less than twice the numerator if the numerator and denominator both are increased by 1 the fraction becomes 3 by 5, Find the fraction.

$\bold {Solution:}$

Let the numerator be x.

Then, the denominator =2x-1

Atq,

$= > \frac{x + 1}{(2x - 1) + 1} = \frac{3}{5} \\ = > \frac{x + 1}{2x - 1 + 1} = \frac{3}{5} \\ = > \frac{x + 1}{2x} = \frac{3}{5 } \\ = > 5(x + 1) = 2x \times 3 \\ = > 5x + 5 = 6x \\ = > 5x - 6x = - 5 \\ = > - x = - 5 \\ = > x = 5$

Numerator =5

Denominator

$= (2\times 5-1)\\=10-1</p><p>\\=9$

$\fbox{\green{Required\:Fraction=\frac{5}{9}}}$

$\bold {Verification:}$

$= > \frac{x + 1}{(2x - 1) + 1} = \frac{3}{5} \\ = > \frac{5 + 1}{(2 \times 5 - 1) + 1} = \frac{3}{5} \\ = > \frac{6}{(10 - 1) + 1} = \frac{3}{5} \\ = > \frac{6}{9 + 1} = \frac{3}{5} \\ = > \frac{6}{10} = \frac{3}{5} \\ = > \frac{3}{5} = \frac{3}{5} \\=>LHS=RHS$

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Answer 2

$\mathtt{ \purple{ \huge{\fbox{ \: Solution : \: \: }}}}$

Let , the numerator be x

So , denominator = 2x - 1

It is given , if the numerator and denominator both are increased by 1 the fraction becomes 3 by 5

Thus ,

$\sf \hookrightarrow \frac{x + 1}{2x - 1 + 1} = \frac{3}{5} \\ \\\sf \hookrightarrow 5x + 5 = 6x \\ \\\sf \hookrightarrow x = 5$

Hence , fraction is 6/10

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