If 11011_2+11101_2+1111_2+1001_2= \X_10 then X=

TS ICET 23 May 2018 Shift 2 Paper

Question : 109 of 200

Marks: +1, -0

11011_2+11101_2+1111_2+1001_2

\X_10

then X=

Solution:

x=11011_{2}+11101_{2}+1111_{2}+1001_{2}

11011_{2}=(1 × 2^{4})+(1 × 2^{3})+(0 × 2^{2})+(1 × 2^{1})+(1 × 2^{0})=27

11101_{2}=(1 × 2^{4})+(1 × 2^{3})+(1 × 2^{2})+(0 × 2^{1})+(1 × 2^{0})=29

1111_{2}=(1 × 2^{3})+(1 × 2^{2})+(1 × 2^{1})+(1 × 2^{0})=15

1001_{2}=(1 × 2^{3})+(0 × 2^{2})+(0 × 2^{1})+(1 × 2^{0})=9

X = 27 + 29 + 15 + 9 = 80

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