MATH 1108 R06 MIDTERM EXAM 2 SOLUTION
FALL 2015 - MOON
• Write your answer neatly and show steps.
• Except calculators, any electronic devices including laptops and cell phones are
not allowed.
• Do not use the graphing function on your calculator.
(1) A box of 10 flashbulbs contains 3 defective bulbs. A random sample of 2 is
selected and tested. Let X be the random variable associated with the number
of defective bulbs in the sample.
(a) (6 pts) Find the probability distribution table of X.
defective bulbs
0
1
7 C2
7 C1 · 3 C1
probability
10 C2
10 C2
• Finding each probability correctly: +2 pts.
2
3 C2
10 C2
(b) (2 pts) Find the expected number of defective bulbs in a sample.
7 C2
7 C1 · 3 C1
3 C2
E(X) = 0 ·
+1·
+2·
10 C2
10 C2
10 C2
21
3
21
+1·
+2·
= 0.6
= 0·
45
45
45
• If the probability distribution table is incorrect, one can get at most 1 pt.
Date: November 19, 2015.
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MATH 1108 Midterm Exam 2
Fall 2015 - Moon
(2) At a price of $1.88 per pound, the supply for cherries in NY is 16,000 pounds.
When the price drops to $1.46 per pound, the supply decreases to 10,000 pounds.
(a) (3 pts) Find the price-supply equation.
For the quantity, we’ll use the thousand pound as the unit. Then when the
price is p1 = 1.88, the supply is q1 = 16. And when the price is p2 = 1.46,
the supply is q2 = 10.
The price-supply equation is p = mq + b.
p2 − p1
1.46 − 1.88
m=
=
= 0.07
q2 − q1
10 − 16
So the equation is p = 0.07q + b.
p1 = 0.07q1 + b ⇒ 1.88 = 0.07 · 16 + b ⇒ b = 1.88 − 0.07 · 16 = 0.76
The price-supply equation: p = 0.07q + 0.76
(b) (4 pts) Suppose that the price-demand equation is
p = −0.2q + 4,
where p is the price (dollars) and q is the quantity (thousand pounds). Find
the equilibrium price and quantity.
p = 0.07q + 0.76
p = −0.2q + 4
0.07q + 0.76 = −0.2q + 4 ⇒ 0.07q + 0.2q = 4 − 0.76
⇒ 0.27q = 3.24 ⇒ q ≈ 12
p ≈ −0.2 · 12 + 4 ≈ 1.6
Equilibrium price: $1.6
Equilibrium quantity: 12000 pounds
• One can obtain at most 2 pts if he/she uses incorrect units.
• If the price-supply equation is incorrect, one can obtain at most 3 pts.
• -1 pt if one doesn’t mention units.
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MATH 1108 Midterm Exam 2
Fall 2015 - Moon
(3) (3 pts) Perform the row operation −4R1 + R2 → R2 on the following matrix
1 −3 2
.
4 −6 −8
−4R1 = −4 12 −8
−4R1 + R2 = −4 12 −8 + 4 −6 −8 = 0 6 −16
1 −3 2
1 −3 2
−4R1 +R2 →R2
−→
4 −6 −8
0 6 −16
(4) To save the college tuition for Emily, her father decides to invest $25,000 for 18
years.
(a) (3 pts) Merrill Lynch offers a money market account that earns 5.2% compounded monthly. If he chooses this account, how much will it be worth in
18 years? Round to the nearest cent.
r
0.052
P = 25000, m = 12, i =
=
, n = mt = 12 · 18 = 216
m
12
216
0.052
n
A = P (1 + i) = 25000 1 +
≈ $63615.28
12
(b) (4 pts) How long does it take to be $50,000?
0.052
i=
12
n
0.052
n
A = P (1 + i) ⇒ 50000 = 25000 1 +
12
n
50000
0.052
⇒2=
= 1+
25000
12
n
0.052
0.052
ln 2 = ln 1 +
= n ln 1 +
12
12
0.693147
0.693147 ≈ 0.004324n ⇒ n ≈
≈ 160.30
0.004324
It takes 161 months, or 13 years and 5 months.
A = 50000,
P = 25000,
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MATH 1108 Midterm Exam 2
Fall 2015 - Moon
(c) (3 pts) JPMorgan Chase offers an investment account that earns 5% compounded continuously. If he chooses this account, how much will it be
worth in 18 years? Round to the nearest cent.
P = 25000,
t = 18,
r = 0.05
A = P ert = 25000 · e0.05·18 ≈ $61490.08
(d) (3 pts) The third option is a saving account which is offered by Barclays.
This account earns 5.3% compounded annually. Find APY’s of these three
accounts and determine which option is the best choice.
12
APY of Merrill Lynch: 1 + 0.052
− 1 ≈ 0.053257
12
0.05
APY of JPMorgan Chase: e
− 1 ≈ 0.051271
1
APY of Barclays: (1 + 0.053) − 1 = 0.053
