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71. Sum of 1(2)+2(2)^2+3(2)^3+.......+100(2)^100
Answer: Option E
Explanation:Here is no explanation for this answer
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72. Find the number that can be put in place of the question mark.
3 7 14 23 36 49 ? 83 104
Answer: Option B
Explanation:7-3=4 4*4=16 16-2=14
36-23=13 13*4=52 52-3=49
83-56=27 27*4=108 108-4=104
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73. There are 20 balls which are red,blue,or green.If 7 balls are green and the sum of red balls and green balls is less than 13, at most how may red balls are ?
Answer: Option A
Explanation:Green ball =7
G + R < 13
So, red can be 1 or 2 or 3 or 4 or 5
For atmost red ball take red = 5
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74. If a publication occurs every seven years and the sum of the years is 13524. Then find the first year if the no. of publication is 7.
Answer: Option D
Explanation:Here a= ? , d=7 , s=13524 , n=7
sum of A.P. = n/2(2a + (n-1)d)
so we have series a+(a+7)+(a+14).......=13524;
=> 13524=7/2(2*a+(7-1)7)
=> a=1911;
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75. 3 mangoes and 4 apples costs Rs 85. 5 apples and 6 peaches costs Rs. 122. 6 mangoes and 2 peaches cost Rs.114. what is the combined price of 1 apple, 1 peach and 1 mango?
Answer: Option A
Explanation:Form the below eq. from the given info. in question.
3m+4a=85 -------- (i)
5a+6p=122 -------- (ii)
6m+2p=114 ----- (iii)
By solving eq. (i), (ii) and (iii)
m=15,a=10,p=12
so 1 apple, 1 peach and 1 mango price = 15+10+12 = Rs. 37
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76. What is the remainder of (16937^30)/31 ?
Answer: Option C
Explanation:Here is no explanation for this answer
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77. In a cricket match,two batsman scores are 96,96 respectively.They require only 5 runs in 3 balls,can both the batsman complete their centuries?
Answer: Option B
Explanation:Here is no explanation for this answer
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78. There are several bags of same weights. A bag is 6 kg plus three fourth of weight of an another bag . What is the weight of the bag?
Answer: Option B
Explanation:Lets assume bags as a,b,c,d....
Given condition 1: A bag is 6 kg plus three fourth of weight of an another bag
so, a=6+(3/4)b ------ (1)
condition 2: weights are same
so, a=b=c=d
=> a=6+(3/4)a
=> a=24
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79. There is a square field of side 10 m. A man runs with different speed 10 kmph, 15 kmph, 20 kmph, 25 kmph on the four sides of the field. What is the average speed of man ?
Answer: Option D
Explanation:Avg. Speed = (Total Distance /Total Time)
Total Distance = 10*4 = 40m 0r 0.04 km.
So, avg. Speed = 0.04/[(0.01/10)+(0.01/15) + (0.01/20) + (0.01/25)]
=15.58kmph
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80. LEADING" arrange it in such a way that atleast two vowels always together ?
Answer: Option A
Explanation:Total possible arrangement = 7! = 5040
Given condition 2 vowel should comes together : {_L_D_N_G_)
So, 3 vowels can be placed at 5 positions = 5p3 * 4! = 60*24 = 1440.
So required arrangements : (5040 - 1440) = 3600
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Here is the list of questions asked in Solved Placement Papers of TCS TCS aptitude questions. Practice TCS Written Test Papers with Solutions and take Q4Interview TCS Online Test Questions to crack TCS written round test. Overall the level of the TCS Online Assessment Test is moderate. Only those candidates who clear the written exam will qualify for the next round, so practic all the questions here and take all the free tests before going for final selection process of TCS