Use the following flowchart to answer the below two questions 1) and 2).

*2 points*

*2 points*

*3 points*

*3 points*

*3 points*

*3 points*

*2 points*

*3 points*

*2 points*

*2 points*

*Computational Thinking week 2 graded assignment Complete Solutions Are Discussed In This Blog. We Hope This Might Help You All In Matching Answers . Or For Some Others Reasons If Not Able To Complete Graded Assignments *

- The following procedure is executed using the “Scores” dataset. At the end of the execution, diff stores the difference between the highest and lowest total marks. The programmer may have made mistakes in one or more steps. Identify all such steps (if any). Assume that all the steps not mentioned in options are free from errors. It is a Multiple Select Question (MSQ).

Step 1. Arrange all cards in a single pile called Pile 1

Step 2. Initialize variables max to 0, min to 300, and diff to 0

Step 3. If Pile 1 is empty then go to step 8

Step 4. Read the top card in Pile 1

Step 5. If*Total*< max then store*Total*in max

Step 6. If*Total*< min then store*Total*in min

Step 7. Move the current card to another pile called Pile 2 and repeat from step 3

Step 8. Update value of diff to (max – min)

Step 2. Maintain two variables A, B and initialize them to 0

Step 3. If Pile 1 is empty then stop the iteration

Step 4. Read the top card in Pile 1

Step 5. Increment variable B

Step 6. If

*Word*does not end with a full stop then execute step 9

Step 7. If

*Word*ends with a full stop and B > A then store B in A

Step 8. Re-initialize the variable B to 0

Step 9. Move the current card to another pile called Pile 2 and repeat from step 3

4. The following procedure is executed using “Words” dataset. What will A and B represent at the end of the execution?

Step 1. Arrange all cards in a single pile called Pile 1

Step 2. Maintain two variables A, B and initialize them to 100000

Step 3. If Pile 1 is empty then stop the iteration

Step 4. Read the top card in Pile 1

Step 5. If *Mathematics Marks* < A, then store A in B and *Mathematics Marks* in A

Step 6. If *Mathematics Marks* > A and *Mathematics Marks* < B, then store *Mathematics Marks* in B

Step 7. Move the current card to another pile called Pile 2 and repeat from step 3

*Physics*which is 37. Let all the cards be arranged in a single pile called Pile 1 in such a way that Kajal’s card is below Rashmita’ s card. What will the values of A and B be at the end of the execution?

Step 1. Maintain two variables A and B and initialize them to 101 and “None” respectively

Step 2. If Pile 1 is empty then stop the iteration

Step 3. Read the top card in Pile 1

Step 4. If

*Physics marks*<= A, then store

*Physics marks*in A and store student’s name in B

Step 5. Move the current card to another pile called Pile 2 and repeat from step 2

7. he following procedure is executed using “Scores” dataset. What will X and Y represent at the end of the execution?

Step 1. Arrange all cards in a single pile called Pile 1

Step 2. Maintain variables A, B, Y and Initialize them to 0

Step 3. Maintain a variable X and initialize it to “None”

Step 4. If Pile 1 is empty then stop the iteration

Step 5. Read the top card in Pile 1

Step 6. If the *Gender *is ‘F’ then add one to A. If A > Y then store A in Y and “Female” in X

Step 7. If the *Gender *is ‘M’ then add one to B. If B > Y then store B in Y and “Male” in X

Step 8. Move the current card to another pile called Pile 2 and repeat from step 4

## Add Your Heading Text Here

# Week-2, Graded

Week-2, Graded

Question 1 [2 Marks]

Statement

Options

(a)

(b)

(c)

(d)

(e)

Answer

Solution

Question-2 [2 Marks]

Statement

Options

(a)

(b)

(c)

(d)

Answer

Solution

Question-3 [3 Marks]

Statement

Options

(a)

(b)

(c)

(d)

Answer

Solution

Question-4 [3 Marks]

Statement

Options

(a)

(b)

(c)

(d)

Answer

Solution

Question-5 [3 Marks]

Statement

Options

(a)

(b)

(c)

(d)

Answer

Solution

Question-6 [3 Marks]

Statement

Options

(a)

(b)

(c)

(d)

Answer

Solution

Question-7 [3 Marks]

Statement

Options

(a)

(b)

(c)

(d)

Answer

Solution

Question-8 [3 Marks]

Statement

Options

(a)

(b)

(c)

(d)

Answer

Solution

Question-9 [3 Marks]

Statement

Options

(a)

(b)

(c)

(d)

Answer

Solution

Question-10 [3 Marks]

Statement

Options

(a)

(b)

(c)

(d)

Answer

Solution

## Question 1 [2 Marks]

### Statement

The following procedure is executed using the "Scores" dataset. At the end of the execution, **diff **stores the

difference between the highest and lowest total marks. The programmer may have made mistakes in one or

more steps. Identify all such steps (if any). Assume that all the steps not mentioned in options are free from

errors. It is a Multiple Select Question (MSQ).

