GATE2014 AG: GA-5

Question

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? 

• A. $3-4$ years
• B. $4-5$ years 
• C. $5-6$ years 
• D. $6-7$ years

Comments

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 $72/20 = 3.6$.

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Can  we do it via Compound interest formula ?

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Obviously Yes  @Chhotu..

Answer 1

Initial population $P=5M$

After $1$ year. $P=5M\times 1.2=6M$

Now $2^{nd}$ year, $P=6M$

Now after increment at end of $2$ years, $P=6m\times 1.2=7.2M$

After $3$ years $P=7.2m\times 1.2=8.64M$

After $4$ years $P=8.64m\times 1.2=10.368M$

So, answer should be $A.$

Comments

FV = PV × (1+r)n

where FV = Future Value
PV = Present Value
r = annual interest rate
n = number of periods

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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 $172.8\times \dfrac{120}{100} = 207.36$

Answer 2

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 $4.$

Answer 3

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)

Answer 4

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