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

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Bunuel
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The population of a city grows at a rate of 5% per annum. If in 2006 [#permalink] New post   31 Dec 2019, 02:53
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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
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E
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Re: The population of a city grows at a rate of 5% per annum. If in 2006 [#permalink] New post   02 Jan 2020, 08:39
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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


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!
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Re: The population of a city grows at a rate of 5% per annum. If in 2006 [#permalink] New post   31 Dec 2019, 11:03
Bunuel wrote:
The population of a city grows at a rate of 5% per annum. If in 2006 its population is 18,52,200, what was its population in 2004?

A. 12,60,000
B. 13,60,000
C. 15,60,000
D. 16,00,000
E. 16,80,000



2006= 1852200
2005 = 18,52,200/105
2004=18,52,200/(1.05)^2 =
IMO E: 16,80,000
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Re: The population of a city grows at a rate of 5% per annum. If in 2006 [#permalink] New post   Updated on: 01 Jan 2020, 11:26
(1.05)^2 * P = 1852200
P - cost in 2004 = (1852200*10^4)/105^2
Answer :E

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Originally posted by Ilishar on 31 Dec 2019, 13:10.
Last edited by Ilishar on 01 Jan 2020, 11:26, edited 1 time in total.
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Re: The population of a city grows at a rate of 5% per annum. If in 2006 [#permalink] New post   31 Dec 2019, 13:11
Option E

\(2006 Population = 1.852.200\)

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

\(2004 Population = \frac{1.852.200}{1,05^2} = 1.680.000\)
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Re: The population of a city grows at a rate of 5% per annum. If in 2006 [#permalink] New post   31 Dec 2019, 16:06
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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
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Re: The population of a city grows at a rate of 5% per annum. If in 2006 [#permalink] New post   02 Jan 2020, 10:40
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
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Re: The population of a city grows at a rate of 5% per annum. If in 2006 [#permalink] New post   02 Jan 2020, 11:06
Bunuel wrote:
The population of a city grows at a rate of 5% per annum. If in 2006 its population is 18,52,200, what was its population in 2004?

A. 12,60,000
B. 13,60,000
C. 15,60,000
D. 16,00,000
E. 16,80,000


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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Re: The population of a city grows at a rate of 5% per annum. If in 2006 [#permalink]
02 Jan 2020, 11:06
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