TCS Placement Paper | MCQ 3

This is a TCS model placement paper for aptitude preparation. This placement paper will cover aptitude questions that are asked in TCS recruitment drives and also strictly follows the pattern of questions asked in TCS interviews. It is recommended to solve each one of the following questions to increase your chances of clearing the TCS interview.

  1. If f(x) = ax^4 – bx^2 + x + 5 and given f(-3) = 2, then f(3) = ? (a^b = a raised to power b)
    a) 3
    b) 8
    c) 1
    d) -2

    Answer: b) 8

    Solution:



    We can directly solve:
    => f(-3) = a(-3)^4 – b(-3)^2 + (-3) + 5 = 2
    => 81a – 9b – 2 = 2
    => 81a – 9b = 0
    Now solving f(3),
    => f(3) = 81a – 9b + 8
    => f(3) = 0 + 8 = 8(Answer)

  2. There is a chocolate factory which distributes chocolates to a class. It supplies chocolates to a class of 50 students for 30 days, keeping in mind that all students get an equal number of chocolates. For the first 10 days, only 20 students were present. How many students be accommodated into the group so that all the chocolates get consumed?
    a) 70
    b) 55
    c) 60
    d) 45

    Answer: d) 45

    Solution:

    Let each student get 1 chocolate each, so the total number of chocolates = 50 * 30 = 1500 chocolates.
    For first 10 days 20 students were present, so total chocolates consumed = 20 * 10 = 200 chocolates.
    Chocolates left = 1300. These are to be distributed for the next 20 days. Therefore in each day 1300 / 20 chocolates were to be consumed which = 65 chocolates per day.
    So the required answer = 65 – 20 = 45 chocolates.



  3. Given, log(0.318) = 0.3364 and log(0.317) = 0.3332, find log(0.319)?
    a) 0.3396
    b) 0.3394
    c) 0.3393
    d) 0..390

    Answer: a) 0.3396

    Solution:

    => log(0.319) = log(0.318) + (log(0.318) – log(0.317))
    = 0.3364 + (0.3364 – 0.3332)
    = 0.3364 + 0.3332
    = 0.3396 (Answer)

  4. There are a set of 20 students out of which 18 are boys and 2 are girls. They are to be seated in a circular manner so that the two girls are always separated by a boy. In how many ways can the students be arranged?
    a) 12
    b) 18!x2
    c) 17×2!
    d) 17!

    Answer: b) 18!x2

    Solution:

    There are in all 20 places out of which if one girl sits in one position then the other girl may sit either to her left or right skipping one place, which is to be filled by a boy. So total number of ways the boys can sit = 18! ways and girls may alternate there sits so the total answer would be = 18! * 2 ways.

  5. Ram appears for an exam. In paper A he scores 18 out of 70. In paper B he scores 14 out of 30. So in which paper did he perform better?
    a) Paper A
    b) Paper B

    Answer: b) Paper B

    Solution:



    We just need to calculate the percentage he scored in each paper.
    In paper A: (18/70) * 100 = 25.7%
    In paper B: (14/30) * 100 = 46.6% (Answer)

  6. A flight takes off at 2 a.m. from a place at 18N 10E and landed at 36N 70W, 10 hours later. What is the local time of the destination?
    a) 6:00 a.m.
    b) 6:40 a.m.
    c) 7:40 a.m.
    d) 7:00 a.m.
    e) 8:00 a.m.

    Answer: b) 6:40 a.m

    Solution:

    Let’ calculate the difference in the number of latitudes = 70 + 10 = 80 degrees towards east.
    We know 1 degree = 4 min, so 80 degrees = 80 * 4 = 320 mins
    320 mins = 5 hr 20 minutes
    Now the plane landed 10 hours later so time of landing = 12 hrs accoding to the starting place
    So time at destination = 12 hrs – 5 hrs 20 min = 6 hr 40 mins(Answer)

  7. An athletic run at 9 km/hr along a railway track. The track is 240m long and ahead of a train 120m long running at 45km/hr, in the same direction. how much time will the train take to completely cross the athlete?
    a) 3.6 sec
    b) 18 sec
    c) 72 sec
    d) 36 sec

    Answer: d) 36 sec 

    Solution:

    Let’s try to find the relative speed = 45 – 9 = 36km/hr
    = 36 * 5/18 = 10m/s
    Now the total distance needed to be covered by the train to completely cross the athlete = 240 + 120 = 360m
    So time = dist/speed = 360/10 = 36 seconds

  8. A takes 3 days to complete a work while B takes 2 days. Both of them finish a work and earn Rs. 150. What is A’s share of money?
    a) Rs. 70
    b) Rs. 30
    c) Rs. 60
    d) Rs. 75



    Answer: c) 60

    Solution:

    A completes 1/3rd of work in one-day and B completes 1/2 of work in one day. So the ratio of there work is:
    A:B = 2:3
    So A’s share = (2/5)*150 = 60 rupees(Answer)

  9. Ram and Shyam salaries are in the ratio of 2:3. If both of there salaries are increased by Rs 4000 each, the new ratio becomes 40:57. What is Shyam’s present salary?
    a) Rs. 17, 000
    b) Rs. 20, 000
    c) Rs. 25, 500
    d) None of these

    Answer: d) None of these

    Solution:

    Let Rams and Shyams salary be ‘x’. The ratio of there salary according to the question is 2x:3y
    According to the question,
    (2x+4000):(3x+4000) = 40:57
    On solving we get 3x = 34000
    Therefore, the present salary of Shyam is Rs. 38000

  10. At what rate per cent per annum will the SI on a sum of money be 2/5 of the amount in 10 years?
    a) 6%
    b) 5 2/3 %
    c) 4%
    d) 6 2/3 %

    Answer: c) 4%

    Solution:

    Let the sum of money be Rs ‘x’. So SI = 2x/5
    So, rate = (SI*100)/(P*Time)
    => (2x*100)/(5*x*10)
    => 4 % (Answer)



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