The population of a city grows at a rate of 5% per annum. If in 2006

Question

The population of a city grows at a rate of 5% per annum. If in 2006 its population is 1,852,200, what was its population in 2004?

A. 1,260,000 
B. 1,360,000 
C. 1,560,000 
D. 1,600,000 
E. 1,680,000

Archit3110 · Answer

2006= 1852200
2005 = 18,52,200/105
2004=18,52,200/(1.05)^2 = 
IMO E: 16,80,000

Ilishar · Answer

(1.05)^2 * P = 1852200
P - cost in 2004 = (1852200*10^4)/105^2 
Answer :E

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PTD1995 · Answer

Option E

2006 Population = 1.852.200

As 2 years have elapsed, we have to discount those years

2004 Population = \frac{1.852.200}{1,05^2} = 1.680.000

EgmatQuantExpert · Answer

Solution

Given  
 • The population of a city grows at a rate of 5% per annum.
• In 2006 its population is 18,52,200

To find  
 • The population of the city in 2004

Approach and Working out  
Let in 2004 city population was x.
 • Hence, in 2006, the city’s population = x * 1.05 *1.05
• x * 1.05 *1.05 =18,52,200
• x = 1680000

Thus, option E is the correct answer.
    Correct Answer: Option E

conormdr203 · Answer

I understand all the answers so forgive me if this is a dumb question - but what is the practical way to calculate 1.05 * 1.05 * 18,52,200. Basic arithmetic isn't my strong-suit, and I'm still wrapping my head around all the basics of being able to handle these calculations without doing it the longest way possible.

Also my first post - woo!

Thanks!

brains · Answer

A 5% increase is equivalent to fraction of 21/20.

So if P be the population in 2004 , then

using multiplication factor,

P* 21/20 * 21/20 = 1852200
P = (18522/21*21)*20*20*100 ( Since 18522 can be reduced to simpler fraction by dividing by 7 or 21)
P = 1680000

Kinshook · Answer

Compounded growth in 2 years @5% per annum = 1.05*1.05 = 1.1025
In 2004, population was = 18,52,200/1.1025 = 16,80,000

IMO E

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bumpbot · Answer

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The population of a city grows at a rate of 5% per annum. If in 2006

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