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A positive number when decreased by $4$ is equal to $21$ times the reciprocal of the number. The number(s) is/are :

  1. $3$
  2. $7$
  3. $3$ and $-7$
  4. $-3$ and $7$
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Let the positive number be $x$.

so $x-4=\frac{21}{x}$

$\implies x(x-4)=21$

$\implies x^2-4x-21=0$

$\implies x^2-7x+3x-21=0$

$\implies x(x-7)+3(x-7)=0$

$\implies (x-7)(x+3)=0$

$\implies x=7,-3$

Option (D) is correct.
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