Question

5. In the first year of a $10,000 investment the interest rate was 6%. All earned interest remained

invested for the second year, and the interest rate increased to 7%. What was the percent increase of
the entire investment after two years?
16.7%
1%
13.4%
70%

Answer 1

The % increase in investment after two years is 13.05 %

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 = $
`A_1`

Let The % increase in investment after two years = x

Now, According to question

From Compounded Interest method

Amount = Principal ×
`(1+(\textrm rate)/( 100))^(\textrm  time)`

Or,
`A_1` = p ×
`(1+(\textrm r)/(100))^(\textrm t)`

Or,
`A_1` = $10,000 ×
`(1+(\textrm 6)/( 100))^(\textrm 1)`

Or,
`A_1` = $10,000 ×
`(1.06)^(1)`

Or,
`A_1` = $10,000 × 1.06

∴
`A_1` = $10,600

So, The Amount paid after 1 year =
`A_1` = $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 = $
`A_2`

Now, According to question

From Compounded Interest method

Amount = Principal ×
`(1+(\textrm rate)/( 100))^(\textrm  time)`

Or,
`A_2` = p ×
`(1+(\textrm r)/(100))^(\textrm t)`

Or,
`A_2` = $11,200 ×
`(1+(\textrm 7)/( 100))^(\textrm 1)`

Or,
`A_2` = $11,200 ×
`(1.07)^(1)`

Or,
`A_2` = $11,200 × 1.07

∴
`A_2` = $11,984

So, The Amount paid after 1 year =
`A_2` = $11,984

Now, Again

% increase in investment after two years =
`(A_2 - A_1)/(A_1)` ×100

Or , x =
`(11,984 - 10,600)/(10,600)` ×100

Or , x =
`(1384)/(10,600)` ×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