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04 May 2021, 08:27
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
41% (02:31) correct
59%
(02:28)
wrong
based on 145
sessions
History
Date
Time
Result
Not Attempted Yet
The population of a country increased by an average of 1% per year from 2016 to 2019 .If the population of the country was 21 million on December 31,2019,then the population of this country on January 1,2016 ,to the nearest thousand had been:
A. 20400
B. 20200
C. 20000
D. 18000
E. 15000
A. 20400
B. 20200
C. 20000
D. 18000
E. 15000
04 May 2021, 22:09
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Jihyo
Let population on January 1, 2016 = x =?, Population increases by 1% per annum, Population on December 31, 2019 = 21 million.
Population on January 1, 2017 = 1.01 * x
Population on January 1, 2018 = 1.01 * (1.01 * x) = \(1.01^2\) * x
Population on January 1, 2019 = \(1.01^3\) * x
Population on January 1, 2020 = Population on December 31, 2019 = \(1.01^4\) * x = 21,000,000
=> x = \(\frac{21,000,000}{1.01^4}\) = 20,180,000 \(\approx\) 20,200 thousand
So, correct answer is option B.
10 Oct 2021, 07:04
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Hey! How to shorten this calculation since we wont be able to use calculator;-;
Posted from my mobile device
Posted from my mobile device
