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The population of a town is 1,76,400. If it increases at the rate of 5% per annum, then what will be its population 2 years hence?
1,94,572
1,94,621
1,93,636
1,94,481
1,94,481
Use the growth multiplier: 5% increase means 105% of the previous value. For 2 years, the multiplier is 1.05×1.05=1.1025. Calculate 1,76,400×1.1025.
Initial population P = 1,76,400, Annual growth rate r = 5%, Time period n = 2 years
A=P×(1+100r)n
Use the growth multiplier: 5% increase means 105% of the previous value. For 2 years, the multiplier is 1.05×1.05=1.1025. Calculate 1,76,400×1.1025.
Students often calculate simple interest instead of compound growth (1,76,400×1.10 = 1,94,040), which is incorrect for population growth.
Identify parameters
Given principal population P=1,76,400, rate r=5%, and time n=2 years.
P=1,76,400,r=5,n=2
Apply growth formula
The population after n years is calculated using the compound increase formula A=P×(1+r/100)n.
A=1,76,400×(1+1005)2
Calculate the multiplier
Simplify the fraction 1+5/100=1.05. Squaring 1.05 gives 1.1025.
1.052=1.1025
Final multiplication
Multiply the initial population by the growth factor: 1,76,400×1.1025=1,94,481.
A=1,76,400×1.1025=1,94,481
D is correct because the population after 2 years of 5% growth is 1,94,481.
This concept is identical to calculating Compound Interest. If you know the formula for amount A=P(1+R/100)n, you can solve for population, depreciating assets (use minus sign), or compound interest problems.
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