Step 1. Arrange all cards in a single pile called Pile 1

Step 2. Initialize variables **max **to 0, **min **to 300, and **diff **to 0

Step 3. If Pile 1 is empty then go to step 8

Step 4. Read the top card in Pile 1

Step 5. If *Total < ***max ***then store Total *in **max**Step 6. If *Total *< **min **then store *Total *in **min**

Step 7. Move the current card to another pile called Pile 2 and repeat from step 3

Step 8. Update value of **diff **to (**max **- **min**)

### Options

#### (a)

Step 2: Incorrect initialization of variable **max**

#### (b)

Step 2: Incorrect initialization of variable **min**

#### (c)

Step 3: The statement should be "If Pile 1 is empty then stop the iteration"

#### (d)

Step 5: Incorrect conditional statement to update **max**

#### (e)

No mistake

### Answer

(d)

### Solution

In the given procedure we want to find the difference between the maximum total marks secured and

minimum total marks secured. For this, the procedure is storing the maximum mark in **max **and minimum

marks in **min**. In step 5, if the *Total *marks are more than **max **(as we see that in step 1, max is initialized to

0), the the value of **max **will be updated. The correct conditional statement should have been, " If *Total >***max ***then store Total *in **max **" Then only we can say that max is storing the maximum total marks. But in the

given procedure, "if *Total *is less than max then *Total *is stored in **max**" which results in never updating the

value of **max**. Therefore, option (d) is correct.

## Question-2 [2 Marks]

### Statement

Match the following expressions in the Column 1 with the appropriate values in column 2.

Column 1 | Column 2 |

a. 4 == 4 or 2 < 6 | 1. Invalid expression |

b. 2 == 2 and 2 > 5 | 2. True |

c. 8 = 3 | 3. False |

d. 45 + '2' | 4. 47 |

e. 2 < 3 | 5. "47" |

### Options

#### (a)

a - (2), b - (3), c - (1), d - (4), e - (2)

#### (b)

a - (2), b - (3), c - (1), d - (5), e - (2)

#### (c)

a - (2), b - (3), c - (1), d - (1), e - (2)

#### (d)

a - (1), b - (3), c - (2), d - (1), e - (1)

### Answer

(c)

### Solution

4 == 4 - True , 2 < 6 - True. Therefore, True or True = True. **(a. maps to 2.)**

2 == 2 - True, 2 > 5 - False. Therefore, True and False = False. **(b. maps to 3.)**

8 = 3 - Invalid expression, Assignment is for variables not the values(like integer,

strings, etc) **(c. maps to 1.)**

45 + '2' - Invalid expression, Addition of two different data types is not possible. **(d. maps to 1.)**2 < 3 - True. **(a. maps to 2.)**

## Question-3 [3 Marks]

### Statement

The following procedure is executed using "Words" dataset. What will **A **represent at the end of the

execution?

Step 1. Arrange all cards in a single pile called Pile 1

Step 2. Maintain two variables **A, B **and initialize them to 0

Step 3. If Pile 1 is empty then stop the iteration

Step 4. Read the top card in Pile 1

Step 5. Increment variable **B**

Step 6. If *Word *does not end with a full stop then execute step 9

Step 7. If *Word *ends with a full stop and **B > A **then store **B **in **A**Step 8. Re-initialize the variable **B **to 0

Step 9. Move the current card to another pile called Pile 2 and repeat from step 3

### Options

#### (a)

Length of the shortest sentence based on the number of words

#### (b)

Length of the longest sentence based on the number of words

#### (c)

Length of the longest sentence based on the number of letters

#### (d)

Length of the shortest sentence based on the number of letters

### Answer

(b)

### Solution

Step 5: **B **stores the sum of the words but how many words?

Step 6: It says if the word does not end with full stop then go to step 9 which is nothing but repeating the

iteration. It means when a word ends with a full stop, then step 7 and 8 will be executed. From step 8 it is

clear that whenever the word ends with a full stop then **B **is being reinitialized to 0. This means **B **stores the

word counts in a sentence.

