TCS Digital Aptitude Questions with Answers For 2021 Pass-Outs PDF

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Lecture notes of TCS digital Questions with answers for 2021 pass outs from university of oxfrd...

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9/9/2020

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TCS Digital Aptitude Questions with Answers | For 2021 Pass-Outs Published on 02 Sep 2020

This article will deal with Aptitude Questions (with Answers) that have been asked in TCS Digital in the previous years. If you're a 2021 pass-out, it's time that you gear up for the TCS Digital Recruitment Process.

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(https://www.faceprep.in/tcs-digital-cracker/?aff_id=1002) TCS Digital Aptitude Questions and Answers Q1. If x = (163 +173 +183 +193 ), then x when divided by 70 leaves a remainder of ? A. 0 B. 1 C. 35 D. 69 Answer: A Explanation: x = (16³+17³+18³+19³) x = (16³+19³+17³+18³) x = (16+19)(16²+19²-16*19) + (17+18)(17²+18²-17*18) x = 35(even + odd - even) + (35)(odd + even - even) x = 35(odd) + 35(odd) x = 35 (odd + odd) x = 35 (even) x = 35 (2n) x = 70n

Remainder when x is divided by 70 will be ‘0’

Q2. Consider the triangle shown in the figure where BC = 12 cm, DB = 9 cm, CD = 6 cm and ∠BCD = ∠BAC, what is the ratio of the perimeter of ΔADC to that of ΔBDC?

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A. 7/9 B. 8/9 C. 6/9 D. 5/9

Answer: A Explanation: Consider △ABC and △BCD ∠BCD = ∠BAC (given) ∠B = ∠B (Common angle) By Angle-Angle property △ABC and △BCD are similar Hence,

Q3. Consider an obtuse-angled triangle with sides 8 cm, 15 cm and x cm. If x is an integer then how many such triangles exist? A. 5 B. 21 C. 10 D. 15 E. 14 Answer: C Explanation: As per side-inequality theorem, 15-8 < x < 15+8 or, 7 17 Possible values of x = 18, 19, 20, 21, 22 = 5 values Thus, there are 10 possible triangles.

Q4: If

where [x] denotes the greatest

integer less than or equal to x, then A. 96 ≤ n < 104 B. 104 ≤ n < 107 C. 107 ≤ n < 111 D. 111 ≤ n < 116 Answer: C Explanation: log₁₀1 = 0 log₁₀10 = 1 log₁₀100 = 2 Since [x] denotes the greatest integer less than or equal to x --> [log₁₀2] = [log₁₀3] = .......... = [log₁₀9] = 0 --> [log₁₀1] + [log₁₀2] + [log₁₀3] + .......... + [log₁₀9] = 0 Also, [log₁₀10] = [log₁₀11] = ................ = [log₁₀99] = 1 --> [log₁₀10] + [log₁₀11] + ................ + [log₁₀99] = 90 Similarly, [log₁₀100] = [log₁₀101] = [log₁₀102] = ..................= [log₁₀999] = 2 [log₁₀1] + [log₁₀2] .... +[log₁₀9] + [log₁₀10] + .... + [log₁₀99] + [log₁₀100] + ...... [log₁₀n] = n --> 0 + 90 + [log₁₀100] + ... [log₁₀n] = n --> 90 + 2(n - 99) = n (Since n cannot be greater than 116 from the options) --> 2n - n - 198 + 90 = 0 --> n = 108

Hence the correct answer is (c) 107 ≤ n < 11.

Q5. A man travels three-fifths of distance AB at a speed of 3a, and the remaining at a speed of 2b, if he goes from A to B and back at a speed of 5c in the same time, then A. 1/a + 1/b = 1/c B. a + b = c C. 1/a + 1/b = 2/c D. None of these Answer: C https://www.faceprep.in/tcs/tcs-digital-aptitude-questions/

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Time taken to cover the distance with speed 3a = ((3x/5))/3a Time taken to cover the distance with speed 2b = ((2x/5))/2b Time taken to cover the distance x from B to A then return = 2x/5c

(https://www.faceprep.in/tcs-digital-cracker/?aff_id=1002) Q6. Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight-member joint family is nearest to: A. 23 years B. 22 years C. 21 years D. 25 years E. 24 years

Answer: E Explanation: Ten years ago, the sum of the ages of 8 members in the family = 231 After 3 years, the age of every member increases by 3; hence the total age increases by 8 x 3 = 24. The total age is 231 + 24 = 255 A member of 60 years age died = 255 – 60 = 195 Similarly, again after 3 years, Sum of the ages of all members = 195 + 24 = 219 Again, a member of age 60 died = 219 – 60 = 159 Now they have asked the sum in the current years, so the remaining years are 4. Hence the sum increases by 8 * 4 = 32 Total sum of the ages of 8 members in the present year = 159 + 32 = 191 Average age = 191 / 8 = 23.8 = 24 (approx)

Q7. George can do some work in 8 hours. Paul can do the same work in 10 hours while Hari can do the same work in 12 hours. All the three of them start working at 9 a.m. while George stops work at 11:00 a.m., the remaining two complete the work, approximately when will the work be finished? A. 11:30 a.m. B. 12:00 p.m. C. 12:30 p.m. D. 1:00 p.m. https://www.faceprep.in/tcs/tcs-digital-aptitude-questions/

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Answer: D

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Explanation: George

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--> 8 hrs

Paul

--> 10 hrs

Hari

--> 12 hrs

Total work

= LCM (8,10,12) = 120 units

George

-->15 units/hr

Paul

--> 12 units/hr

Hari

--> 10 units/hr

Combined efficiency = 15 + 12 + 10 = 37 units/hr Work done by all of them in 2 hrs(by 11am) Remaining work

= 120 – 74

= 37 x 2 = 74 units

= 46 units

Combined efficiency of Paul & Hari Time taken to complete remaining work

= 22 units/hr = 46/22 = Approx 2 hours

Thus, the final time = 11 am + approx 2 hours = 1 pm.

Q8. The 260th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, f, f, f, f, f, f, ....is: A. U B. W C. T D. V Answer: B Explanation: Total number of terms in the series = 26(26+1)/2 = 351 351 – 26 = 325 --> (326 to 351 will be z) 325 – 25 = 300 --> (301 to 325 will be y) 300 – 24 = 276 --> (277 to 300 will be x) 276 – 23 = 253 --> (254 to 276 will be w) So, the 260th element in the series is w.

Q9. In the diagram below, the areas of the triangles are as follows: A1 = 1024, A2 =1016, A3=1057. What is the area of A4?

A. 1023 B. 1036 C. 1020 D. 1065

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Q10.

TCS Digital Cracker

Rs. 3000 is distributed among A, B and C such that A gets 2/3rd of what B and C together get and C gets 1/2 of what A and B together get. Find C’s share?

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A. Rs. 1500 B. Rs. 1000 C. Rs. 1200 D. Rs. 1800

Answer: B Explanation: C get ½ of what A and B together get. --> C = ½(A + B) --> A + B = 2C ---(1) Given, A + B + C = 3000

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From (1) (2C) + C = 3000

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C’s share = 1000

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