March 28, 2010

TCS Recruitment Rounds | Placement Papers -II |TCS Technical Interview Questions


TCS Recruitment Rounds PART-2
1.Written Test
2.Technical Interview
3.MR ( Managerial)
4. HR Interview
WRITTEN TEST :( ONLINE TEST)

Contains 3 sections

1) Verbal (Synonyms - Antonyms - Comprehension Passages)
2) Quantitative Aptitude
3) Critical Reasoning

SECTION-1(Verbal- 30 questions - 20 min)
1.Synonyms (Refer In GRE BARRONS 12th Edition )
2.Antonyms (Refer In GRE BARRONS 12th Edition (page no -126))
3. Passage completion


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31) A =0 0 0 0 1 1 1 1
B =0 0 1 1 0 0 1 1
C =0 1 0 1 0 1 0 1
( A U B ) n C Find the fourth row, having the bit pattern as an integer in an 8-bit computer, and express the answer in its decimal value.
Ans: 29

32) A, B and C are 8 bit nos. They are as follows:
A 1 1 0 1 1 0 1 1
B 0 1 1 1 1 0 1 0
C 0 1 1 0 1 1 0 1
Find ( (A-B) u C )=?
Hint: 109 A-B is {A} - {A n B}
Ans: 0 1 1 1 1 1 1 1 (DB)

33) If A, B and C are the mechanisms used separately to reduce the wastage of fuel by 30%, 20% and 10%. What will be the fuel economy if they were used combined.
Ans: 20%

34) In the class of 40 students, 30 speak Hindi and 20 speak English. What is the lowest possible number of students who speak both the languages?
(a) 5 (b) 20 (c) 15 (d) 10 (e) 30

35) In a two-dimensional array, X (9, 7), with each element occupying 4 bytes of memory, with the address of the first element X (1, 1) is 3000, find the address of
X (8, 5).
Sol: [HINT~ Formula=Base Add + Byte reqd {N (i-1) + (j-1)}
Where,
Base Add=3000;
Byte reqd=4;
N=no of columns in array=7;
i=8; j=5;
IN ROW MAJOR ORDER]
Ans: 3212
36) If the vertex (5, 7) is placed in the memory. First vertex (1, 1),s address is 1245 and then address of (5, 7) is ----------
Ans: 1279

37) A 2D array is declared as A [9, 7] and each element requires 2 byte. If A [1, 1] is stored in 3000. Find the memory of A [8, 5]?
Ans: 3106

38) One circular array is given (means the memory allocation takes place like a circular fashion) dimension (9X7) .starting address is 3000.find the address of (2, 3)
Ans: 555

39) The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If the size of the program is increased by 1% then how much memory now occupied?
Sol: N is increased by 1%
Therefore new value of N=N + (N/100)
=101N/100
M=sqrt (100 * (101N/100))
Hence, we get
M=sqrt (101 * N)
Ans: 0. 5 %( =SQRT 101N)

40) A bus started from bus stand at 8.00a m and after 30 min staying at destination, it returned back to the bus stand. The destination is 27 miles from the bus stand. The speed of the bus 50 percent fast speed. At what time it retur4ns to the bus stand.
Sol: (this is the step by step solution :)
A bus cover 27 mile with 18 mph in =27/18= 1 hour 30 min.
And it wait at stand =30 min.
After this speed of return increase by 50% so 50%of 18 mph=9mph
Total speed of returning=18+9=27
Then in return it take 27/27=1 hour
Then total time in journey=1+1:30+00:30 =3 hour
So it will come at 8+3 hour=11 a.m.
So Ans==11 a.m
41) A Flight takes off at 2 A.M from northeast direction and travels for 11 hours to reach the destination which is in North West direction. Given the latitude and longitude of source and destination. Find the local time of destination when the flight reaches there?
Ans: 7 AM or 1.00 PM

42) My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed?
a) 6:00 am b) 6:40am c) 7:40 d) 7:00 e) 8:00
(Hint: Every 1 deg longitude is equal to 4 minutes. If west to east add time else subtract time)
Ans: 8:00

43) A moves 3 kms east from his starting point. He then travels 5 kms north. From that point he moves 8 kms to the east. How far is A from his starting point?
Ans: 13 kms

44) A plane moves from 9°N40°E to 9°N40°W. If the plane starts at 10 am and takes 8 hours to reach the destination, find the local arrival time.

45) In Madras, temperature at noon varies according to -t^2/2 + 8t + 3, where t is elapsed time. Find how much temperature more or less in 4pm to 9pm. (May be we can solve it by Definite Integration. Check any way}
Ans: at 9 pm 7.5 more or 385.8 (DB)

46) For Temperature a function is given according to time: ((t**2)/6) + 4t +12 what is the temperature rise or fall between 4.AM TO 9 AM
Sol: In equation first put t=9,
We will get 34.5........................... (1)
Now put t=4,
We will get 27.............................. (2)
So Ans=34.5-27
=7.5

47) For Temperature a function is given according to time: ((t**2)/6) + 4t +12 what is the temperature rise or fall between 5 PM to 8 PM

48) Low temperature at the night in a city is 1/3 more than 1/2 high as higher temperature in a day. Sum of the low tem. And highest temp is 100 degrees. Then what is the low temp?
Sol: Let highest temp be x
So low temp=1/3 of x of 1/2 of x plus x/2 i.e. x/6+x/2
Total temp=x+x/6+x/2=100
Therefore, x=60
Lowest temp is 40
Ans :( 40 deg.)

