Aptitude Questions Y5

31. What was the day of the week on 12th January, 1979?



Ans: Friday



Sol: Number of odd days in (1600 + 300) years = (0 + 1) = 1 odd day.

78 years = (19 leap years + 59 ordinary years) = (38 + 59) odd days = 6 odd days

12 days of January have 5 odd days.

Therefore, total number of odd days= (1 + 6 + 5) = 5 odd days.

Therefore, the desired day was Friday.



32. Find the day of the week on 16th july, 1776.



Ans: Tuesday



Sol: 16th july, 1776 means = 1775 years + period from 1st january to 16th july

Now, 1600 years have 0 odd days.

100 years have 5 odd days.

75 years = 18 leap years + 57 ordinary years

= (36 + 57) odd days = 93 odd days

= 13 weeks + 2 odd days = 2 odd days

Therefore, 1775 years have (0 + 5 + 2) odd days = 0 odd days.

Now, days from 1st Jan to 16th july; 1776

Jan Feb March April May June July

31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days

= (28 weeks + 2 days) odd days

Therefore, total number of odd days = 2

Therefore, the day of the week was Tuesday


33 .Find the angle between the minute hand and hour hand of a click when the time is

7.20?

Ans: 100deg

Sol: Angle traced by the hour hand in 12 hours = 360 degrees.

Angle traced by it in 7 hrs 20 min i.e. 22/3 hrs = [(360/12) * (22/3)] = 220 deg.

Angle traced by minute hand in 60 min = 360 deg.

Angle traced by it in 20 min = [(360/20) * 60] = 120 deg.

Therefore, required angle = (220 - 120) = 100deg.









34.The minute hand of a clock overtakes the hours hand at intervals of 65 min of the correct time. How much of the day does the clock gain or lose?



Ans: the clock gains 10 10/43 minutes



Sol: In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60

minutes.

To be together again, the minute hand must gain 60 minutes over the hour hand.

55 minutes are gained in 60 min.

60 min. are gained in [(60/55) * 60] min == 65 5/11 min.

But they are together after 65 min.

Therefore, gain in 65 minutes = (65 5/11 - 65) = 5/11 min.

Gain in 24 hours = [(5/11) * (60*24)/65] = 10 10/43 min.

Therefore, the clock gains 10 10/43 minutes in 24 hours.



35.A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours. What will be the true time when the clock indicates 1 p.m. on the following day?



Ans. 48 min. past 12.



Sol: Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours.

24 hours 10 min. of this clock = 24 hours of the correct clock.

145/6 hrs of this clock = 24 hours of the correct clock.

29 hours of this clock = [24 * (6/145) * 29] hrs of the correct clock

= 28 hrs 48 min of the correct clock.

Therefore, the correct time is 28 hrs 48 min. after 8 a.m.

This is 48 min. past 12.


36. At what time between 2 and 3 o’ clock will the hands 0a a clock together?



Ans: 10 10/11 min. past 2.


Sol: At 2 o’ clock, the hour hand is at 2 and the minute hand is at 12, i.e. they are 10

min space apart.

To be together, the minute hand must gain 10 minutes over the other hand.

Now, 55 minutes are gained by it in 60 min.

Therefore, 10 min will be gained in [(60/55) * 10] min = 10 10/11 min.

Therefore, the hands will coincide at 10 10/11 min. past 2.



37. A sum of money amounts to Rs.6690 after 3 years and to Rs.10035 after 6 years on compound interest. Find the sum.



Ans: Rs. 4460



Sol: Let the Sum be Rs. P. Then

P [1 + (R/100)]^3 = 6690………..(i)

P [1 + (R/100)]^6 = 10035………..(ii)

On dividing, we get [1 + (R/100)]^3 = 10035/6690 = 3/2.

P * (3/2) = 6690 or P = 4460.

Hence, the sum is Rs. 4460.



38. Simple interest on a certain sum is 16/25 of the sum. Find the rate percent and time, if both are numerically equal.



Ans: Rate = 8% and Time = 8 years



Sol: Let sum = X. Then S.I. = 16x/25

Let rate = R% and Time = R years.

Therefore, x * R * R/100 = 16x/25 or R^2 = 1600/25, R = 40/5 = 8

Therefore, Rate = 8% and Time = 8 years.





39. Find

i. S.I. on RS 68000 at 16 2/3% per annum for 9 months.

ii. S.I. on RS 6250 at 14% per annum for 146 days.

iii. S.I. on RS 3000 at 18% per annum for the period from 4th Feb 1995 to 18th April 1995.



