TCS Placement Paper | MCQ 9

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. When a + b is divided by 12 the remainder is 8, and when a – b is divided by 12 the remainder is 6. If a > b, what is the remainder when ab divided by 6?
    a) 3
    b) 1
    c) 5
    d) 4

    Answer: b) 1

    Solution:
    According to the question,
    a + b = 12k + 8
    => $(a+b)^2 = 144k^2 + 64 + 192k$
    a – b = 12l + 6
    => $(a-b)^2 = 144l^2 + 36 - 144l$
    Subtracting both the equations we get,
    ab = $36(k^2 - l^2) + 48k - 36l + 7$
    Now all the terms of ab is divisible by 6, except 7. So the remainder left is 1.



  2. There is a set of 26 questions. For each wrong answer, five marks were deducted and eight points were added for each correct answer. Assuming that all of the questions were answered, and the score was 0, how many questions were answered correctly?
    a) 12
    b) 10
    c) 11
    d) 13

    Answer: b) 10

    Solution:
    This can be easily solved using hit and trial method.
    Let’s consider the first option. If 12 questions in all are answered correctly, then the total score = 12 * 8 = 96 marks.
    If 12 questions are answered correctly, then 14 questions were wrongly answered. So total deductions = 14 * 5 = 70 marks.
    So total score = 96 – 70 = 26 which is not correct.
    Let’s consider the second option. If 10 questions in all are answered correctly, then the total score = 10 * 8 = 80 marks.
    If 10 questions are answered correctly, then 16 questions were wrongly answered. So total deductions = 16 * 5 = 80 marks.
    So total score = 80 – 80 = 0
    Hence 10 is the correct option.

  3. One day, Ramesh started 30 minutes late from home and driving at 25% slower than the usual speed, reached the market 50 minutes late. How much time in minutes does Ramesh usually take to reach the market from home?
    a) 20
    b) 40
    c) 60
    d) 80

    Answer: a) 60

    Solution:
    Let the usual speed of Ramesh be ‘s’
    Let the distance between home and market be ‘d’
    So usual time took = d/s
    Time took on that particular day = d/(3s/4)
    So according to the question,
    d/s(4/3 – 1) = 20
    or, d/s = 60



  4. Three containers A, B and C are having mixtures of milk and water in the ratio of 1:5, 3:5, 5:7 respectively. If the capacities of the containers are in the ratio 5:4:5, find the ratio of milk to water, if all the three containers are mixed together.
    a) 54:115
    b) 53:113
    c) 53:115
    d) 54:113

    Answer: c) 53:115

    Solution:
    Using the weighted average formula we can calculate the weight of milk,
    => [5*(1/6) + 4*(3/8) + 5*(5/12)]/(5+4+5) = 53/168
    So weight of water = 168 – 53 = 115
    So the ratio of milk to water = 53:115

  5. Aman participates in an orange race. In the race, 20 oranges are placed in a line of intervals of 4 meters with the first orange 24 meters from the starting point. Aman is required to bring the oranges back to the starting place one at a time. How far would he run in bringing back all the oranges?
    a) 1440
    b) 2440
    c) 1240
    d) 2480

    Answer: d) 2480 

    Solution:
    Since every orange is placed at a difference of 4 meters and the first potato is placed at 24 meters from the starting position. Every orange is placed at 24m, 28m, 32m, 36m, ….20 terms.
    Now to bring ever orange one at a time, Aman needs to cover the double of the distance = 48, 56, 64, …20 terms.
    So putting the values in the sum of AP formula, a = 48, d= 8, n = 20.
    Total distance travelled = 20/2 [2 * 48 + (20-1)*8]
    = 2480 meters

  6. There are two decks of cards each deck containing 20 cards, with numbers from 1 to 20 written on them. A card is drawn at random from each deck, getting the numbers x and y What is the probability that log x + log y is a positive integer. (Logs are taken to the base 10.)
    a) 7/400
    b) 29/100
    c) 3/200
    d) 1/80

    Answer: a) 7/400

    Solution:
    We know that log x + log y = log xy
    for log xy to be positive, we have the following choices:
    (1, 10), (10, 1), (10, 10), (5, 20), (20, 5), (2, 5), (5, 2)
    So the probability = 7/400

  7. There is a conical tent which can accommodate 10 persons. Each person requires 6 sq.meter space to sit and 30 cubic meters of air to breathe. What will be the height of the cone?
    a) 72 m
    b) 15 m
    c) 37.5 m
    d) 155 m

    Answer: b) 15 m

    Solution:
    All the persons are to sit on the ground forming the base of the cone.
    Total base covered = pi * $r^2$ = 6*10 = 60 sq-meter.
    The total volume of the tent will be equal to the total air to breathe by the 10 people = 30*10 = 300 cubic meter
    So, 1/3(pi * $r^2$ * h) = 300
    => h = 15 meters.

  8. Find the greatest power of 143 which can divide 125! exactly.
    a) 11
    b) 8
    c) 9
    d) 7

    Answer: c) 9

    Solution:
    We can write 143 = 11 × 13.
    So the highest power of 13 should be considered in 125!, which is 9 (13 * 9 = 117)
    The highest power of 11 in 125! is 12 (11 * 11 = 121 and remaining 1).
    That means, 125! = 11^12×13^9×…
    So only nine 13’s are available. So we can form only nine 143’s in 125!. So maximum power of 143 is 9.

  9. A car starts at 6:00 pm. from the starting point at a speed of 18 m/s, reached its destination. There it waited for 40 minutes and returned back at the speed of 28 m/s. Find the time taken to reach the destination.
    a) 9:44 pm
    b) 8:32 pm
    c) 7:30 pm
    d) 9:30 pm

    Answer: a) 9:44 pm

    Solution:
    Let the distance covered be D m
    The time to cover the starting distance = D/18 secs.
    The time taken for the reverse journey = D/28 secs.
    According to the quesiton,
    D/18 – D/28 = (40 × 60)
    On solving this we get,
    D = 2400 × 252/5 = 120960 m
    No the total time taken = (D/18) + (D/28) + 2400 = 13440 seconds
    = 3 hours and 44 minutes
    Therefore, the bus reaches back at 9:44 PM

  10. The value of a house depreciates each year, by 3/4 of its initial value at the beginning of the year. If the initial value of the scooter is Rs. 40, 000. What will be the value at the end of 3 yrs?
    a) Rs. 19000
    b) Rs. 16875
    c) Rs. 17525
    d) Rs. 18000

    Answer:  b) 16875

    Solution:
    This is the question of succession depreciation.
    the starting amount = Rs. 40000
    This reduces by 3/4 th of its initial value every year = (40, 000) * (3/4)^3 = 16875



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