Question

Nina made two investments: Investment \text{A}A has a value of \$50$50 at the end of the first year and increases by 8\%8% per year. Investment \text{B}B has a value of \$60$60 at the end of the first year and increases by \$3$3 per year. Nina checks the value of her investments once a year, at the end of the year. What is the first year in which Nina sees that investment \text{A}A's value exceeded investment \text{B}B's value?

Answer 1

Answer:

6 years

Step-by-step explanation:

Given,

The amount of investment A  = $ 50,

Annual rate of interest = 8 %,

Thus, her amount after x years,

$A_1=50(1+\frac{8}{100})^x$

$=50(1+0.08)^x$

$\implies A_1=50(1.08)^x$

Now, the amount of investment B = $ 60,

Added amount per year = $ 3,

⇒ Added amount in x years = 3x,

Thus, Total amount after x years,

$A_2=60+3x$

Since, Investment A's value exceeded investment B's value,

If,

$A_1>A_2$

$\implies 50(1.08)^x > 60+3x$

By the below graph,

We found that,

x > 5.552

Hence, the first year at which the Investment A's value exceeded investment B's value is 6 years ( approx )

Answer 2

Answer: 7 years

Step-by-step explanation: