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The population of a certain country increases at the rate of 30,000

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The population of a certain country increases at the rate of 30,000 [#permalink]

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02 Jul 2017, 01:50
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25% (medium)

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78% (01:25) correct 22% (02:01) wrong based on 111 sessions

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The population of a certain country increases at the rate of 30,000 people every month. The population of the country in 2012 was 360 million. In which year would the population of the country be 378 million?

(A) 2060
(B) 2061
(C) 2062
(D) 2063
(E) 2064

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Re: The population of a certain country increases at the rate of 30,000 [#permalink]

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02 Jul 2017, 02:35
1
Bunuel wrote:
The population of a certain country increases at the rate of 30,000 people every month. The population of the country in 2012 was 360 million. In which year would the population of the country be 378 million?

(A) 2060
(B) 2061
(C) 2062
(D) 2063
(E) 2064

Population Increase Rate = 30,000 Per Month

Population in 2012 = 360 Million

Population in x year = 378 Million

378 Million - 360 Million = 18 Million

We want to calculate the year when the population increases by 18 Million

$$= \frac{18,000,000}{30,000} * \frac{1}{12}$$ ====> 1/12 as we are converting per month to per year

$$= 50$$ Years

Start Year $$= 2012$$

We will add 50 Years $$= 2012 + 50 = 2062$$

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Re: The population of a certain country increases at the rate of 30,000 [#permalink]

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02 Jul 2017, 05:13
Imo C
Rate of increase is 30000 per month , so per year it would be 12*30000=3,60,000 people
In 2012 population =360*10^6
In what year the population would be 378*10^6
378*10^6-360*10^6 =18*10^6
Now we have total increment dividing it by by the rate we will get the year
18*10^6/(3,60,000 )=50 years
so 2012+50=2062
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The population of a certain country increases at the rate of 30,000 [#permalink]

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03 Jul 2017, 03:29
Bunuel wrote:
The population of a certain country increases at the rate of 30,000 people every month. The population of the country in 2012 was 360 million. In which year would the population of the country be 378 million?

(A) 2060
(B) 2061
(C) 2062
(D) 2063
(E) 2064

1. Increase per year: 30,000 per month * 12 mos. = 360,000 per year --->
$$3.6 * 10^5$$

2. Total increase needed: 378 million - 360 million = 18 million --->
$$18 * 10^6$$

3. # of years = Total increase divided by increase per year:

$$\frac{18 * 10^6}{3.6 * 10^5}$$ =

5 x 10 = 50 years

4. Year achieved: 2012 + 50 = 2062

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Re: The population of a certain country increases at the rate of 30,000 [#permalink]

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03 Jul 2017, 09:22
Bunuel wrote:
The population of a certain country increases at the rate of 30,000 people every month. The population of the country in 2012 was 360 million. In which year would the population of the country be 378 million?

(A) 2060
(B) 2061
(C) 2062
(D) 2063
(E) 2064

Population in 2012 is $$360000000$$
Desired population is $$378000000$$

Net increase is $$378000000 - 360000000 = 18000000$$

Increase in population per year is $$360000$$

So, Number of years required is $$\frac{18000000}{360000} = 50$$

Thus, 50 years from 2012 , ie in 2062 Population will be 378 million, answer will be (C)
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Re: The population of a certain country increases at the rate of 30,000 [#permalink]

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20 Sep 2017, 10:49
Population Increase Rate = 30,000 Per Month

Population in 2012 = 360 Million

Population in ? year = 378 Million

378 Million - 360 Million = 18 Million

18 Million / 30,000=600

in years = 600/12 = 50

Start Year =2012

Re: The population of a certain country increases at the rate of 30,000   [#permalink] 20 Sep 2017, 10:49
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