Question Paper for TCS Placement Exam

Will you please provide here the Tata Consultancy Services placement aptitude question paper?

You need the Tata Consultancy Services placement aptitude question paper, i am giving here:

1) If log 0.317=0.3332 and log 0.318=0.3364 then find log 0.319 ?
Sol) log 0.317=0.3332 and log 0.318=0.3364, then
log 0.319=log0.318+(log0.318-log0.317) = 0.3396

2) A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg packets are there ?
Sol) x= 2 kg Packs
y= 1 kg packs
x + y = 150 .......... Eqn 1
2x + y = 264 .......... Eqn 2
Solve the Simultaneous equation; x = 114
so, y = 36
ANS : Number of 2 kg Packs = 114.

3) 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?
6:00 am b) 6:40am c) 7:40 d) 7:00 e) 8:00
Sol) The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360).
When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours).
Hence, the time at the destination place is 12 noon - 5:20 hours = 6: 40 AM

4) 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 ?
Sol) Since it is moving from east to west longitide we need to add both
ie,40+40=80
multiply the ans by 4
=>80*4=320min
convert this min to hours ie, 5hrs 33min
It takes 8hrs totally . So 8-5hr 30 min=2hr 30min
So the ans is 10am+2hr 30 min
=>ans is 12:30 it will reach

5) The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent? Please tell me the answer with explanation. Very urgent.
Sol) see it doesn't matter that wat the time is being taken to fill the bucket.after reaching program it waits there for 10ms and back to the programmer in 20 ms.then total time to get the response is 20ms +10 ms=30ms...it's so simple....

6) A file is transferred from one location to another in 'buckets'. The size of the bucket is 10 kilobytes. Each bucket gets filled at the rate of 0.0001 kilobytes per millisecond. The transmission time from sender to receiver is 10 milliseconds per bucket. After the receipt of the bucket the receiver sends an acknowledgement that reaches sender in 100 milliseconds. Assuming no error during transmission, write a formula to calculate the time taken in seconds to successfully complete the transfer of a file of size N kilobytes.
(n/1000)*(n/10)*10+(n/100)....as i hv calculated...~~!not 100% sure

8) Y catches 5 times more fishes than X. If total number of fishes caught by X and Y is 42, then number of fishes caught by X?
Sol) Let no. of fish x catches=p
no. caught by y =r
r=5p.
r+p=42
then p=7,r=35

9) Three companies are working independently and receiving the savings 20%, 30%, 40%. If the companies work combinely, what will be their net savings?
suppose total income is 100
so amount x is getting is 80
y is 70
z =60
total=210

10) The ratio of incomes of C and D is 3:4.the ratio of their expenditures is 4:5. Find the ratio of their savings if the savings of C is one fourths of his income?
Sol) incomes:3:4
expenditures:4:5
3x-4y=1/4(3x)
12x-16y=3x
9x=16y
y=9x/16
(3x-4(9x/16))/((4x-5(9x/16)))
ans:12/19

11) If G(0) = -1 G(1)= 1 and G(N)=G(N-1) - G(N-2) then what is the value of G(6)?
ans: -1
bcoz g(2)=g(1)-g(0)=1+1=2
g(3)=1
g(4)=-1
g(5)=-2
g(6)=-1

12) If A can copy 50 pages in 10 hours and A and B together can copy 70 pages in 10 hours, how much time does B takes to copy 26 pages?
Sol) A can copy 50 pages in 10 hrs.
A can copy 5 pages in 1hr.(50/10)
now A & B can copy 70 pages in 10hrs.
thus, B can copy 90 pages in 10 hrs.[eqn. is (50+x)/2=70, where x--> no. of pages B can copy in 10 hrs.]
so, B can copy 9 pages in 1hr.
therefore, to copy 26 pages B will need almost 3hrs.
since in 3hrs B can copy 27 pages.

