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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?

1. $3-4$ years
2. $4-5$ years
3. $5-6$ years
4. $6-7$ years
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+3

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 $72/20 = 3.6$.

0
Can  we do it via Compound interest formula ?

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 $A.$
by Boss (38.7k points)
edited by
0

FV = PV × (1+r)n

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

0
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 $172.8\times \dfrac{120}{100} = 207.36$

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)

by Boss (17.1k points)
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

by (41 points)
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 $4.$
by Veteran (432k points)
–1 vote
When population 100 then 20 men increase in 1 year

"         "             1    "      20  "       "          " 1⨉100  years

"         "             1    "       1  "       "          " 1⨉100 /20 years

"         "         5000000   "   1  "       "        "   1⨉100  /(20⨉5000000)

"         "         5000000  "   5000000 "   "    "   1⨉100 ⨉5000000 /(20⨉5000000) = 5 years

Ans will (C) [As min 5 years needed for doubling growth rate]
by Veteran (119k points)
+2
"growing at 20% annually"

means every year the population increases by 20% or becomes 1.2 times that at the start of the year. In your approach you have considered it to be 20% every year increase, from the initial value at start of year 1.