use Option elimination technique,u will get A

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On dividing $50$ into two parts such that the sum of their reciprocals is $\dfrac{1}{12}$, we get the parts as:

- $20,30$
- $24,26$
- $28,22$
- $36,14$

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Answer: A (20, 30)

Lets say 50 divided into two parts x and 50-x.

Now, reciprocals are 1/x and 1/50-x

1/x + 1/(50-x)= 1/12

Solve, you’ll get x= 20 and 30.

Lets say 50 divided into two parts x and 50-x.

Now, reciprocals are 1/x and 1/50-x

1/x + 1/(50-x)= 1/12

Solve, you’ll get x= 20 and 30.

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$\textrm{let’s assume one part of 50 is $x$,then other will be $(50-x)$}$

$\textrm{Now according to the question sum of reciprocal is $\frac{1}{12}$}$

$\implies$ $\frac{1}{x}+\frac{1}{(50-x)}=\frac{1}{12}$

$\implies$ $\frac{50-x+x)}{x*(50-x)}=\frac{1}{12}$

$\implies$ $50*12=x*(50-x)$

$\implies$ $600=50x-x^2$

$\implies$ $x^2-50x+600=0$

$\implies$ $x^2-30x-20x+600=0$

$\implies$ $x(x-30)-20(x-30)=0$

$\implies$ $(x-30)(x-20)$

$\implies$ $x=30,20$

$\textrm{option a is correct.}$

$\textrm{Now according to the question sum of reciprocal is $\frac{1}{12}$}$

$\implies$ $\frac{1}{x}+\frac{1}{(50-x)}=\frac{1}{12}$

$\implies$ $\frac{50-x+x)}{x*(50-x)}=\frac{1}{12}$

$\implies$ $50*12=x*(50-x)$

$\implies$ $600=50x-x^2$

$\implies$ $x^2-50x+600=0$

$\implies$ $x^2-30x-20x+600=0$

$\implies$ $x(x-30)-20(x-30)=0$

$\implies$ $(x-30)(x-20)$

$\implies$ $x=30,20$

$\textrm{option a is correct.}$