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MATH 166 Exam 3

1. (5 points) If and ,and C = BA, find c12:

Answer:

d) The element in the first row, second column of the product is equal to the matrix product
of the first row of B times the second column of A. This gives us 2 * e + 5 * d.

2. (5 points) Alice borrows $15000 at 3% (compounded monthly) to buy a car, and makes
no payments on the loan for 5 years. How much does she owe (rounded to two decimal
places)?

Answer:

c) $17424.25172 is the answer given by the TVM solver when we enter n = 60, I% = 3,
PV = 15000, and C/Y = 12.
3. (5 points) Find the inverse of

Answer:

c) If we enter the matrix in the problem as A, then A-1 gives us answer b)

4. (5 points)

Bob borrowed $750 five months ago. If he now owes $776.25, what simple annual interest
rate did he borrow at?

Answer:

b) Bob owes $26.25 in interest for five months. A month’s interest would be
so a year’s interest would be $63. gives us the interest rate of 8.4%.

5. (5 points) If A is a 2× 3 matrix, B is a 2× 3 matrix, and C is a 3× 2 matrix, which of
the following is not a valid operation between A,B, and C?)

Answer:

d) Since the number of columns in the first matrix must equal the number of rows in the
second matrix, the B * A is not valid, since B has 3 columns and A has 2 rows.

6. (5 points) Using the matrices from the previous problem, which of the following is a
false statement?)

Answer:

b) Since A and C have different orders, it is not possible to add them .

7. (5 points) Paul buys a savings bond for $75 and in 5 years, it will mature to a value
of $100. Assuming quarterly compounded interest , what interest rate is he getting on his
savings bond?

Answer:

a) If we plug in n = 20,PV = 75, FV = −100, and C/Y = 4, we get 5.79522% as the
interest rate.

8. (12 points) Francine is investing $1000 from her paycheck each year into real estate,
stocks, treasury bonds, and a bank account. She wants to puts half as much into her
savings account as she does into the treasury bonds and stocks combined. She estimates the
annual return on the investments to be 5% for real estate and savings, 3.5% for treasury
bonds, and 8% for stocks. She wishes to achieve a 6.5% average annual return on her
investments.

(a) Give a system of equations that represents the above information. (Use the variables r
for real
estate, s for stocks, t for treasury bonds, and a for the bank account.)

(b) Solve the system of equations, and parameterize the answer in terms of a , the amount
in the bank account.

(c) Using the information above, what is the minimum and maximum she can put into her
bank account?

(d) if she chooses to put $200 into her bank account each year, how much does she put into
the other investments?

Answer:

(a) The information above can be summarized by writing equations (letting r= real estate,
s = stocks, t = treasury bonds, and a = bank account):
r + t + s + a = 1000
a = .5t + .5s
.05r + .08s + .035t + .08a = .065 * 1000

(b) . r s t a
We type this in as the following matrix:
Using rref on this matrix, we get the following:

This tells us that and that

(c) Since Additionally, 1000 − 3a ≥ 0, so a ≤ 333.33, giving us
250 ≤ a ≤ 333.33. Once we know boundaries for a, we can plug them to get boundaries
for the other variables:

(d) Plugging in a = 200, we get r = $400, s = 466.66 and t = −66.66 . The negative value
for t illustrates what happens when we give an a outside of our acceptable boundaries
above.

9. (6 points) Using the following system of equations, either state that there are no
solutions, that there are infinitely many solutions (with simplified equations limiting the
solutions), or state the solution. Write all numbers as fractions in your answer.

x − 27 = 3z
4z + 5 = 6y
7x = 8z

Answer:

Putting the equations in standard form, we get x − 3z = 27,−6y + 4z = −5, and

7x − 8z = 0. Entering these in as the matrix
, and using rref, we get the
following answers: and

10. (8 points)
A single trip for a certain fishing boat consists of going to the fishing area,
fishing for a number of days, and then returning to dock to sell the fish. The owner of the
boat has calculated the gas cost for each trip to be $300 to travel to the fishing area, $100
per day of fishing, and $1 per pound of fish carried plus $300 to return to the dock. If the
owner can sell fish at $3 per pound and catches 200 pounds of fish per day spent in the
fishing area, how many days must the boat remain in the fishing area to pay for its gas?

