ISRO CSE 2011 | Question: 40

Question

Consider the following pseudocode

x:=1;
i:=1;
while ( x <= 500)
begin
x:=2^x;
i:=i+1;
end

What is the value of $\textsf{i}$ at the end of the pseudocode?

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

$\underline{\textbf{Answer:}\Rightarrow}\mathbf{B)\;5}$

$\underline{\textbf{Explanation:}\Rightarrow}$

$\underline{\text{Step:1}}$

$\mathrm{x = 1,\;i = 1}$

$\underline{\text{Step:2}}$

$\mathrm{x = 2,\;i = 2}$

$\underline{\text{Step:3}}$

$\mathrm{x = 4,\;i = 3}$

$\underline{\text{Step:4}}$

$\mathrm{x = 16,\;i = 4}$

Now, $\mathrm{x = 16}\le500$

$\underline{\text{Step:5}}$

$\mathrm{x = 2^{16},\;i = 5}$

$\therefore \;\mathbf 5$ is the right answer.

Answer 2

ans is B  value of i=5  initially x=1<1000 so i becomes 2 then x=2 , i =3 ,then x=4 , i=4 then x=16, i=5 now x=2^16>1000 so exit finally i=5

Comments

Here the condition given is (x<=500), then the answer will be A) 4, Right?

Answer 3

ans is (b)

Answer 4

B

because when x =16, the while loop was (16<=500)executed then i= 5 and x=2^16, next  in whlie(2^16<=500) it is false . so ans is 5

Answer 5

before while loop

                x=1           i=1

in while loop

               x=2            i=2

               x=4            i=3

               x=16           i=4

               x= 65536       i=5

now,while condition fails and terminates.hence,i=5