Cognizant Placement papers with solutions

Cognizant Placement papers with solutions for on campus and off campus drives are discussed here. Even before you start practicing the CTS placement papers, you need to know Cognizant test pattern for the role you have applied. Check here to know – Cognizant Test Pattern

cognizant placement papers with solutions

How to Prepare for the Cognizant drive

The best way to easily prepare for the Cognizant drive is through our CTS online course. This course covers all the previous year placement papers. Our CTS online course will also include 10 hours of Video Classes and 9 full length Cognizant mock tests. Click here to check our Cognizant Online Course

Cognizant Placement Papers – Quantitative Aptitude

Cognizant Aptitude section is a part of Cognizant online Test for Programmer Analyst role. Cognizant hiring for the Associate role does not have this section in the online test.

1) In an election between two candidates, one got 55% of the total valid votes and got 20% invalid votes. At the end of the day when the total number of votes were counted, the total number was found to be 7500. So what was the total number of valid votes that the winning candidate got, was:
a) 2400
b) 3100
c) 3400
d) 2700

Answer:  d
Explanation:
Since 20% of the votes were invalid, 80% of the votes were valid = 80% of 7500 = 6000 votes were valid. One candidate got 55% of the total valid votes, then the second candidate must have 45% of the votes = 0.45 * 6000 = 2700 votes.

2) A whole number n which when divided by 4 gives 3 as remainder. What will be the remainder when 2n is divided by 4?
a) 0
b) 1
c) 4
d) 2

Answer:  d
Explanation:
According to the question, n = 4q + 3. Therefore, 2n = 8q + 6 or 2n = 4(2q + 1) + 2. Thus, we get when 2n is divided by 4, the remainder is 2.

Cognizant Placement Papers – Logical Reasoning

Cognizant Logical reasoning questions are easy when compared to the remaining sections. Cognizant Logical reasoning placement papers with solutions are here. If you practice these carefully, then you can easily score well.

1) Statements: Some envelops are gums. Some gums are seals. Some seals are adhesives.
Conclusions:
1) Some envelopes are seals.
2) Some gums are adhesives.
3) Some adhesives are seals.
4) Some adhesives are gums.
a) Only (3)
b) Only (1)
c) Only (2)
d) Only (4)

Answer: a

2) Ankit is the son of Zubin. Manju is the daughter of Anil. Sheela is the mother of Manju. Mohan is the brother of Manju. How is Mohan related to Sheela?
a) Brother
b) Father
c) Son
d) Cannot be determined

Answer: c
Explanation:
Manju is the daughter of Anil. Sheela is the mother of Manju. Therefore, Mohan is the son of Sheela.

Cognizant Placement Papers – Verbal Ability

Cognizant placement papers or previous year papers analysis shows that the verbal ability section is the toughest of all. You need to prepare well for this section to crack the exam.

1) Choose the best antonym for ‘Exodus’ 
a) Influx
b) Homecoming
c) Return
d) Restoration
Answer: a

2) Choose the best antonym for ‘Amused’
a) Jolted
b) Frightened
c) Saddened
d) Astonished
Answer:  c

Cognizant Placement Papers – Automata Fix

This a new section introduced in Cognizant on-campus recruitment Process. For Automata Fix syllabus, sample questions  – Please check here.

Cognizant Placement Papers – Coding or Programming

This round is only for Cognizant off campus drive via Amcat. It is not applicable for Cognizant on campus hirings. This is a very crucial section. You need to answer at least 1 out of the 2 given coding questions in the Cognizant online test. Else your chances of getting selected to the next round are very less.

Find the occurrence of a substring in a parent string

Input:

aAbcDefabcAdf
Substring : abc

In C

#include
#include
char str[100], sub[100];
int count = 0, count1 = 0;
int main()
{
int i, j, l, len1, len2;
printf(“\nEnter a string : “);
scanf(“%[^\n]s”, str);
len1 = strlen(str);
printf(“\nEnter a substring : “);
scanf(” %[^\n]s”, sub);
len2 = strlen(sub);
for (i = 0; i < len1;)
{
j = 0,  count = 0;
while ((str[i] == sub[j]))
{
count++;   i++;  j++;
}
if (count == len2)
{            count1++;
count = 0;
}
}
printf(“%d”, count1);
return 0; }

Output: 1

 

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