Move Zeroes to I
move 5 to j
perform para1 varying I from 10 by -2 until I = 0
display j.
para1.
Add 5 to j.
What’ll be the value after execution of display stmt.
A) 35
B) 40
C) 30
D) 25
please explain how?
Answers were Sorted based on User's Feedback
Answer / gayathri
The para1 will be performed only 5 times for the values of
I = 10, 8, 6, 4, and 2 respectively. Para1 can not be
executed for the value I = 0 as it uses Perform until
condition.
so the value of J is 5+(5*5) equals 30
correct answer is c) 30
Please correct me if I am wrong.
| Is This Answer Correct ? | 13 Yes | 1 No |
Answer / sudeshna majumdar
Correct answer is c) 30.
The para1 will be performed 6 times where the values of I
will be 10, 8, 6, 4, 2, and 0 respectively.
After each processing of para1, value of J will be
incremented by 5 and value of I will be decremented by 2.
Now, the para1 is called starting from I = 10 and ends when
I = 0. Thus, after 6 time para1 processing,J value will be
5*6 = 30.
| Is This Answer Correct ? | 10 Yes | 4 No |
Answer / shivakumar deshpande
Gayathri is correct ie Answer -3 is correct
| Is This Answer Correct ? | 4 Yes | 0 No |
Answer / suputhru
Gaayathri is right!!!
para1 will be performed only 5 times
J is 5+(5*5) equals 30
correct answer is c) 30
| Is This Answer Correct ? | 4 Yes | 0 No |
Answer / avi
The value of J before the Para call is 5.
Now,the loop executes for 6 times, so 5*6 is added to the
existing value of J.
J=5+(5*6)
=35.
The correct option is 35.
| Is This Answer Correct ? | 7 Yes | 12 No |
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