Therefore Merrill Lynch is the best potion.
(5) A couple purchased a house 12 years ago for $279,000. The house was financed
by paying 20% down and signing 30 years mortgage at 8.4% on the unpaid
balance (The interest is compounded monthly.).
(a) (3 pts) Find the monthly mortgage payment. Round to the nearest cent.
The amount of loan: 279000 · 0.8 = 223200
0.084
r
=
= 0.007, n = mt = 12 · 30 = 360
P V = 223200, m = 12, i =
m
12
1 − (1 + i)n
1 − (1 + 0.007)360
P V = P MT
⇒ 223200 = P M T
≈ P M T · 131.261561
i
0.007
223200
≈ $1700.42
⇒ P MT ≈
131.261561
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MATH 1108 Midterm Exam 2
Fall 2015 - Moon
(b) (3 pts) Find the current unpaid loan. Round to the nearest cent.
The couple has to make 12 × 18 = 216 monthly payments of $1700.42.
P M T = 1700.42,
P V = P MT
m = 12,
i = 0.007,
n = 12 · 18 = 216
1 − (1 + 0.007)−216
1 − (1 + i)−n
= 1700.42 ·
≈ $189078.70
i
0.007
• One can get at most 2 pts if he/she uses a wrong monthly payment.
(c) (2 pts) Compute the total interest they have paid in last 12 years. Round to
the nearest cent.
Actual total payment: 1700.42 × 12 × 12 = 244860.48
Reduced loan: 223200 − 189078.70 = 34121.30
Paid interest: 244860.48 − 34121.30 = $210738.70
• Getting actual payment: 1 pt.
(d) (3 pts) The couple wishes to borrow money using the equity in its home
for collateral. A loan company will loan the couple up to 80% of their equity. The current market value of the house is $315,000. How much can the
couple borrow from the loan company? Round to the nearest cent.
Equity = 315000 − 189078.70 = 125921.30
Loan = 0.8 × 125921.30 = $100737.04
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MATH 1108 Midterm Exam 2
Fall 2015 - Moon
(6) A 45-year-old man puts $2500 in a retirement account at the end of each quarter for 15 years, until he reaches the age of 60. The annual interest rate is 6%,
compounded quarterly.
(a) (3 pts) How much will it be in the account? Round to the nearest cent.
r
0.06
P M T = 2500, m = 4, i =
=
= 0.015, n = mt = 4 · 15 = 60
m
4
(1 + i)n − 1
(1 + 0.015)60 − 1
F V = P MT
= 2500 ·
≈ $240536.63
i
0.015
(b) (3 pts) For the next five years, he makes no further deposits. When he
reaches the age of 65, he decides to retire. How much will it be in the
account? Round to the nearest cent.
Because he doesn’t make any further deposits, the future value is computed
by ordinary future value formula.
P = 240536.63,
i = 0.015,
n = mt = 4 · 5 = 20
A = P (1 + i)n = 240536.63 · (1 + 0.015)20 ≈ $323967.96
• One can get at most 2 pts if he/she uses a wrong present value.
(c) (4 pts) He decides to withdraw equal quarterly payments for 20 years, and
at the end of which time (he will be 85) the account will have a zero balance.
How much should he withdraw each quarter? Round to the nearest cent.
P V = 323967.96,
m = 4,
i = 0.015,
n = mt = 4 · 20 = 80
1 − (1 + i)−80
1 − (1 + 0.015)−80
⇒ 323967.96 = P M T
i
0.015
323967.96
⇒ 323967.96 ≈ P M T · 46.407323 ⇒ P M T ≈
≈ $6980.97
46.407323
P V = P MT
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MATH 1108 Midterm Exam 2
Fall 2015 - Moon
(7) (8 pts) By using Gauss-Jordan method, solve the system of linear equations
x + y + 2z = 18
3x + y + 2z = 34
x + 3y + 2z = 22.



1 1
2
18
1 1 2 18
+R2 →R2
 0 −2 −4 −20 
 3 1 2 34  −3R1−→
1 3
2
22
1 3 2 22




1 1
2
18
1 1 2 18
1
− R2 →R2
−R1 +R3 →R3
 0 −2 −4 −20  2−→
 0 1 2 10 
−→
0 2
0
4
0 2 0 4




1 0 0 8
1 0 0
8
−R2 +R1 →R1
+R3 →R3
 0 1 2 10  −2R2−→
 0 1 2
−→
10 
0 2 0 4
0 0 −4 −16




1 0 0 8
1 0 0 8
− 14 R3 →R3
−2R3 +R2 →R2
 0 1 0 2 
−→  0 1 2 10 
−→
0 0 1 4
0 0 1 4
⇒ x = 8, y = 2, z = 4

• Finding the correct answer x = 8, y = 2, z = 4 with appropriate transformations: 8 pts.
• Performing an incorrect transformation: -2 pts each.
• Without using Gauss-Jordan method, you can get at most 5 pts.
7