Question

divide the sum of 65/12 and 8/3 by their difference and add 1/2 for resultant​

Answer 1

Answer:

First we have to find the sum (65/12) + (8/3)

LCM of 12 and 3 is 12

Then,

(65 × 1)/(12 × 1) = (65/12) and (8 x 4)/(3 × 4) = (32/12)

Now,

(65 +32)/12

= (97/12)

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Now find the difference of (65/12) - (8/3)

LCM of 12 and 3 is 12

Then, (65 × 1)/(12 × 1) = (65/12)

(8 × 4)/(3 × 4) = (32/12)

Now,

= (65-32)/12

= (33/12)

Now divide (97/12) = (33/12)

(97 × 12) × (12/33)

= (97 × 12)/(12 × 33)

= (97 × 1)/(1 × 33)

= (97/33)

Answer 2

To solve : divide the sum of $\frac{65}{12}$  and $\frac{8}{3}$  by their difference and add $\frac{1}{2}$ for resultant​ .

Solution :

• According to question we have to divide the sum of $\frac{65}{12}$  and $\frac{8}{3}$  by their difference and add $\frac{1}{2}$ for resultant​ .

• Therefore , firstly we find sum of $\frac{65}{12}$  and $\frac{8}{3}$    by cross multiplying both terms .
• $\frac{65}{12} +\frac{8}{3} = \frac{195 + 96}{36 }$

                  $=\frac{291}{36}$

                 $= \frac{97}{12}$

• Now , we find difference between  $\frac{65}{12}$  and $\frac{8}{3}$ by cross multiplying both terms .
• $\frac{65}{12} - \frac{8}{3} = \frac{195 -96}{36}$

                   $= \frac{99}{36} \\\\= \frac{11}{4}$

• Now dividing sum of  $\frac{65}{12}$  and $\frac{8}{3}$  by their difference and adding $\frac{1}{2}$ for resultant​ .  

       $(\frac{97}{12}\div \frac{11}{4} )+ \frac{1}{2}$

        $=( \frac{97}{12} \times\frac{4}{11} ) +\frac{1}{2} \\\\= (\frac{97}{3\times 11}) + \frac{1}{2} \\\\= \frac{97}{33} +\frac{1}{2} \\\\= \frac{194 + 33 }{66} \\\\= \frac{227}{66}$

Hence , resultant answer on adding $\frac{1}{2}$ is $\frac{227}{66}$ .