Ans: i. RS 8500.

ii. RS 350.

iii. RS 108.



Sol:

i. P = 68000, R = 50/3% p.a. and T = 9/12 year = ¾ years

Therefore, S.I. = (P * Q * R/100)

= RS (68000 * 50/3 * ¾ * 1/100) = RS 8500.

ii. P = RS 6265, R = 14% p.a. and T = (146/365) year = 2/5 years.

Therefore, S.I. = RS (6265 * 14 * 2/5 *1/100) = RS 350.

iii. Time = (24 + 31 + 18) days = 73 days = 1/5 year

P = RS 3000 and R = 18% p.a.

Therefore, S.I. = RS (3000 * 18 * 1/5 * 1/100) = RS 108



40. A sum at simple interest at 13 ½% per annum amounts to RS 2502.50 after 4 years. Find the sum.



Ans: sum = RS 1625



Sol: Let sum be x. Then,

S.I. = (x * 27/2 * 4 * 1/100) = 27x/50

Therefore, amount = (x + 27x/50) = 77x/50

Therefore, 77x/50 = 2502.50 or x = 2502.50 * 50 / 77 = 1625

Hence, sum = RS 1625

Aptitude Questions Y4

21. If a man walks at the rate of 5kmph, he misses a train by only 7min. However if he walks at the rate of 6 kmph he reaches the station 5 minutes before the arrival of the train.Find the distance covered by him to reach the station.



Ans:6km.

Sol: Let the required distance be x km.

Difference in the times taken at two speeds=12mins=1/5 hr.

Therefore x/5-x/6=1/5 or 6x-5x=6 or x=6km.

Hence ,the required distance is 6 km



22. Walking 5/6 of its usual speed, a train is 10min late. Find the usual time to cover the journey?

Ans:50 min



Sol: New speed = 5/6 of usual speed

New time = 6/5 of usual time

Therefore, (6/5 of usual time) – usual time = 10min

Therefore Usual time = 50min



23.A train running at 54 kmph takes 20 seconds to pass a platform. Next it takes 12 seconds to pass a man walking at 6 kmph in the same direction in which the train is going. Find the length of the train and the length of the platform.



Ans. length of the train=160m

length of the platform=140 m.



Sol: Let the length of the train be x meters and length of the platform be y meters.

Speed of the train relative to man=(54-6) kmph =48 kmph.

=(48*5/18) m/sec =40/3 m/sec.

In passing a man, the train covers its own length with relative speed.

Therefore, length of the train=(Relative speed *Time)

=(40/3 * 12) m =160 m.

Also, speed of the train=(54 * 5/18) m/sec=15 m/sec.



Therefore, x+y/2xy=20 or x+y=300 or y=(300-160 m=140 m.

Therefore, Length of the platform=140 m.



24. A man is standing on a railway bridge which is 180m long. He finds that a train crosses the bridge in 20seconds but himself in 8 seconds. Find the length of the train and its speed.



Ans: length of train=120m

Speed of train=54kmph



Sol: Let the length of the train be x meters

Then, the train covers x meters in 8 seconds and (x + 180) meters in 20 seconds.

Therefore x/8 = (x+180)/20 ó 20x = 8(x+180) ó x = 120

Therefore Length of the train = 120m

Speed of the train = 120/8 m/sec = 15 m/sec =15 * 18/5 kmph = 54kmph

25. A man sells an article at a profit of 25%. If he had bought it at 20 % less and sold it for Rs.10.50 less, he would have gained 30%. Find the cost price of the article?

Ans. Rs. 50.



Sol: Let the C.P be Rs.x.

1st S.P =125% of Rs.x.= 125*x/100= 5x/4.

2nd C.P=80% of x. = 80x/100 =4x/5.

2nd S.P =130% of 4x/5. = (130/100* 4x/5) = 26x/25.

Therefore, 5x/4-26x/25 = 10.50 or x = 10.50*100/21=50.

Hence, C.P = Rs. 50.

26. A grosser purchased 80 kg of rice at Rs.13.50 per kg and mixed it with 120 kg rice at Rs. 16 per kg. At what rate per kg should he sell the mixture to gain 16%?



Ans: Rs.17.40 per kg.



Sol: C.P of 200 kg of mix. = Rs (80*13.50+120*16) = Rs.3000.

S.P = 116% of Rs 3000= Rs (116*3000/100) = Rs.3480.

