TCS Placement Paper Part - I ( 2015 - 2016 Batch) - PrepareInterview.com

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Thursday, December 17, 2015

TCS Placement Paper Part - I ( 2015 - 2016 Batch)


tcs
Company Name: TCS

*Written exam there were two sections one is verbal and other is Arithmetic portions

*In verbal sections there we had to write an email writing and in arithmetic part there were several problems like series problem, Probability, train, profit & loss, Allegation ,etc...and some tricky problems

1. Here is 15 dots. If you select 3 dots randomly, what is the probability that 3 dots make a triangle?

a. 440/455.
b. 434/455.
c. 449/455.
d. 438/455.

Answer: 
Total ways of selecting 3 dots out of 15 is 15C3 = 455 If 3 dots are collinear then triangle may not be formed. Now look at the above diagram. 

If we select any 3 dots from the red lines they may not form a triangle. They are 5 x 5C3 = 50.  If we select the three letters from blue lines, they may not form a triangle.

They are in total 5 ways. Also there are 6 others lines which don't form a triangle.  Total = 50 + 5 + 6 = 61.  So we can form a triangle in 455 - 61 = 394. So answer could be 394/455. 

2. In how many ways a team of 11 must be selected from 5  men and 11 women such that the team must comprise of not more than 3 men?

Answer: The team may consist of 0 men + 11 women, 1 men + 10 women, 2 men + 9 women, or 3 men + 8 women.

So Number of ways are = 11C11+5C1x11C10+5C2x11C9+5C3x11C8 = 2256.

3. Eesha bought 18 sharpeners for Rs.100.  She paid 1 rupee more for each white sharpener than for each brown sharpener.  What is the price of a white sharpener and how many white sharpener did she buy ?

a. Rs.5, 10
b. Rs.6, 10
c. Rs.5, 8
d. Rs.6, 8

Sol: B.

Explanation:
Just check the options. If she bought 10 white sharpeners at Rs.6 per piece, She has spent Rs.60 already.

And with the remaining Rs.40, she bought 8 brown sharpeners at 40/8 = Rs.5 which is Rs.1 less than White sharpener.

4. On a 26 question test, five points were deducted for each wrong answer and eight points were added for each correct answer. 

If all the questions were answered, how many were correct, if the score was zero ?
a. 10
b. 12
c. 11
d. 13

Ans: A

Explanation:
Take options and check. If 10 are correct, his score is 10 x 8 = 80.  But 16 are wrong. So total negative marking is 16 x 5 = 80. So final score is zero.

5. If f(f(n))+f(n)=2n+3 and f(0)=1, what is the value of f(2012)?
a) 2011
b) 2012
c) 2013
d) 4095

Ans: Option C.

6.1(1!)+2(2!)+3(3!)....2012(2012!) = ?
Ans: 2013!-1

     1(1!)=1  => 2!-1
     1(1!)+2(2!)=1+4=5 => 3!-1
     1(1!)+2(2!)+3(3!)=1+4+18=23 => 4!-1
      ........................
      .......................
     1(1!)+2(2!)+3(3!)+........+2012(2012!)=>2013!-1.

7.A cow and horse are bought for Rs.2,00,000. The cow is sold at a profit of 20% and the horse is sold at a loss of 10%.  The overall gain is Rs.4000, the Cost price of cow?

a) 130000.
b) 80000.
c) 70000.
d) 120000.

Ans: Overall profit = 4000/200000x100=2%.

By applying alligation rule, we get
So cost price of the cow = 2/5 x 200000 = 80,000.

8.Ahmed, Babu, Chitra, David  and Eesha each choose a large different number. Ahmed says, " My number is not the largest and not the smallest".

Babu says, "My number is not the largest and not the smallest". Chitra says, "My number is the largest".

David says, " My number is the smallest". Eesha says, " My number is not the smallest".  Exactly  one of the five children is lying. The others are telling the truth. Who has the largest number?

a) Eesha.
b) David.
c) Chitra.
d) Babu.

Ans: A.

9.A beaker contains 180 liters of alcohol. On 1st day, 60 l of alcohol is taken out and replaced by water.  2nd day, 60 l of mixture iss taken out and replaced by water and the process continues day after day. What will be the quantity of alcohol in beaker after 3 days.

