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This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License. math exams: November 2011

Sunday, November 27, 2011

Question FM 11/27/2011 11:25

To accumulate 4000 at the end of 3n years, deposits of 49 are made at the end of each of the first n years and 98 at the end of each the next 2n years. The annual effective rate of interest is i. You are given $(1+i)^{n}=2$.
Determine i.

Ans: $i=0.1225$

Question FM 11/27/2011 10:47

Lea deposits 100 into an account at the beginning of each 3-year period for 30 years. The account credits interest at an annual effective interest rate of i.
The accumulated amount in the account at the end of 30 years is X, which is 5 times the accumulated amount in the account at the end of 15 years.
Calculate X.

Ans: 6,194.72

Question FM 11/27/2011 10:34

An annuity pays 1 at the end of each year for 20 years. Using an annual effective interest rate of i, the accumulated value of the annuity at time 2n+1 is 12.967. It is also known that $(1+i)^{n}=2.351$.
Calculate n.

Ans: $n=\frac{\ln(2.351)}{\ln(1.53641)}$

Question FM 11/27/2011 10:26

An annuity pays 100 at the end of each of the next 10 years and 200 at the end of each of the next 10 years following. If 1=0.08, find the present value of the annuity.

Ans : $100[a_{\bar{20}}+a_{\bar{10}}\nu^{10}]$

Saturday, November 26, 2011

Question FM 11/21/2011 11:12

Find the force of interest at which $\bar{s}_{\bar{20}}=4\bar{s}_{\bar{10}}$.

Ans: $\delta=\frac{ln(3)}{10}$.

Question FM 11/21/2011 10:27

A student wishes to accumulate $\$100,000$ in education fund at the end of 20 years. If he deposit $\$1000+X$ in the fund at the end of each of the first 10 years and $\$1000+2X$ in the fund of each of the second 10 years, find X if the fund earns $6\%$ effective per year.

Ans: $\frac{100,000-1000s_{\bar{20}}}{s_{\bar{20}}+s_{\bar{10}}}$

Monday, November 21, 2011

Question FM 11/21/2011 03:52

Determine the accumulated value at time 10 years of payments that are received continuously over each year. The payment is $\$105$ during the first year, $\$110$ during the second year, $\$115$ during the third year, and so on, up to the last payment of $\$150$ in year 10. The annual effective interest rate is $6\%$.

Ans : $100\bar{s}_{\bar{10}}+5(\bar{Is})_{\bar{10}}$

Question FM 11/21/2011 03:19

You invest $\$300$ at time 8 years, $\$600$ at time 9 years, $\$900$ at time 10 years, and so on, up to the last payment at time 20 years. What is the accumulated value of these payments at time 31 years using an annual effective rate of interest of $3\%$.

Ans: $300(Ia)_{\bar{13}}(1.03)^{24}$

Question FM 11/21/2011 02:56

You invest X now, the bank will give you 13000 at the time 7 years, 12000 at time 8 years, 110000 at time 9 years, and so on, with the last payment being at time 14 years, using an annual effective rate of interest of $8\%$, determine X.

Ans: $X=500a_{\bar{8}}+1000(Da)_{\bar{8}}$

Question FM 11/21/2011 02:22

Lina receives $\$200$ in 1 year, $\$250$ in 2 years, $\$300$ in 3 years, and so on until the final payment of $\$700$. Using an annual effective rate of interest of $4\%$. Find the present value of these payments at time 0.

Ans:$50.(Ia)_{\bar{11}}+150.a_{\bar{11}}$

Sunday, November 20, 2011

Question FM 11/20/2011 05:53

Calculate the present value at t=0 of a continuous payment stream made at the constant rate of $\rho_{t}=\$6$ per year from time 3 to time 6 years. The force of interest is given by $\delta_{t}=0.04$

Ans: $150[e^{-0.24}+e^{-0.12}]$

Question FM 11/20/2011 04:53

Calculate the present value at t=0 of a continuous payment stream made at the constant rate of $\rho_{t}=\$10$ per year from time 0 to time 20 years. The force of interest is given by $\delta_{t}=0.05$

Ans: $\frac{10}{0.05}[1-e^{-1}]$

Saturday, November 19, 2011

Question FM 11/19/2011 08:36

John receives payments at the start of each of the next 10 years. The first payment is 1000, which is paid now.
The payments from then on increase at a rate effective rate of $15\%$ each year.
Using an annual effective rate on interest of $23.05\%$, find the present value of these payments at time 0.

