RedpineLatest (2011)Test Pattern and latest questions asked
Three Sections
Three sections mentioned comprises total of 25 questions (weightage -not uniform)
and they carry 100 marks, time given will be 90 minutes..........!
Section: 1(ELECTRONICS)
Section: 2(COMPUTING)
Section: 3(APTITUDE)
Time Limit: 90 Minutes
Section:1(ELECTRONICS)
Simple realization of logic gates given, we have 2 find d output?
The maximum value of signed number that can be fit into 2 byte register?
Section:2(COMPUTING)
C,C++
Program to obtain value of ,k, and to obtain ,k, in d given program code u should have d knowledge of modular division
Section:3(APTITUDE)
Topics to concentrate
Equations
Average and age
Train problems
Probablity
Direction sense
Some questions asked
Section:1(ELECTRONICS)
1.Simple realization of logic gates given, we have 2 find d output?
Sol: Have to know the properties of AND,NAND, EX-OR functions
2.The maximum value of signed number that can be fit into 2 byte register?
Sol: We know that byte consists of 8 bits and the left most bit consist of sign hence only seven bits represent the magnitude similarly for 2 byte register d left most bit i.e.,16th bit represent sign hence the max value can be obtained by placing all 1,s in remaining 15 bit positions.
Therefore d value is: (2^15)-1=32767
3. Ideal op-amp sum was given and we have 2 find d output voltage?
Sol: It was difficult to draw d diagram i will explain d procedure so that u can b able 2 interpret d diagram easily
At terminal1: (2-V1)/5=(V1-Vout)/10;
At terminal2: (0-V2)/10=(V2-Vout)/100; as given it is ideal op-amp V1=V2;
By solving d above three equations we can get Vout=-5.5V
4.Given a series of three sources connected in a ckt with load resistor(R) and power delivered by them individually is given by 18W,50W,98W.What is the total power delivered when all the three sources r active?
Sol: E1^2/R=18W;
E2^2/R=50W; Equations formed by interpreting the given data
E3^2/R=98W;
Hence total power delivered when all the 3 sources r active is E^2/R;
where E=E1+E2+E3;(as they r connected in series)
By multiplying 1st&2nd eqs we get (E1*E2)/R=sqrt(18*50)=30W;
By multiplying 2nd&3rd eqs we get (E3*E2)/R=sqrt(98*50)=70W;
By multiplying 1st&3rd eqs we get (E3*E1)/R=sqrt(18*98)=42W;
Therefore total power delivered is:P=(E1^2+E2^2+E3^2+2E1E2+2E2E3+2E1E3)/R
P=18+50+98+2*30+2*70+2*42=68+98+60+140+84=166+200+84
Hence total power delivered P=450W
5.A 26Kbyte memory, there is memory it contains 12 address lines and 4 bit data bus, the number of these type of memories required to design 26Kbyte memory?
Sol: 26K byte=26*1024=26624 bytes
Address lines=12;memory occupied=2^12=4096 bytes
4 bit data bus memory can be neglected as it is very small
Hence 26624 bytes=N*4096 bytes;
=>N=26624/4096=6.5
Hence 7(6.5)type of memories r required to design 26Kbyte memory
6.The no of 2-input XOR gates required to design 19-input XOR gate?
Sol: Lengthy procedure........!
7.This questions was given based on rising n falling edges of a flip flop?
Sol: Have a brief look on the theory of flip flops
8.Conversion of given multiplexers to AND gates
9.Convert d following:
a)73(in decimal) to binary
b)octal to binary
c)decimal to hexadecimal.........!
