Question

The population of cattle in a farm increases so that the difference between the population in year $n$ and $n+2$ is $n+1$. If the populations in year $2010$, $2011$ and $2013$ were $39,60$ and $123$, respectively, then the population in $2012$ was:

• A. $81$
• B. $84$
• C. $87$
• D. $90$

Answer 1

Answer: (B)

$84$

Let the population in $2012$ be $x$.

It is given that population difference between the years $n$ and $n+2$ is proportional to population in the year $n+1$

Thus, $\frac{p_{2013}-p_{2011}}{p_{2012}}=\frac{p_{2012}-p_{2010}}{p_{2011}}$

$\Rightarrow \frac{123-60}{x}=\frac{x-39}{60}\Rightarrow x^2-39x=63\times 60$

$\Rightarrow x^2-39x-3780=0$

$\Rightarrow x=45,84$

But, cattle population increases. Thus $x=84$