CPSC_103_Midterm_2021W1_Solution-1
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103
Subject
Accounting
Date
Apr 3, 2024
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12
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2021W1 Midterm Answer Key
Question 1
numbers = [2, 3, 2, 7, 4, 3]
tot = 0
for i in numbers:
if is_even(i): # assume is_even(i) is a function that is
# complete and correct (and available for use)
# is_even(i) returns True if i is even
tot = tot + i
tot
8
Question 2
def foo1(x):
return x/2
def foo2(x, y):
y = 20
x == y
x = foo1(y)
y = 3
return x
foo2(6, 5)
10
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2021W1 Midterm Answer Key
Question 3
The Beverage data below (similar to the one you saw in Worksheet 4) records information about
a beverage offered at the cafe, including:
●
Its name, as it appears in the menu
●
Its price
●
Whether or not it is dairy-free
Even though this data definition will run with no errors, there are some errors in the data design.
Identify the line number where each error is found and describe how you would fix each error.
For example, if you believe a line should be removed, write its number and SHOULD BE
REMOVED after it (capitalization does not matter) (i.e., Line 7 should be removed or LINE 7
SHOULD BE REMOVED).
01.
from
typing
import
NamedTuple
02.
Beverage
=
NamedTuple(
'Beverage'
, [(
'name'
,
str
),
03.
(
'price'
,
float
),
04.
(
'dairy_free'
,
str
)])
05.
06.
#interp. a beverage available in the cafe’s menu,
07.
# including its name, price, and whether or not
it is dairy free.
08.
09.
BLACK_COFFEE
=
Beverage(
"Black coffee"
,
2.50
,
"Yes"
)
10.
EARL_GRAY
=
Beverage(
"Earl Gray Tea"
,
3.25
,
"Yes"
)
11.
LONDON_FOG
=
Beverage(
"London Fog"
,
4.75
,
"No"
)
12.
13.
# template based on compound (3 fields)
14.
@typecheck
15.
def
fn_for_drink
(b: Beverage)
-> ...
:
16.
return
...
(b
.
name,
17.
b
.
price)
Line 3: an interval comment is needed to show that price cannot be a negative number
Line 4:
dairy_free
should be a bool, not a str
Lines 9-11: The
“Yes”
should be changed to
True
and the
“No”
should be
False
Line 15: It should be
fn_for_beverage
, not
fn_for_drink
Line 18: Missing a
b.dairy_free
in the data template.
Page 2 of 12
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2021W1 Midterm Answer Key
Question 4
Time to work on a classic RPG game! Design a data type for a character in a game. A
character’s information is comprised of:
●
Its name
●
Its race (one of: human, elf, dwarf, ogre)
●
Its health points (between 0 and 300)
●
Its mana points (between 0 and 300)
The data definition for race is already given to you:
from enum import Enum
Race = Enum("Race", ["HUMAN", "ELF", "DWARF", "OGRE"])
# interp. The possible races for a character in the game
# examples are redundant for enumerations
@typecheck
# Template based on Enumeration (4 cases)
def fn_for_race(r: race) -> ...:
if r == Race.HUMAN:
return …
elif r == Race.ELF:
return …
elif r == Race.DWARF:
return ...
elif r == Race.OGRE:
return ...
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2021W1 Midterm Answer Key
from typing import NamedTuple
Character = NamedTuple(‘Character’, [(‘name’, str),
(‘race’, Race),
(‘health’, int) # in range [0,300]
(‘mana’, int) # in range [0,300]])
# interp. information about a character in the RPG game,
# including name (name), race (race), health points
# (health), and mana points (mana).
CH1 = Character(“Raghard”, Race.Human, 150, 100)
CH2 = Character(“Wargo”, Race.Ogre, 0, 300)
CH3 = Character(“Oggy”, Race.Dwarf, 300, 0)
# template based on compound (4 fields) and reference rule
@typecheck
def fn_for_character(c: Character) -> ...:
return ...(c.name,
fn_for_race(c.race),
c.health
c.mana)
Page 4 of 12
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2021W1 Midterm Answer Key
Data Definitions Used in Questions 5 and 6
Type = Enum("Type", ["APARTMENT", "HOUSE"])
# interp. Property type. It can be an apartment or a house.
# Examples are redundant for enumerations
# template based on Enumeration
@typecheck
def fn_for_type(t: Type) -> ...:
if t == Type.APARTMENT:
return ...
elif t == Type.HOUSE:
return ...
Property = NamedTuple("Property", [("type", Type),
("square_feet", int), # in range (0,...)