Ans: $1000.\frac{1-(1.07)^{-10}}{\frac{0.07}{1.07}}$

Question FM 11/19/2011 08:21

Amelie receives a payments at the end of each yera for 20 years. The first payment is $\$2000$. The remaining payments increase by $5\%$, compounded each time.
Calculate the prsent value of the payments at time 0, using an annual effective rate of interest of $10\%$.

Ans: $\frac{2000}{1.05}[\frac{1-(1.047619)^{-20}}{0.047619}]$

Thursday, November 17, 2011

Question FM 11/17/2011 11:07

Jade invests X at time 10 in order to receive 100 at the end of the 15 th year. Find X,
using an annual effective rate of interest of $\%5$.

Ans: $100\frac{1-(1.05)^{-10}}{0.05}.(1.05)^{-4}$

Question FM 11/17/2011 10:40

Payments of $\$10$ are received at the end of each year for 5 years, after which payments of $\$100$
are received at the end of each year forever. The annual effective interest rate is $\%10$.
Determine the present value of these payments.

Ans: $10\frac{1-(1.1)^{-5}}{0.1}+\frac{100}{0.1}.(1.1)^{-5}$

Question FM 11/17/2011 10:15

David receives payments of $\$X$ at end of the end of each year, starting 6 years from today, forever. Karen receives payments of $\$10$
at the beginning of each year, including today, forever.
The present values of their payments are the same at an annual effective interest rate of $\%10$. Calculate X.

Ans: $\$17.7$

Monday, November 14, 2011

Question FM 11/14/2011 10:16

A bank makes payments continuously at a rate of 100 per year. The payments are made between 3 and 13 years. Find the accumulated value of these payments at time 20 years using an annual interest rate of $7\%$.

Ans: $100\frac{(1.07)^{10}-1}{ln(1.07).(1.07)^{7}}$

Question FM 11/15/2011 12:08

Find the accumulated value at time 30 years of payments of $\$100$ at time 11 years, 12 years and so on, with the last payment at time 17 years. Use an annual effective interest rate of $6\%$.

Ans: $100\frac{1.06^{7}-1}{0.06}1.06^{13}$

Question FM 11/14/2011 11:47

Determine the present value of payments of $\%10$ payable continuously each year, starting now and continuing indefinitely. The annual effective interest rate is $6\%$.

Ans: $\$171.62$

Question FM 11/14/2011 11:33

Determine the present value of payments of $\%10$ to be made at the beginning of each year, starting now. The payments continue forever. The annual effective interest rate is $6\%$.

Ans: $\$176.6$

Question FM 11/14/2011 10:54

Determine the present value of payments of $\$10$ to be made at the end of each year starting this year. The payments continue forever. The annual effective interest rate is $6\%$.

Ans: $\$166.67$.

Question FM 11/14/2011 4:09

A bank makes payments continuously at a rate of 1000 per year. The payments are made between times 7 and 9 years. Find the present value of these payments at time 5 years using an annual effective rate of discount of $\%4$

Ans: $1000\frac{1-0.96^{2}}{\ln(1/0.96)}(0.96)^{2}$

Question FM 11/14/2011 3:37

Find the present value at time 0 of regular payments of 100 at time 15 years, 16 years, and so on, with the last payment at time 20 years. Use an annual effective interest rate of $8\%$.

Ans: $\$157.39$

Sunday, November 13, 2011

Question FM 11/13/2011 9:43

Show that the ratio of the accumulated value of $\$1$ invested ta rate i for n periods, to the accumulated value of $\$1$ invested at rate j for n periods, is equal to the accumulated value of 1 invested for mn periods at rate r. Find an expression for r at as a function of i,j and m.

