Open In App

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.

Answer: b) 1

Solution: According to the question, a + b = 12k + 8 => a – b = 12l + 6 => Subtracting both the equations we get, ab = Now all the terms of ab is divisible by 6, except 7. So the remainder left is 1.



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.

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



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

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

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

Answer: b) 15 m

Solution: All the persons are to sit on the ground forming the base of the cone. Total base covered = pi * = 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 * * h) = 300 => h = 15 meters.

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.

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 question, 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

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

Article Tags :