Question

The population of a city at previous consecutive census years were 4,00,000, 5,58,500, 7,76,000 and 10,98,500. The anticipated population at the next census to the nearest 5,000 would be_______

• A. 1415000

Answer 1

The correct option is A 1415000
The problem is solved by incremental increase method since method is not mentioned. Incremental increase method gives results somewhere between arithmetic increase method and geometric increase method. Moreover, it is considered best for any city whether old or new.

$$\begin{array}{cccc}\text{Census year}& \text{Population}& \text{Population Increase (x)}& \text{Incremental increase (y)}\\ \text{1.}& \text{4,00,000}& & \\ \text{2.}& \text{5,58,500}& \text{158500}& \\ \text{3.}& \text{7,76,000}& \text{217500}& \text{59000}\\ \text{4.}& \text{10,98,500}& \text{322500}& \text{1,05,000}\\ & & \overline{x}=\frac{\sum x}{3}=232833.3& \overline{y}=\frac{\sum y}{2}=8200\end{array}$$


$P_n=P_0+n\overline{x}+\frac{n(n+1)}{2}\overline{y}$
$P_5=P_0+\overline{x}\times 1+\left[1\right(1+1\left)/2\right]\times \overline{y}$
$P_5=10,98,500+232833.33+82000$
$P_5=1413333.33$
$\simeq 1415000$ (to the nearest $5000$)