Quantitative Aptitude - A 0 0 0 0 1 1 1 1B 0 0 1 1 0 0 1 1C 0 1 0 1 0 1 0 1Find ( A u B ) n C

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Quantitative Aptitude | Question 27 of 49

A 0 0 0 0 1 1 1 1

B 0 0 1 1 0 0 1 1

C 0 1 0 1 0 1 0 1

Find ( A u B ) n C

Total Answers and Comments: 8

Last Update: June 29, 2009 Asked by: Purnima

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October 09, 2006 13:00:46

manoj

RE: A 0 0 0 0 1 1 1 1B 0 0 1 1 0 0 1 1

tell me the procedure to solve this problem

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October 10, 2006 00:58:29

Francis

RE: A 0 0 0 0 1 1 1 1B 0 0 1 1 0 0 1 1

Hi I think you can try this way. Consider U (Union) as OR gate and Intersection as AND gate. Then you can come at the solution

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February 01, 2007 23:02:29

jai ailani

RE: A 0 0 0 0 1 1 1 1B 0 0 1 1 0 0 1 1

ans is 00010101

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March 05, 2007 14:14:28

ayush

RE: A 0 0 0 0 1 1 1 1B 0 0 1 1 0 0 1 1

A: 1 1 0 1 1 0 1 1

B: 0 1 1 1 1 0 1 0

by binary sub. a-b 01100001 (1-0 1 1-1 0 0-0 0 now (a-b)uc 01100001

or 01101101

gives 1101101... convert to decimal equals 109

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May 20, 2008 09:42:41

spandu

Member Since: May 2008 Contribution: 4

RE: A 0 0 0 0 1 1 1 1B 0 0 1 1 0 0 1 1C 0 1 0 1 0 1 0 1Find ( A u B ) n C

aub union means add 00001111+00110011 01000010
((aub)nc) intersection means and operation 01000000(64)

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May 27, 2008 05:08:13

chandradeepika

Member Since: May 2008 Contribution: 36

RE: A 0 0 0 0 1 1 1 1B 0 0 1 1 0 0 1 1C 0 1 0 1 0 1 0 1Find ( A u B ) n C

To find A-C We will find 2's compliment of C and them add it with A
That will give us (A-C)
2's compliment of C 1's compliment of C+1
10101010+1 10101011
A-C 10101011+00001111
10111010
Now (A-C) U B is .OR. logic operation on (A-C) and B
10111010.OR . 00110011
The answer is 10111011
Whose decimal equivalent is 187.

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June 26, 2009 10:26:23

sagarpdy

Member Since: June 2009 Contribution: 2

RE: A 0 0 0 0 1 1 1 1B 0 0 1 1 0 0 1 1C 0 1 0 1 0 1 0 1Find ( A u B ) n C

It is one of the representation of sets

i.e.

A {0 1 2 3}

likewise

hence A U B A|B 00111111

(AUB)nC (AUB)&C 00010101

Ans {0 2 4}

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June 29, 2009 06:30:35

Mekaladevi

Member Since: June 2009 Contribution: 2

RE: A 0 0 0 0 1 1 1 1B 0 0 1 1 0 0 1 1C 0 1 0 1 0 1 0 1Find ( A u B ) n C

A u B 00111111
(A u B) n C 00010101

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