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13 Dec 2021, 03:15
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
79% (02:45) correct
21%
(02:35)
wrong
based on 47
sessions
History
Date
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If the population of a city increases by \(\frac{100}{9}\%\) per year and if in 2013, the population of the city is 10530 then find the difference between the population of the city in 2015 and 2014.
A. 1300
B. 1400
C. 1700
D. 1800
E. 2000
A. 1300
B. 1400
C. 1700
D. 1800
E. 2000
13 Dec 2021, 07:08
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Population in 2014 increases by = 10530 * (100/9) = 1170
Population in 2014 = 1170+10530 = 11700 ............1
Similarly
Population in 2015 increases by = 11700 * (100/9) = 1300
Population in 2015 = 1300+11700 = 13000 ..........2
Subtract 1 from 2
Ans 1300
Population in 2014 = 1170+10530 = 11700 ............1
Similarly
Population in 2015 increases by = 11700 * (100/9) = 1300
Population in 2015 = 1300+11700 = 13000 ..........2
Subtract 1 from 2
Ans 1300
14 Dec 2021, 06:30
Kudos
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100/9% = 100/9/100/1
100/9*1/100= 1/9 (reciprocal)
amount increased in 2014 from 2013= 10530*1/9=1,170
population at the end of 2014 = 10530+1170=11,700
population increase in 2015 = 11700*1/9=1,300
population at the end of 2015 = 11,700+1300=13,000
change in the amount between 13,000-11,700=1,300 so (A)
100/9*1/100= 1/9 (reciprocal)
amount increased in 2014 from 2013= 10530*1/9=1,170
population at the end of 2014 = 10530+1170=11,700
population increase in 2015 = 11700*1/9=1,300
population at the end of 2015 = 11,700+1300=13,000
change in the amount between 13,000-11,700=1,300 so (A)