13) what's the answer for that :
A, B and C are 8 bit no's. They are as follows:
A -> 1 1 0 0 0 1 0 1
B -> 0 0 1 1 0 0 1 1
C -> 0 0 1 1 1 0 1 0 ( - =minus, u=union)
Find ((A - C) u B) =?

To find A-C, We will find 2's compliment of C and them add it with A,
That will give us (A-C)
2's compliment of C=1's compliment of C+1
=11000101+1=11000110
A-C=11000101+11000110
=10001001
Now (A-C) U B is .OR. logic operation on (A-C) and B
10001001 .OR . 00110011
The answer is = 10111011,
Whose decimal equivalent is 187.

14) One circular array is given(means memory allocation tales place in circular fashion) diamension(9X7) and sarting add. is 3000, What is the address of (2,3)........
Sol) it's a 9x7 int array so it reqiure a 126 bytes for storing.b'ze integer value need 2 byes of memory allocation. and starting add is 3000
so starting add of 2x3 will be 3012.

15) 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) initial x (1,1) = 3000 u hav to find from x(8,1)so u have x(1,1),x(1,2) ... x(7,7) = so u have totally 7 * 7 = 49 elementsu need to find for x(8,5) ? here we have 5 elements each element have 4 bytes : (49 + 5 -1) * 4 = 212 -----( -1 is to deduct the 1 element ) 3000 + 212 = 3212

16) Which of the following is power of 3 a) 2345 b) 9875 c) 6504 d) 9833

17) 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) M=sqrt(100N)
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)

18)
1)SCOOTER --------- AUTOMOBILE--- A. PART OF
2.OXYGEN----------- WATER ------- B. A Type of
3.SHOP STAFF------- FITTERS------ C. NOT A TYPE OF
4. BUG -------------REPTILE------ D. A SUPERSET OF
1)B 2)A 3)D 4)C

19) A bus started from bustand at 8.00a m and after 30 min staying at destination, it returned back to the bustand. the destination is 27 miles from the bustand. the speed of the bus 50 percent fast speed. at what time it returns to the bustand
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 returnig=18+9=27
Then in return it take 27/27=1 hour
then total time in joureny=1+1:30+00:30 =3 hour
so it will come at 8+3 hour=11 a.m.
So Ans==11 a.m

20) In two dimensional array X(7,9) each element occupies 2 bytes of memory.If the address of first element X(1,1)is 1258 then what will be the address of the element X(5,8) ?
Sol) Here, the address of first element x[1][1] is 1258 and also 2 byte of memory is given. now, we have to solve the address of element x[5][8], therefore, 1258+ 5*8*2 = 1258+80 = 1338 so the answer is 1338.

21) The temperature at Mumbai is given by the function: -t2/6+4t+12 where t is the elapsed time since midnight. What is the percentage rise (or fall) in temperature between 5.00PM and 8.00PM?

22) 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 temperature 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

23) 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. Ans. At 9pm 7.5 more
Sol) In equestion 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

24) A person had to multiply two numbers. Instead of multiplying by 35, he multiplied by 53 and 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

25) How many positive integer solutions does the equation 2x+3y = 100 have?
a) 50 b) 33 c) 16 d) 35
Sol) There is a simple way to answer this kind of Q's 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)

26) The total expense of a boarding house are 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

27) 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 erarsers = 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 byamal)
i.e 5/8 = 500/8% = 62.5% is the answer

28) 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

29) 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
to me the chances are
when n=3
(i) nC1= 3C1 =3 for 10 boxes .. 10*3=30
(ii) 10C1=10 for 10 boxes ....10*10=100
(iii)9C1=9 for 10 boxes .....10*9=90

30) (1-1/6) (1-1/7).... (1- (1/ (n+4))) (1-(1/ (n+5))) = ?
leaving the first numerater and last denominater, all the numerater and denominater will cancelled out one another. Ans. 5/(n+5)

31) 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

32) 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

33) wind blows 160 miles in 330min.for 80 miles how much time required
Sol) 160 miles= 330 min
1 mile = 330/160
80 miles=(330*80)/160=165 min.

