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Indiawidejobs.com is publishing Inautix placement papers specially for the candidates those who are preparing for written test in Inautix company. Our website is strictly committed for providing all the related information which will be required for candidates those who are trying to attempt for jobs in several companies like Inautix and failing with a small reason like lack of awareness about Inautix placement papers. Our intention is to find out the way and pattern of questions which were asked in written tests conducted in those companies. These Inautix placement papers were vividly framed by keeping in mind the possibilities of job seekers and difficulties they are facing at written tests. Inautix placement papers with answers are prepared by us in a specific way so that job seekers those who are attempting to qualify in written test conducted at Inautix company will learn quickly with simple reference.
These Inautix placement papers will definitely filter candidates for better results in written tests. These written tests are intended to check out talent of jobseekers based on their skills which will definitely zoom out the capabilities of candidates those who are willing to work at Inautix. Complicacy in questions can be clearly confirmed by analyzing placement paper pattern which was followed in written test. Candidates those who refer Inautix placement papers with answers are having better chances of cracking written tests at this company. There will be a chance of repeating questions in Inautix placement papers year to year. So please concentrate more on referring placement papers with solutions by downloading Inautix placement papers pdf from our website. By preparing Inautix placement papers Indiawidejobs.com is paving an easy way to succeed at written tests conducted in Inautix. Indiawidejobs.com will always try to provide complete information regarding Inautix placement papers up to the date. So keep on visiting this website for Inautix placement papers with answers. In the below post you can refer sample questions that are commonly appeared in Inautix placement papers.
About the Company: Inautix Technologies India Private Limited is one of the major client based software business enterprise in India. Inautix Technologies based in Chennai and Pune and its parent organization BNY Mellon and other subsidiaries have track record of our more than 5000 consultants, analysts, and technologists. Inautix offers endless opportunities to grow in knowledge and skill while gaining invaluable insights into the processes of a global organization.
Sample Aptitude Questions:
1) Of all the numbers whose literal representations in capital letters consists only of straight line segments (for example, FIVE), only one number has a value equal to the number of segments used to write it. Which number has this property?
This is the only solution that satisfies the requirement that the capital letters shall consist only of straight line segments.
2) Greengrocer C. Carrot wants to expose his oranges neatly for sale. Doing this he discovers that one orange is left over when he places them in groups of three. The same happens if he tries to place them in groups of 5, 7, or 9 oranges. Only when he makes groups of 11 oranges, it fits exactly. How many oranges does the greengrocer have at least?
Assume the number of oranges is A. Then A-1 is divisible by 3, 5, 7 and 9. So, A-1 is a multiple of 5×7×9 = 315 (note: 9 is also a multiple of 3, so 3 must not be included!). We are looking for a value of N for which holds that 315×N + 1 is divisible by 11. After some trying it turns out that N = 3. This means that the greengrocer has 946 oranges.
3) A number is called a palindrome when it is equal to the number you get when all its digits are reversed. For example, 2772 is a palindrome. We discovered a curious thing. We took the number 461, reversed the digits, giving the number 164, and calculated the sum of these two numbers: 461 164 + ——- 625 We repeated the process of reversing the digits and calculating the sum two more times: 625 526 + ——- 1151 1511 + ——- 2662 To our surprise, the result 2662 was a palindrome. We decided to see if this was a pure coincidence or not. So we took another 3-digit number, reversed it, which gave a larger number, and added the two. The result was not a palindrome. We repeated the process, which resulted in another 3-digit number which was still not a palindrome. We had to repeat the process twice more to finally arrive at a 4-digit number which was a palindrome. What was the 3-digit number we started with the second time?
Because the reverse of the starting number is greater than the starting number itself, the first digit of the starting number must be less than the last digit. Therefore, the starting number must be at least 102. Secondly, we know that after two summations, the result has still only 3 digits.
We know that def is not a palindrome. Therefore, d differs from f. This is only possible if d=f+1 (d can only be one greater than f, because b is at most 9). Since abc is at least 102, def is at least 403, so d+f will be at least 7. Since ghi is still a 3-digit number but not a palindrome, i can be at most 8, so d+f can be at most 8. Since d=f+1, d+f can only be 7, from which we conclude that a=1 and c=2. Now we have:
To make the first digit of 4e3 a 4, b must be 5, 6, 7, 8, or 9. Now calculate the sum of 4e3 and 3e4:
Because the first digit of the sum must be 8, e must be at least 5. Therefore, the only remaining candidates for b are 8 (8+8=16) and 9 (9+9=18). Now it can easily be found that b must be 9 and the starting number we are looking for is 192:
291 + (291 is greater than 192)
867 (still a 3-digit number)
6996 (the 4-digit palindrome).
4) The legendary king Midas possessed a huge amount of gold. He hid this treasure carefully: in a building consisting of a number of rooms. In each room there were a number of boxes; this number was equal to the number of rooms in the building. Each box contained a number of golden coins that equaled the number of boxes per room. When the king died, one box was given to the royal barber. The remainder of the coins had to be divided fairly between his six sons. Is a fair division possible in all situations?
A fair division of Midas’ coins is indeed possible. Let the number of rooms be N. This means that per room there are N boxes with N coins each. In total there are N×N×N = N3 coins. One box with N coins goes to the barber. For the six brothers, N3 – N coins remain. We can write this as: N(N2 – l), or: N(N – 1)(N + l). This last expression is divisible by 6 in all cases, since a number is divisible by 6 when it is both divisible by 3 and even. This is indeed the case here: whatever N may be, the expression N(N – 1)(N + l) always contains three successive numbers. One of those is always divisible by 3, and at least one of the others is even. This even holds when N=1; in that case all the brothers get nothing, which is also a fair division!
5) On a sunny morning, a greengrocer places 200 kilograms of cucumbers in cases in front of his shop. At that moment, the cucumbers are 99% water. In the afternoon, it turns out that it is the hottest day of the year, and as a result, the cucumbers dry out a little bit. At the end of the day, the greengrocer has not sold a single cucumber, and the cucumbers are only 98% water. How many kilograms of cucumbers has the greengrocer left at the end of the day?
In the morning, the 200 kilograms of cucumbers are 99% water. So the non-water part of the cucumbers has a mass of 2 kilograms. At the end of the day, the cucumbers are 98% water. The remaining 2% is still the 2 kilograms of non-water material (which does not change when the water evaporates). If 2% equals 2 kilograms, then 100% equals 100 kilograms. So, the greengrocer has 100 kilograms of cucumbers left at the end of the day.
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