49) A person had to multiply two numbers. Instead of multiplying by 35, he multiplied by 53and the product went up by 540. What was the raised product?
a) 780
b) 1040
c) 1590
d) 1720
Sol: x*53-x*35=540=> x=30 therefore, 53*30=1590
Ans: 1590

50) How many positive integer solutions does the equation 2x+3y = 100 have?
a) 50
b) 33
c) 16
d) 35
Sol: Given 2x+3y=100, take l.c.m of ,x, coeff and ,y, coeff i.e. l.c.m of 2,3 ==6then divide 100 with 6 , which turns out 16 hence answer is 16short cut formula--- constant / (l.cm of x coeff and y coeff)

51) The total expense of a boarding house is partly fixed and partly variable with the number of boarders. The charge is Rs.70 per head when there are 25 boarders and Rs.60 when there are 50 boarders. Find the charge per head when there are 100 boarders.
a) 65
b) 55
c) 50
d) 45
Sol: let a = fixed cost and
k = variable cost and n = number of boarders
Total cost when 25 boarders c = 25*70 = 1750 i.e. 1750 = a + 25k
Total cost when 50 boarders c = 50*60 = 3000 i.e. 3000 = a + 50k
Solving above 2 eqns, 3000-1750 = 25k i.e. 1250 = 25k i.e. k = 50
Therefore, substituting this value of k in either of above 2 eqns we get
a = 500 (a = 3000-50*50 = 500 or a = 1750 - 25*50 = 500)
So total cost when 100 boarders = c = a + 100k = 500 + 100*50 = 5500
So cost per head = 5500/100 = 55

52) Amal bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and 14 pencils for an amount which was half more than what Amal had paid. What % of the total amount paid by Amal was paid for pens?
a) 37.5%
b) 62.5%
c) 50%
d) None of these
Sol: Let, 5 pens + 7 pencils + 4 erasers = x rupees
So 10 pens + 14 pencils + 8 erasers = 2*x rupees
Also mentioned, 6 pens + 14 pencils + 8 erasers = 1.5*x rupees
So (10-6) = 4 pens = (2-1.5) x rupees
So 4 pens = 0.5x rupees => 8 pens = x rupees
So 5 pens = 5x/8 rupees = 5/8 of total (note x rupees is total amt paid by
Amal) i.e. 5/8 = 500/8% = 62.5%
Ans: 62.5%

53) I lost Rs.68 in two races. My second race loss is Rs.6 more than the first race. My friend lost Rs.4 more than me in the second race. What is the amount lost by my friend in the second race?
Sol: x + x+6 = rs 68
2x + 6 = 68
2x = 68-6
2x = 62
x=31
x is the amt lost in I race
x+ 6 = 31+6=37 is lost in second race
Then my friend lost 37 + 4 = 41 Rs
Ans: 41 Rs

54) A face of the clock is divided into three parts. First part hours total is equal to the sum of the second and third part. What is the total of hours in the bigger part?
Sol: The clock normally has 12 hr
Three parts x, y, z
x+y+z=12
x=y+z
2x=12
x=6
So the largest part is 6 hrs
Ans: 6 hrs

55) (1- 1/6) (1-1/7).... (1- (1/ (n+4))) (1-(1/ (n+5))) = ?
Sol: Leaving the first numerator and last denominator, all the numerator and denominator will cancelled out one another.
Ans: 5/ (n+5)

56) Ten boxes are there. Each ball weighs 100 gms. One ball is weighing 90 gms.
i) If there are 3 balls (n=3) in each box, how many times will it take to find 90 gms ball? ii) Same question with n=10
iii) Same question with n=9
Sol: The chances are
When n=3
(i) nC1= 3C1 =3 for 10 boxes.. 10*3=30
(ii) nC1=10C1=10 for 10 boxes ....10*10=100
(iii) nC1=9C1=9 for 10 boxes.....10*9=90

57) With 4/5 full tank vehicle travels 12 miles, with 1/3 full tank how much distance travels?
Sol: 4/5 full tank= 12 mile
1 full tank= 12/ (4/5)
1/3 full tank= 12/ (4/5)*(1/3) = 5 miles
Ans: 5 miles

58) Wind flows 160 miles in 330min.for 80 miles how much time required
160 miles?
Sol: 1 mile = 330/160
80 miles= (330*80)/160=165 min.
Ans: 165 min.

59) A person was fined for exceeding the speed limit by 10mph.another person was also fined for exceeding the same speed limit by twice the same if the second person was traveling at a speed of 35 mph. find the speed limit
Sol :( x+10) =(x+35)/2
Solving the eqn we get x=15
Ans: 15

60) A sales person multiplied a number and get the answer is 3 instead of that number divided by 3.what is the answer he actually has to get.
Sol: Assume 1
1* 3 = 3
1*1/3=1/3
So he has to got 1/3
Ans: 1/3

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