Question

Let $X =\{1,2,3\}$. A number is selected from set $X$ with replacement and this process is done 1100 times. For each $i , 1 \leq i \leq 3$. Let $f$ (i) represent the number of times the number $i$ is selected. Clearly $f(1)+f(2)+f(3)=1100$.
Also, let $S$ denote the total sum of 1100 numbers selected. If $S^3=162 f(1) f(2) f(3)$.
The value of harmonic mean of $f(1), f(2)$ and $f(3)$ is

• A. 100
• B. 200
• C. 300
• D. 400

Answer 1

Answer: (C)

$\Theta S=f(1)+2 f(2)+3 f(3) $
$\Theta A M \geq G M \Rightarrow \frac{S}{3} \geq(6 f(1) f(2) f(3))^{1 / 3} $
$\Rightarrow S^3 \geq 162 f(1) f(2) f(3) $
$\text { but given } S^3=162 f(1) f(2) f(3) $
$\therefore f(1)=2 f(2)=3 f (3)$
$\& f(1)+f(2)+f(3)=1100$
$\therefore f(1)=600, f(2)=300 \text { and } f(3)=200 $
$\therefore H M=\frac{1}{\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}}=\frac{300}{\frac{1}{6}+\frac{1}{3}+\frac{1}{2}}=300$