Question

The present population of a village is 130,000. If the population increases each year by $4\% $ of what it had been at the beginning of each year, then the population of the village after four years will be x. Find the sum of the digits of x.

Answer 1

Hint:
Given the present population of a village and the rate at which it increases every year of what it had been at the beginning of each year. After 4 years by assuming that the population be P then we can calculate it by using the formula
 $P = 130,000 \times {(1 + \dfrac{r}{{100}})^n}$ Where r is the rate at which the population increases and n is the number of years.

Complete step by step solution:
Let us assume that the population of the village after four years be P
Since we know that if the population increase each year by $4\% $ of what it had been at the beginning of each year so $r = 4$ here
And we have to calculate the population of the village after four years so $n = 4$
Now by using the formula $P = 130,000 \times {(1 + \dfrac{r}{{100}})^n}$ and putting the values of r and n we get
 $ \Rightarrow P = 130,000 \times {(1 + \dfrac{4}{{100}})^4}$
On simplifying the bracket, we get,
 $ \Rightarrow P = 130,000 \times {(1 + 0.04)^4}$
On adding terms inside the bracket, we get,
 $ \Rightarrow P = 130,000 \times {(1.04)^4}$
On simplifying the last term, we get,
 $ \Rightarrow P = 130,000 \times 1.16986154$
On multiplication we get,
 $ \Rightarrow P = 152,082$
Now the population is $x = 152,082$
Therefore, the sum of digits of is $x = 1 + 5 + 2 + 0 + 8 + 2 = 18$

Hence the answer is 18.

Note:
In this question, we assume the value of the population after 4 years ago with any of the variables and then use it to find the original value. Also, we will use the formula for calculating the present population
 $P = {P_i} \times {(1 + \dfrac{r}{{100}})^n}$
Where P is the population after n years, ${P_i}$ is the initial population and r is the rate of increment every year.