Ans:$(\frac{1+i}{1+j})^{\frac{1}{m}}-1$

Question FM 11/13/2011 9:32

It is known that 800 invested for two years will earn 100 in interest.
Find the accumulated value of 1000 invested at the same rate of compound interest for three years.

Ans: 1193.24

Question FM 11/13/2011 9:16

a) Show that $d^{(m)}=i^{(m)}\nu^{\frac{1}{m}}$
b) Verbally interpret the result in a)

Question FM 11/13/2011 9:02

Show that $m=\frac{d^{(m)}i^{(m)}}{i^{(m)}-d^{(m)}}$

Question FM 11/13/2011 8:43

a) Find the accumulated value of $\$1$ at the end of n periods where the effective rate of interest for the k th period, k=1,2,...n, is defined by
$i_{k}=(1+r)^{k^{2}}(1+i)-1$
b)Show that the answer to a) can be written in the form $(1+j)^{n}$. Find j.

Ans:
a) $a(n)=(1+r)^{\frac{1}{6}(n)(n+1)(2n+1)}(1+i)^{n}$
b) $j=(1+r)^{\frac{1}{6}(n+1)(2n+1)}(1+i)-1$

Friday, November 11, 2011

Question FM 11/11/2011 11:47

The sum of the accumulated value of $\$1$ at the end of four years at a certain effective rate of interest i, and the present value of $\$1$ to be paid at the end of four years at an effective rate of discount equal to i is 1.0042. Find the rate i.

Ans: $2.645\%$

Question FM 11/11/2011 10:26

It is known that an investment of 100 will accumulate to 680 at the end of 20 years. If it is assumed that the investment earns simple interest at rate i during the 1st year, 2i during the 2 nd years,...,20i during the 20 th year, find i.

Ans: $2.76\%$

Question FM 11/11/2011 9:55

It is known that an amount of money will triple in 20 years at a varying force of interest $\delta_{t}=kt+1$. Find an expression for k.

Ans: $\frac{ln3-20}{200}$

Question FM 11/11/2011 9:32

You can receive one of the following two payment streams:

i) 100 at time 0, 100 at time 5, 200 at time n, and 300 at time 2n.
ii) 600 at time 10.

At an annual effective interest rate of i, the present values of the two streams are equal.
Given $\nu=0.7$, determine i.

Ans: $2.34\%$

Question FM 11/11/2011 9:24

Find the exact effective rate of interest at which payments of $\$300$ at the present, $\$300$ at the end of one year, and $\$300$ at the end of two years will accumulate to $\$900$ at the end of two years.

Ans: $\frac{\sqrt{13}-1}{2}$

Question FM 11/11/2011 3:03

Fund A accumulates at a nominal rate of discount convertible quarterly of $\%8$. Fund B accumulates with a force of interest $\delta_{t}=\frac{t}{100}$. Find the next time that the two funds are equal.

Ans: 2 years

Question FM 11/11/2011 2:30

In return for payments of $\$3000$ at the end of six years and $\$6000$ at the end of eight years, an investor pay $\$2000$ immediately and agree to make an additional payment at the end of four years. Find the amount of the additional payment if $i^{6}=0.06$.

Ans: 4848.31

Question FM 11/11/2011 2:06

An investor makes four deposits into account, at end of 2, 6, 9, and 11 years. The amount of the deposits at time t is $300(1.025)^{t}$. Find the size of the account at the end of 13 years, if the nominal of discount convertible bi-monthly is 6/41.

Ans: 3796.47

Question FM 11/11/2011 1:56

Find how long $\$100$ should be left to accumulate at $8\%$ effective in order that it will amount to triple the accumulated value of $\$100$ deposited at the same time at $6\%$ effective.

Ans: 58.8 years

Thursday, November 10, 2011

Question FM 11/10/2011 11:50

If $r=\frac{i^{(m)}}{d^{(m)}}$ express $\nu$ in terms of $r$

Ans: $r^{-m}$

Question FM 11/10/2011 11:30

Suppose :

$1+\frac{i^{(m)}}{m}=\frac{1+\frac{i^{(5)}}{5}}{1+\frac{i^{(6)}}{6}}$

Find m?

Ans : m=30.