Step 7: For every sentence, **A **is compare with **B**. The value of **A **gets updated with the value of **B **whenever **B**is greater than **A**.

As **A **is initialized with 0, and **B **stores the number of words in a sentence, **A **stores the maximum words in a

sentence i.e., length of longest sentence in terms of number of words.

## Question-4 [3 Marks]

### Statement

The following procedure is executed using "Words" dataset. What will **A **and **B **represent at the end of the

execution?

Step 1. Arrange all cards in a single pile called Pile 1

Step 2. Maintain two variables **A, B **and initialize them to 100000

Step 3. If Pile 1 is empty then stop the iteration

Step 4. Read the top card in Pile 1

Step 5. If *Mathematics Marks *< **A**, then store **A **in **B **and *Mathematics Marks *in **A**

Step 6. If *Mathematics Marks *> **A **and *Mathematics Marks *< **B**, then store *Mathematics Marks *in **B**Step 7. Move the current card to another pile called Pile 2 and repeat from step 3

### Options

#### (a)

A = Highest mark in Mathematics and B = Lowest mark in Mathematics

#### (b)

A = Lowest mark in Mathematics and B = Highest mark in Mathematics

#### (c)

A = Highest mark in Mathematics and B = Second highest mark in Mathematics

#### (d)

A = Lowest mark in Mathematics and B = Second lowest mark in Mathematics

### Answer

(d)

### Solution

Let us take the first 10 *Mathematics *marks from dataset. From step 5 and 6 it is clear that we only need the*Mathematics *marks of students (cards).

The Mathematics marks of first five students are 68, 62, 57, 42 and 87.

First card:

**Step 5: **If we pick the first card with *CardNo *0, the *Mathematics *mark is 68. As 68 is less than 100000 (the

current value of **A**), **B **will store the value of **A **therefore, the value of **B **will be 100000 and **A **will be updated

with 68. Therefore, the current value of **A **and **B **are 68 and 100000 respectively.

Step 6: *Mathematics *mark is not greater than A, step 6 will not be executed.

Second Card:

**Step 5: **The *Mathematics *mark is 62 which is less than 68 therefore, step 5 will be executed. **A **will be

updated with current *Mathematics *marks and **B **will be updated with **A**. Therefore, the current value of **A**and **B **are 62 and 68 respectively.

Step 6: *Mathematics *mark is not greater than A, step 6 will not be executed.

Third card

**Step 5: **The *Mathematics *mark is 57 which is less than 62 therefore, step 5 will be executed. **A **will be

updated with current *Mathematics *marks and **B **will be updated with **A**. Therefore, the current value of **A**and **B **are 57 and 62 respectively.

Step 6: *Mathematics *mark is not greater than A, step 6 will not be executed.

Fourth card

**Step 5: **The *Mathematics *mark is 42 which is less than 57 therefore, step 5 will be executed. **A **will be

updated with current *Mathematics *marks and **B **will be updated with **A**. Therefore, the current value of **A**and **B **are 42 and 57 respectively.

Step 6: *Mathematics *mark is not greater than A, step 6 will not be executed.

From the first three cards having *Mathematics *marks 68, 62, 57 and 42, it is clear that **A **stores the lowest

marks in *Mathematics *whereas **B **stores the second lowest marks in *Mathematics*

## Question-5 [3 Marks]

### Statement

The given procedure is executed using the “Scores” dataset. Kajal and Rashmita both have scored the lowest

mark in *Physics *which is 37. Let all the cards be arranged in a single pile called Pile 1 in such a way that

Kajal's card is below Rashmita' s card. What will the values of **A **and **B **be at the end of the execution?

Step 1. Maintain two variables **A **and **B **and initialize them to 101 and "None" respectively

Step 2. If Pile 1 is empty then stop the iteration

Step 3. Read the top card in Pile 1

Step 4. If *Physics *marks <= **A**, then store *Physics *marks in **A **and store student's name in **B**Step 5. Move the current card to another pile called Pile 2 and repeat from step 2

### Options

#### (a)

A = 37, B = "Rashmita"

#### (b)

A = 37, B = "Kajal"

#### (c)

A = "Kajal", B = 37

#### (d)

A = "Rashmita", B = 37

### Answer

(b)

### Solution

Step 4: If the *Physics *marks of the picked card is less than or equal to **A**, **A **will be updated with the current*Physics *marks of the picked card and simultaneously the student ' s name will be stored in **B**. Now let after a

certain iterations,

Let us assume that, **A **= z where and **B **= "xyz"

As Kajal's card is below Rashmita 's card, Rashmita's card will be picked up first. As 37 is less than or equal to

the current value of **A **(as it is mentioned in the question that 37 is the minimum marks), **A **will be updated

with 37 and **B **will be updated with "Rashmita".

