Question

Joseph has $10,000 to invest. He can go to Yankee Bank that pays 5% simple interest or Met Bank that pays 4% interest compounded annually. At how many years will Met Bank be the better choice? ​

Answer 1

Answer: After 12 years ( approx )

Explanation:

Let after t years the amount in met bank will exceed to the amount in Yankee bank,

For Yankee bank,

The principal amount, P = $ 10,000

Simple annual interest rate, r = 5 %

Hence, the amount in this bank after x years,

`A= P +(P* r* t)/(100)`

`A=10000+ (10000* 5* x)/(100)`

`A=10000 + 500x`

Now, For Met bank,

The principal amount, P = $ 10,000

Compound annual interest rate, r = 4 %

Hence, the amount in this bank after x years,

`A= P(1+(r)/(100))^x`

`A=10000(1+(4)/(100))^x`

`A=10000(1+0.04)^x`

`A=10000(1.04)^x`

⇒For exceeding,

`10000(1.04) > 10000+500x`

By the below graph, the above condition occurs after intersection of
`10000(1.04)^x` and
`10000+50x` at (11.919, 15959.341)

Hence, after x = 11.919 ≈ 12 years the amount in Met bank will be greater than that of Yankee bank.

⇒ For 12 years Met Bank will be the better choice. ​