Peer Instruction: Iterations Multiple Choice Questions

Q-1: Which of the following prints the numbers i through n?

• for i in range(1,i):
  

• Incorrect! Here the range printed is 1 to i-1

• for i in range(1,n):
  

• Incorrect! Here the range printed is 1 to n-1

• for i in range(1,n+1):
  

• Incorrect! Here the range printed is 1 to n

• for i in range(i,n):
  

• Incorrect! Here the range printed is i to n-1

• None of the above

• Correct! Use range(i, n+1) to print numbers through i to n

Q-2: Which of the following adds up the numbers 1 through 4?

• for i in range(1,5):
     sum = 0
     sum = sum + i
  

• Incorrect! Here the sum will be set to 0 each time the loop runs from 1 to 4.

• sum = 0
  for i in range(1,5):
     sum = sum + 1
  

• Incorrect! sum needs to be added by i and not 1.

• sum = 0
     for i in range(1,5):
     sum = sum + sum
  

• Incorrect! sum needs to added by i and not sum.

• sum = 0
  for i in range(1,5):
     sum = sum + i
  

• Correct! The code will keep adding numbers from 1 to 4 with each iteration.

• I don’t know

• Incorrect! Define a variable, say sum, and set it to 0. Run a loop from 1 to 4 and keep adding i to sum.

Q-3: What does the following code print?

n = 4
for i in range(1,n):
   print(i,n,end='')
   n = 6

• 1 4 2 4 3 4
• Incorrect! n has been reset to 6 inside the loop. So, the code should print 6 from the second iteration. However, the n inside range will not be affected by this reset.
• 1 4 2 6 3 6
• Correct! n has been reset to 6 inside the loop. So, the code should print 6 from the second iteration. However, the n inside range will not be affected by this reset.
• 1 4 2 6 3 6 4 6 5 6
• Incorrect! n has been reset to 6 inside the loop. So, the code should print 6 from the second iteration. However, the n inside range will not be affected by this reset.
• This will cause an error
• Incorrect! n has been reset to 6 inside the loop. So, the code should print 6 from the second iteration. However, the n inside range will not be affected by this reset.
• I don't know
• Incorrect! n has been reset to 6 inside the loop. So, the code should print 6 from the second iteration. However, the n inside range will not be affected by this reset.

Q-4: Which of the following generates this pattern for n = 5?

*****
*****
*****
*****
*****

• for i in range(0,n):
     print("*" * i)
  

• Incorrect! Although there will be 5 iterations, it will not print 5 rows and columns of asterisks. With each iteration, i columns of asterisks will be printed. It will generate a staircase pattern.

• for i in range(0,n):
     print("*" * n)
  

• Correct! This will print 5 rows and columns of asterisks. There will be 5 iterations. With each iteration, n=5 columns of asterisks will be printed.

• for i in range(1,n):
     print("*" * i)
  

• Incorrect! This code will run 4 iterations, starting from 1 through n-1=4. Also, with each iteration, i columns of * will be printed. It will generate a staircase pattern.

• for i in range(1,n):
     print("*" * n)
  

• Incorrect! This will print 4 rows and 5 columns of asterisks. There will be 4 iterations, starting from 1 through n-1=4. With each iteration, n=5 columns of asterisks will be printed.

• I don’t know

• Incorrect! Print n=5 columns of asterisks. Run a for loop to print n=5 rows.

Q-5: For n = 5, which of the following is the number of frontspaces and stars printed?

 for r in range(1, n + 1):
     print(' ' * frontspaces, '*' * stars)


    *
   ***
  *****
 *******
*********

• frontspaces = (n-r) / 2, stars = r
• Incorrect! (n - r) / 2 can output a non-integer. This will cause an error.
• frontspaces = n - r, stars = 2 * r
• Incorrect! This will not print the correct pattern. For instance, take n = 5 and r = 1. To print the first row, we need 4 frontspaces and 1 star. But here, n-r = 4 and 2 * r = 2
• frontspaces = n - r, stars = 2 * r - 1
• Correct! This will print the correct pattern. For instance, take n = 5 and r = 1. To print the first row, we need 4 frontspaces and 1 star. Here, n-r = 4 and 2 * r - 1 = 1
• frontspaces = r, stars = n - r
• Incorrect! This will not print the correct pattern. For instance, take n = 5 and r = 1. To print the first row, we need 4 frontspaces and 1 star. But here, r = 1 and n - r = 4.
• I don't know
• Incorrect! For instance, to print the first row we need 4 frontspaces and 1 star. So, n - r = 5 - 1 = 4 and 2 * r - 1 = 2 * 1 - 1 = 1.

Q-6: What does the following code print?

for i in range(1, 4):
   for j in range(1, 4):
      print(i,j,end=‘ ')

• 1 1 2 2 3 3
• Incorrect! The nested for loop will run through j = 1 to 3 for every i.
• 1 2 3 1 2 3 1 2 3
• Incorrect! The nested for loop will run through j = 1 to 3 for every i.
• 1 1 1 2 1 3 2 1 2 2 2 3 3 1 3 2 3 3
• Correct! The nested for loop will run through j = 1 to 3 for every i.
• 1 1 2 1 3 1 2 1 2 2 2 3 3 1 3 2 3 3
• Incorrect! The nested for loop will run through j = 1 to 3 for every i.
• I don't know
• Incorrect!The nested for loop will run through j = 1 to 3 for every i.

Q-7: Which of the following code generates the times table for any n as shown below?

For n = 4,

1 2 3 4 2 4 6 8 3 6 9 12 4 8 12 16

• for i in range(0, n):
     for j in range(0, n):
        print(i * j, end=‘ ')
  print()
  

• Incorrect! This will print a row of 0s in the first row and a row of n-1 times in the last row.