Rate of S.P of the mixture = Rs.3480/200.per kg. = Rs.17.40 per kg.



27. Two persons A and B working together can dig a trench in 8 hrs while A alone can dig it in 12 hrs. In how many hours B alone can dig such a trench?



Ans:24hours.



Sol: (A+B)’s one hour’s work =1/8, A’s one hour’s work =1/12

Therefore, B’s one hour’s work = (1/8-1/12) =1/24.

Hence, B alone can dig the trench in 24 hours.



28. A and B can do a piece of work in 12 days ; B and C can do it in 20 days. In how many days will A, B and C finishes it working all together?

Also, find the number of days taken by each to finish it working alone?



Ans:60 days



Sol: (A+B)’s one day’s work=1/12; (B+C)’s one day’s work=1/15 and (A+C)’s one day’s

work=1/20.

Adding, we get: 2(A+B+C)’s one day’s work = (1/12+1/15+1/20)=1/5.

Therefore, (A+B+C)’s one day’s work=1/10.

Thus, A, B and C together can finish the work in 10 days.

Now, A’s one day’s work

= [(A+B+C)’s one day’s work] – [(B+C)’s one day’s work]

= 1/10-1/15)

= 1/30.

Therefore, A alone can finish the work in 30 days.

Similarly, B’s 1 day’s work = (1/10 -1/20) = 1/20.

Therefore, B alone can finish the work in 20 days.

And, C’s 1 day’s work= (1/10-1/12) = 1/60.

Therefore, C alone can finish the work in 60 days.



29. A is twice as good a workman as B and together they finish a piece of work in 18

days.In how many days will A alone finish the work?



Ans:27 days.



Sol: (A’s 1 day’s work): (B’s 1 day’s work) = 2:1.

(A + B)’s 1 day’s work = 1/18.

Divide 1/18 in the ratio 2:1.

Therefore A’s 1 day’s work = (1/18 * 2/3) = 1/27.

Hence, A alone can finish the work in 27 days.



30. 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work?



Ans: 12 ½ days.



Sol: Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work =y.

Then, 2x+3y=1/10 and 3x+2y=1/8.

Solving, we get: x=7/200 and y=1/100.

Therefore (2 men +1 boy)’s 1 day’s work = (2*7/200 + 1*1/100) = 16/200 = 2/25.

So, 2 men and 1 boy together can finish the work in 25/2 =12 ½ days.

Aptitiude Questions Y2

11. Vivek travelled 1200km by air which formed 2/5 of his trip.One third of the whole trip , he travelled by car and the rest of the journey he performed by train. The distance travelled by tain was???

Ans.800km



Sol: Let the total trip be x km.

Then 2x/5=1200

x=1200*5/2=3000km

Distance travelled by car =1/3*3000=1000km

Journey by train =[3000-(1200+1000)]=800km.









12. In a college ,1/5 th of the girls and 1/8 th of the boys took part in a social camp.What of the total number of students in the college took part in the camp?

Ans: 2/13



Sol: Out of 5 girls 1 took part in the camp

out of 8 boys 1 took part in the camp

so, out of 13 students 2 took part in the camp.

So, 2/13of the total strength took part in the camp.



13. On sports day,if 30 children were made to stand in a column,16 columns could be formed. If 24 children were made to stand in a column , how many columns could be formed?

Ans. 20


Sol: Total number of children=30*16=480

Number of columns of 24 children each =480/24=20.



14. Two trains 200mts and 150mts are running on the parallel rails at this rate of 40km/hr and 45km/hr.In how much time will they cross each other if they are running in the same direction.



Ans: 252sec



Sol: Relative speed=45-40=5km/hr=25/18 mt/sec

Total distance covered =sum of lengths of trains =350mts.

So, time taken =350*18/25=252sec.



15. 5/9 part of the population in a village are males. If 30% of the males are married, the percentage of unmarried females in the total population is:



Ans: (250/9)%



Sol: Let the population =x Males=(5/9)x

Married males = 30% of (5/9)x = x/6

Married females = x/6

Total females = (x-(5/9)x)=4x/9

Unmarried females = (4x/9 – x/6) = 5x/18

Required percentage = (5x/18 * 1/x * 100) = (250/9)%



16. From height of 8 mts a ball fell down and each time it bounces half the distnace back. What will be the distance travelled



Ans.: 24

Sol. 8+4+4+2+2+1+1+0.5+0.5+ and etc .. =24



17. First day of 1999 is sunday what day is the last day

Ans.: Monday

18. Increase area of a square by 69% by what percent should the side be incresed

Ans.: 13

Sol:Area of square=x2

Then area of increase=100+69=169

square root of 169 i.e 13 .



