Question

duvide the sum of 65/12and 8/3 by their difference​

Answer 1

Answer:

Let A=\frac{65}{12}\: and \:</p><p>B = \frac{8}{3}

i ) Sum =A+B = \frac{65}{12}+\frac{8}{3}

=\frac{65+32}{12}

=\frac{97}{12} ---(1)

ii) Difference = A-B=\frac{65}{12}-\frac{8}{3}

=\frac{65-32}{12}

=\frac{33}{12} ---(2)

According to the problem given,

\frac{A+B}{A-B}

= \frac{\frac{97}{12}}{\frac{33}{12}}

= \frac{97}{33}

Therefore,

Divide the sum of 65/12 and 8/3 by their difference =\frac{97}{33}

Answer 2

Answer:

Let A=\frac{65}{12}\: and \:</p><p>B = \frac{8}{3}

i ) Sum =A+B = \frac{65}{12}+\frac{8}{3}

=\frac{65+32}{12}

=\frac{97}{12} ---(1)

ii) Difference = A-B=\frac{65}{12}-\frac{8}{3}

=\frac{65-32}{12}

=\frac{33}{12} ---(2)

According to the problem given,

\frac{A+B}{A-B}

= \frac{\frac{97}{12}}{\frac{33}{12}}

= \frac{97}{33}

Therefore,

Divide the sum of 65/12 and 8/3 by their difference =\frac{97}{33}