("year", int),
("price", int),
("pending", bool)])
# interp. information about a property, including its type #
(apartment or house), footage, the year it was built,
# price and whether or not it has a pending offer.
P1 = Property(Type.APARTMENT, 625, 2020, 450000, False)
P2 = Property(Type.HOUSE, 1100, 1980, 1350000, False)
P3 = Property(Type.HOUSE, 925, 1995, 1100000, True)
# template based on Compound and reference rule
@typecheck
def fn_for_property(p: Property) -> ...:
return ...(fn_for_type(p.type),
p.square_feet,
p.year,
p.price,
p.pending)
Page 5 of 12
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2021W1 Midterm Answer Key
# List[Property]
# interp. a list of properties in the agency’s inventory
L0 = []
L1 = [P1, P2, P3]
# template based on arbitrary-sized and reference rule
@typecheck
def fn_for_lop(lop: List[Property]) -> ...:
# description of the accumulator
acc = ... # type: …
for p in lop:
acc = ...(fn_for_property(p), acc)
return ...(acc)
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2021W1 Midterm Answer Key
Question 5
The data definitions on the previous two pages are for a Real Estate agency that deals in
private homes. Each property has a type (apartment or house), square feet, year built, price,
and a status to indicate whether or not an offer is pending on it.
The following function examines a list of properties to determine whether or not it contains a
house that costs at most $1000000 and is of a given footage or greater. Complete the helper
functions and add new ones if needed. You can assume has_cheap_house_min_footage() is
complete and correct.
Be sure to follow the HtDF recipe. You are allowed to assume that the tests for all functions in
this question are present, complete, and correct.
from cs103 import *
@typecheck
def has_cheap_house_min_footage(lop: List[Property], sf: int) -> bool:
"""
Returns True if there is at least one house of footage
equal or greater than sf that costs at most one
million dollars in the list; False otherwise.
"""
# return True # stub
# template copied from List[Property] with 1
# additional parameter
for p in lop:
if is_property_house(p) and has_min_footage(p,
sf) and costs_at_most_one_million(p):
return True
return False
# Complete the helper functions below and add new functions if needed
# Be sure to follow the FULL HtDF recipe
Page 7 of 12
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2021W1 Midterm Answer Key
@typecheck
def is_property_house(p: Property) -> bool:
"""
Returns True if p is a house; False otherwise.
"""
#
return True # stub
# template based on Property
# return is_house(p.type)
@typecheck
def is_house(type: Type) -> bool:
"""
Returns True if type is Type.HOUSE; False otherwise.
"""
# return True # stub
# template based on Type
if type == Type.APARTMENT:
return False
elif type == Type.HOUSE:
return True
@typecheck
def has_min_footage(p: Property, square_footage: int) -> bool:
"""
Returns True if p's square footage (i.e., size) is at
least square_footage large; False otherwise.
"""
#
return True # stub
# template based on Property with additional parameter
return p.square_feet >= square_footage
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2021W1 Midterm Answer Key
@typecheck
def costs_at_most_one_million(p: Property) -> bool:
"""
Returns True if p costs one million dollars or less;
False otherwise.
"""
#
return True # stub
# template based on Property
return p.cost <= 1000000
# You are allowed to assume that all tests required for the
# functions above are present, complete, and correct
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2021W1 Midterm Answer Key
Question 6
Consider the Real Estate agency data definitions given earlier.
We want to design a function to return all apartments built after a given year and below a given
price.
Here are some functions which may or may not prove useful. It is possible that your answer will
not call every helper function listed in this question. You cannot write any other function for this
question. Only use what you are given.
You do NOT need to provide a complete implementation of the helpers. You only need to
complete find_apartments_by_year_and_price(). You do not need to provide tests for
find_apartments_by_year_and_price().
@typecheck
def find_all_apartments(lop: List[Property]) -> List[Property]:
"""
Returns all apartments in lop.
"""
return []
# stub
@typecheck
def is_property_apartment(p: Property) -> bool:
"""
Returns True if p is an apartment; False otherwise.
"""
# return True # stub
@typecheck
def is_apartment(t: Type) -> bool:
"""
Returns True if t represents an apartment; False
otherwise.
"""
# return True # stub
Page 10 of 12
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2021W1 Midterm Answer Key
@typecheck
def find_properties_in_budget(lop: List[Property], p: int) ->
List[Property]:
"""
Returns all properties with price equal to or below p.
"""
return []
# stub
@typecheck
def is_property_in_budget(p: Property, pr: int) -> bool:
"""
Returns True if the property p has price equal to or
below pr; False otherwise.
"""
# return True # stub
@typecheck
def find_properties_built_after_year(lop: List[Property], y: int) ->
List[Property]:
"""
Returns all properties built strictly after year y.
"""
return []
# stub
@typecheck
def is_property_built_after_year(p: Property, y: int) -> bool:
"""
Returns True if the property p was built strictly
after year y; False otherwise.
"""
# return True # stub
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2021W1 Midterm Answer Key
@typecheck
def find_apartments_by_year_and_price(lop: List[Property], y: int, p:
int) -> List[Property]:
"""
Returns a list of all apartments built after a given
year (y) and below a given price (p).
"""
# return [] # stub
# template based on composition
# 1. filter all apartments
# 2. filter all apartments built after year y
# 3. filter all apartments built after year y with
# price equal or less than price p
# 4. return final list
# Complete the rest of the
# find_apartments_by_year_and_price function
all_apartments = find_all_apartments(lop)
apartments_after_year =
find_properties_built_after_year(all_apartments, y)
apartments_at_most_price =
find_properties_in_budget(apartments_after_year, p)
return apartments_at_most_price
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.
