Cognizant Quantitative Aptitude Questions And Answers
Cognizant Quantitative Aptitude MCQs : This section focuses on "Quantitative Aptitude" for Cognizant Exam. These Quantitative Aptitude MCQs are asked in previous Cognizant placements/recruitment exams and will help you to prepare for upcoming Cognizant drives.
1. The LCM of three different numbers is 1024. Which one of the following can never be there HCF?
A. 8
B. 32
C. 124
D. 256
View Answer
Ans : C
Explanation: All numbers except 124 is factors of 1024 , so answer is 124
2. The sum of three consecutive natural numbers each divisible by 3 is 72. What is the largest among them?
A. 28
B. 27
C. 24
D. 23
View Answer
Ans : B
Explanation: Let x , x+3 , x+6 are three consecutive numbers divisible by 3
According to question
Substitute the value of x in consecutive numbers
x=21
x+3=21+3=24
x+6=21+6=27
Therefore, The largest number among them is 27.
3. Convert 0.0155 into fractions?
A. 31/2000
B. 35/3000
C. 20/4000
D. 30/2500
View Answer
Ans : A
Explanation: Multiply 10000 to both numerator & denominator
(0.0155 x 10000)/(1 x 10000) = 155/10000
It can be written as 1.55% = 1.55/100 or 155/10000
find LCM (Least Common Multiple) for 155 & 10000.
5 is the LCM for 155 & 10000
divide numerator & denominator by 5
155/10000 = (155 / 5) / (10000 / 5)
= 31/2000
4. In an election between two candidates, one got 70% of the total valid votes and got 10% invalid votes. At the end of the day when the total number of votes were counted, the total number was found to be 10000. So what was the total number of valid votes that the winning candidate got, was:
A. 1500
B. 2000
C. 2500
D. 3000
View Answer
Ans : D
Explanation: Since 10% of the votes were invalid, 90% of the votes were valid = 90% of 10000 = 9000 votes were valid.
One candidate got 70% of the total valid votes, then the second candidate must have 30% of the votes
=> 0.30 * 9000 = 3000 votes.
5. 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. 2
B. 3
C. 4
D. 5
View Answer
Ans : A
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.
6. Three cubes of edges 6 cms, 8 cms and 10 cms are meted without loss of metal into a single cube. The edge of the new cube will be:
A. 8 cms
B. 12 cms
C. 16 cms
D. 20 cms
View Answer
Ans : B
Explanation: Volume of new cube = Volume of cube 1 + cube 2 + cube 3
=> 6^3 + 8^3 + 10^3
=> 216 + 512 + 1000
a^3 = 1728
a = (1728)^(1/3) = 12
7. If 378 coins consist of rupee, 50 paise and 25 paise coins, whose values are proportional to 13 :11 : 7, the number of 50 paise coins will be :
A. 132
B. 136
C. 140
D. 150
View Answer
Ans : A
Explanation: If values are proportional to 13 : 11 : 7, then the number of coins will be proportional to 13/1 : 11/0.50 : 7/0.25
=> 13 : 22 : 28.
Now from this the number of coins of 50 paise will be 378 × 22/63 = 132.
8. What is the greatest number that will divide 964, 1238 and 1400 and leave a remainder of 41, 31 and 51 respectively?
A. 64
B. 71
C. 58
D. 79
View Answer
Ans : B
Explanation: To reach to the solution we just need to find the HCF of (964 – 41), (1238 – 31), (1400 – 51) = 923, 1207, 1349
The HCF of 923, 1207 and 1349 = 71
9. There are 6 cities, and every city is connected to each other. How many different routes can one trace from A to B, such that no city is touched more than once in any one route?
A. 40
B. 60
C. 65
D. 78
View Answer
Ans : C
Explanation: There must be 1 direct route.
There are 4 ways to cover 1 city
There are 4 * 3 = 12 ways to cover 2 cities
There are 4 * 3 * 2 ways to cover 3 cities
There are 4 * 3 * 2 * 1 ways to cover 4 cities
Total ways = 65 ways
10. If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is:
A. 80 Km
B. 45 Km
C. 70 Km
D. 50 Km
View Answer
Ans : D
Explanation: Let us assume, the actual distance traveled be x km.
Then, x/10 = (x + 20)/14
14x = 10x + 200
4x = 200
x = 50 km.
Discussion