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# The population of a city is 200,000. If annual birth rate and death ra

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Joined: 18 Jul 2018
Posts: 374
Location: India
Concentration: Finance, Marketing
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The population of a city is 200,000. If annual birth rate and death ra  [#permalink]

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09 Sep 2018, 18:20
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35% (medium)

Question Stats:

68% (01:35) correct 32% (01:04) wrong based on 25 sessions

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The population of a city is 200,000. If annual birth rate and death rate are 6% and 3% respectively, then what is the population of the city after 2 years?

1) 206,090.
2) 212,000.
3) 212,090.
4) 212,180.
5) 215,000.

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Joined: 11 Mar 2018
Posts: 85
Re: The population of a city is 200,000. If annual birth rate and death ra  [#permalink]

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09 Sep 2018, 18:50
Afc0892 wrote:
The population of a city is 200,000. If annual birth rate and death rate are 6% and 3% respectively, then what is the population of the city after 2 years?

1) 206,090.
2) 212,000.
3) 212,090.
4) 212,180.
5) 215,000.

Annual Birth Rate = 6% of current population
Annual Death Rate = 3% of current population
Total annual percent addition in current population = (6% - 3%) of current population = 3% of current population

Year 1 population =
3% of current population + current population
$$\frac{3}{100} * 200,000 + 200,000$$
=206,000

Year 2 population =
3% of previous year population + previous year population
$$\frac{3}{100} * 206,000 + 206,000$$
=212,180

Hence D
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Re: The population of a city is 200,000. If annual birth rate and death ra  [#permalink]

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10 Sep 2018, 06:46
Afc0892 wrote:
The population of a city is 200,000. If annual birth rate and death rate are 6% and 3% respectively, then what is the population of the city after 2 years?

1) 206,090.
2) 212,000.
3) 212,090.
4) 212,180.
5) 215,000.

Annual Birth Rate = 6% of current population
Annual Death Rate = 3% of current population
Total annual percent addition in current population = (6% - 3%) of current population = 3% of current population

Year 1 population =
3% of current population + current population
$$\frac{3}{100} * 200,000 + 200,000$$
=206,000

Year 2 population =
3% of previous year population + previous year population
$$\frac{3}{100} * 206,000 + 206,000$$
=212,180

Hence D

No need to calculate twice, after finding out effective interest rate go for compound interest calculation to save time.

$$200000( 1 + \frac{3}{100})^2 = 212,180$$
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Re: The population of a city is 200,000. If annual birth rate and death ra  [#permalink]

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10 Sep 2018, 08:11
Afc0892 wrote:
The population of a city is 200,000. If annual birth rate and death rate are 6% and 3% respectively, then what is the population of the city after 2 years?

1) 206,090.
2) 212,000.
3) 212,090.
4) 212,180.
5) 215,000.

Since the answer is not an estimate and exact, the units digit should help us in avoiding calculations...

Effective increase is 6-3=3%

So end of first year 200000*103/100
Next year.. $$200000*\frac{103}{100}*\frac{103}{100}$$
U
The first non zero digit will be the units digit of 3*3*2=18, so 8

Only D left..

Otherwise specially in compound interest, Vedic maths is helpful..
103 * 103
Both are 3 away from 100..
number.... Distance from 100
103………… +3
103…………+3
Add diagonals for first 3 digit so 103+3=106
For next two digits as 100 has two zeroes, multiply the right numbers, so 3*3=9, thus 09
Number =10609
So 10609*200000/100*100=10609*20=212180

D

Say the numbers were 102*105..
102.....+2
105......+5
So (102+5)__(5*2)=10710
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Re: The population of a city is 200,000. If annual birth rate and death ra  [#permalink]

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10 Sep 2018, 08:21
For percentages like 6% or 3%, one way is to use 1.06 or 1.03.

Here.
Birth rate = 1.06
Death rate = 1.03

So effectively the increase in population is 1.06-1.03 = 1.03
For two years, the total population is 200000*1.03*1.03 = 200000*1.0609 = 212180
Re: The population of a city is 200,000. If annual birth rate and death ra &nbs [#permalink] 10 Sep 2018, 08:21
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