.Atrustfundfora10yearoldchildissetupbyasinglepaymentofGHS14000.Atwhat agewillthechild’sfundaccumulateuptoGHS27000giventhatinterestratestandsat 13%perannum.

To solve this problem, we will use the formula for compound interest which is A = P(1 + r/n)^(nt).

Here, A = the amount of money accumulated after n years, including interest. P = the principal amount (the initial amount of money) r = annual interest rate (in decimal) n = number of times that interest is compounded per year t = the time the money is invested for in years

Given in the problem: P = GHS 14,000 A = GHS 27,000 r = 13% or 0.13 n = 1 (since it's compounded annually)

We need to find t.

Rearranging the formula to find t, we get:

t = log(A/P) / n * log(1 + r/n)

Substituting the given values:

t = log(27000/14000) / log(1 + 0.13)

After calculating the above expression, we get the value of t. This will give us the number of years it will take for the child's fund to accumulate up to GHS 27,000.

Please note that the child's current age is 10, so you will need to add 10 to the calculated t to find the child's age when the fund will accumulate to GHS 27,000.

To solve this problem, we will use the formula for compound interest which is A = P(1 + r/n)^(nt).

Here, A = the amount of money accumulated after n years, including interest. P = the principal amount (the initial amount of money) r = annual interest rate (in decimal) n = number of times that interest is compounded per year t = the time the money is invested for in years

Given in the problem: P = GHS 14,000 A = GHS 27,000 r = 13% or 0.13 n = 1 (since it's compounded annually)

We need to find t.

Rearranging the formula to solve for t, we get:

t = log(A/P) / n[log(1 + r/n)]

Substituting the given values:

t = log(27000/14000) / 1[log(1 + 0.13/1)] t = log(1.92857) / log(1.13) t = 0.2857 / 0.0539 t = 5.3 years

So, the child will be 10 + 5.3 = 15.3 years old. Since age is usually not expressed in decimals, the child will be 16 years old when the trust fund accumulates up to GHS 27,000.