Now Kajal's card will be picked up. 37 is equal to the current value of **A **which is 37, Step 4 will be executed

and the values of **A **and **B **will change to 37 and "Kajal" respectively.

Therefore, **A **= 37 and **B **= "Kajal".

## Question-6 [3 Marks]

### Statement

The following pseudocode is executed using the “Scores” dataset. What will **count **represent at the end of

the execution of pseudocode?

count = 0

while(Pile 1 has more cards){

Read the top card X from Pile 1

C = True

if(X.Mathematics < 56){

C = False

}

if(X.Physics < 56){

C = False

}

if(X.Chemistry < 56){

C = False

}

if(C == True){

count = count + 1

}

Move X to Pile 2

}

### Options

#### (a)

Number of students who scored more than 56 marks in all three subjects

#### (b)

Number of students who scored less than 56 marks in all three subjects

#### (c)

Number of students who scored less than 55 marks in all three subjects

#### (d)

Number of students who scored more than 55 marks in all three subjects

### Answer

(d)

### Solution

The question is about finding what **C **represents? From the condition given in the line 14 it is clear that**count **totally depends on the value of **C**. Whenever **C **is true, **count **is being incremented. So it is clear that**count **represents the number of times **C **is being true.

Now let us focus on when **C **is true? **C **is initialized to True at the start. In the line 5, it is said that **C **will be

false whenever the Mathematics mark is less than 56. It means **C **will be true if Mathematics marks is

greater than 55. From line 8, it is clear that **C **is also being false whenever Physics marks are less than 56.

That means if **C **is true, then the student in the selected card has scored more than 55 in Physics. Similarly

from line 11, it is also clear that **C **is true whenever students scores more than 55 in Chemistry. So if any of

the conditions become true, that is student has scored less than 56 in either Mathematics or Physics or

Chemistry, **C **will be False. Otherwise, it is true. Which means **count **is the number of Students who have

scored more than 55 in all the subjects.

## Question-7 [3 Marks]

### Statement

The following procedure is executed using "Scores" dataset. What will **X **and **Y **represent at the end of the

execution?

Step 1. Arrange all cards in a single pile called Pile 1

Step 2. Maintain variables **A**, **B**, **Y **and Initialize them to 0

Step 3. Maintain a variable **X **and initialize it to “None”

Step 4. If Pile 1 is empty then stop the iteration

Step 5. Read the top card in Pile 1

Step 6. If the *Gender *is 'F' then add one to **A**. If **A > Y **then store **A **in **Y **and “Female" in **X**Step 7. If the *Gender *is 'M' then add one to **B**. If **B > Y **then store **B **in **Y **and “Male” in **X**Step 8. Move the current card to another pile called Pile 2 and repeat from step 4

### Options

#### (a)

X : "Male"

Y : 17

#### (b)

X : "Female"

Y : 17

#### (c)

X : "Female"

Y : 13

#### (d)

X : "Male"

Y : 13

### Answer

(a)

### Solution

Here in step 2, variables **A**, **B **and **Y **are initialized to 0 whereas in step 3, variable **X **is initialized to ”None”

which indicates that this variable is used to store value of string datatype. If we observe steps 6 and 7 then

we can notice that the computation in these steps is identical except it is for different gender.

Variables **A **and **B **store the number of cards from 'F' and 'M' respectively. Variable **Y **stores maximum

between **A **and **B**. This step also stores the gender which corresponds to the maximum in the variable **X**.

Here is the full list of cards which belong to the respective cities.

