Top 14 Capgemini Pseudocode Questions

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[@VikasMauryaAcademy]

Question 3 full Explanation: As X is call by reference, last updated value of X will be assigned to the variable for all the recursive calls. When c value becomes 0, x value will be 9. So, f(5,5) ->f(6,4) * 9 -> f(7,3) * 9 -> f(8,2) * 9 -> f(9,1) * 9 -> c == 0 & returns 1 Ans => 1 * 9 * 9 * 9 * 9 = 6561

[@shreyanshbhagare1267]

for question 4 answer will be - output will be 6 try by doing code

[@VikasMauryaAcademy]

@@shreyanshbhagare1267 Explanation: Here i = 0, x = 3. x = x++( x = 3) Post increment will assign value before gets updated. I value will increment by 1 and loop will run 3 times. When I becomes 3, while loop fails(3 != 3 returns false) X value remains 3 because always value gets reassigned before the updation.

[@VikasMauryaAcademy]

In question 5, the option was incorrect, so I selected a new one after solving the entire question. check full Explanation: Here i=0,s=0,n=127 (127>0) while condition true r = 127 % 10 = 7 p = 8 ^ 0 = 8 s = 0 + 8 * 7 = 56 i = 1 n = 127/10 = 12 (12>0) while condition true r = 12 % 10 = 2 p = 8 ^ 1 = 9 s = 56 + 9 * 2 = 74 i = 2 n = 12/10 = 1 (1>0) while condition true r = 1 % 10 = 1 p = 8^2 = 10 s = 74 + 1 * 10 = 84 I = 3 N = 1/10 = 0 (0>0) While condition fails S = 84

[@coursetotal6570]

shi hai question 5 me option nhi mil raha toh ek option apna likh kar vhi correct krdo, nikal jayega capgemini

[@VikasMauryaAcademy]

Explanation: Here i=0,s=0,n=127 (127>0) while condition true r = 127 % 10 = 7 p = 8 ^ 0 = 8 s = 0 + 8 * 7 = 56 i = 1 n = 127/10 = 12 (12>0) while condition true r = 12 % 10 = 2 p = 8 ^ 1 = 9 s = 56 + 9 * 2 = 74 i = 2 n = 12/10 = 1 (1>0) while condition true r = 1 % 10 = 1 p = 8^2 = 10 s = 74 + 1 * 10 = 84 I = 3 N = 1/10 = 0 (0>0) While condition fails S = 84

[@VikasMauryaAcademy]

bro jab option wrong hai or jab maine solve kiye toh option match nhi kar reh the toh maine ek new option likha hai by mistake option c mein 84 hona chaiye then 87 hogya tha

[@yadavalka6829]

Question 3 : 9 kaise aaya

[@VikasMauryaAcademy]

Explanation: As X is call by reference, last updated value of X will be assigned to the variable for all the recursive calls. When c value becomes 0, x value will be 9. So, f(5,5) ->f(6,4) * 9 -> f(7,3) * 9 -> f(8,2) * 9 -> f(9,1) * 9 -> c == 0 & returns 1 Ans => 1 * 9 * 9 * 9 * 9 = 6561

[@madhavaggarwal8328]

i have my capgemini exam tomorrow. is this the exact level of pseudocodes that we will get?

[@VikasMauryaAcademy]

Yes

[@VikasMauryaAcademy]

All solutions questions 1 Explanation: goto label is expected to give at any line except, conditional areas. In the given code, compiler fails to identify I1 as label & it will consider I1 as an undeclared identifier. Questions 2 Explanation: Here, main() is a recursive function which will keep on executing till if condition is false. Initial value of static variable i = 5. i--(Pre-decrement) will update the i value as 4. As long as value inside if brackets are Non-zero numbers, condition will be true. As I is a static variable, re-initialization in the next function call will not affect I value. i value will keep on decrement and print 4 3 2 1. After 1, i becomes 0 and the if condition fails followed by ending recursive calls. Questions 3 Explanation: As X is call by reference, last updated value of X will be assigned to the variable for all the recursive calls. When c value becomes 0, x value will be 9. So, f(5,5) ->f(6,4) * 9 -> f(7,3) * 9 -> f(8,2) * 9 -> f(9,1) * 9 -> c == 0 & returns 1 Ans => 1 * 9 * 9 * 9 * 9 = 6561 Questions 4 Explanation: Here i = 0, x = 3. x = x++( x = 3) Post increment will assign value before gets updated. I value will increment by 1 and loop will run 3 times. When I becomes 3, while loop fails(3 != 3 returns false) X value remains 3 because always value gets reassigned before the updation. Questions 5 Explanation: Here i=0,s=0,n=127 (127>0) while condition true r = 127 % 10 = 7 p = 8 ^ 0 = 8 s = 0 + 8 * 7 = 56 i = 1 n = 127/10 = 12 (12>0) while condition true r = 12 % 10 = 2 p = 8 ^ 1 = 9 s = 56 + 9 * 2 = 74 i = 2 n = 12/10 = 1 (1>0) while condition true r = 1 % 10 = 1 p = 8^2 = 10 s = 74 + 1 * 10 = 84 I = 3 N = 1/10 = 0 (0>0) While condition fails S = 84

[@coursetotal6570]

i didn't understand question 3, how is coming 9, in the entire question 9 is not present and no any operation will be performed so how will be coming 9

[@VikasMauryaAcademy]

Explanation: As X is call by reference, last updated value of X will be assigned to the variable for all the recursive calls. When c value becomes 0, x value will be 9. So, f(5,5) ->f(6,4) * 9 -> f(7,3) * 9 -> f(8,2) * 9 -> f(9,1) * 9 -> c == 0 & returns 1 Ans => 1 * 9 * 9 * 9 * 9 = 6561

[@sparshartist01]

bro x is call by refrence mean x value is same throughout the recursion calls, so first we reach base case than as we return ans we use last updated x value eg- dry run should be f(5,5) ->f(6,4) * x -> f(7,3) * x -> f(8,2) * x -> f(9,1) * x (see here x current value is 9 ) agar call by value normal hota tho tum jo soch rahe the whi hota simple 5*6*7*8*9, yaas rakhna bas ki call by refrence mai variable k saath recursion lagana -> c == 0 & returns 1

[@VikasMauryaAcademy]

Question 6 Explanation: The above pseudocode will print Fibonacci numbers from 1 to 13 without any space (1