Ans: 53.3.

These are the questions which I remembered some questions also coming from Geometry portion also (2 questions).

10. The value of diamond varies directly as the square of its weight. If a diamond falls and breaks into two pieces with weights in the ratio 2:3. what is the loss percentage in the value?

Sol: Let weight be “x”
the cost of diamond in the original state is proportional to x2

when it is fallen it breaks into two pieces 2y and the 3y
x = 5y
Original value of diamond = (5y)2 = 25y2
Value of diamond after breakage = (2y)2+(3y)2=13y2

so the percentage loss will be = 25y2−13y225y2×100=48%

11.Five college students met at a party and exchanged gossips. Uma said, “Only one of us is lying”. David said, “Exactly two of us are lying”. Thara said, “Exactly 3 of us are lying”. Querishi said, “Exactly 4 of us are lying”. Chitra said “All of us are lying”. Which one was telling the truth?

a)David
b)Querishi
c)Chitra
d)Thara

Sol:  As all are contradictory statements, it is clear that ONLY one of them is telling the truth. So remaining 4 of them are lying. Querishi mentioned that exactly 4 are lying. So, he is telling the truth.
Explanation: Let us 1st assume that Uma is telling the truth. Then according to her only one is lying. But if only one is lying then all the others’ statements are contradicting the possibility. In the same way all the other statements should be checked. If we assume the Querishi is telling the truth, according to him exactly 4 members are lying. So all the others are telling lies and he is the one who is telling the truth. This case fits perfectly.

12.Cara, a blue whale participated in a weight loss program at the biggest office. At the end of every month, the decrease in weight from original weight was measured and noted as 1, 2, 6, 21, 86, 445, 2676. While Cara made a steadfast effort, the weighing machine showed an erroneous weight once. What was that.

a) 2676
b) 2
c) 445
d) 86

Sol: This is a number series problem nothing to do with the data given.
1x 1+1=2
2 x 2+2=6
6 x 3+3=21
21 x 4+4=88 and not 86
88 x 5+5 = 445
445*6+6 = 2676

13.The letters in the word ADOPTS are permuted in all possible ways and arranged in alphabetical order then find the word at position 42 in the permuted alphabetical order?

a) AOTDSP
b) AOTPDS
c) AOTDPS
d) AOSTPD

Sol:
In alphabetical order : A D O P S T
A _ _ _ _ _ : the places filled in 5! ways = 120, But we need a rank less than 120.  So the word starts with A.
A D _ _ _ _ : empty places can be filled in 4!=24
A O _ _ _ _ : the places filled with 4! ways = 24.  If we add 24 + 24 this total crosses 42. So We should not consider all the words starting with AO.
A O D _ _ _ : 3!= 6
A O P _ _ _ : 3!=6
Till this 36 words are obtained, we need the 42nd word.
AOS _ _ _ : 3!= 6
 Exactly we are getting the sum 42. So last 3 letters in the descending order are TPD.
So given word is AOSTPD

14.A man who goes to work long before sunrise every morning gets dressed in the dark. In his sock drawer he has 6 black and 8 blue socks. What is the probability that his first pick was a black sock, but his second pick was a blue sock?

Sol:  This is a case of without replacement.  We have to multiply two probabilities.  1. Probability of picking up a black sock, and probability of picking a blue sock, given that first sock is black.
6C114C1×8C113C1=2491

15.There are 6 red balls,8 blue balls and 7 green balls in a bag. If 5 are drawn with replacement, what is the probability at least three are red?
Sol: At least 3 reds means we get either : 3 red  or  4 red or 5 red. And this is a case of replacement.
case 1 : 3 red balls : 6/21 x 6/21 x 6/21 x 15/21 x 15/21
case 2 : 4 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 15/21
case 3 : 5 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 6/21

Total probability =  = (6/21 x 6/21 x 6/21 x 15/21 x 15/21)+(6/21 x 6/21 x 6/21 x 6/21 x (15 )/21)+ (6/21 x 6/21 x 6/21 x 6/21 x 6/21)
 = 312/16807

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