34) 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 travelling at a speed of 35 mph. find the speed limit
Sol)
(x+10)=(x+35)/2
solving the eqn we get x=15

35) 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
this is the exact answer

36) A person who decided to go weekend trip should not exceed 8 hours driving in a day average speed of forward journey is 40 mph due to traffic in Sundays the return journey average speed is 30 mph. How far he can select a picnic spot.

37) Low temperature at the night in a city is 1/3 more than 1/2 hinge as higher temperature in a day. Sum of the low temp and high temp is 100 c. then what is the low temp.
ans is 40 c.
Sol) let x be the highest temp. then,
x+x/2+x/6=100.
therefore, x=60 which is the highest temp
and 100-x=40 which is the lowest temp.

38) car is filled with four and half gallons of oil for full round trip. Fuel is taken 1/4 gallons more in going than coming. What is the fuel consumed in coming up.
Sol) let feul consumed in coming up is x. thus equation is: x+1.25x=4.5ans:2gallons

39) A work is done by the people in 24 min. One of them can do this work alone in 40 min. How much time required to do the same work for the second person
Sol) Two people work together in 24 mins.
So, their one day work is
(1/A)+(1+B)=(1/24)
One man can complete the work in 40mins
one man's one day work (1/B)= (1/40)
Now,
(1/A)=(1/24)-(1/40)
(1/A)=(1/60)
So, A can complete the work in 60 mins.

40) In a company 30% are supervisors and 40% employees are male if 60% of supervisors are male. What is the probability? That a randomly chosen employee is a male or female?
Sol) 40% employees are male if 60% of supervisors are male so for 100% is 26.4%so the probability is 0.264

41) In 80 coins one coin is counterfeit what is minimum number of weighing to find out counterfeit coin
Sol) the minimum number of wieghtings needed is just 5.as shown below
(1) 80->30-30
(2) 15-15
(3) 7-7
(4) 3-3
(5) 1-1

42) 2 oranges, 3 bananas and 4 apples cost Rs.15. 3 oranges, 2 bananas, and 1 apple costs Rs 10. What is the cost of 3 oranges, 3 bananas and 3 apples?
2x+3y+4z=15
3x+2y+z=10 adding
5x+5y+5z=25
x+y+z=5 that is for 1 orange, 1 bannana and 1 apple requires 5Rs.
so for 3 orange, 3 bannana and 3 apple requires 15Rs.
i.e. 3x+3y+3z=15

43) In 8*8 chess board what is the total number of squares refers
Sol) odele discovered that there are 204 squares on the board We found that you would add the different squares - 1 + 4 + 9 + 16+ 25 + 36 + 49 + 64.
Also in 3*3 tic tac toe board what is the total no of squares
Ans 14 ie 9+4(bigger ones)+1 (biggest one)
If you ger 100*100 board just use the formula
the formula for the sum of the first n perfect squares is
n x (n + 1) x (2n + 1)
______________________
6
if in this formula if you put n=8 you get your answer 204

44) One fast typist type some matter in 2hr and another slow typist type the same matter in 3hr. If both do combinely in how much time they will finish.
Sol) Faster one can do 1/2 of work in one hourslower one can do 1/3 of work in one hourboth they do (1/2+1/3=5/6) th work in one hour.so work will b finished in 6/5=1.2 hour i e 1 hour 12 min.

45) If Rs20/- is available to pay for typing a research report & typist A produces 42 pages and typist B produces 28 pages. How much should typist A receive?
Here is the answer Find of 42 % of 20 rs with respect to 70 (i.e 28 + 42) ==> (42 * 20 )/70 ==> 12 Rs

46) An officer kept files on his table at various times in the order 1,2,3,4,5,6. Typist can take file from top whenever she has time and type it.What order she cannt type.?

47) In some game 139 members have participated every time one fellow will get bye what is the number of matches to choose the champion to be held?
the answer is 138 matches
Sol) since one player gets a bye in each round,he will reach the finals of the tournament without playing a match.
therefore 137 matches should be played to detemine the second finalist from the remaining 138 players(excluding the 1st player)
therefore to determine the winner 138 matches shd be played.

