Question

Of the three numbers, the first is twice the second and the second is twice the third. The average of the recriprocal of the numbers is $\frac{7}{72}$. The numbers are:

• A. $16,8,4$
• B. $20,10,5$
• C. $24,12,6$
• D. $36,18,9$

Answer 1

Answer: (C)

$24,12,6$

Let the three numbers be $x,y$ and $z$

Given

$x=2y$

So,

$x=4z$

$y=2z$

$z=z$

The average of the reciprocal numbers is $\frac{7}{72}$

$\Rightarrow \frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}{3}=\frac{7}{72}$

$\Rightarrow \frac{yz+xz+xy}{3xyz}=\frac{7}{72}$

$\Rightarrow \frac{2z^2+4z^2+8z^2}{3\times 4z\times 2z\times z}=\frac{7}{72}$

$\Rightarrow \frac{14z^2}{24z^3}=\frac{7}{72}$

$\Rightarrow 84z=504$

$\Rightarrow z=6$

So, $x=4z=4\times 6=24$

$y=2z=2\times 6=12$

Thus the numbers are $24,12$ and $6$.