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Posted By: Ramkumar Member Level: Diamond Posted Date: 12 May 2008 Category: Computer Basics
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Consider the pseudocode:
Name="Suzy"
display "Enter name"
accept "Name"
display Name
Identify the error:
Select Answer:
Variable Name already has a value therefore, another value cannot be assigned to it.
The statement Name="Suzy" does not store the value in the variable because it is specified in double quotes.
The statement accept "Name" is incorrect. Name should not be in double quotes.
The last display statement should have been written as display "Name" to display the value of the variable.
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Your Test Score: 0 / 0
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Next Question:
The following pseudocode has been written to display a series of numbers:
begin
numeric nNum1, nNum2, nNum3, nCounter = 2
nNum1 = 1
nNum2 = 1
display nNum1, nNum2
while nCounter < 10
begin
nNum3 = nNum1 + nNum2
display nNum3
nNum1 = nNum2
nNum2 = nNum3
nCounter = nCounter + 1
end
end
What is the output of the above pseudocode?
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Previous Question:
The following pseudocode has been written to display a series of numbers:
begin
numeric nNum1, nNum2, nNum3, nCounter = 2
nNum1 = 1
nNum2 = 1
display nNum1, nNum2
while nCounter < 10
begin
nNum3 = nNum1 + nNum2
display nNum3
nNum1 = nNum2
nNum2 = nNum3
nCounter = nCounter + 1
end
end
How many times would the while loop get executed if in the declaration statement, the variable nCounter was initialized to 20 instead of 2?
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p=x;
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sum = sum + *p;
i++, p++;
}
printf("%d\n",sum);
}
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