48) One rectangular plate with length 8inches, breadth 11 inches and 2 inches thickness is there. What is the length of the circular rod with diameter 8 inches and equal to volume of rectangular plate?
Sol) Vol. of rect. plate= 8*11*2=176
area of rod=(22/7)*(8/2)*(8/2)=(352/7)
vol. of rod=area*length=vol. of plate
so length of rod= vol of plate/area=176/(352/7)=3.5

49) One tank will fill in 6 minutes at the rate of 3cu ft /min, length of tank is 4 ft and the width is 1/2 of length, what is the depth of the tank?
3 ft 7.5 inches

50) A man has to get air-mail. He starts to go to airport on his motorbike. Plane comes early and the mail is sent by a horse-cart. The man meets the cart in the middle after half an hour. He takes the mail and returns back, by doing so, he saves twenty minutes. How early did the plane arrive?
ans:10min:::assume he started at 1:00,so at 1:30 he met cart. He returned home at 2:00.so it took him 1 hour for the total jorney.by doing this he saved 20 min.so the actual time if the plane is not late is 1 hour and 20 min.so the actual time of plane is at 1:40.The cart travelled a time of 10 min before it met him.so the plane is 10 min early.

51) Ram singh goes to his office in the city every day from his suburban house. His driver Mangaram drops him at the railway station in the morning and picks him up in the evening. Every evening Ram singh reaches the station at 5 o'clock. Mangaram also reaches at the same time. One day Ram singh started early from his office and came to the station at 4 o'clock. Not wanting to wait for the car he starts walking home. Mangaram starts at normal time, picks him up on the way and takes him back house, half an hour early. How much time did Ram singh walked?

52) 2 trees are there. One grows at 3/5 of the other. In 4 years total growth of the trees is 8 ft. what growth will smaller tree have in 2 years.
Sol) THE BIG TREE GROWS 8FT IN 4 YEARS=>THE BIG TREE GROWS 4FT IN 2 YEARS.WHEN WE DIVIDE 4FT/5=.8*3=>2.4
ans: 1.5 mt 4 (x+(3/5)x)=88x/5=2x=5/4 after 2 years x=(3/5)*(5/4)*2 =1.5

53) There is a six digit code. Its first two digits, multiplied by 3 gives all ones. And the next two digits multiplied by 6 give all twos. Remaining two digits multiplied by 9 gives all threes. Then what is the code?
sol) Assume the digit xx xx xx (six digits)
First Two digit xx * 3=111
xx=111/3=37
( first two digits of 1 is not divisible by 3 so we can use 111)
Second Two digit xx*6=222
xx=222/6=37
( first two digits of 2 is not divisible by 6 so we can use 222)
Thrid Two digit xx*9=333
xx=333/9=37
( first two digits of 3 is not divisible by 9 so we can use 333)

54) There are 4 balls and 4 boxes of colours yellow, pink, red and green. Red ball is in a box whose colour is same as that of the ball in a yellow box. Red box has green ball. In which box you find the yellow ball?
ans is green...
Sol) Yellow box can have either of pink/yellow balls.
if we put a yellow ball in "yellow" box then it wud imply that "yellow" is also the colour of the box which has the red ball(becoz acordin 2 d question,d box of the red ball n the ball in the yellow box have same colour)
thus this possibility is ruled out...
therefore the ball in yellow box must be pink,hence the colour of box containin red ball is also pink....
=>the box colour left out is "green",,,which is alloted to the only box left,,,the one which has yellow ball..