19. Ten years ago, chandrawathi’s mother was four times older than her daughter. After 10years, the mother will be twice older than daughter. The present age of Chandrawathi is:



Ans.20 years


Sol: Let Chandrawathi’s age 10 years ago be x years.
Her mother’s age 10 years ago = 4x

(4x+10)+10=2(x+10+10)

x=10

Present age of Chandrawathi = (x+10) = 20years



20. Finding the wrong term in the given series

7, 28, 63, 124, 215, 342, 511

Ans:28



Sol: Clearly, the correct sequence is

2^3 – 1, 3^3 – 1, 4^3 – 1, 5^3 – 1, ……….

Therefore, 28 is wrong and should be replaced by (3^3 – 1) i.e, 26.

Aptitude Questions Y1

1.A three digit number consists of 9,5 and one more number . When these digits are reversed and then subtracted from the original number the answer yielded will be consisting of the same digits arranged yet in a different order. What is the other digit?



Sol. Let the digit unknown be n.

The given number is then 900+50+n=950+n.



When reversed the new number is 100n+50+9=59+100n.

Subtracting these two numbers we get 891-99n.

The digit can be arranged in 3 ways or 6 ways.

We have already investigated 2 of these ways.

We can now try one of the remaining 4 ways. One of these is n 95

100n+90+5=891-99n

or 199n =796

so, n=4

the unknown digit is 4.

2.A farmer built a fence around his 17 cows,in a square shaped region.He used 27 fence poles on each side of the square. How many poles did he need altogether???



Ans.104 poles



Sol. Here 25 poles Must be there on each side .And around four corners 4 poles will be

present. 4*25+4=100+4=104 poles.





3.On the first test of the semester, kiran scored a 60. On the last test of the semester, kiran scored

75%

By what percent did kiran's score improve?



Ans: 25%



Sol. In first test kiran got 60

In last test he got 75.

% increase in test ( 60(x+100))/100=75

0.6X+60=75

0.6X=15

X=15/0.6=25%





4.A group consists of equal number of men and women. Of them 10% of men and 45% of women are unemployed. If a person is randomly selected from the group. Find the probability for the selected person to be an employee.

Ans:29/40

Sol: Assume men=100,women=100 then employed men & women r (100-10)+(100-45)=145



So probability for the selected person to be an employee=145/200=29/40



5. Randy's chain of used car dealership sold 16,400 cars in 1998. If the chain sold 15,744 cars in

1999, by what percent did the number of cars sold decrease?



Ans: 4%



Sol. Let percentage of decrease is x , then

16400(100-x)/100=15744

16400-15744=164x

x=656/164=4%





6. A radio when sold at a certain price gives a gain of 20%. What will be the gain percent, if sold for thrice the price?



A) 260%



B) 150%



C) 100%



D) 50%



E) None of these

Ans: 260%



Sol. Let x be original cost of the radio.

The solding price = (100+20)x=120x

If , it is sold for thrice the price ,then 3*120x=360x



So, gain percent is (360-100)=260%.





7. Find the perimeter of the shape below.







1cm





















Ans: 24cm





Sol: 2+4+5+1+7+5=24



8.If the Arithmetic mean is 34 and geometric mean is 16 then what is greates number in that series of numbers?



Ans. 64



Sol. Let two numbers be x, y;

Arthmetic mean=34=>( x+y)/2=34

x+y=68

geometric mean=16=>(xy)pow 1/2=16

xy=16*16=256



By trail and error 16*16=64*4



And 64+4/2=34

So the greatest number int hat series is 64.

9. The diameter of the driving wheel of a bus is 140cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 kmph?

Ans. 250



Sol. Distance to be covered in 1 min=(66*1000)/60 m=1100m

Circumference of the wheel =(2*22/7*0.70)m=4.4m.

So, Number of revolutions per min=1100/4.4=250.





10. The boys and girls in a college are in the ratio 3:2. If 20% of the boys and 25% of the girls are adults, the percentage of students who are not adults is:??

Ans.78%



Sol: Suppose boys = 3x and girls = 2x

Not adults = (80*3x/100) + (75*2x/100) = 39x/10

Required percentage = (39x/10)*(1/5x)*100 = 78%

Aptitude..

