Question

Drew wants to save $2,500 to go to the next World Cup. To the nearest dollar, how much will he need to invest in an account now with 6.25% APR, compounding daily, in order to reach his goal in 4 years

Answer 1

Drew will need to invest approximately $1,949 in an account with a 6.25% APR, compounding daily, to reach his $2,500 goal in 4 years.

Step-by-step explanation:

Drew needs to calculate how much to invest right now to achieve his goal of having $2,500 in 4 years with a 6.25% APR compounding daily. To find this, he can use the formula for compound interest P = A / (1 + r/n)^(nt), where P is the principal investment amount (what Drew needs to find), A is the future value of the investment, r is the annual interest rate (decimal), n is the number of times that interest is compounded per year, and t is the number of years the money is invested for.

First, we convert 6.25% APR to a decimal by dividing by 100, which gives us 0.0625. The interest is compounded daily, which means n = 365. Drew wants to have $2,500 in 4 years, so A = $2,500 and t = 4.

Using the formula:

P = 2500 / (1 + 0.0625/365)^(365*4)

Calculating the value inside the parentheses first:

P = 2500 / (1 + 0.0001712328767)^1460

Then, calculate the exponent:

P = 2500 / (1.283035621)

And finally, divide $2,500 by this number:

P = 2500 / 1.283035621

P = $1948.55 (rounded to the nearest dollar)

So, Drew will need to invest approximately $1,949 to reach his goal of $2,500 in 4 years with a 6.25% APR compounded daily.