Correct Sum = (36 * 50 + 48 - 23) = 1825.

Correct mean = = 1825/50 = 36.5

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 .

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

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.

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.

TRUTH TABLE FOR HALF ADDER

A    B  SUM  CARRY

0    0    0         0

0    1    1         0

1    0    1         0

1    1    0         1

INPLEMENTATION:

For SUM, The two inputs A and B are given to XOR gate.

For Carry, The two inputs A and B are given to AND gate.

 Clearly, to find the average, we ought to know the number of boys, girls or students in the class, neither of which has been given.

So, the data provided is inadequate.

Total age of 4 members, 10 years ago = (24 x 4) years = 96 years.

Total age of 4 members now = [96 + (10 x 4)] years = 136 years.

Total age of 6 members now = (24 x 6) years = 144 years.

Sum of the ages of 2 children = (144 - 136) years = 8 years.

Let the age of the younger child be x years.

Then, age of the elder child = (x+2) years.

So, x+(x+2) =8 <=> x=3

Age of younger child  =  3 years.