Question

Colin invests £2350 into a savings account. The bank gives 4.2% compound interest for the first 4 years and 4.9% thereafter. How much will Colin have after 10 years to the nearest pound?

Answer 1

To solve this, we are going to use the compound interest formula: $A=P(1+ \frac{r}{n} )^{nt}$
where
$A$ is the final amount after $t$ years 
$P$ is the initial investment 
$r$ is the interest rate in decimal form 
$n$ is the number of times the interest is compounded per year

For the first 4 years we know that: $P=2350$, $r= \frac{4.2}{100} =0.042$, $t=4$, and since the problem is not specifying how often the interest is communed, we are going to assume it is compounded annually; therefore, $n=1$. Lest replace those values in our formula:
$A=P(1+ \frac{r}{n} )^{nt}$
$A=2350(1+ \frac{0.042}{1} )^{(1)(4)}$
$A=2350(1+0.042)^{4}$
$A=2770.38$

Now, for the next 6 years the intial investment will be the final amount from our previous step, so $P=2770.38$. We also know that: $r= \frac{4.9}{100} =0.049$, $t=6$, and $n=1$. Lets replace those values in our formula one more time:
$A=P(1+ \frac{r}{n} )^{nt}$
$A=2770.38(1+ \frac{0.049}{1})^{(1)(6)$
$A=2770.38(1+0.049)^6$
$A=3691.41$

We can conclude that Collin will have £3691.41 in his account after 10 years.