The population of cattle in a farm increases | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $P_n$ be the population in year $n$. According to the problem,$P_{n+2} - P_n = k P_{n+1}$,where $k$ is a constant.Given populations:$P_{2010}

Vedclass · Accepted Answer

$84$

Vedclass · Answer

(B) Let $P_n$ be the population in year $n$. According to the problem,$P_{n+2} - P_n = k P_{n+1}$,where $k$ is a constant.Given populations:$P_{2010} = 39$$P_{2011} = 60$$P_{2013} = 123$Let $P_{2012} = x$.For $n = 2010$:$P_{2012} - P_{2010} = k P_{2011}$ $\Rightarrow x - 39 = k(60)$ $\Rightarrow k = \frac{x - 39}{60} \quad (i)$For $n = 2011$:$P_{2013} - P_{2011} = k P_{2012}$ $\Rightarrow 123 - 60 = k(x)$ $\Rightarrow 63 = kx$ $\Rightarrow k = \frac{63}{x} \quad (ii)$Equating $(i)$ and $(ii)$:$\frac{x - 39}{60} = \frac{63}{x}$$x(x - 39) = 60 	imes 63$$x^2 - 39x - 3780 = 0$Solving the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{39 \pm \sqrt{(-39)^2 - 4(1)(-3780)}}{2}$$x = \frac{39 \pm \sqrt{1521 + 15120}}{2}$$x = \frac{39 \pm \sqrt{16641}}{2}$$x = \frac{39 \pm 129}{2}$Since population must be positive,$x = \frac{39 + 129}{2} = \frac{168}{2} = 84$.Thus,the population in $2012$ was $84$.

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

Similar Questions

Given the equation $4x^2 + 4(a - 1)x + (1 - 2a) = 0$ has roots $\sin \theta$ and $\cos \theta$ $(0 < \theta < \frac{\pi}{2})$,then the maximum value of $(a + \sin \theta)$ is-

If $\alpha$ is a root of the equation $x^2-x+1=0$,then $\left(\alpha+\frac{1}{\alpha}\right)^3+\left(\alpha^2+\frac{1}{\alpha^2}\right)^3+\left(\alpha^3+\frac{1}{\alpha^3}\right)^3+\left(\alpha^4+\frac{1}{\alpha^4}\right)^3=$

The number of solutions of the equation $x^2 + y^2 = a^2 + b^2 + c^2$,where $x, y, a, b, c$ are all prime numbers,is

If the quadratic equation formed by eliminating $x$ from $x^2+\alpha x+\beta=0$ and $xy+l(x+y)+m=0$ has the same roots as that of the given quadratic equation,then the set of values of $\beta$ is

The number of pairs of real numbers $(x, y)$ such that $x = x^2 + y^2$ and $y = 2xy$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module