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Out of three given numbers, the first one is twice the second and three times the third. If the average of these numbers is $88,$ then the difference between the second and third is :

  1. $24$
  2. $48$
  3. $72$
  4. $96$
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Let’s assume that first number is $x$, second number is $x/2$ third number is $x/3$.

Now according to the given questions

$\implies\frac{x+\frac{x}{2}+\frac{x}{3}}{3}=88$

$\implies \frac{6x+2x+3x}{6}=88*3$

$\implies 11x=88*18$

$\implies x=144$

therefore

  1. first number =$144$
  2. second number =$72$
  3. third number= $48$

So the difference between the second and third numbers is =$72-48=24$

Option $(A)$ is correct.

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