All Information about CDAC's CCAT Examination

I have given the examination on 30th June'13 and I want to share my experience with my readers :) .

I applied for the DAC and DITISS courses so I have given only Section A and B exams so there are details of these two sections only. First, I would like say something about CDAC for those reader who doesn't know anything about this examination.

Centre for Development of Advanced Computing (C-DAC) is the premier R&D organization of the Department of Electronics and Information Technology (DeitY), Ministry of Communications & Information Technology (MCIT) for carrying out R&D in IT, Electronics and associated areas. Different areas of C-DAC, had originated at different times, many of which came out as a result of identification of opportunities.

They takes an entrance examination called CCAT. The examination held two times in a year. One in the month of June/July and second in the month of December/January. The duration of this courses are six months. After completion of any course you will get a certification PG Diploma.

Syllabus of two sections and it's duration are given below : 

Section A :

• English - There were five questions from a very small passage. Five questions of jumbled sentences and five questions from prepositions (Read English Grammar book and  any Competitive English book ).
• Quantitative Aptitude - Questions from Problem on trains, Ages, Profit and loss, Pipes, Mensuration, Time, distance and work etc. (Read  Quants by RS Agarwal ).
• Logical Reasoning - Questions from Analytical reasoning, Coding decoding, Situations Determining, Blood relations, Analogy, Series, etc. (Read Verbal and non verbal by RS Agarwal). 

 
These were the questions of Section A. I suggest you that don't panic during examination just focus on questions, read questions carefully and maintain your speed and accuracy.

Section B :

 

• C : Logical programming questions so focus on your c programming skills and else basic c questions. 
• C++ : Very easy questions, only basic theoretical questions were there. 
• DS : Questions from Search, Sorting, Stack Queue and Linklist only. 
• Networking : Read only first two chapters of any Computer Network book. 
• OS : Read Galvin's book and focus on Deadlock, Memory Management, Segmentation etc. 

 
These are the two sections which was asked in 30th June'13 CCAT's paper. I hope it might be helpful for upcoming examination. If you like my efforts then give me some feedback for sure. 

Thanks guys for reading this...

Download : CDAC Materials

TCS new pattern Question 2013

My name is PREET.But my son accidentally types the by interchanging a pair of letters in my name.What is the probability that despite this interchange, the name remains unchanged? a.5% b.20% c.25% d.12.5%

Ans:

10%

there are 10 possible pairs

out of which 1 will not change my name.

so probability that name remains unchanged = 1/10 = 10%

Roy is now 4 year older than Erik and half of that amount than Lewis.If in two years Roy will be twice as old as Erick,then in two year what would be Roy age multiplied by Lewis age??

Ans:ans is 48
firstly two equation from 1st sentence ..
suppose roy=x, erik=y, lewis=z
so x=y+4....(1)
2nd is x=z+2...(2) (Roy is now 4 year older than Erik and half of that amount than Lewis so 4/2=2)
also after 2 years erik and roy relation wil b
x+2=2(y+2).. (3)
so put this into 1st equ. u will get y=2..
and then x=6,z=4(by substituting other)
nw asking multiplication after two years of roy n lewis is (6+2)*(4+2)=48..

a man cut small cubes of 3 cubic cms each. Which he joined to form a cube wit 10 cubes length, 3 cubes in depth and 3 cubes wide. How many more small cubes does she require to form a perfect cube?

 Ans:

At present, the number of cubes used are 10*3*3=90. For a perfect cube, the cube should be 10 cubes length, 10 cubes in depth, and 10 cubes wide, i.e. 10*10*10=1000. So, additional number of cubes required is 1000-90= 910

one person has no siblings and says,"the guy in the photo is the only son of my fathers's son". what is the relation of the guy to the person

Ans:

His own son.

in a shopping mall with staff of 5 members, the average is 45.After 5 years a person joined them and the average is again 45. what is the age of 6th person?

Ans:

Since the average age is 45 yrs for the 5 members.
sum total of their ages = 45 X 5=225 yrs.
after 5 years, sum total of their ages = 225 + ( 5 X 5 ) = 250 yrs
now , let x be the age of the 6th person, so
the sum total of age is now = (250 + x ) yrs.
given the average is 45 yrs.
so,
( 250 + x )/ 6 = 45
or, x = 20 yrs.
so, the age of the 6th person is 20 years.

49 members attend the party.in that 22 are males,17 females.the shake hands between males,females,male,female.total 12 people given shake hands.how many such kinds of such shake hands are possible?

If you mean only 12 persons shook hands .
then 12C2 = 66 possible shake hands

what is the reminder when 6^17+17^6 is divided by 7?

First break the sum into two parts..
(i) 6^17 (ii) 17^6

sol-(i)- By remainder theorem there are always two remainders for any number, i.e. one is in negative and another is positive. In this case when we divide 6 by 7 we will get two remainders that is 6 and (-1). We can take any of these but for simplicity we will take (-1) is the remainder. Now (-1)^17 = -1.

sol-(ii)-- we have 17^6.. Now. 17^6 /7 = 3^6/7= (3^3)^2/7 = 27^2/7 = (-1)^2/7 = 1 is the remainder.