55) A bag contains 20 yellow balls, 10 green balls, 5 white balls, 8 black balls, and 1 red ball. How many minimum balls one should pick out so that to make sure the he gets at least 2 balls of same color.
Ans:he should pick 6 ball totally.
Sol) Suppose he picks 5 balls of all different colours then when he picks up the sixth one, it must match any on of the previously drawn ball colour.
thus he must pick 6 balls

56) What is the number of zeros at the end of the product of the numbers from 1 to 100
Sol) For every 5 in unit palce one zero is added
so between 1 to 100 there are 10 nos like 5,15,25,..,95 which has 5 in unit place.
Similarly for every no divisible by 10 one zero is added in the answer so between 1 to 100 11 zeros are added
for 25,50,75 3 extra zeros are added
so total no of zeros are 10+11+3=24

57) 10 Digit number has its first digit equals to the numbers of 1's, second digit equals to the numbers of 2's, 3rd digit equals to the numbers of 3's .4th equals number of 4's..till 9th digit equals to the numbers of 9's and 10th digit equals to the number of 0's. what is the number?.(6marks)
ans:2100010006
2---shows that two 1's in the ans
1---shows that one 2 in ans
0---shows no 3 in the ans
0---shows no 4 in the ans
0---shows no 5 in the ans
1---shows one 6 in the ans
0---shows no 7 in the ans
0---shows no 8 in the ans
0---shows no 9 in the ans
6---shows six 0's in the ans

58) There are two numbers in the ratio 8:9. if the smaller of the two numbers is increased by 12 and the larger number is reduced by 19 thee the ratio of the two numbers is 5:9. Find the larger number?
sol) 8x:9x initialy
8x+ 12 : 9x - 19 = 5x:9x
8x+12 = 5x
-> x = 4
9x = 36 not sure about the answer ..

59) There are three different boxes A, B and C. Difference between weights of A and B is 3 kgs. And between B and C is 5 kgs. Then what is the maximum sum of the differences of all possible combinations when two boxes are taken each time
A-B = 3
B-c = 5
a-c = 8
so sum of diff = 8+3+5 = 16 kgs

60) A and B are shooters and having their exam. A and B fall short of 10 and 2 shots respectively to the qualifying mark. If each of them fired atleast one shot and even by adding their total score together, they fall short of the qualifying mark, what is the qualifying mark?
ans is 11
coz each had atleast 1 shot done so 10 + 1 = 11
n 9 + 2 = 11
so d ans is 11

61) A, B, C, and D tells the following times by looking at their watches. A tells it is 3 to 12. B tells it is 3 past 12. C tells it is 12:2. D tells it is half a dozen too soon to 12. No two watches show the same time. The difference between the watches is 2,3,4,5 respectively. Whose watch shows maximum time?
sol) A shows 11:57, B shows 12:03, C shows 12:02, and D shows 11:06 therefore, max time is for B

62) Falling height is proportional to square of the time. One object falls 64cm in 2sec than in 6sec from how much height the object will fall.
Sol) The falling height is proportional to the squere of the time.
Now, the falling height is 64cm at 2sec
so, the proportional constant is=64/(2*2)=16;
so, at 6sec the object fall maximum (16*6*6)cm=576cm;
Now, the object may be situated at any where.
if it is>576 only that time the object falling 576cm within 6sec .Otherwise if it is situated<576 then it fall only that height at 6sec.

63) Gavaskar average in first 50 innings was 50. After the 51st innings his average was 51 how many runs he made in the 51st innings
Ans) first 50 ings.- run= 50*50=2500
51st ings.- avg 51. so total run =51*51=2601.
so run scored in that ings=2601-2500=101 runs.

64) Anand finishes a work in 7 days, Bittu finishes the same job in 8 days and Chandu in 6 days. They take turns to finish the work. Anand on the first day, Bittu on the second and Chandu on the third day and then Anand again and so on. On which day will the work get over?
a) 3rd b) 6th c) 9th d) 7th
Ans is d) 7th day
Sol) In d 1st day Anand does 1/7th of total work
similarly,
Bithu does 1/8th work in d 2nd day
hence at d end of 3 days, work done = 1/7+1/8+1/6=73/168
remaining work = (168-73)/168 = 95/168
again after 6 days of work, remaining work is = (95-73)/168 = 22/168
and hence Anand completes the work on 7th day.(hope u understood.)

