Question

In 2001 the world’s population was 250000. With a growth rate of approximately \[14\% \]. What was the population in 2011?

Answer 1

Hint: Here we write the initial population, growth rate and number of years after which we have to find the population. Substitute the values in the formula to find the population in 2011.
* If A is the initial population and B is the population after n period of time (in years) with a growth rate of r% then we can write the formula for population as \[B = A \times {\left[ {1 + \dfrac{r}{{100}}} \right]^n}\]

Complete step-by-step answer:
We are given the initial population in 2001 as 250000
\[A = 250000\]
Rate of growth is given \[14\% \]
\[r = 14\% \]
The time period from 2001 and 2011 is \[2011 - 2001 = 10\]years.
\[n = 10\]
Substitute the values of\[A = 250000\],\[r = 14\% \] and\[n = 10\]in the formula\[B = A \times {\left[ {1 + \dfrac{r}{{100}}} \right]^n}\]
\[ \Rightarrow B = 250000 \times {\left[ {1 + \dfrac{{14}}{{100}}} \right]^{10}}\]
Calculate the value inside the bracket by taking LCM
\[ \Rightarrow B = 250000 \times {\left[ {\dfrac{{100 + 14}}{{100}}} \right]^{10}}\]
\[ \Rightarrow B = 250000 \times {\left[ {\dfrac{{114}}{{100}}} \right]^{10}}\]
Cancel the same factors from numerator and denominator.
\[ \Rightarrow B = 250000 \times {\left[ {\dfrac{{57}}{{50}}} \right]^{10}}\]
Use the formula \[{\left( {\dfrac{a}{b}} \right)^n} = \dfrac{{{a^n}}}{{{b^n}}}\]to open the values in the bracket
\[ \Rightarrow B = 250000 \times \left[ {\dfrac{{{{57}^{10}}}}{{{{50}^{10}}}}} \right]\]
Use the formula \[{(pq)^m} = {p^m}{q^m}\]to open the denominator.
\[ \Rightarrow B = 250000 \times \left[ {\dfrac{{{{57}^{10}}}}{{{5^{10}} \cdot {{10}^{10}}}}} \right]\]
Use the formula \[{p^{m + n}} = {p^m} \cdot {p^n}\]to break the terms in the denominator.
\[ \Rightarrow B = 250000 \times \left[ {\dfrac{{{{57}^6} \cdot {{57}^4}}}{{{5^2} \cdot {5^8} \cdot {{10}^6} \cdot {{10}^4}}}} \right]\]
\[ \Rightarrow B = 250000 \times \left[ {\dfrac{{{{57}^6} \cdot {{57}^4}}}{{250000 \cdot {5^8} \cdot {{10}^6}}}} \right]\]
Cancel same terms from numerator and denominator.
\[ \Rightarrow B = \dfrac{{{{57}^6} \cdot {{57}^4}}}{{{5^8} \cdot {{10}^6}}}\]
\[ \Rightarrow B = \dfrac{{{{57}^6}}}{{{5^8}}} \times \dfrac{{{{57}^4}}}{{{{10}^6}}}\]
Open the powers with multiplication.
\[ \Rightarrow B = \dfrac{{34296447249}}{{390625}} \times \dfrac{{10556001}}{{1000000}}\]
Solve the first fraction
\[ \Rightarrow B = 87798.90495744 \times \dfrac{{10556001}}{{1000000}}\]
Multiply the numerator with the term obtained
\[ \Rightarrow B = \dfrac{{926805328529.64}}{{1000000}}\]
Shift the decimal to six places to the left from its original place
\[ \Rightarrow B = 926805.32852964\]

So, the population in 2011 is 926805 approximately.

Note: Students are likely to make mistakes in substituting the value of r as 0.14 as it is given in percentage form. Keep in mind when the formula has \[\dfrac{r}{{100}}\] then we substitute only the numeric part of the percentage, else we convert the percentage into fraction or decimal and then substitute. Also, while performing the complex calculations, move step by step to avoid calculation mistakes.