Gender | Card No. | Total no. of cards |

F | 3,4,5,10,12,14,15,18,21,22,23,25,28 | 13 |

M | 0,1,2,6,7,8,9,11,13,16,17,19,20,24,26,27,29 | 17 |

## Question-8 [3 Marks]

### Statement

The given pseudocode is executed using the “Words” dataset. At the end of the execution, **count **finds the

number of sentences with least number of words. Choose the correct option to complete the pseudocode.

count = 0, A = 0, B = 100000

while(Pile 1 has more cards){

Read the top card X from Pile 1

A = A + 1

if(X.Word ends with full stop){

if(A == B){

*** Statement 1 ***

}

if(A < B){

B = A

*** Statement 2 ***

}

A = 0

}

Move X to Pile 2

}

### Options

#### (a)

Statement 1: **count **= 1, Statement 2: **A **= **A **+ 1

#### (b)

Statement 1: **count **= 1, Statement 2: **count **= **count **+ 1

#### (c)

Statement 1: **count **= **count **+ 1, Statement 2: **count **= 1

#### (d)

Statement 1: **count **= **count **+ 1, Statement 2: **A **= 1

### Answer

(c)

### Solution

In the question, **count **is supposed to find the number of sentences with least number of words in it.

Therefore, **count **should be initialized to 0. Along with **count**, we need two more other variables, one to

keep track of the number of words in the current sentence. This is being done by the variable **A**. Second

variable to keep track of the number of least words in a sentence seen till now. From the pseudocode it is

clear that this job is being done by the variable **B**.

Remember, there could be more sentences having the least number of words. Therefore, whenever we find

the next least, **B **should be updated with the minimum number of words in a sentence. Next, if we find the

same number of words in the different sentence (i.e., equal to the current minimum) **count **should be

incremented. In Line 10, **B **is being updated whenever the next minimum is being found. Simultaneously,**count **should also be initialized to one. Therefore, Statement 2: **count = **1. For counting number of

sentences with same minimum number of words **count **should be incremented whenever we find the

sentence with same min words and therefore, Statement 1: **count **= **count **+ 1

## Question-9 [3 Marks]

### Statement

The following pseudocode is executed using the “Words” dataset. At the end of the execution, **A **captures

the minimum letter count of an adjective. Choose the correct code fragment(s) to complete the

pseudocode. It is Multiple Select Question (MSQ).

1. A = 100

while(Pile 1 has more cards){

Read the Top card X in Pile 1

4. *********************

5. * Fill the code *

6. *********************

7. Move X to Pile 2

8. }

### Options

#### (a)

if(X.PartOfSpeech == “Adjective” and X.LetterCount > A){

A = X.LetterCount

3. }

#### (b)

if(X.PartOfSpeech == “Adjective” and X.LetterCount < A){

A = X.LetterCount

3. }

#### (c)

if(X.PartOfSpeech == “Adjective”){

if(X.LetterCount > A){

A = X.LetterCount

4. }

5. }

#### (d)

if(X.PartOfSpeech == “Adjective”){

if(X.LetterCount < A){

A = X.LetterCount

4. }

5. }

### Answer

(b), (d)

### Solution

From the question, it is clear that **A **stores the minimum number of letters in an Adjective. At the start of

pseudocode, **A **is initialized to 0. To find the minimum letter count of an adjective, we know that **A **should be

compared to the number of letters in the word. Whenever the part of speech of a word is adjective and the

number of letters in the word is less than **A**, **A **should be updated with the letter count. Therefore, there are

two conditions to be followed. First, the picked word should be an adjective and second, the letter count of

this adjective should be less than **A **(i.e., is the current value of **A**). These two conditions are found in options

in (b) and (d).

## Question-10 [3 Marks]

### Statement

The following pseudocode is executed using the “Olympics” dataset. At the end of the execution, **A **captures

the number of players who have won a "Gold" medal but are not Indians. Choose the INCORRECT code

fragment(s) that cannot be used to complete the pseudocode. It is Multiple Select Question (MSQ).

1. A = 0

while(Pile 1 has more cards){

Read the Top card X in Pile 1

4. *********************

5. * Fill the code *

6. *********************

7. Move X to Pile 2

8. }

### Options

#### (a)

1. if(X.Medal == “Gold” and X.Nationality != "Indian"){

2. A = 1

3. }

#### (b)

1. if(X.Medal == “Gold” and X.Nationality != "Indian"){

2. A = A + 1

3. }

#### (c)

if(X.Medal == “Gold”){

if(X.Nationality != "Indian"){

3. A = A + 1

4. }

5. }

#### (d)

if(X.Nationality != "Indian"){

if(X.Medal == “Gold”){

3. A = A + 1

4. }

5. }

### Answer

(a)

### Solution

From the question, it is clear that **A **stores the number of players in who have won a "Gold" medal and are

not from India. From the options, it is clear that (b), (c), and (d) are the correct code fragments. In option (a),

the condition to check is correct , however the value of **A **is wrongly updated to one always. Hence, option

(a) has incorrect code fragment. Therefore, option (a) is the correct answer.

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