(a) We can use the formula for compound interest to find the rate of interest per annum: A = P * (1 + r/n)^(n*t), where A is the final amount, P is the principal amount, r is the annual interest rate, n is the number of times interest is compounded per year, and t is the time in years.
Using the table, we can see that the loan increases by a factor of 1.058 after 1 year and by a factor of 1.058/1.045 = 1.0124 after 2 years. Thus, we have:
1.058 = (1 + r/n)^(n*1)
1.0124 = (1 + r/n)^(n*2)
Dividing the second equation by the first, we get:
1.0124/1.058 = (1 + r/n)^(n*1)
Taking the nth root of both sides, we get:
(1 + r/n) = (1.0124/1.058)^(1/n)
Solving for r, we get:
r = n * [(1.0124/1.058)^(1/n) - 1]
We don't know the value of n, but we can make an educated guess that it's either 1 or 2, since the loan is increasing by roughly 5.8% per year. If we try n = 1, we get:
r = 1 * [(1.0124/1.058)^(1/1) - 1] = -0.0427
This is clearly not the correct answer, since the interest rate cannot be negative. If we try n = 2, we get:
r = 2 * [(1.0124/1.058)^(1/2) - 1] = 0.0574
This is a reasonable answer, since it's close to the observed increase in the loan value of 5.8% per year. Therefore, the rate of interest per annum is 5.7% to 1 d.p.
(b) Using the formula for compound interest, we can find the value of the loan 10 years after it starts:
A = P * (1 + r/n)^(n*t) = 2500 * (1 + 0.0574/1)^(1*10) = £4,745.60 (to the nearest 1p