Question

Answer this Question !! ​

Answer 1

Answer:

$\frac{1}{3}$

Step-by-step explanation:

$\frac{m}{2n} = \frac{3}{4} \\ \\ \frac{3m}{2n} = \frac{3}{4} \\ \\ \frac{3m}{n} = \frac{6}{4} \\ \\ \\ \frac{3m - n}{n} = \frac{6 - 4}{4} \\ \\ \frac{3m - n}{n} = \frac{2}{4} \\ \\ \frac{3m - n}{n} = \frac{1}{2} \: \: \: \: \: equation \: \: \: \: (i) \\ \\ \\ \\ \\ \\ \frac{m}{2n} = \frac{3}{4} \\ \\ \\ \frac{m}{n} = \frac{6}{4} \\ \\ \frac{m}{3n} = \frac{6}{12} \\ \\ \frac{m}{3n} = \frac{1}{2} \\ \\ \\ \frac{m + 3n}{3n} = \frac{1 + 2}{2} \\ \\ \frac{m + 3n}{3n} = \frac{3}{2} \\ \\ \\ \frac{m + 3n}{3n} = \frac{3}{2} \: \: \: \: \: \: equation \: \: \: \: \: \: (ii) \\ \\ \\ \\ from \: (i) \: \: \: and \: \: \: \: (ii) \\ \\ \\ \\ \frac{3m - 2n}{m + 3n} = \frac{1}{2} \times \frac{2}{3} \\ \\ \\ \frac{3m - 2n}{m + 3n} = \frac{ \frac{1}{3} }{1} = \frac{1}{3}$

Answer 2

Answer:

$\sf \dashrightarrow \ \dfrac{3m - n}{m + 3n} = \dfrac{7}{9}$

Step-by-step explanation:

We've been given that;

$\sf \dashrightarrow \ \dfrac{m}{2n} = \dfrac{3}{4}$

On transposing 2n we get;

$\sf \dashrightarrow \ \dfrac{m}{1} = \dfrac{3 \times 2n}{4}$

On dividing 2n and 4 we get;

$\sf \dashrightarrow \ \dfrac{m}{1} = \dfrac{3 \times n}{2}$

$\sf \dashrightarrow \ m = \dfrac{3n}{2}$

On substituting this value in the fraction given in the question we get;

$\sf \dashrightarrow \ \dfrac{3m - n}{m + 3n}$

$\sf \dashrightarrow \ \dfrac{3\left(\dfrac{3n}{2}\right) - n}{\left(\dfrac{3n}{2}\right) + 3n}$

On taking LCM we get;

$\sf \dashrightarrow \ \dfrac{\bigg(\dfrac{9n}{2} - n\bigg)}{\bigg(\dfrac{3n}{2} + 3n\bigg)}$

$\sf \dashrightarrow \ \dfrac{\bigg(\dfrac{9n - 2n}{2}\bigg)}{\bigg(\dfrac{3n + 6n}{2} \bigg)}$

$\sf \dashrightarrow \ \bigg(\dfrac{9n - 2n}{2}\bigg) \times \bigg(\dfrac{2}{3n + 6n} \bigg)$

$\sf \dashrightarrow \ \bigg(\dfrac{9n - 2n}{3n + 6n} \bigg)$

$\sf \dashrightarrow \ \bigg(\dfrac{7n}{9n} \bigg)$

$\sf \dashrightarrow \ \dfrac{7}{9}$

Therefore;

$\sf \dashrightarrow \ \boxed{\bold{\dfrac{3m - n}{m + 3n} = \dfrac{7}{9}}}$

Hence solved.