Answer: After 12 years ( approx )
Step-by-step 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,
[tex]A= P +\frac{P\times r\times t}{100}[/tex]
[tex]A=10000+ \frac{10000\times 5\times x}{100}[/tex]
[tex]A=10000 + 500x[/tex]
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,
[tex]A= P(1+\frac{r}{100})^x[/tex]
[tex]A=10000(1+\frac{4}{100})^x[/tex]
[tex]A=10000(1+0.04)^x[/tex]
[tex]A=10000(1.04)^x[/tex]
⇒For exceeding,
[tex]10000(1.04) > 10000+500x[/tex]
By the below graph, the above condition occurs after intersection of [tex]10000(1.04)^x[/tex] and [tex]10000+50x[/tex] 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.