Answer:

Let f be the pounds of fish caught, and d be the number of days in the fishing area. Then
we can represent this problem as the following system of equations: f = 200d and 3f =
300 + 300 + f + 100d, meaning that $3 times the pounds of fish (the ship’s income from
fishing) must equal the amount of gas used ($300 for going out, $300 plus f for coming back,
and $100 per day). Putting these equations in simplified form gives us f − 200d = 0 and
2f − 100d = 600, giving us the following matrix:

Solving via rref gives us that f = 400, d = 2, so the ship must spend two days fishing to
break even.

Alternatively, we can substitute f = 200d as we make the equations, giving the single
equation 600d = 300 + 300 + 200d + 100d, which gives us the same answer.

11. (8 points) Solve for x,y, and z in the following equation:

Answer:

Adding the matrices together gives us the following equations: x+1+24 = 27, 2+4z = 14,
8y + 8 = 16, which gives us x = 2,y = 1, z = 3.

12. (8 points) A car manufacturer calculates their cost per car manufactured in France
and India, and the number produced in each location in 2007 and 2008. The production
matrix,P is given by:

The cost matrix, C, which gives the cost per unit produced in each country, is given by:



(a) Find a meaningful product of the given matrices. (Be sure to clearly show how you
multiplied them .)

(b) What information do the elements of the product give us? Be precise for each element
in the product.

Answer:

(a) In order to have the labels match up correctly in the multiplication, we need to change
C into a column matrix, giving us

Then we multiply, getting

(b) The first row will give us total costs for 2007, covering both France and India. The
second row will give this same value for 2008. (The item in row 1 is French production
for 2007 times unit cost plus Indian production for 2007 times unit cost. The item in
the second row is the same, but for 2008 production.)

13. (11 points) A printing business takes out a 7 year loan at 4.9% interest (compounded
monthly) to purchase a new printer worth $220,000. The company paid 15% of the value
up-front and borrowed the rest, making monthly payments.

Answer:

Entering in N = 12* 7, I = 4.9%, PV = 187000, FV = 0,P/Y = C/Y = 12 and solving for
PMT gives us 2634.263095 as our monthly payment.

(a) Knowing our monthly payment, we set N = 12*5 and solve for FV to find the remaining
principle after 5 years, which gives us 60106.46392

(b) We take the value owed from above (at the end of the fifth year is the same as the
beginning of the 6th year). We need to calculate our interest each month and subtract
that from our monthly payment to get the principal paid. Once we know the principal
paid, we can find the principal remaining by subtraction.

Month Principal at Start Interest Paid Principal Paid Principal Remaining
1 $ 60106.46392 $ 245.43472 $ 2388.82836 $ 57717.63555
2 $ 57717.63555 $ 235.58034 $ 2398.58275 $ 55319.05280
3 $ 55319.05280 $ 225.88613 $ 2408.37696 $ 52910.67584

14. (12 points) Joe is saving for retirement. He deposits $200 every month in a bank
account which gives 3.5% interest, compounded monthly.

(a) What is the effective annual rate of interest on the account?

(b) After 20 years, how much is the account worth?

(c) After 20 years, how much interest has Joe earned?

Answer:

(a) There are 12 interest periods in the year, so the effective interest rate is found by calculating
so the effective interest rate is 3.55669%.

(b) Using the TVM solver, we put in N = 12 * 20, I = 3.5%, PV = 0, PMT = −100, and
P/Y = C/Y = 12. Solving for future value gives us 69373.85382.

(c) For an investment, Interest = FV - PV - PMT * N. Since the future value is $69373.85382

and Joe has put in 12 * 200 * 20 = $48000, the total interest is the difference of these, or
$21373.85382.

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