Hi friends welcome to India wide jobs here we are providing the information of TCS placement papers with solutions. Actually Lot of candidates are suffering in the written exam failure in Software companies like TCS, Wipro etc., For the students sake we are providing some TCS Question Papers based on TCS latest placement papers. By using these candidates can get the good knowledge on the written test. By Practicing these placement papers of TCS students can get the clear information related to entrance test format and it is useful for the practice. In these Paper students will get specific category of TCS aptitude test papers with answers in pdf format.
By using these TCS previous papers students can assume what types of question we can get in written test. The TCS model papers are given below if any student want to download the TCS sample paper please click on below link of the TCS previous year question papers and they are downloaded in to your devices. The TCS previous year question paper are given in PDF Format candidates can take the print out of TCS sample papers and practice well and get good knowledge on TCS question papers with their solutions. The TCS placement papers download process is by clicking each links of TCS test papers provided below. If want to take print out of TCS previous papers with solutions pdf click on the print button in below showing images.
Actually in TCS they main concentrate on written test because based on written test marks they can easily filter the students and forward them interview, So these TCS placement papers are very useful to all the students please do the practice these questions. After completion of written test they will conduct the Group discussion the selected candidates in the GD They call for the Technical Round and then HR Round.! Keep Practice of the TCS placement papers and get job Easily.
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Tata Consultancy Services (TCS) is a global leader in IT, digital and business solutions that partners with its clients to simplify, strengthen and transform their businesses. Tata Consultancy Services has been recognized by Brand Finance as one of the Big 4 Global IT Services Brands. Our continued industry-leading growth is a testament to the certainty our clients experience every day.
Some Sample TCS aptitude test papers with answers
- An old man and a young man are working together in an office and staying together in a nearby apartment. The old man takes 30 minutes and the young 20 minutes to walk from apartment to office. If one day the old man started at 10.00 AM and the young man at 10:05AM from the apartment to office, when will they meet?
Let the distance be 12 km. So the old man speed = 12 km12hr = 24 kmph
The young man speed = 12 km13hr = 36 kmph
As the old man started 5 minutes earlier, he covers 24×560 = 2 km in 5 minutes.
Now the time taken to the young man to meets him = 236−24×60 = 10 min.
So the time at which young man meet the old man = 10.05 + 10 = 10. 15 min.
2.There are 16 teams divided in 4 groups. Every team from each group will play with each other once. The top 2 teams will go to the next round and so on the top two teams will play the final match. Minimum how many matches will be played in that tournament?
In each group, total matches played = 4C2 = 6.
So total matches played in the first round = 6 × 4 = 24
Now top two teams from each group progress to the next round. Now these 8 teams are pooled into 2 groups. Total matches played in the second round = 6 × 2 = 12
Now 4 teams progress to the next round. Total matches played in the third round = 6
From this round, 2 teams progress to the next round. And final will be played between them.
Total matches = 24 + 12 + 6 + 1 = 43
- A sealed envelope contains a card with a single digit written on it. Three of the following statements are true and one is false.
- The digit is 1.
- The digit is not 2.
III. The digit is not 9.
- The digit is 8.
Which one of the following must necessarily be correct?
- II is false
- III is true*
- IV is false
- The digit is even.
- I is true
Three of the given statements are true. So both II and III are correct, and the given number is one of 1 or 8. So option b is correct.
- How many 2’s are there between the terms 112 to 375?
Let us calculate total 2’s in the units place. (122, 132, 142 … 192), (201, 212, 222, … 292), (302, 312, … 372) = 8 + 10 + 8 = 26
Total 2’s in tenth’s place, (120, 121, 122, …, 129) + (220, 221, …, 229) + (320, 321, …, 329) = 30
Total 2’s in hundred’s place = (200, 201, … 299) = 100.
Total 2’s between 112 and 375 = 26 + 30 + 100 = 156
- Ram and Shakil run a race of 2000 meters. First, Ram gives Shakil a start of 200 meters and beats him by one minute. If, Ram gives Shakil a start of 6 minutes Ram is beaten by 1000 meters. Find the time in minutes in which Ram and Shakil can run the races separately.
