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# If a town of 25,000 people is growing at a rate of approximately 1

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Re: If a town of 25,000 people is growing at a rate of approximately 1 [#permalink]
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1% is quite small and hence even simple interest calculation with it'll yield result quite close to comp interest. So ans A
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Re: If a town of 25,000 people is growing at a rate of approximately 1 [#permalink]
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NoHalfMeasures wrote:
1% is quite small and hence even simple interest calculation with it'll yield result quite close to comp interest. So ans A

But how far can we assume SI=CI? 1%? 2%? 3%?
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Re: If a town of 25,000 people is growing at a rate of approximately 1 [#permalink]
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NoHalfMeasures wrote:
If a town of 25,000 people is growing at a rate of approx. 1% per year, the population of the town in 5 years will be closest to?
a.26000
b.27000
c.28000
d.29000
e.30000

Let population in the beginning = 100
at the end of 1st year population = 100+1% of 100 = 100+1 = 101
at the end of 2nd year population = 101+1% of 101 = 101+1 approx = 102 approx (on the lower side because 1% of 101 is greater than 1)
at the end of 3rd year population = 102+1% of 102 = 102+1 approx = 103 approx (on the lower side because 1% of 102 is greater than 1)
at the end of 4th year population = 103+1% of 103 = 103+1 approx = 104 approx (on the lower side because 1% of 103 is greater than 1)
at the end of 5th year population = 104+1% of 104 = 104+1 approx = 105 approx (on the lower side because 1% of 104 is greater than 1)

i.e. 5% greater than 25000 = 25000+5% of 25000 = 25000+1250 = 26250 (on the lower side)

i.e. Population must be greater than 26250 but since the closest option is 26000 so

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Re: If a town of 25,000 people is growing at a rate of approximately 1 [#permalink]
NoHalfMeasures wrote:
If a town of 25,000 people is growing at a rate of approx. 1% per year, the population of the town in 5 years will be closest to?
a.26000
b.27000
c.28000
d.29000
e.30000

After 1 year, the population will be:

25,000 + 250 = 25,250

After 2 years, the population will be:

25,250 + 252 = 25,502

After 3 years, the population will be:

25,502 + 255 = 25,757

So essentially, we see that the population growth is a little more than 250 people a year.

25,000 + 250 x 5 = 26,250, so about 26,000.

Alternate solution:

The population of the town in 5 years is:

25,000 x (1 + 0.01)^5 = 25,000 x 1.01^5 = 26,275, which is closest to 26,000.

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Re: If a town of 25,000 people is growing at a rate of approximately 1 [#permalink]
NoHalfMeasures wrote:
If a town of 25,000 people is growing at a rate of approx. 1% per year, the population of the town in 5 years will be closest to?
a.26000
b.27000
c.28000
d.29000
e.30000

1% of 25,000 = 250

Since 1% compounded is really insignificant, we can use simple interest: $$250 * 5 = 1,250$$

$$25,000 + 12,250 = 16,250$$

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Re: If a town of 25,000 people is growing at a rate of approximately 1 [#permalink]
We can use Simple Interest formula - SI = PTR/100. Here , P = 25000, T=5, R=1 . therefore, PTR/100 = 1250. Therefore, the population of the town after 5 years would be closest to 26000. (25000+1250* = 26000*)
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Re: If a town of 25,000 people is growing at a rate of approximately 1 [#permalink]
We can solve this question using Binomial Expansion Formula
(1+x)^n= 1 + n x + [n(n - 1)/2!] x^2 + [n(n - 1)(n - 2)/3!] x^3 +...
In this question, x=1/100, which is a very small number. So, we can avoid higher order terms (x^2, x^3,....)
So,
25000(1+1/100)^5=25000(1+5/100)
=25000+25000*5/100
=25000+1250
=26250 Ans(A)
Re: If a town of 25,000 people is growing at a rate of approximately 1 [#permalink]
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