FMS TEST BANK 2026 QUESTIONS WITH
SOLUTIONS GRADED A+
◉ Sum of geometric series. Answer: [(first term) - (first omitted term)] /
[1 - (first term)]
◉ s-angle-n formula. Answer: [(1+i)^n - 1] / i
◉ a-angle-n formula. Answer: [1 - v^n] / i
◉ Amount of SIMPLE interest in ANY period. Answer: X(0) * i, where
i is the eir for the period
◉ If annual effective simple interest rate is i, what is the effective
interest rate for 6 month? Answer: i/2
◉ What is a sinking fund? Answer: Borrower only pays interest on the
loan each period (the payment each period is thus the same)
Then, at the end, the borrower pays total principal amount in one
payment
Each year, the borrower deposits money into a sinking fund, and uses the
amount in this fund at the end of the term to pay the balance off
,◉ Dollar-weighted rate of return. Answer: Approximates annual
effective return rate i using simple interest approximations for each
withdrawl/deposit
E.g.:
If fund starts at 75, gets deposits of 10 at end of each month, has
withdrawls of 5 at end of Feb, 25 at end of June, 80 halfway through
Oct, and 25 at end of Oct, and ends with 60, the formula for i is:
60 = 75(1+i) + 10[(1 + 11/12i) + (1 + 10/12i) + .. (1+0/12i)] -5(1 +
10/12i) - 25(1+6/12i) - 80(1 + 5/24i) - 35(1 + 2/12i)
◉ Unit increasing annuity formula. Answer: [(a double dot angle n) -
nv^n] / i
◉ Present value of perpetuity. Answer: X/i (where X is the payment
amount each period)
◉ v * a double dot angle n, is the same as. Answer: a angle n
◉ A perpetuity costs 77.1 and makes end-of-year payments. The
perpetuity pays 1 at the end of year 2, 2 at the end of year 3, .., n at the
end of year (n+1). After year (n+1), the payments remain constant at n.
The annual effective interest rate is 10.5%. Answer: From 1 to n, it is a
,unit increasing annuity immedaite, which must then be discounted one
period to get to time 0.
From n+1 on, it is a perpetuity with payments of n, which must be
discounted n+1 periods to get to time 0.
77.1 = v(Ia)angle-n + (v^(n+1) * n/i)
(Ia)angle-n = (a double dot angle n - n*v^n) / i
77.1 = v[(a double dot angle n - n*v^n) / i] + (v^(n+1) * n/i)
v(a double dot angle n) is the same as a-angle-n
◉ Accumulated value of unit decreasing annuity. Answer: (Ds)angle-n =
[(n(1+i)^n - s-angle-n) / i]
◉ 1000 is deposited into Fund X, which earns an annual effective rate of
6%. At the end of each year, the interest earned plus an additional 100 is
withdrawn from the fund. At the end of the tenth year, the fund is
depleted.
The annual withdrawals of interest and principal are deposited into Fund
Y, which earns an annual effective rate of 9%.
, Calculate the accumulated value of Fund Y at the end of year 10.
Answer: THe interest deposits are 6 times a decreasing unit annuity of
10 years
The 100 deposits are 100 for 10 years
So, accumulated value is:
6[(Ds)-angle-10] + 100s-angle-10, with i = .09
6[(n(1+i)^n - s-angle-n)/i] + 100s-angle-10
◉ If loan payments are 150% of interest due, and interest rate is 10%,
what percent of that payment is used to pay down pricinpal? Answer:
5%
Why?
150% of interest due is 1.5(.1)(Principal)
Interest due is .1(Principal)
Amount of payment applied to principal is thus 1.5(.1)(Principal) -
.1(Principal) = .05(Principal)
SOLUTIONS GRADED A+
◉ Sum of geometric series. Answer: [(first term) - (first omitted term)] /
[1 - (first term)]
◉ s-angle-n formula. Answer: [(1+i)^n - 1] / i
◉ a-angle-n formula. Answer: [1 - v^n] / i
◉ Amount of SIMPLE interest in ANY period. Answer: X(0) * i, where
i is the eir for the period
◉ If annual effective simple interest rate is i, what is the effective
interest rate for 6 month? Answer: i/2
◉ What is a sinking fund? Answer: Borrower only pays interest on the
loan each period (the payment each period is thus the same)
Then, at the end, the borrower pays total principal amount in one
payment
Each year, the borrower deposits money into a sinking fund, and uses the
amount in this fund at the end of the term to pay the balance off
,◉ Dollar-weighted rate of return. Answer: Approximates annual
effective return rate i using simple interest approximations for each
withdrawl/deposit
E.g.:
If fund starts at 75, gets deposits of 10 at end of each month, has
withdrawls of 5 at end of Feb, 25 at end of June, 80 halfway through
Oct, and 25 at end of Oct, and ends with 60, the formula for i is:
60 = 75(1+i) + 10[(1 + 11/12i) + (1 + 10/12i) + .. (1+0/12i)] -5(1 +
10/12i) - 25(1+6/12i) - 80(1 + 5/24i) - 35(1 + 2/12i)
◉ Unit increasing annuity formula. Answer: [(a double dot angle n) -
nv^n] / i
◉ Present value of perpetuity. Answer: X/i (where X is the payment
amount each period)
◉ v * a double dot angle n, is the same as. Answer: a angle n
◉ A perpetuity costs 77.1 and makes end-of-year payments. The
perpetuity pays 1 at the end of year 2, 2 at the end of year 3, .., n at the
end of year (n+1). After year (n+1), the payments remain constant at n.
The annual effective interest rate is 10.5%. Answer: From 1 to n, it is a
,unit increasing annuity immedaite, which must then be discounted one
period to get to time 0.
From n+1 on, it is a perpetuity with payments of n, which must be
discounted n+1 periods to get to time 0.
77.1 = v(Ia)angle-n + (v^(n+1) * n/i)
(Ia)angle-n = (a double dot angle n - n*v^n) / i
77.1 = v[(a double dot angle n - n*v^n) / i] + (v^(n+1) * n/i)
v(a double dot angle n) is the same as a-angle-n
◉ Accumulated value of unit decreasing annuity. Answer: (Ds)angle-n =
[(n(1+i)^n - s-angle-n) / i]
◉ 1000 is deposited into Fund X, which earns an annual effective rate of
6%. At the end of each year, the interest earned plus an additional 100 is
withdrawn from the fund. At the end of the tenth year, the fund is
depleted.
The annual withdrawals of interest and principal are deposited into Fund
Y, which earns an annual effective rate of 9%.
, Calculate the accumulated value of Fund Y at the end of year 10.
Answer: THe interest deposits are 6 times a decreasing unit annuity of
10 years
The 100 deposits are 100 for 10 years
So, accumulated value is:
6[(Ds)-angle-10] + 100s-angle-10, with i = .09
6[(n(1+i)^n - s-angle-n)/i] + 100s-angle-10
◉ If loan payments are 150% of interest due, and interest rate is 10%,
what percent of that payment is used to pay down pricinpal? Answer:
5%
Why?
150% of interest due is 1.5(.1)(Principal)
Interest due is .1(Principal)
Amount of payment applied to principal is thus 1.5(.1)(Principal) -
.1(Principal) = .05(Principal)
