TCS NQT Input/Output Quiz- 2

In the given pseudo code we can see that the code will gives the output of the maximum element present in the given array, so the correct missing code will be :
If array[i]>max
max=array[i]
End if

In the given pseudo code x=3, y=4, z=4

Associativity of the relational operator is left to right,

So in the given expression, ((z>=y>=x)) => ((4>=4>=3)) evaluation will be done from left to right, i.e,

=> 1>=3 => 0 (as 4>=4 return 1 and 1>=3 return 0)

So, the code will print 0.

Array c[10] is initialize with value {11, 22, 33, 44, 55, 66, 77, 88, 99, 100}
For loop will terminate when a>=7,
So, initially b =0,
Inside for loop when a=0,
As, c[0]%2==1 i.e, (11%2 =1)
b = 11,
when a = 1
As, c[1]%2!=1 i.e, (22%2 =0)
So, b = 11 (remains same),
when a = 2
As, c[2]%2==1 i.e, (33%2 =1)
So, b = 11+33 = 44,
Similarly…
When a = 3, 5 b remains same
But when a=4, 6
b = 44+55+77=176.
When a = 7 condition of for loop will become false, and loop will not execute further.

Nothing will be printed on the screen

Explanation : Inside for loop before the print statement continue keyword is used so, it skips the print statement in every iteration of the loop.
So, nothing will get printed on the output screen.

Since, in the given code digit = 0
So the if condition becomes false hence else part will execute and print digit i.e, 0

In the given code rose =1, lotus =2
Associativity to the assignment operator is left to right,
So the in the given expression : rose += lotus = rose
First, lotus is assign with 1 (value of rose), so lotus =1
Then rose = 2 (i.e, rose + lotus)
So we get output as 21.

Loop will execute from 0 to dest-1 so in reverse format, from the given options, for(mile=dest-1; mile>-1;mile--)
Is same as for(mile=dest-1; mile>=0; mile--)

Continue statement is used to skip the part,

In the given code the for loop will executes for skip_num values 1, 2, 3, 4, 5 when it becomes 6 the loop will terminate

Inside the loop, for the value 2 and 4 continue statement is used, it means that when skip_num =2 or skip_num = 4, cout<<skip_num will not execute,

Hence we get output : 1 3 5

In the given code, function problem is call with argument 1,
Means num =1
Since if condition becomes false, as num i.e 1 is not greater than 5
Hence function will return 2 (as it return ++num means it increment the value of num by 1 then return it)
So, we get output 2.

Initially a=3, b=5
Now, Since b-- (i.e, 5--) return true and b becomes, 4 i.e, (b=4)
a++; // it will increment the value of a by 1 i.e, (a=4)
Again, b-- (i.e, 4--) return true and b becomes, 3 i.e, (b=3)
a++; // it will increment the value of a by 1 i.e, (a=5)
Again, b-- (i.e, 3--) return true and b becomes, 2 i.e, (b=2)
a++; // it will increment the value of a by 1 i.e, (a=6)
Again, b-- (i.e, 2--) return true and b becomes, 1 i.e, (b=1)
a++; // it will increment the value of a by 1 i.e, (a=7)
Again, b-- (i.e, 1--) return true and b becomes, 0 i.e, (b=0)
a++; // it will increment the value of a by 1 i.e, (a=8)
Now, b-- (i.e, 0--) return false and loop gets terminated.

So, we get output 8. (because a=8).