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If the population of a city was 42,500 and it grew at a rate of 2% per annum compounded annually, what would the population be after 2 years?
40,800
40,400
44,217
41,616 Question ID : 4410092028075 Option 1 ID : 4410097985943 Option 2 ID : 4410097985940 Option 3 ID : 4410097985941 Option 4 ID : 4410097985942 Status : Answered Chosen Option : C
44,217
Use the formula for two years successives: r1тАЛ+r2тАЛ+100r1тАЛ├Чr2тАЛтАЛ. Here, 2+2+1002├Ч2тАЛ=4.04%. Calculate 4.04% of 42,500, which is 425├Ч4.04=1,717. Add this to 42,500 to get 44,217.
Initial population P = 42,500, Rate of growth r = 2% per annum, Time period n = 2 years
A=P(1+100rтАЛ)n
Use the formula for two years successives: r1тАЛ+r2тАЛ+100r1тАЛ├Чr2тАЛтАЛ. Here, 2+2+1002├Ч2тАЛ=4.04%. Calculate 4.04% of 42,500, which is 425├Ч4.04=1,717. Add this to 42,500 to get 44,217.
Students often calculate simple interest instead of compound interest, resulting in an incorrect answer of 44,200 (42,500+4% of 42,500).
Define variables
Identify the principal amount P=42500, growth rate r=2%, and time n=2 years.
P=42500,r=2,n=2
Apply population growth formula
Use the compound growth formula to determine the population after 2 years.
A=42500(1+1002тАЛ)2
Simplify the expression
Simplify the fraction inside the parenthesis: 1+0.02=1.02. Now calculate the square: (1.02)2=1.0404.
A=42500├Ч1.0404
Calculate final value
Perform the multiplication to find the final population count.
A=44217
C is correct because the compounded growth of 42,500 at 2% for 2 years yields exactly 44,217.
This concept is identical to compound interest calculations and is frequently tested in banking exams and CSAT under 'Successive Percentage Changes'.