Sol: To convert decimal number 2 binary divide d given decimal number by 2
To convert decimal number 2 hexadecimal/octal divide d given number by 16/8
To convert binary number 2 decimal multiply d digits with powers of 2
Section:2(COMPUTING)
10.Simple C programs(3 questions were given)
11.Program to obtain value of ,k, and to obtain ,k, in d given program code u should have d knowledge of modular division
Eg: 16 mod 7=remainder obtained when 16 is divided by 7 i.e.,2
12.A(m,n)=n+1,if m=0;
=A(m-1,A(m,n)),if m>0,n=0;
=A(m-1,A(n,1)),if m>0,n>0; then find A(2,2)
Sol: Looks simpler but takes lot of time 2 answer its based on mainly RECURSIVE function used in C language
Section:3(APTITUDE)
13.Six monkeys take 4 minutes to eat 6 bananas
a)How many minutes r required for 2 monkeys to eat 2 bananas
b)How many monkeys r required to eat 15 bananas in 15 minutes
Sol:a)From given data we can interpret as follows:
1 monkey in 1 minute eats 6/(4*6)=1/4th of a banana--------eq1
Therefore 1 monkey takes 8 minutes to eat 2 bananas which means that 2 monkeys can eat 2 bananas in just 4 minutes
b)Let us assume that ,N, monkeys r required to eat 15 bananas in 15 minutes
Therefore 1 monkey in 1 minute can eat 15/(15*N)=1/Nth of a banana-------eq2
By equating 1 and 2 we get 1/N=1/4;
=> N=4;Hence no.of monkeys required are 4
14.A man walking along a railway bridge heard d whistle sound of a train when he already 5/13th distance of a bridge. Then he runs n can be escaped from making accident with d train. If he had ran back from there to starting point he could be escaped. If the velocity of man is 12kmph.What is velocity of man?
Sol: By given data it is clear that man has to cover total distance of d bridge and 5/13th
distance of d bridge as he walked back. Let d distance of d bridge=,x,
Hence total distance man has to cover=x+(5/13)x=(18/13)x
But train requires only ,d, distance to cover and also time taken by both must b same
=> Velocity of train=x/t;
Velocity of man=(18/13)x/t=12kmph;
=>x/t=(12*13)/18=26/3kmph
15.In a party 12 members had a board meeting and hands were shaken between them before and after d party. Therefore total no.of handshakes possible?
Sol:1 st person can shook his hand with other 11 persons=>handshakes possible=11
2nd person can shook his hand with other 10 persons=>handshakes possible=10
11th person can shake hand with 1 person only=>handshakes possible=1
Hence total no.of handshakes possible=11+10+9+8+.......+1=66
But hands had done it twice their work in d party. Therefore total no.of handshakes possible=2*66=132
16.The average age of 10 members of a given committee= average age of 10 members 4 yrs. ago because older member is replaced by a younger member. What is d age difference?
Sol: Present average of ages=(a1+a2+......+x)/10-------eq1;
Average of ages before 4 yrs ago=(a1+a2....-9*4+y)/10-----eq2;
By equating eqs 1&2 we get
x=y-36;
y-x=36=>the age difference is 36 yrs........!
17.Abbreviations from CN like.....
Sol: CDMA-Code division multiple access
FTP- File transfer protocol
IEEE-Institute for electrical and electronics engineers
I have covered d total test paper and make practice of these type of questions for getting into RPS(Red Pine Signals) wishing u all d best for your future........!
NOTE: These are the solutions written by me in my test hope they will be helpful for u I will be happy if any one of my friend get benefitted by this.....!
Digital:
It includes both STLD ,VHDL and Microprocessors.
1)Design 3:1 multiplexer using one tri-state buffer, AND gates and NOT gates.
2)The no of 2-input XOR gates required to design 19-inprt XOR gate?
3)A 26Kbyte memory, there is memory it contains 12 address lines and 4 bit data bus, the number of these type of memories required to design 26Kbyte memory?
4)Write a VHDL or Verilog HDL code for
input:a,clock,reset
output :out is assigned to 1 when a is ,1, for two clock cycles.
5)what is the output of following fig.100ps is the delay for XOR gate and 50ps for AND gate .all +ve and -ve edges start at boundaries of nanoseconds.(actually the output of fig is A(B(notC)+(not B)C) ,and the waveforms are given).
6)design forwhich the output is 10MHz clock,input to that circuit is 30MHz.Communications:
7) What is shorans theorem?
8)X is Gaussianly distributed signal
a)p(Xb)
swap(a,b);
printf("a=%d,b= %d",a,b);
}
a)a=5 b=6 b)a=6 b=5 c)a=0 b=6 d)None Ans : a a=5 b=6
3)what is output for the following program.
#include
main()
{
unsigned char i;
for( i=0;i
