TATA CONSULTANCY SERVICES
Paper Type : Whole Testpaper
Test Date : 11 March 2008
Test Location : Integral University Lucknow
TCS PAPER ON 11th MARCH 2008 AT LUCKNOW
Hi friends i am RAHUL from INTEGRAL UNIVERSITY . ..i have got placed
with TCS , thru campus placements......out of 207 .students 98 cleared
written and finally 32 selected. n i am thankful to each one of u for
being so informative thru this group n specially to all those who
provided me with info abt TCS in last few days... thr selection
procedure.
1. u need to hv first class throughtout(10th,12th,grad)
2. engineering student then 60% n above
3.if post grad(MCA) 60% n above
Paper pattern
1 set has three sections
a> English(10 antonyms n 10 syno..)...( 32 marks 20 mins)
b> Reading comprehension passages and comleting sentence(6 + 6)marks
2nd section is Quantitative...(40 mins 32 ques)..
3rd section is Critical reasoning..(30 mins 12 ques..)
FOR ANTONYMS N SYNONYMS
just go through the barrons 12 edition .but i think thats too much so
its better not to worried about the antonyms and synonyms .they are too
tough and dont waste ur time in that becoz the time alloted for this
section is just 20 minutes.......and you have to complete the reading
comprehension and sentence completion as well..............
REMEMBER there is no sectional cut off and no upper or lower cut off
neither any type of negative marking......so attempt all the
question.....
QUANTITATIVE SECTION
they are just repeating these section every time some of the questions
are TCS PAPER ON 19th JANUARY 2008
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+(log(0.3180.317)) = 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 20min
It takes 8hrs totally . So 85hr 20 min=2hr 40min
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 1010/5=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=10eq4
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
so amount x is getting is 80
y is 70
z =60
total=210
but total money is 300
300210=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
3x4y=1/4(3x)
12x16y=3x
9x=16y
y=9x/16
(3x4(9x/16))/((4x5(9x/16)))
ans:12/19
11) If G(0) = 1 G(1)= 1 and G(N)=G(N1)  G(N2) 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 AC, We will find 2's compliment of C and them add it with A,
That will give us (AC)
2's compliment of C=1's compliment of C+1
=11000101+1=11000110
AC=11000101+11000110
=10001001
Now (AC) U B is .OR. logic operation on (AC) 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 twodimensional 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=34.527
=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*53x*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, 30001750 = 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 = 300050*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 (106) = 4 pens = (21.5)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 = 686
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) (11/6) (11/7).... (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 100x=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 times 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>3030
(2) 1515
(3) 77
(4) 33
(5) 11
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 cann t 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 airmail. He starts to go to airport on his
motorbike. Plane comes early and the mail is sent by a horsecart. 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
2shows that two 1's in the ans
1shows that one 2 in ans
0shows no 3 in the ans
0shows no 4 in the ans
0shows no 5 in the ans
1shows one 6 in the ans
0shows no 7 in the ans
0shows no 8 in the ans
0shows no 9 in the ans
6shows 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
AB = 3
Bc = 5
ac = 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=26012500=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 = (16873)/168 = 95/168
again after 6 days of work, remaining work is = (9573)/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 ( (AB) u C )==?
Hint : 109
AB 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. ab = 01100001 (10=1, 11=0,00=0, n for the 1st 3
digits 110011=011)
now (ab)uc= 01100001
or 01101101
gives 1101101... convert to decimal equals 10
CRITICAL REASONING
this section is very important there are 12 question and u have to solve
it in 30 minutes.. for critical reasoning ....go thru the barrons 5
model test paper....just mug up all the answer but beware .....they can
change the character names so be careful..............
after clearing my written they called us for interview on 12th march
TECHNICAL INTERVIEW
after waiting for more than seven hours they called me for the interview
.look just be confident and have eye contact with them .......actually
in the starting i was very much confident and answered all his question
but after some time dont knw wat he asked to me and and my mind was just
blocked.....dont knw y......even this is not the end ...when i finished
my tech interview and came out of the hall .....i forgot my file
there.....[:d].......well after few minutes i called for the H.R.
H.R.
same things ...introduce urself, strengths weakness,future plan,where do
u see urself after 5 years.....i told them that i am interested in
embedded ..they asked if we send u u.s. will u go?.....and some more
h.r. question..
i advice u guys dont even try to make them fool ...if u dont knw any
answer ....just tell them 'u dont knw'
M.R.
same h.r. questions introduce urself......wagerah wagerah finally result
announced and thank god! i was one of the guys who have
selected....yeppiieeeeeeee
all the best ..... 
