Question

4. The population of a town increases by 10% annually. If the present population is 22000 find population 1 year ago

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Answer 1

Answer:

Heya,

Here is your answer :-

Present population of town = 22000

Rate of increase in population annually = 10%

Let the population 1 year ago be x .

Now,

(x + 10% of x) = 22000

(x + x/10) = 22000

11x/10 = 22000

x = 22000 × 10/11

x = 20000

So,

The population 1 year ago will be 20000.

Hope my answer will be helpful to you.

Answer 2

Answer:

The answer given below is 100% correct. Mark as brainliest if you prefer.

Have a phenomenal future!

Step-by-step explanation:

Amount(A)= Present Population of The Town=22000

Rate of Increase(R)= 10% annually

Time(T)=1 year

Principal= Population one year ago=P

                     ATQ,

  A=P+$\frac{PRT}{100}$    [Simple Interest Formula]  

⇒22000= P+$\frac{P(10)(1)}{100}$

⇒22000=P+$\frac{P}{10}$

⇒22000=$\frac{10P+P}{10}$

⇒22000=$\frac{11P}{10}$

⇒P=$\frac{22000*10}{11}$

⇒P=2000×10

⇒P=20000

Therefore, population one year ago=20000

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