Verbal

1) STAR : CLUSTER:: ? :?

a) Orange : rind b) ???????? c) trees : clump d) mirror : reflection

2) GIBBLE : SENSE

3) GAUFFAW : LAUG

a) Sneeze : cough b) Whimper : cry …… c) ???????????????

4) PISTON : CYLINDER::?:?

a) SHAFT: ELEVATOR b) BULLET: REVOLVER c)???? d)???

Aptitude

1) abcD+abcd+aBCd+aBCD then the simplified function is ( Capital letters are copliments of corresponding

letters A=compliment of a)

[a] a [b] ab [c] abc [d] a(bc)* [e] none (bc)*=compliment of bc

2) A 12 address lines maps to the memory of

[a] 1k bytes [b] 0.5k bytes [c] 2k bytes [d] none

3) In a processor these are 120 instructions. Bits needed to implement this instructions

[a] 6 [b] 7 [c] 10 [d] none

4) In 8085 microprocessor READY signal does.which of the following is incorrect statements

[a]It is input to the microprocessor

[b] It sequences the instructions

5) Return address will be returned by function to

[a] Pushes to the stack by call

6) n=7623

{

temp=n/10;

result=temp*10+ result;

n=n/10

}

7) If A>B then

F=F(G);

else B>C then

F=G(G);

In this , for 75% times A>B and 25% times B>C then, is 10000 instructions are there ,then the ratio of F to G

[a] 7500:2500 [b] 7500:625 [c] 7500:625 if a=b=c else

8) In a compiler there is 36 bit for a word and to store a character 8bits are needed. IN this to store a

character two words are appended. Then for storing a K characters string, How many words are needed

[a] 2k/9 [b] (2k+8)/9 [c] (k+8)/9 [d] 2*(k+8)/9 [e] none

Technical

1) C program cod

int zap(int n)

{

if(n<=1)then zap=1;

else zap=zap(n-3)+zap(n-1);

}

then the call zap(6) gives the values of zap

[a] 8 [b] 9 [c] 6 [d] 12 [e] 15

2) Virtual memory size depends on

[a] address lines [b] data bus [c] disc space [d] a & c [e] none

3) Critical section is

[a]

[b] statements which are accessing shared resourses

4) Load a, mul a, store t1, load b, mul b, store t2, mul t2, add t1, then the content in accumulator is

5) question (3) in old paper

6) q(4) in old paper

7) The little indian representetion of 84 will be

a) 00001000 00000100 b) 00010000 00000100 c) 00001000 00000010

8) The standard for HTML is

a) HTML b) STML c) CGI d) DHTML

9) The standard for exchanging mails and communicat in files is

a) SMTP b) FTP c) HTTP d) ISP

10 The equivalent of NAND gate is

NAD with NOT b) OR with NOT c)NOT with OR

11) The operation to be performed on a stack using a linked list can be realised using the function

(A) delete front , instert rear (B) insert front , delete rear

(C) delete front , insert front (D) delete rear , insert rear

12) The binary equivalent of 188 is

13) what world wide network connects the largest number of servers and
computers

(A) WWW (B) HTTP (C) ARPANET (D) INTERNET

14. The network connecting all the networks are called : (A) INTERNET
15. The layer not present in TCP/IP model is:
(a) security layer (b)presentation layer (c)
16. Making use of objects with diferent ............. -polymorphism
17. Tree is a type of data structure

a) Primitive b) non- primitive
20. Unix is a _______ type of OS tree is a type of data structure
(a) multi user (b)multi tasking time sharing (c)single user
21. The protocol used to transfer web pages
(a)HTTP (b)FTP (c)TCP/IP .........
22. The data transfer ............memory ...............
(a)DMA
23. Mode of operation of stack............
(a)LIFO (b)FIFO (c)FCFS.......
24. What is ROM
(a) read only memory
25. address on the net is .......
(a)URL
26. Language for queries used in RDBMS
a)SQL ............

Logical reasoning

Verbal
1. antonym of SAGE:ans FOOL(GRE Barrons)
2. antonym of orthodox:ans uncoventionality(GRE Barrons)
3. analogy Excessive : Moderation::impusive:reflection(GRE barrons)
4. analogy Construct:Renovate::write:delete
Logical reasoning
1 odd man out --tautology:oncology:repetition:redundancy ans oncology
2 odd man out -- enroll:capitulate:enlist:conscript ans capitulate
3 few questions to check senetences if gramatically right
4. the following letters are assigned the codes like
a=1,c=2,d=3,r=7,y=9,o=4,i=5,t=6,n=8 then the word dictionary is
ans a> 3526548179 note :numbers may vary
5. If the nos 8 & 7 are interchanged as well as the signs +&/ are interchanged which of the following is valid

equation
ans c>8+8/7=9(it is 7/7+8=9)
6. whichof the following options is related to cricket,tennis,football
a)carom b)chess c)volleyball d)all ans c