On adding case -(i) and (ii)... we will get final remainder = 0.

or

1   6^17 mod 7=(7-1)^17 mod 7
so 7^17 divide by 7 and remainder will 0,so we only take (-1)^17 mod 7=-1

2) 17^6 mod 7=(7*2+3)^6
so now (7*2)^17 divide by 7 and remainder will 0,so we only take (3)^6 mod 7=
729 mod 7=1
now add (1)+(2)
-1+1=0(ans)

What is the greatest possible positive integer n if 8^n divides (44)^44 without leaving a remainder?
a)14 b)28 c)29 d)15

Ans:

44^44 = 2^88 *11^44
= 8^29 * 2 * 11^44
so ans is 29

Complete the series 2, 7, 24, 77,__
Ans: 240

Ans:

(1st no.*3)+1=(2*3)+1=7
(2nd no.*3)+3=(7*3)+3=24
(3rd no.*3)+3=(24*3)+5=77
(4th no.*3)+1=(77*3)+7=238

or

ist*12=3rd 2*12=24
2nd*11=4th 7*11=77
3rd*10=5th 24*10=240

or

(2*3)+1=7
(7*3)+3=24
(24*3)+5=77
so ans is (77*3)+7=240

 

There is a circular pond and there is a road around the pond. The road is 4ft wide. The area of the pond is 11/25 of the area of the road. What is the radius of the pond?

Ans:let radius of pond=r
then,radius of road=r+4
since area of pond=11/25(area of road)
pi r^2=11/25 pi[(r+4)^2-r^2]
after solving this equa we get
25r^2-88r-176=0
r=4.94

There are two container. The volume of the second container is twice than the first. Half of the first cont. is filled with Wine & The one quarter of the second container is filled with wine. Now the remaining of the both container is filled with water. There is a third container which is enough to have the total liquid of the first two containers. What will be the ratio of the total volume and wine of the third container? Answer will be 1:3

Ans:consider A(first container)=100 for A wine filled half 100/2= 50
consider B(second container)=200 for B wine filled 1/4(200)=50
this two container mixed in 3rd container let as C=300
the ratio of wine to the volume of C is 100/300=1/3(1:3)

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Posted By Sundeep aka SunTechie

Sundeep is a Founder of Youth Talent Auzzar, a passionate blogger, a programmer, a developer, CISE and these days he is pursuing his graduation in Engineering with Computer Science dept.

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Placement Paper : TCS : Pattern 7 (Pattern Wise)

Pattern 7:

1. (40*40*40 – 31*31*31)/(40*40+40*31+31*31)=?
a)8 b)9 c)71 d)51

Solution:a³  -b³ =(a-b)*(a²+a*b+b²) so from this formula we will find (a-b) value

2. (98*98*98 – 73*73*73)/( 98*98*98 – 73*73*73)=?
a).171 b).4 c).420 d).415

3. (209*144)^2 + (209*209)+(209*144)+(144*144) = ?
a)905863729 b)905368729 c)905729368 d)65

To Previous Pattern :(click on link below)
Pattern 1 :http://youthtalentauzzar.blogspot.com/2011/11/placement-papers-tcs-tata-consultancy.html
Pattern 2 : http://youthtalentauzzar.blogspot.com/2011/11/placement-papers-tcs-tata-consultancy_15.html
Pattern 3 : http://youthtalentauzzar.blogspot.com/2011/11/placemnt-papertcs-pattern-3.html
Pattern 4 : http://youthtalentauzzar.blogspot.com/2011/11/here-is-patten-4-for-tcs-placement.html
Pattern 5: http://youthtalentauzzar.blogspot.com/2011/11/placement-paper-tcs-pattern-5.html
Pattern 6: http://youthtalentauzzar.blogspot.com/2011/11/placement-paper-tcs-pattern-6-pattern.html

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Placement Paper : TCS : Pattern 4

Here is the Patten 4 for the TCS placement papers...

Pattern 4:
1. In the reading room of a library, there are 23 reading spots. Each reading spot consists of a round table with 9 chairs placed around it. There are some readers such that in each occupied reading spot there are different numbers of readers. If in all there are 36 readers, how many reading spots do not have even a single reader?
a) 8 b) none c) 16 d) 15

Solution: 23 reading spots, Each reading spot consists of 9 chairs placed around it so There are some readers such that in each occupied reading spot there are different numbers of readers.

For each table different no of persons are sat,so for first table 1 person is sit,2^nd table 2 persons are sit 36

readers means(1+2+3+4+5+6+7+8 so 8 tables are filled so 23-8=15 reading spots does not have single reader.

