# please help me for midterm ON GRADE 11 boolean algebra

Discussion in 'Homework Help' started by joker_tony, May 28, 2008.

1. ### joker_tony Thread Starter New Member

May 18, 2008
3
0
i really need someone to me verify my answer. i will be having my exam next week and iam stuck . please iam really stuck on some of this question.
' this is not

'' this is not, not

1. F = B.(A''. B' . C')

2. F = A. (A''+ B' + C')

3. F = A . (B''.C') + A . (B + C')

4. F = (A'' + B'') + A . B

5. F = A'.B'+C'+A.(B'+C')

6. F= A'. B+ D.(C+D') + A. B' . C' . D

7. F = A. C'.D'+(A''.B'.C') + (A''.B')

8. F = A.B'.C' + A'.B.C.D + A'.B.C'+A.B'C.D

MY ANSWERS

1. F=BA'.BC'
2 F=A.AB.AC'
3. F= AB + AC
4. F=A.B

Last edited: May 28, 2008
2. ### joker_tony Thread Starter New Member

May 18, 2008
3
0
step by step solution

1. AB'+BB'+BC' (use complements laws B.B'=0)
= AB'+BC'

2. AA.AB'.AC' ( Use idempotency A.A= A)
=A.AB'.AC'

3. AB+AC+AB+AC(use idempotency A+A=A)
=AB+AC

4. (A.B)+A.B ( i think this one is wrong)
=A.AB.AB.B
=A.B

Last edited: May 28, 2008
3. ### Caveman Senior Member

Apr 15, 2008
471
1
This is incorrect. B AND B' = 0.

This is incorrect as well.

This is incorrect.
Incorrect.

I'll explain 1, and you should take it from there:
1. F = B.(A''. B' . C')
First, get rid of the silly '' things A'' = A.
1. F = B.(A . B' . C')
Notice that they are all AND operations, so you can just drop the parenthesis.
1. F = B. A''. B' . C'
You can also swap the order (AND is commutative)
1. F = A''. B. B' . C'
B.B' = 0
1. F = A''. 0 . C'
Anything AND 0 = 0
1. F = 0

4. ### joker_tony Thread Starter New Member

May 18, 2008
3
0
y do you have 1. F = B. A''. B' . C'

A'' = A but you have put it back there after you have drop it

you have A''

Last edited: May 28, 2008
5. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
218
There is some helpful material on Boolean Algebra in the AAC ebook section. There are also some worked examples that may be worth reviewing.

hgmjr

6. ### sejan New Member

May 24, 2008
7
0
your no 1 is incorrect,it is 0.
about no 2-it is A+A.B'+A.C'
reduction-A(A"+B'+c')
=AA"+AB'+AC'
A+AB'+AC'
no-3,it seems to be ok although my technique is different,i checked with K-map
no-4-
(A"+B")+A.B
=A+B+A.B
=A(1+B)+B
=A+B (1+X=1)
no 5-
A'B'+C'+A(B'+c')
=A'B'+C'+AB'+AC'
=B'(A+A')+C'(1+A)
=B'+C'

I think you can carry on now-if confused use k-map to check

7. ### Caveman Senior Member

Apr 15, 2008
471
1
Made a mistake there (cut-and-paste laziness), but the result is the same.