Answer:
The % increase in investment after two years is 13.05 %
Step-by-step explanation:
Given as :
The principal investment = p = $10,000
The rate of interest = r = 6%
The time period t = 1 year
Let The Amount paid after 1 year = $[tex]A_1[/tex]
Let The % increase in investment after two years = x
Now, According to question
From Compounded Interest method
Amount = Principal × [tex](1+\dfrac{\textrm rate}{ 100})^{\textrm time}[/tex]
Or, [tex]A_1[/tex] = p × [tex](1+\dfrac{\textrm r}{100})^{\textrm t}[/tex]
Or, [tex]A_1[/tex] = $10,000 × [tex](1+\dfrac{\textrm 6}{ 100})^{\textrm 1}[/tex]
Or, [tex]A_1[/tex] = $10,000 × [tex](1.06)^{1}[/tex]
Or, [tex]A_1[/tex] = $10,000 × 1.06
∴ [tex]A_1[/tex] = $10,600
So, The Amount paid after 1 year = [tex]A_1[/tex] = $10,600
Now, Interest earn = Amount - Principal
Or, I = $10,600 - $10,000
i.e I = $600
So, This interest earn is invested for second year
So, Principal for second year = $10,600 + $600
i.e Principal for second year = $11,200
The rate of interest = r = 7%
The time period t = 1 year
Let The Amount paid after 1 year = $[tex]A_2[/tex]
Now, According to question
From Compounded Interest method
Amount = Principal × [tex](1+\dfrac{\textrm rate}{ 100})^{\textrm time}[/tex]
Or, [tex]A_2[/tex] = p × [tex](1+\dfrac{\textrm r}{100})^{\textrm t}[/tex]
Or, [tex]A_2[/tex] = $11,200 × [tex](1+\dfrac{\textrm 7}{ 100})^{\textrm 1}[/tex]
Or, [tex]A_2[/tex] = $11,200 × [tex](1.07)^{1}[/tex]
Or, [tex]A_2[/tex] = $11,200 × 1.07
∴ [tex]A_2[/tex] = $11,984
So, The Amount paid after 1 year = [tex]A_2[/tex] = $11,984
Now, Again
% increase in investment after two years = [tex]\dfrac{A_2 - A_1}{A_1}[/tex] ×100
Or , x = [tex]\dfrac{11,984 - 10,600}{10,600}[/tex] ×100
Or , x = [tex]\dfrac{1384}{10,600}[/tex] ×100
∴ x = 13.05 %
So, % increase in investment after two years = x = 13.05 %
Hence,The % increase in investment after two years is 13.05 % Answer