65) A man, a women and a child can do a piece of work in 6 days,man can do it in 14 days, women can do it 16 days, and in how many days child can do the same work?
The child does it in 24 days

66)
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}
A: 1 1 0 1 1 0 1 1
B: 0 1 1 1 1 0 1 0
by binary sub. a-b = 01100001 (1-0=1, 1-1=0,0-0=0, n for the 1st 3 digits 110-011=011)
now (a-b)uc= 01100001
or 01101101

__________________
Answered By StudyChaCha Member

Hi I am about to completed my bachelor degree and preparing for the placement exam for the TCS please provide me placement question paper for TCS placement exam?

As you want to known about the question paper of the TCS placement exam here I am providing you the same


1. A manufacturer of chocolates makes 6 different flavors of chocolates. The chocolates are sold in boxes of 10. How many “different” boxes of chocolates can be made?

Sol:
If n similar articles are to be distributed to r persons, x1+x2+x3......xr=n each person is eligible to take any number of articles then the total ways are n+r−1Cr−1
In this case x1+x2+x3......x6=10
in such a case the formula for non negative integral solutions is n+r−1Cr−1
Here n =6 and r=10. So total ways are 10+6−1C6−1 = 3003

2. In a single throw with two dice, find the probability that their sum is a multiple either of 3 or 4.
a. 1/3
b. 1/2
c. 5/9
d. 17/36

Sol: Their sum can be 3,4,6,8,9,12
For two dice, any number from 2 to 7 can be get in (n-1) ways and any number from 8 to 12 can be get in (13 - n) ways.
Then possible ways are 2 + 3 + 5 + 5 + 4 + 1 = 20 possible cases.
So probability is (20/36)=(5/9)

3. B alone can do piece of work in 10 days. A alone can do it in 15 days. If the total wages for the work is Rs 5000, how much should B be paid if they work together for the entire duration of the work?
a. 2000
b. 4000
c. 5000
d. 3000

Sol:
Time taken by A and B is in the ratio of = 3:2
Ratio of the Work = 2 : 3 (since, time and work are inversely proportional)
Total money is divided in the ratio of 2 : 3 and B gets Rs.3000

4. 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

5. Arun makes a popular brand of ice cream in a rectangular shaped bar 6cm long, 5cm wide and 2cm thick. To cut costs, the company had decided to reduce the volume of the bar by 19%. The thickness will remain same, but the length and width will be decreased by some percentage. The new width will be,
a. 5.5
b. 4.5
c. 7.5
d. 6.5

\Sol:
Volume =l×b×h = 6×5×2 = 60 cm3
Now volume is reduced by 19%.
Therefore, new volume = (100−19)100×60=48.6
Now, thickness remains same and let length and breadth be reduced to x%
so, new volume: (x100×6)(x100×5)2=48.6
Solving we get x =90
thus length and width is reduced by 10%
New width = 5-(10% of 5)=4.5

6. If all the numbers between 11 and 100 are written on a piece of paper. How many times will the number 4 be used?

Sol: We have to consider the number of 4's in two digit numbers. _ _
If we fix 4 in the 10th place, unit place be filled with 10 ways. If we fix 4 in units place, 10th place be filled with 9 ways (0 is not allowed)
So total 19 ways.
Alternatively:
There are total 9 4's in 14, 24, 34...,94
& total 10 4's in 40,41,42....49
thus, 9+10=19.

7. If twenty four men and sixteen women work on a day, the total wages to be paid is 11,600. If twelve men and thirty seven women work on a day, the total wages to be paid remains the same. What is the wages paid to a man for a day’s work?

Sol: Let man daily wages and woman daily wages be M and W respectively
24M+16W=11600
12M+37W=11600
solving the above equations gives M=350 and W=200

8. The cost price of a cow and a horse is Rs 3 lakhs. The cow is sold at 20% profit and the horse is sold at 10% loss. Overall gain is Rs 4200. What is the cost price of the cow?

