Page 149 - FUNDAMENTALS OF COMPUTER

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NPP Number System, Boolean Algebra and Logic Circuits 149

Solution: hc :
(a) The given decimal number 53 is an (a) {X`m J`m Zå~a 53 EH$ nyUmªH$ h¡ Ÿ& AV… Bgo
integer. Therefore divide it by 8, write 8 go ^mJ Xmo Ÿ& eof\$b {bImo VWm A§V _| eof\$bm| H$mo
remainder and read upward :
ZrMo go D$na n‹T>mo&
Remainder eof\$c
8 53 5
8 6 6
0
53
65
Thus, ( ) → ( ) 8
10
(b) The given decimal number is a fraction. (b) Xr JB© g§»`m nyU©V… Am§{eH$ g§»`m h¡ Ÿ& Bgo 8
Now multiply by 8, write integer part and read go JwUm H$amo, nyUmªH$ {bImo VWm nyUmªH$m| H$mo D$na go ZrMo
downward. We can approximate up to three
places of octal point : H$s Amoa n‹T>mo& h_ Am°ŠQ>b nm°BÝQ> Ho$ VrZ ñWmZm| VH$
AZw_mZ bJm gH$Vo h¢Ÿ&
.61 × 8 = 4.88 4
.88 × 8 = 7.04 7
.04 × 8 = 0.32 0
Thus, (.61 ) → (.470 ) 8
10
(c) The given decimal number 14.52 is a (c) Xr JB© g§»`m 14.52 EH$ {_{lV g§»`m h¡
mixed number containing both integer and {Og_| nyUmªH$ d Am§{eH$ ^mJ XmoZm| h¢ Ÿ& Bgr{bE XmoZm|
fractional part. Therefore consider both the parts
separately. First convert integer 14 into octal : {hñgm| H$mo AbJ- AbJ H$aHo$ n[adV©Z H$a| Ÿ& nyUmªH$ 14
H$mo bo …
8 14 6
8 1 1
0

16
14
Thus, ( ) → ( ) 8
10
Now, consider fractional part : A~ Am§{eH$ ^mJ H$mo bo …
. .52 × 8 = 4. 16 4
. 16 × 8 = 1.28 1
.28 × 8 = 2.24 2

52
Thus, ( ) → (.412 ) 8
10
Combining both the results we get : XmoZm| n[aUm_m| H$mo {_bmZo na …
(14.52 ) → (16.412 ) 8
10