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Q) Let the binary sum after BCD addition is stored in K, Z8, Z4, Z2, and Z1. Then the condition for a correction and output carry can be expressed as C =

A) K + Z8 Z4 + Z8 Z2  B)  K + Z8 Z4 + Z4 Z2  C) K + Z8 Z2 + Z8 Z1  D) K + Z4 Z2 + Z2 Z1
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Is it A ?
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Yes. Could you please explain the logic?
+1 Since it will be very time consuming to draw a K-map of 5 variables, you can do it in this way i.e. by dividing Z8Z4Z2Z1 in two groups -> one in which K=0 and other in which K=1.

So for instance, the cell 3 in 2nd K-map corresponds to 1 0011 i.e. KZ8'Z4'Z2Z1.

We don't have any row in the Truth Table for numbers>19. So we put don't care in the cells where Z8Z4Z2Z1>0011. For eg.Z8Z4Z2Z1=0100 i.e.  K Z8Z4Z2Z1=1 0100 = (20)10

we add bcd correction if there is

 a final carry     1001 +  1001 ---------------------   1 0010 here K=1 hence we add 0110 for correction the result of sum of individual bcd digit is greater than 9          0110 +       0011 ------------------------        1010 here the result is 1010 which is 10 in binary but not in bcd. to correct it add 0110

hence for correction to be added either K = 1 or result out of range

for the following cases we'll get out of range

$z_8$$z_4$$z_2$$z_1 1010 1011 11xx which is z_8$$z_2$ + $z_8$$z_4 option A) K + z_8$$z_2$ + $z_8$$z_4$ is correct

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