Sol:
Profit = 4200
Profit =SP - CP
4200=SP - 300000 therefore SP=304200
x+y = 300000
1.2x + 0.9y = 304200
Solving for x = 114000 = CP of cow.

9. 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4......
In the above sequence what is the number of the position 2888 of the sequence.
a) 1
b) 4
c) 3
d) 2

Sol: First if we count 1223334444. they are 10
In the next term they are 20
Next they are 30 and so on
So Using n(n+1)2×10≤2888
For n = 23 we get LHS as 2760. Remaining terms 128.
Now in the 24th term, we have 24 1's, and next 48 terms are 2's. So next 72 terms are 3's.
The 2888 term will be “3”.

10. How many 4-digit numbers contain no.2?

Sol: Total number of four digit numbers =9000 (i.e 1000 to 9999 )
We try to find the number of numbers not having digit 2 in them.
Now consider the units place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Tens place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Hundreds place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
Thousands place can be selected in 8 ways (i.e 1,3,4,5,6,7,8,9) here '0' cannot be taken
Total number of numbers not having digit 2 in it =9 x 9 x 9 x 8 =5832
Total number of numbers having digit 2 in it = 9000-5832 =3168

11. 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%

12. 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.

13. 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

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

16. Total number of 4 digit number do not having the digit 3 or 6.

Sol:
consider 4 digits _ _ _ _
1st blank can be filled in 7C1 ways (0,3,6 are neglected as the first digit should not be 0)
2st blank can be filled in 8C1 ways (0 considered along with 1,2,4,5,7,8,9)
3st blank can be filled in 8C1 ways
4st blank can be filled in 8C1 ways
Therefore total 4 digit number without 3 or 6 is 7 x 8 x 8 x 8=3584

17. Find the missing in the series: 70, 54, 45, 41,____.

Sol: 40
70-54 = 16 = 42
54-45 = 9 = 32
45-41 = 4 = 22
41-40 = 1 = 12

18. A school has 120, 192 and 144 students enrolled for its science, arts and commerce courses. All students have to be seated in rooms for an exam such that each room has students of only the same course and also all rooms have equal number of students. What is the least number of rooms needed?

Sol: We have to find the maximum number which divides all the given numbers so that number of roots get minimized. HCF of 120,192 & 144 is 24. Each room have 24 students of the same course.
Then rooms needed 12024+19224+14424 = 5 +8 + 6 = 19

19. A farmer has a rose garden. Every day he picks either 7,6,24 or 23 roses. When he plucks these number of flowers the next day 37,36,9 or 18 new flowers bloom. On Monday he counts 189 roses. If he continues on his plan each day, after some days what can be the number of roses left behind? (Hint : Consider number of roses remaining every day)
a)7
b)4
c)30
d)37

Sol:
let us consider the case of 23. when he picks up 23 roses the next day there will be 18 new, so in this case., 5 flowers will be less every day. So when he counts 189, the next day 184, 179,174,169,................
finally the no. of roses left behind will be 4.

20. What is the 32nd word of "WAITING" in a dictionary?
Sol: Arranging the words of waiting in Alphabetical Order : A,G,I,I,N,T,W
Start with A_ _ _ _ _ _ This can be arranged in 6!/2! ways=720/2=360 ways
so can't be arranged starting with A alone as it is asking for 32nd word so it is out of range
AG_ _ _ _ _then the remaining letters can be arranged in 5!/2! ways so,120/2=60 ways. Out of range as it has to be within 32 words.
AGI_ _ _ _ Now the remaining letters can be arranged in 4! ways =24
AGN _ _ _ _ can be arranged in 4!/2! ways or 12 ways
so,24+12 =36th word so out of range. So we should not consider all the words start with AGN
now AGNI_ _ _can be arranged in 3! ways =6 ways
so 24+6=30 within range
Now only two word left so, arrange in alphabetical order.
AGNTIIW - 31st word
AGNTIWI - 32nd word

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Answered By StudyChaCha Member