Question

The population of a town increases by 6 percent every year . If the present population is 15900 ,find its population a year ago​

Answer 1

$\bold{\underline{\underline{\huge{\sf{AnsWer:}}}}}$

Population of town a year ago was 14946

$\bold{\underline{\underline{\large{\sf{StEp\:by\:stEp\:explanation:}}}}}$

GiVeN :

• The population of a town increases by 6 percent every year
• Present population is 15900

To FiNd :

• Population of town a year ago

SoLuTiOn :

Let the population of the town a year ago be x.

Present population is given as 15900

And the population increases by 6 % every year.

We will calculate 6% of the present population.

$\longrightarrow$$\sf{15900\:\times\:{\dfrac{6}{100}}}$

$\longrightarrow$ $\sf{159\:\times\:6}$

$\longrightarrow$ $\sf{954}$

$\sf{\therefore{6\:percent\:of\:initial\:population\:=\:954}}$

We conclude that the 954 people adds up to the population every year.

So, to calculate the population a year ago, we just have to subtract the present population with the increase in population every year.

$\longrightarrow$ $\sf{15900\:-\:954}$

$\longrightarrow$ $\sf{14946}$

$\sf{\therefore{Population\:of\:town\:a\:year\:ago\:was\:14946}}$

Answer 2

SOLUTION:-

Given:

•The population of a town increases by 6% every year (annually).

•If the present population is 15900.

To find:

The population of town a year ago.

Explanation:

Let the population of the town a year ago be R.

We have,

• Present population is 15900.
• The population increases by 6% per year.

°•°Present population= 106% of R.

$= > 15900 = \frac{106}{100} \times R \\ \\ = > 1590000 = 106R\\ \\ = > R = \frac{1590000}{106} \\ \\ = > R = 15000$

Thus,

The population of the town a year ago was R= 15000.