7. A sports discussion is taking place in a meeting. All those in the meeting who enjoy the batting of Sachin also

enjoy the batting of Lara. Those who enjoy the batting of Lara despise the batting of Rahul. Some of those

who enjoy the batting of Lara also enjoy the batting of Saurav; those who enjoy the batting of Rahul also

enjoy the batting of Ricky. Some of those who enjoy the batting or Ricky also enjoy the batting of Lara.

Some of those who enjoy the batting of Saurav despise the batting of Rahul. All those who enjoy the batting

of Ricky also enjoy the batting of Astle. Miss Sujatha enjoys the batting of Rahul. Which of the following

must be false?
A)She does not enjoy the batting of Saurav
B)She enjoys the batting of Ricky
C)She may enjoy the batting of Lara
D)She enjoys the batting of Astle
6. Seven students-F, G, H, I, J, K, and L-are being considered for a scholarship. All of them had to take an
examination which was based on their technical knowledge. The scores obtained in the examination were used

for evaluating these students. In analyzing this information regarding the scores, the Principal made the following
observations: None of the students has exactly the same score,as any other student. K's score is higher than L's,
but lower than H's.I has a higher score than J.both F and G have higher scores than H. If J has a higher score

than, how many different rankings, by score, of the students are possible?
A)2 8)1 C)8D)4


Technical


1. binary equalent of 455 ?
2. whats the o/p of the following code ?
int main ()
{
int i = 0 ;
switch ( printf ("k") , printf ("ku"))
{
case 1 : printf (" %d" , i ) ;
break ;
case 2 : printf (" %d " , ++i ) ;
break ;
}
}
ans : a) k ku 1 also ans d) ( both are same )
3. Whats used to connect the computer with terminals ?
ans : a) Data Terminal equipment
4. a question about unix & linux
ans : b) [ unix from Bell labs ]
5 .which is used to convert the assmbly code into binary
ans : assembler
6. whats the o/p of the following code ?
int main ()
{
int i = -1 ; j = 1 , k = 0 , m ;
m = i++ j++ && k++ ;
printf ("\n\n%d %d %d " , i , j , k , m ) ;
}
7. Expansion of CDMA code divison multiple access .
8. A question from Microprocessor ( which segment is not used by Microprocessor ? )
a. ) Data segment b) extended segment c) information segment d) none of these
9. Which of the following layer is not in Tcp/ip ?
a) transport b) application layer c) Network layer d) none of the above
10. If A= 10000011 and B=1010111 and c= 11000011 then whats the string represent 100001100011 .. (some thing

like that )
11. a question regarding PCM
12. the best case comparison of Quick sort ?
13. the usage of static function ?
ans : to access static members
14.ACL stands for
a. access central limit
b.access control limit
c access control list
d.both a&b
15.Choose a valid statement: in c++ "this" is
ans c references the object that has invoked the method
3 buffer registers are found in
a.data bus b main memory c. I/O devices d.cpu
ans c(check)
16 An expression like ((A+B)*(D+E)/f) write the postfix expression
ans d AB+DE+*...
17. The depth of a tree is
ans maximum level of a node
18.IEEE standard of wireless protocol ans 802.11
19. main()
{
i=10;
printf("%d",++i++);
printf("%d",++i);
}
a. 11&13 b.12&13 c.error d.? ans C (check)

20 What does the func return
int fun()
{
i=3;
i=6;
if(i=6)
i=7;
return i;
}
ans 7
21.
main()
{
int a[]={0 1 2 3};
int *b;
b=&a[2];
printf("%d",b[-1]);
}
ans 1(check)
note: the elements of array may vary
Technical Interview
1. What are the layers in OSI stack & Explain them ?
2. What u know about trees ? & tell the types of tree ?
3. Give the syntax of binary tree ?
4. Write the program of factorial using recursion ?
5. Write a Sql Query for ......
6. Some questions from stack & Queue .
7. Write program for Single, doubly linked lists ?
8. Tell about ur future projects & plan ?

HR interview
1. Tell about ur family background .
2. Why do u like to join in Wipro ?
3.Why should I hire u ?
4. Ur achievement ?
5. Do u have any qustion to ask ?
6. . What do u know about wipro ?