Answer / khamuruddin

05 A PIC S9(4). = 4
05 B PIC XXXBXXX. = 7
05 C PIC ____9.99. = 4
05 D PIC S9(5) COMP-3. = 3
05 E PIC 9(3) COMP. = 2
-------
20
-------

Answer / ezhil_18

@ above

COMP3 data require always requrie n+1/2 bytes. here 5+1/2
takes only 3 bytes not 4 bytes.

Answer / elite group

20 Bytes

Answer / raj

05 A PIC S9(4) -> 4
05 B PIC XXXBXXX -> 7
05 C PIC ____9.99 -> 4
05 D PIC S9(5) COMP-3 -> 3
05 E PIC 9(3) COMP -> 2

SO TOTAL OF 20 BYTES

Answer / vijay

05 A PIC S9(4) -> 5 (BY CONSIDERING ONE SIGN BIT)

05 B PIC XXXBXXX -> 7 ( AS YOU ALL KNOWS ABOUT THIS)

05 C PIC ____9.99 -> 4 (SPACE AND 9V99=9(3))I THINK

05 D PIC S9(5) COMP-3 -> 3 ((5/2)+0.5)FOR ODD NUM WE HAVE
CONSIDER ONE 1/2 BYTE ONLY=0.5)

05 E PIC 9(3) COMP -> 2 (9(1) TO 9(4)=HALFWORD=2BYTES)

(9(5) TO 9(9)= WORD= 4 BYTES)
(9(10) TO 9(18)= FULLWORD=8BYTES)

Answer / suresh kumar

sorry to post the last answer, I would like to go for 24
bytes.

Answer / guest

y there is soo much confusion.. plz mention correct ans..
else it will confuse the readers.. :(

Answer / priyanka

I think t will be 25 since

05 A PIC S9(4). - 5 :4 + 1 sign byte
05 B PIC XXXBXXX. - 7
05 C PIC ____9.99. - 8
05 D PIC S9(5) COMP-3. - 3
05 E PIC 9(3) COMP. - 2
__________________________________________
Total = 25 bytes

Answer / madhu

A - 2
B - 7
C - 4
D - 3
E - 2

Total 18 bytes.

Answer / raj

05 A PIC S9(4) -> 4
05 B PIC XXXBXXX -> 7
05 C PIC ____9.99 -> 4
05 D PIC S9(5) COMP-3 -> 4
05 E PIC 9(3) COMP -> 2

SO TOTAL OF 21 BYTES

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