Question

Let $X=\{1,2,3\}$. A number is selected from set $X$ with replacement and this process is done 1100 times. For each $i,1\le i\le 3$. 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 $S^3=162f(1)f(2)f(3)$.
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)

$\Theta S=f(1)+2f(2)+3f(3)$
$\Theta AM\ge GM\Rightarrow \frac{S}{3}\ge (6f(1)f(2)f(3){)}^{1/3}$
$\Rightarrow S^3\ge 162f(1)f(2)f(3)$
$\text{ but given }{S}^3=162f(1)f(2)f(3)$
$\therefore f(1)=2f(2)=3f(3)$
$\&f(1)+f(2)+f(3)=1100$
$\therefore f(1)=600,f(2)=300\text{ and }f(3)=200$
$\therefore HM=\frac{1}{\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}}=\frac{300}{\frac{1}{6}+\frac{1}{3}+\frac{1}{2}}=300$