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Required information
[The following information applies to the questions displayed below.]
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The management team of Wickersham Brothers Incorporated is preparing its annual financial statements. The statements are
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Current Year
Prior Year
Balance Sheet
Assets
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$ 91,500
$ 111,000
Accounts Receivable
Inventory
Property and Equipment
Less: Accumulated Depreciation
97,500
00006
000'06
(48,800)
$ 422,700
Total assets
$ 378,500
Liabilities:
$ 15,000
$ 18,000
Accounts Payable
Salaries and Wages Payable
Notes Payable, Long-Term
Stockholders' Equity:
000'e
000 sL
00s T
000 '06
Common Stock
120,000
Retained Earnings
189,700
000 6tT
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$ 378,500
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Test: Multiple Functions
FILE UPLOAD 1
Question 12
Listen
Use the function f(x)=2x-5
.
•
•
Find the inverse of f(x).
Graph f(x) and f1(x) and state the domain of each function
Prove that f(x) and f¹(x) are inverses, both graphically and
algebraically.
ATTACHMENTS
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red
Question 11
4) Listen
Use the graphs below to categorize the statements to describe either Graph A
or Graph B.
3
Graph A
Graph B
0
#12
"
3/2
2x
0
*/2
3/2
2
-1
T
graph a
graph b
Add an answer item!
Add an answer item!
Answer Bank
amplitude = 1
midline is y=3
f(x)= cos(x)+3
amplitude =3
f(x)=3sin(x)+1
midline is y=1
All Change
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On June 30, Sharper Corporation's stockholders' equity section of its balance sheet appears as follows before any stock
dividend or split. Sharper declares and immediately distributes a 50% stock dividend.
Common stock-$10 par value, 60,000 shares issued and
outstanding
Paid-in capital in excess of par value, common stock
$ 600,000
250,000
685,000
$ 1,535,000
Retained earnings
Total stockholders' equity
(1) Prepare the updated stockholders' equity section after the distribution is made.
(2) Compute the number of shares outstanding after the distribution is made.
Complete this question by entering your answers in the tabs…
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C
unctions
Question 3
Listen
Given
11π
0 =
3
which of the below statements are true?
Press on an item in the answer box, and then press on the box under the corresponding
category. To remove an item, you can press on the box and the trash icon to the right of the
item you want to remove.
True
False
Add an answer item!
Add an answer item!
Answer Bank
11π
tan(1)=-√3
cos()=sin()=√
11π
3
11T is in Quadrant III.
11π
is in Quadrant IV.
cos (13)=-
3
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PLEASE, WRITE THE SOLUTIONS ON PAPER, EXPLAINING THE ENTIRE PROCESS, THE ONLY AND CORRECT ANSWERS ARE FOR (i) V(t) = exp ( -2e^0.02t + 2 ) for 0 < t <10, V(t) = exp( -0.0928 - 0.03t - 0.0005t^2) for 10 < t <15, V(t) = exp ( 0.0197 - 0.045t ) for 15 < t.
AND (ii) C3 = £2500 x 0.5877 = £1,469.16
AND (iii) i^2 = 4.48% pa
AND (iv) PV = £988.38
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PLEASE, WRITE THE SOLUTIONS ON PAPER, EXPLAINING THE ENTIRE PROCESS, THE
ONLY AND CORRECT ANSWERS ARE FOR (i) V(t) = exp (-2e^0.02t + 2 ) for 0 15
(i)
Derive, and simplify as far as possible, expressions in terms of t for V(t), where V(t) is
the present value of a unit sum of cash flow made at time t. You should derive separate
expressions for the three sub-intervals.
(ii)
Hence, making use of the result in part (i), calculate the value at time t = 3 of a payment of
£2,500 made at time t = 15.
(iii) Calculate, to the nearest 0.01%, the constant nominal annual rate of interest convertible
half-yearly implied by the transaction in part (ii).
(iv) Making use of the result in part (i), calculate the present value of a payment stream p(t)
paid continuously from time t = 15 to t = 20 at a rate of payment at time t given by:
p(t) = 300e 0.02t
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Exercise 1-13 (Algo) Identifying effects of transactions on the accounting equation LO P1
The following table shows the effects of transactions 1 through 5 on the assets, liabilities, and equity of Mulan's Boutique. Match each
transaction with its probable description.
Assets
= Liabilities +
Accounts
Equity
Accounts
Receivable
+
+ Supplies
$0 + $ 3,800
Cash +
$ 29,000 +
-4, 800
Land
+ $ 23,000
4,800
=
=
Common
Stock
$ 55,800 +
Payable
$0
Revenues
$0
+
1.
+
2.
+
1,080
+1,080
3.
2,300
4.
-1,080
-1,080
5.
+2, 300
+ $ 4,880
$0
$ 25,420 +
$ 55,800 + $ 2,300
Transaction
Description
1.
2.
3.
4.
5.
2,300
2,300
$0
+
$ 27,800 =
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