Question

16. The present population of a city is 1,80,000.
If it increases at the rate of 8% per annum, its
population after 2 years will be
(a) 2,07,800
(b) 2,09,952
(c) 2,05,992
(d) 2,08,700​

Answer 1

$\huge{\underline{\boxed{\boxed{\red{\mathcal{QUESTION:}}}}}}$

The present population of a city is 1,80,000.  If it increases at the rate of 8% per annum, its  population after 2 years will be

:

1. 2,07,800
2. 2,09,952
3. 2,05,992
4. 2,08,700

$\huge{\underline{\boxed{\boxed{\red{\mathcal{SOLUTION:}}}}}}$

$\star{\bf{\underline{\blue{Given:}}}}$

• Present population of a city (P) = 1,80,000
• Rate of increase (r) = 8%
• Time (t) = 2 years

$\star{\bf{\underline{\blue{To\;Find:}}}}$

• Population after 2 years.

$\star{\bf{\underline{\blue{Formula\;used:}}}}$

• $\sf{Population\;after\;2\;years = P\Bigg(1+\dfrac{r}{100}\Bigg)^{t}}$

Now, put the values in the formula,

$\implies{\sf{Population\;after\;2\;years = P\Bigg(1+\dfrac{r}{100}\Bigg)^{t}}} \\ \\ \\ \implies{\sf{Population\;after\;2\;years=1,80,000\Bigg(1+\dfrac{8}{100}\Bigg)^{2}}}\\ \\ \\ \implies{\sf{Population\;after\;2\;years=1,80,000\Bigg(\dfrac{108}{100}\Bigg)^{2}}}\\ \\ \\ \implies{\sf{Population\;after\;2\;years=1,80,000\Bigg(\dfrac{19440}{10000}\Bigg)}}\\ \\ \\ \implies{\sf{Population\;after\;2\;years=18 \times 19440}}\\ \\ \\ \implies{\sf{Population\;after\;2\;years=3,49,920}}$

Hence, Population after 2 years = 3,49,920.