- 12, 18
- 10, 12
- 11, 18
- 8, 10
Let the speeds of Ram and Shakil = r and s respectively. Always remember, to solve problems involving races, try to equate the ratio of the speeds because how ever the two contestents run, the ratio of their speed won’t change.
In the first race, Ram gives shakil a start of 200 meters. So Ram runs the entire 2000 mts, but shakil runs (1800−60s) as Ram has beaten him by 1 min or 60 seconds.
Therefore, 20001800−60s=rs – – – (1)
In the second race, Ram has given shakil a start of 6 min. So shakil start his race 6 min before Ram. In 6 minutes, shakil runs 360×s = 360s mts. As Ram was beaten by 1000 mts, by that time shakil completes his race, Ram has covered 1000 mts only.
the ratio of the speeds = 2000−360s1000=sr – – – (2)
Equating s/r in both equations, 2000−360s1000=1800−60s2000
Solving we get s = 10/3. So time taken by shakil to cover the distance = 2000/(10/3) = 10 minutes.
- Three generous friends, each with some money, redistribute the money as follows: Sandra gives enough money to David and Mary to double the amount of money each has. David then gives enough to Sandra and Mary to double their amounts. Finally, Mary gives enough to Sandra and David to double their amounts. If Mary had 11 rupees at the beginning and 17 rupees at the end, what is the total amount that all three friends have?
Answer: 10 (71)
Let Sandra, David and Mary each has s, d and 11 respectively.
After the first distribution,
David has d + d = 2d, Mary has 11 + 11 = 22 and Sandra has s – d – 11.
After the second distribution,
Sandra has 2×(s – d – 11) , mary has 2×22 = 44 and david has 2d – (s – d – 11) – 22=3d – s –11.
After the third distribution,
Sandra has 2×2(s – d – 11), david has 2×(3d – s – 11) and mary has 44 – 2(s – d – 11) – (3d – s – 11) = 77 – s – d
It is given that finally Mary has Rs.17. So, 77 – s – d=17
⇒ s + d = 60
⇒ s + d + 11 = 60 + 11 = 71.
- George walks 36 kms partly at a speed of 4 kms per hour and partly at 3 km per hour If he had walked at a speed of 3km per hour when he had walked at 4 and 4 km per when he had walked at 3 he would have walked only 34 kms. The time (in hours) spent by George in walking was
Let George walked “a” hours at 4 kmph, and “b” hours at 3 kmph.
Given, 4a + 3b = 36 – – – (1)
3a + 4b = 34 – – – (2)
Adding the above two equations and simplifying them, a + b = 10.
- The sum of the four consecutive two digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can be one of these four numbers?
Let the numbers are 2a +1, 2a + 3, 2a + 5, 2a + 7 and their sum = 8a + 16
Given that if this sum is divided by 10, results in a perfect square.
As k2 is a perfect square and has to be divided by 4, only even numbers should be considered for k.
For k = 2, we get, a = 3, but 2a + 1 is not a two digit number.
For k = 4, a = 18 for which the given condition is satisfying.
So the numbers are 37, 39, 41, 43.
For k = 6, a = 43. The number are 87, 89, 91, 93. But none of the option contains these numbers.
- Consider the sequence of numbers 0, 2, 2, 4,… Where for n > 2 the nth term of the sequence is the unit digit of the sum of the previous two terms.
Let sn denote the sum of the first n terms of this sequence. What is the smallest value of n for which sn >2771?
[0, 2, 2, 4, 6, 0, 6, 6, 2, 8, 0, 8, 8, 6, 4, 0, 4, 4, 8, 2], 0, 2, 2…this series repeats after every 20 terms.
Sum of these 20 terms = 80
So 2771 =34×80 + 51
Sum of 13 terms = 52
So we have to use 34 times 20 terms = 34×20 = 680
680+13 = 693
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