# Placement Questions & Answers :: TCS

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**TCS Practice Q&A**

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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**Tags:** TCS

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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**Tags:** TCS

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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**Tags:** TCS

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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**Tags:** TCS

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