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TCS APTITUDE PAPER WITH
SOLUTIONS Dear Chetanaites 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 7) A fisherman's day is rated as
good if he catches 9 fishes, fair
if 7 fishes and bad if 5 fishes. He
catches 53 fishes in a week n
had all good, fair n bad days in
the week. So how many good, fair n bad days did the fisher
man had in the week
Ans:4 good, 1 fair n 2 bad days
Sol) Go to river catch fish
4*9=36
7*1=7 2*5=10
36+7+10=53...
take what is given 53
good days means --- 9 fishes so
53/9=4(remainder=17) if u
assume 5 then there is no chance for bad days.
fair days means ----- 7 fishes so
remaining 17 ---
17/7=1(remainder=10) if u
assume 2 then there is no
chance for bad days. bad days means -------5 fishes
so remaining 10---10/5=2days.
Ans: 4 good, 1 fair, 2bad. ====
total 7 days. x+y+z=7--------- eq1
9*x+7*y+5*z=53 -------eq2
multiply eq 1 by 9,
9*x+9*y+9*z=35 -------------
eq3
from eq2 and eq3 2*y+4*z=10-----eq4
since all x,y and z are integer i
sud put a integer value of y
such that z sud be integer in eq
4 .....and ther will be two value
y=1 or 3 then z = 2 or 1 from eq 4 for first y=1,z=2 then from eq1
x= 4
so 9*4+1*7+2*5=53.... satisfied
now for second y=3 z=1 then
from eq1 x=3
so 9*3+3*7+1*5=53 ......satisfied so finally there are two solution
of this question
(x,y,z)=(4,1,2) and (3,3,1)... 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 http://www.ChetanaS.org so amount x is getting is 80
y is 70
z =60
total=210 but total money is 300
300-210=90
so they are getting 90 rs less
90 is 30% of 300 so they r
getting 30% discount 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. http:// www.ChetanaS.org 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 Ch eta naS 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
gives 1101101... convert to
decimal equals 109

hello friend you will find the recent papers in the following links

http://placementqanda.blogspot.com/2011/01/tcs-placement-papers-2011-tcs-placement.html
http://www.indiastudychannel.com/resources/Category20.aspx
http://placementpapers.net/helpingroot/Paper/TCS-Job-Interview-Placement-Paper-January-2011
http://ezinearticles.com/?TCS-Solved-Placement-Papers-2011&id=5655988
http://www.chetanasinterview.com/articles/2126/1/TCS-APTITUDE-PAPER-WITH-SOLUTIONS/Page1.html

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