Question

Mary invests £12000 in a saving account. The account pays 1.5% compound interest per year. Work out the value of her investment after 2 years.

Answer 1

To find the value of Mary's investment after 2 years in a saving account with compound interest, we can use the compound interest formula. Substituting the given values into the formula gives us £12363.61.

Step-by-step explanation:

To calculate the value of Mary's investment after 2 years, we can use the compound interest formula: A = P(1 + r/n)^(nt), where:

• A is the final amount
• P is the principal amount (initial investment)
• r is the annual interest rate (as a decimal)
• n is the number of times the interest is compounded per year
• t is the number of years

In this case, the principal amount is £12000, the annual interest rate is 1.5% (0.015 as a decimal), and the number of times the interest is compounded per year is 1. Plugging these values into the formula, we get:

A = 12000(1 + 0.015/1)^(1*2)

Simplifying this, we get:

A = 12000(1.015)^2

Calculating further, we get:

A ≈ 12000(1.030225) ≈ £12363.61

Therefore, the value of Mary's investment after 2 years is approximately £12363.61.

Answer 2

`\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &12000\\ r=rate\to 1.5\%\to (1.5)/(100)\to &0.015\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per year, thus once} \end{array}\to &1\\ t=years\to &2 \end{cases} \\\\\\ A=12000\left(1+(0.015)/(1)\right)^(1\cdot 2)\implies A=12000(1.015)^2`