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Q:

If the mean of a, b, c is M and ab + bc + ca = 0, then the mean of a2+b2+c23 is :

A) M B) 3 M^2
C) 2 (M/3) D) M^2
 
Answer & Explanation Answer: B) 3 M^2

Explanation:

We have : ( a + b + c) / 3 = M   or (a + b + c) = 3M.

Now. a+b+c2=3M2=9M2 .

 

 

 

<=>a2+b2+c2+2ab+bc+ca=9M2

 

 

 

a2+b2+c2=9M2   ,(ab+bc+ca=0)     

 

 

 

Required mean =a2+b2+c23=9M23=3M2 .

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Q:

A student was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What should be the number in place of x ?

A) 6 B) 7
C) 8 D) 9
 
Answer & Explanation Answer: B) 7

Explanation:

Clearly, we have (3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + x) / 12 = 12.

or  137 + x = 144 or x = 144 - 137 = 7.

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Q:

The average price of 10 books is Rs. 12 while the average price of 8 of these books is Rs. 11.75. Of the remaining two books, if the price of one book is 60% more than the price of the other, what ia the pice of each of these two books ?

A) 16,12 B) 10,16
C) 23,9 D) 8,7
 
Answer & Explanation Answer: B) 10,16

Explanation:

Total pice of the two books = Rs. [(12 x 10) - (11.75 x 8)]

 

                                          = Rs. (120 - 94) = Rs. 26.

 

let the price of one book be Rs. x

 

Then, the price of other book = Rs. (x + 60% of x) = Rs.(x+(3/5)x) = Rs. (8/5)x

 

So,  x+(8/5)x =26 <=> x =10

 

The prices of the two books are Rs. 10 and Rs. 16

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Q:

The average price of three items of furniture is Rs. 15000. If their prices are in the ratio 3:5:7, the price of the cheapest item is :

A) 6000 B) 7000
C) 8888 D) 9000
 
Answer & Explanation Answer: D) 9000

Explanation:

Let their prices be 3x,  5x and 7x.

Then, 3x + 5x + 7x = (15000 * 3) or x = 3000.

Cost of cheapest item = 3x = Rs. 9000.

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Q:

What are the different Adder circuits you studied?

Answer

Adders are generally of five types:


1.Ripple Carry Adder:


The Ripple carry adder(RCA) consists of a building block named Half Adder(HA) which is cascaded to form a Full Adder(FA). These buildingblocks HAs and FAs are also the building blocks of all types of adders.The n full adders are cascaded to form n bit RCA.


The full adder has three input pins(input Ai,input Bi,carryin Ci) and two output pins(Sum and Ci+1).Its equations are:


Sum=Ai^Bi^Ci


Ci+1=Ai.Bi+Bi.Ci+Ai.Ci


 


2. Carry Lookahead Adder:


The Carry Lookahead Adder(CLA) reduces the delay as that in RCA. Let Gi=Ai.Bi, and Pi=Ai^Bi, then Ci+1=Gi+Pi.Ci.


The expressions for Sum and Ci+1 is then defined completely in terms of input pins rather wait for input carry to appear.


 


3. Carry Select Adder:


The carry select adder uses duplicate modules for each combination of input carry(i.e. 1 and 0).The multiplexers then select the appropriate sum and carry output according to the carry output of the preceding stages.


 


4. Carry Skip Adder:


The carry skip adder are as fast as carry lookahead adders which are the fastest adders but its spped decreases to about 20-30% if input operands are 64-bit or more. In these adders we divide the input bit stream into various blocks and make use of two observations:


-if each element of the two bit streams are unequal,i.e. Ai!=Bi than the carry input of the block is equal to the carry input.


-if each element of the two bit streams are equal,i.e. Ai=Bi than the carry input of the block is opposite of the carry input.


 


5. Carry Save Adder:


The carry save adder reduces the addition of three elements into addition of two elements,i.e. if you want sum of nine numbers it reduces it into sum of six numbers. In first step the sum of three numbers is calculated without bothering for the carry.During the second step only carry is calculated which is then added to the generated sum to give the required sum. 

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Subject: Hardware

Q:

What are set up time & hold time constraints? What do they signify? Which one is critical for estimating maximum clock frequency of a circuit?

Answer

Suppose your flip-flop is positive edgetriggered. time for which data should be stable prior to positive edge clock is called setup is called setup time constraint.


Time for which data should be stable after the positive edge of clock is called as hold time constraint.


If any of these constraints are violate d then flip-flop will enter in meta stable state, in which we cannot determine the output of flip-flop.


there are two equation:


1. Tcq + Tcomb > Tskew + Thold


2. Tcq + Tcomb > Tskew + T - Tsetup


Tcq is time delay when data enters the flip flop and data comes at output of flip flop.


Tcomb is the logic delay between two flip flop.


Tskew is the delay of clock to flip flop: suppose there are two flip flop, if clock reaches first to source flip flop and then after some delay to destination flip flop, it is positive skew and if vice versa then negative  skew. 


so if you take 2 eq you will see that setup time is the determining factor of clock's time period.

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Subject: Hardware

Q:

1.Water boils at 100 degree C

2.Water freezes at 0 degree C

Conclusion: At low pressure ,water boils at lower temperature

The conclusion drawn is 

A) True B) false
C) either probably true or probably false D) Irrelavent
 
Answer & Explanation Answer: D) Irrelavent

Explanation:

It is irrelavent

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Q:

The average temperature of the town in the first four days of a month was 58 degrees. The average for the second, third, fourth and fifth days was 60 degrees. If the temperatures of the first and fifth days were in the ratio 7 : 8, then what is the temperature on the fifth day ?

A) 62 degrees B) 64 degrees
C) 65 degrees D) 66 degrees
 
Answer & Explanation Answer: B) 64 degrees

Explanation:

Sum of temperatures on 1st, 2nd, 3rd and 4th days = (58 * 4) = 232 degrees ... (1)

Sum of temperatures on 2nd, 3rd, 4th and 5th days - (60 * 4) = 240 degrees  ....(2)

Subtracting (1) From (2), we get :

Temp, on 5th day  - Temp on 1st day  = 8 degrees.

Let the temperatures on 1st and 5th days be 7x and 8x degrees respectively.

Then, 8x - 7x = 8 or x = 8.

Temperature on the 5th day = 8x = 64 degrees.

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