This can also be solved very easily by using the rule of 72 (http://www.investopedia.com/ask/answers/04/040104.asp) the time taken by a quantity to get doubled at a constant growth rate is $72/rate$. so for this case answer is $72/20 = 3.6$.

5 votes

The population of a new city is $5$ million and is growing at $20\%$ annually. How many years would it take to double at this growth rate?

- $3-4$ years
- $4-5$ years
- $5-6$ years
- $6-7$ years

3

12 votes

Best answer

0

FV = PV × (1+r)n

where **FV** = Future Value

**PV** = Present Value

**r** = annual interest rate

**n** = number of periods

1

Let us suppose the initial population is $100$

- After $1$ year Population become $100\times \dfrac{120}{100} = 120$
- After $2$ year Population become $120\times \dfrac{120}{100} = 144$
- After $3$ year Population become $144\times \dfrac{120}{100} = 172.8$
- After $4$ year Population become $172.8\times \dfrac{120}{100} = 207.36$

7 votes

We can use the Compounding formula like below

$2P = P (1.2)^T,$ where $T$ is the time in years and $1.2$ is the effective value after every year $(1 + r/100).$

So, $\log 2 = T \log 1.2 \implies T = \frac{\log 2}{\log 1.2} \approx 3.8 $

So, option A is the answer here.

If the question was for years after which population gets doubled we will take ceil of $3.8$ and get answer as $4.$

$2P = P (1.2)^T,$ where $T$ is the time in years and $1.2$ is the effective value after every year $(1 + r/100).$

So, $\log 2 = T \log 1.2 \implies T = \frac{\log 2}{\log 1.2} \approx 3.8 $

So, option A is the answer here.

If the question was for years after which population gets doubled we will take ceil of $3.8$ and get answer as $4.$

5 votes

Ans Should be A) 3-4 yr

initial population was 5 million

growing rate = 20 % per annum

now after end of 1 yr population will be = 5(1+20/100) = 6million , ( A= p(1+r/100)$^T$ )

after end of 2nd yr population will be = 6(120/100) = 72/10= 7.2 million

after end of 3rd year population will be = 7.2(120/100)= 43.2/5 = 8.65 million

after end of 4th year population will be = 8.65(120/100) = 10 .38 million ..which is required so it should be** option A)**

3 votes

Formula:. A=P(1+R/100)^T

Upon substitute values and solving we get

2=(1.2)^T.

T=log 2 base 1.2

T=3.81

So answer 1

Upon substitute values and solving we get

2=(1.2)^T.

T=log 2 base 1.2

T=3.81

So answer 1