Respuesta :
2.5 years required for an investment of 5000 dollars to grow to 6000 dollars at an interest rate of 7.5 percent per year, compounded quarterly.
Step-by-step explanation:
The given is,
Initial investment - $5000
Future amount - $6000
Interest rate - 7.5% (compounded quarterly)
Step:1
Formula to calculate the Future amount with compound interest,
[tex]F = P(1+\frac{r}{n} )^{nt}[/tex]...................................(1)
Where, F - Future amount
P - Initial amount
r - Rate of interest
n - No. of compounding in a year
t - Time period
From given,
F = $6000
P = $5000
r = 7.5%
n = 4 (compounded quarterly)
Equation (1) becomes,
[tex]6000=5000(1+\frac{0.075}{4} )^{(t)(4)}[/tex]
[tex]\frac{6000}{5000} =(1+0.01875)^{4t}[/tex]
[tex]1.2 = (1.01875)^{4t}[/tex]
Take log on both sides,
[tex]log 1.2 = 4(t) log 1.01875[/tex]
Substitute log values,
0.07918 = 4(t) (0.0080676)
= (t) (0.0322705)
[tex]t = \frac{0.07918}{0.0322705}[/tex]
= 2.45
t ≅ 2.5 years
Result:
2.5 years required for an investment of 5000 dollars to grow to 6000 dollars at an interest rate of 7.5 percent per year, compounded quarterly.
The time required for an investment of 5000 dollars to grow to 6000 dollars at an interest rate of 7.5 percent per year is 2.06 months
The formula for calculating the compound interest is expressed as:
[tex]A = P(1+\frac{r}{n} )^{nt[/tex]
A is the future value = $6000
P is the amount invested = $5000
r is the rate (in %) = 7.5% = 0.075
t is the time taken (in years)
n is the time of compounding = 4
Substitute the given parameters into the formula:
[tex]6000 = 5000(1+\frac{0.075}{4} )^{4t}\\\frac{6}{5} =(1+0.3)^{4t}\\1.2=(1.3)^{4t}\\log 1.2 = 4tlog1.4\\4t =\frac{log1.2}{log1.4}\\4t= \frac{0.07918}{0.1139}\\4t = 0.6951\\t = \frac{0.6951}{4}\\t= 2.08 months[/tex]
This shows that the time required for an investment of 5000 dollars to grow to 6000 dollars at an interest rate of 7.5 percent per year is 2.06 months
Learn more here:https://brainly.com/question/18442810