• for i in range(1, n + 1):
     for j in range(1, n + 1):
        print(i * j, end=‘ ')
  

• Incorrect! This will print the times table but not in the format given above. There will be no break after the first line.

• for i in range(1, n + 1):
     for j in range(1, n + 1):
        print(i * j, end=‘ ')
     print()
  

• Correct! This will print the times table in the right format due to an additional print in the end. The range in both loops in right and there will be a break after each line of the nested for loop.

• for i in range(1, n + 1):
     for j in range(1, n + 1):
        print(i * j, end=‘ ')
  print()
  

• Incorrect! This will print the times table but not in the format given above. There will be no break after the each line but only after end of the last line.

• I don’t know

• Incorrect! Use a nested for loop and use the * operator.

Q-8: What does the following code print?

x = 5
if (x < 3):
    x = 1
    print("A")
    if(x > 100):
        print("B")
    else:
        print("C")
    print("D")
print("E")

if (x > 2)
    print("F")
    if(x % 3 == 2)
        print("G")
    if (x % 3 == 1)
        print("H")
    else:
        print("I")

• C D E F G I
• Incorrect! Since x=5, x<3 is False and x%3==2 is True. So, E F G I will print.
• D E F G
• Incorrect! Since x=5, x<3 is False and x%3==2 is True. So, E F G I will print.
• E F G I
• Correct! Since x=5, x<3 is False and x%3==2 is True. So, E F G I will print.
• E F H
• Incorrect! Since x=5, x<3 is False and x%3==2 is True. So, E F G I will print.
• I don't know
• Incorrect! Since x=5, x<3 is False and x%3==2 is True. So, E F G I will print.

Q-9: What does the following code print?

x = 6
while(x > 4)
 print(x, end=' ')
 x = x - 1

• 6 5
• Correct! Each time the loop runs, value of x decrements by 1. So, when its value gets down to 4, the loop condition is no longer satisfied.
• 6 5 4
• Incorrect! Each time the loop runs, value of x decrements by 1. So, when its value gets down to 4, the loop condition is no longer satisfied.
• 6 5 4 3
• Incorrect! Each time the loop runs, value of x decrements by 1. So, when its value gets down to 4, the loop condition is no longer satisfied.
• 5 4 3
• Incorrect! Each time the loop runs, value of x decrements by 1. So, when its value gets down to 4, the loop condition is no longer satisfied.
• I don't know
• Incorrect! Each time the loop runs, value of x decrements by 1. So, when its value gets down to 4, the loop condition is no longer satisfied.

Q-10: What does the following code print?

i=0

while(i < 3)
     print(i, end=' ')

• 0 0 0
• Incorrect! The value of i never changes from 0. So, the loop condition is always true and it will keep printing i=0.
• 0 1 2
• Incorrect! The value of i never changes from 0. So, the loop condition is always true and it will keep printing i=0.
• 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . . . .
• Correct! The value of i never changes from 0. So, the loop condition is always true and it will keep printing i=0.
• 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 . . . .
• Incorrect! The value of i never changes from 0. So, the loop condition is always true and it will keep printing i=0.
• I don't know
• Incorrect!The value of i never changes from 0. So, the loop condition is always true and it will keep printing i=0.

Q-11: Which of the following correctly translates the for loop below to a while loop?

for i in range(n):

<body>

• i = 0
  while(i < n)
     <body>
  

• Incorrect! This will be an infinite loop as the value of i never changes.

• i = 0
  while(i < n)
     <body>
     i = i + 1
  

• Correct! The value of i increments by 1 in each iteration till it becomes equal to n at which point the loop condition won’t be satisfied.

• i = 0
  while(i < n)
     <body>
     n = n + 1
  

• Incorrect! This is not the right implementation of the given for loop as the value of i remains the same and the value of n keeps increasing with each iteration.

• i = 1
  while(i < n)
     <body>
     i = i + 1
  

• Incorrect! This is not the right implementation of the given for loop as the value of i remains the same and the value of n keeps increasing with each iteration.

• I don’t know

• Incorrect! The value of i should increment by 1 with each iteration of while loop.

Q-12: Which of these numbers will stop the loop?

valid = False
while not valid:
  x = eval(input ("Enter a number: "))
  valid = (x % 2 == 1 and x % 3 == 0)

• 2
• Incorrect! To get out of the loop, valid should be True. According to the condition provided, an odd number which is a multiple of 3 should work.
• 9
• Correct! To get out of the loop, valid should be True. According to the condition provided, an odd number which is a multiple of 3 should work.
• 6
• Incorrect! To get out of the loop, valid should be True. According to the condition provided, an odd number which is a multiple of 3 should work.
• None of the above
• Incorrect! To get out of the loop, valid should be True. According to the condition provided, an odd number which is a multiple of 3 should work.
• I don't know
• Incorrect! To get out of the loop, valid should be True. According to the condition provided, an odd number which is a multiple of 3 should work.

Q-13: Which of these will exit on 9?

• x = eval(input ("Enter a number: "))
   while (x % 2 == 1 and x % 3 == 0):
     x = eval(input ("Enter a number: "))
  

• Incorrect! Incorrect! 9 is an odd multiple of 3. So, the condition would always hold true and will not exit the loop. There’s no break statement in this option.

• x = eval(input ("Enter a number: "))
   while True:
     if (x % 2 == 1 and x % 3 == 0):
         break;
     x = eval(input ("Enter a number: "))
  

• Correct! This will exit due to the break statement.

• Both!

• Incorrect! There’s no break statement in option A.

• Neither

• Incorrect! There’s a break statement in option B.

• I don’t know

• Incorrect! A break statement can be used to exit the loop.

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