TCS Placement Paper | MCQ 2

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. There is a set of 30 numbers. The average of first 10 numbers is equal to the average of last 20 numbers. What is the sum of last 20 numbers?
    a) Twice the sum of the first ten numbers
    b) Sum of first 10 numbers.
    c) Twice the sum of the last ten numbers
    d) Cannot be determined.

    Answer: a) Twice the sum of the first ten numbers

    Solution:

    Let the sum of the first 10 numbers is equal to ‘x’
    Let the sum of the last 20 numbers is equal to ‘y’
    According to the question:
    x/10 = y/20
    Therefore, y = 2x

  2. There is a town called Metron, where wheels of the front and rear of vehicles are of different size. The measurement unit followed in the town is the metre. The circumference of the front wheel of the car is 133 metres and that of rear wheels is 190 metres. So what is the distance travelled by the cart in metres when the front wheel has done nine more revolutions than the rear wheel?
    a. 1330
    b. 572
    c. 399
    d. 3990



    Answer: d) 3990

    Solution:

    At first, we calculate the LCM of 133 and 190 which is 1330. So, the front wheels take 10 rounds to cover 1330 metres and the rear wheels take 7 rounds to cover the same. So to take 9 extra revolutions the vehicle would have travelled 1330 * 3 = 3990 metres.

  3. Let a number ‘x’ when divided by 406 leaves a remainder 115. What will be the number when the number is divided by 29?
    Answer: 28

    Solution:

    According to the question, the number is equal to 406x + 115.
    Since 406 is divisible completely by 29, therefore any multiple of 406 that is 406x when divided by 29 leaves remainder 0. Now 115, when divided by 29, leaves remainder 28.

  4. A sequence of an alpha-numeric is to be formed. The sequence consisting of two alphabets followed by two numbers is to be formed with no repetitions. In how many ways can it be formed?
    a. 65000
    b. 64320
    c. 58500
    d. 67600

    Answer: c) 58500

    Solution:

    The first can be filled in 26 ways.
    The second place can be filled in 25 ways.
    The third place can be filled in 10 ways.
    The last digit can be filled in 9 ways

    .

  5. According to a particular code language, A=0, B=1, C=2, …, Y=24, Z=25 then can will ONE+ONE (in the form of alphabets only) be coded?
    a) DABI
    b) CIDA
    c) BDAI
    d) ABDI

    Answer: c) BDAI

    Solution:

    This is a 26 base question. Just like there is the Decimal system consisting of 10 digits from 0 to 9, the Base 26 system consist of 26 alphabets where A = 0, B = 1, Z = 25 and so on.
    Let’s calculate, O N E + O N E
    For E(4),
    => E + E
    => 4 + 4
    => 8
    => I
    For N(13),
    => 13 + 13
    => 26
    On converting 26 to Base 26 we get 1 0. Keeping 0(A) and taking 1 as carry
    For O(14),
    => O + O + 1
    => 29
    Dividing 29 by 26 we get 1(B) 3(D)
    So answer is BDAI

  6. In the sequence of problemsolvingproblemsolvingproblemsolving… what is the 2015th alphabet?
    a) p
    b) g
    c) r
    d) n

    Answer: d) n

    Solution:

    ‘problemsolving’ consist of 14 letters. On dividing 2015 by 14 we get 13. So the 13th letter is n and hence the answer.

  7. What is the remainder when the number 101102103104105106107…148149150 is divided by 9?
    Answer: 2

    Solution:

    The divisibility rule for 9 is that the sum of all digits of a number should be divisible by 9. Let’s calculate the sum of the digits:
    There are 50 1’s (unit place) = 50
    There are 10 1’s (tens place) = 10
    There are 10 2’s (tens place) = 20
    There are 10 3’s (tens place) = 30
    There are 10 4’s (tens place) = 40
    There is one 5 (tens place) = 5
    For each number 1 to 9, there are 5 sets of sum 45(1+2+…+9) = 225
    => So sum of all digits = 380
    => 380 / 9 = 2 (Answer)

  8. 30 litres of 78% of a concentrated acid solution is to be prepared. How many litres of 90% concentrated acid needs to be mixed with 75% solution of concentrated acid to get the result?
    a) 10
    b) 6
    c) 3
    d) 4

    Answer: b) 6

    Solution:

    Let’s apply the weighted-average formula.
    Let there be n1 litre of 90% acid solution and n2 litre of 75% solution
    Therefore,
    => 78 = ((90 * n1) + (75 * n2))/(n1 + n2)
    We get,
    => n1/n2 = 1 / 4
    So 30 litres needed to be divided in the ratio of 1:4, which gives us 6 litre as the answer.

  9. Ram will be 32 years old in eight years from now. In 4 years, Ram’s fathers age will be twic as Ram’s age and two years ago, his mother’s age will be twice as his age. What will be the present age of Ram’s father and mother?
    Answer: Father's age = 52, Mother's age = 46

    Solution:

    Ram will be 32 years old in next 8 years. So his present age is 32 – 8 = 24 years old
    After 4 years Ram will be 28 years old. So his father will be 28 * 2 = 56 years old.
    Therefore, fathers present age is 56 – 4 = 52 years old
    Two years ago Ram was 22 years old. So his mothers age the was 22 * 2 = 44 years old
    Therefore mothers present age is 44 + 2 = 46 years old.

  10. In a class, the number of boys is equal to the number of girls. What was the total number of students if twice the number of boys as girls remain when 12 girls entered out?
    Answer: 48

    Solution:

    Let ‘b’ be the number of boys and ‘g’ be the number of girls. According to the question:
    => b / (g – 12) = 2 / 1
    Since b = g;
    we get g = 24.
    So the total number of students = 24 + 24 = 48



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