2. In the reading room of a library, there are 10 tables, 4 chairs per table. In each table there are different numbers of people seated. How many tables will be left out without at least 1 person?
a) 8 b) 6 c) 2 d) 7

3. In the reading room of a library, there are 10 tables, 4 chairs per table. In each table there are different numbers of people seated. How many ways they will sit in the library so that no chair would be blank?
a) 8 b) 6 c) 2 d) 7


To Previous Pattern :(click on link below)
Pattern 1 :http://youthtalentauzzar.blogspot.com/2011/11/placement-papers-tcs-tata-consultancy.html
Pattern 2 : http://youthtalentauzzar.blogspot.com/2011/11/placement-papers-tcs-tata-consultancy_15.html
Pattern 3 : http://youthtalentauzzar.blogspot.com/2011/11/placemnt-papertcs-pattern-3.html

hope you all are liking our posts on YTA ...
seed feedback on youthtalentauzzar@gmail.com or you can join us on facebook : http://www.facebook.com/youthtalentauzzar!
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Placemnt Paper:TCS- pattern 3

TCS Placemnt Paper: pattern 3

Pattern 3:

1. 6 persons standing in queue with different age group, after two years their average age will be 43 and seventh person joined with them. Hence the current average age has become 45. Find the age of seventh person?
a) 43 b) 69 c) 52 d) 31

Solution:

Total age of 6 persons is x hours,after two years total age of 6 persons is x+12

Average age of 6 persons is after two years is 43

So (x+12)/6=43,then solve x,

After 7^th person is added then (x+7^th person age)/7=45

So we will get 7^th person age easily

2. In a market 4 men are standing. The average age of the four before 4years is 45, after some days one man is added and his age is 49. What is the average age of all?
a) 43 b) 45 c) 47 d) 49

3. In a shopping mall with a staff of 5 members the average age is 45 years. After 5 years a person joined them and the average age is again 45 years. What’s the age of 6th person?
a) 25 b)20 c)45 d)30

4. In a market 4 men are standing .The average age of the four before 2 years is 55, after some days one man is added and his age is 45. What is the average age of all?
a) 55 b) 54.5 c) 54.6 d) 54.7


Pattern 4 : comming soon ! 


to check previous pattern ...check this link http://youthtalentauzzar.blogspot.com/p/apti-kit.html


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Placement Papers- TCS (tata consultancy services) : Pattern 2

Pattern 2 of TCS Placement Papers..

Pattern 2:

1. There are two water tanks A and B, A is much smaller than B. While water fills at the rate of 1 liter every hour in A, it gets filled up like, 10, 20, 40, 80, 160 in tank B. (At the end of first hour, B has 10 liters, second hour it has 20 liters and so on). If tank B is 1/32 filled of the 21 hours, what is total duration of hours required to fill it completely?
a) 26 B)25 c)5 d)27

Solution: for every hour water in tank in B is doubled,

Let the duration to fill the tank B is x hours.

x/32 part of water in tank of B is filled in 21 hours,

Next hour it is doubled so,

2*(x/32) part  i.e (x/16) part  is filled in 22 hours,

Similarly (x/8)th part in 23 hours,(x/4)th part is filled in 24 hours,

(x/2)th part is filled in 25 hours, (x)th part is filled in 26 hours

So answer is 26 hours.

2. There are two pipes A and B. If A filled 10 liters in an hour, B can fill 20 liters in same time. Likewise B can fill 10, 20, 40, 80, 160. If B filled in 1/16 of a tank in 3 hours, how much time will it take to fill the tank completely?
a) 9 B) 8 c) 7 d) 6

3. There are two water tanks A and B, A is much smaller than B. While water fills at the rate of 1 liter every hour in A, it gets filled up like, 10, 20, 40,80, 160…..in tank B. 1/8 th of the tank B is filled in 22 hours. What is the time to fill the tank fully?
a) 26 B) 25 c) 5 d) 27

4. A tank is filled with water. In first hour 10 liters, second hours 20 liters, and third hour 40 liters and so on. If time taken to fill ¼ of the tank if 5 hours. What is the time taken to fill up the tank?
a) 5 B) 8 c) 7 d) 12.5

5. If a tank A can be filled within 10 hours and tank B can be filled ¼ in 19 hours then, what is the time taken to fill up the tank completely?
a) 21 B) 38 c) 57 d) 76

Pattern 3 : coming soon ! 

thank you

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YTA

Placement Papers- TCS (tata consultancy services)

Pattern Wise TCS Placement Papers...

Pattern 1:

1. (1/2) of a number is 3 more than the (1/6) of the same number?
a) 6 b)7 c)8 d)9

Solution:

Let the number be x,

((1/2)*x)=3+(1/6)*x,

Then solve x

2. (1/3) of a number is 3 more than the (1/6) of the same number?
a) 6 b)16 c)18 d)21

3. (1/3) of a number is 6 more than the (1/6) of the same number?
a) 6 b)18 c)36 d)24

4. (2/3) of a number is 4 more than the (1/6) of the same number?
a) 6 b)8 c)36 d)24

5. (1/3) of a number is 5 more than the (1/6) of the same number?
a) 6 b)36 c)30 d)72

Pattern 2 : coming soon !

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YTA