Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.5 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5

Question 1.

Express the following circuits in the symbolic form of logic and write the input-output table.

(i)

Solution:

Let p : the switch S_{1} is closed

q : the switch S_{2} is closed

r : the switch S_{3} is closed

~p : the switch S_{1}‘ is closed or the switch S_{1}is open

~q : the switch S_{2}‘ is closed or the switch S_{2} is open

~r : the switch S_{3}‘ is closed or the switch S_{3} is open

l : the lamp L is on

(i) The symbolic form of the given circuit is : p ∨ (q ∧ r) = l

l is generally dropped and it can be expressed as : p ∨ (q ∧ r).

(ii)

Solution:

The symbolic form of the given circuit is : (~ p ∧ q) ∨ (p ∧ ~ q).

(iii)

Solution:

The symbolic form of the given circuit is : [p ∧ (~q ∨ r)] ∨ (~q ∧ ~ r).

(iv)

Solution:

The symbolic form of the given circuit is : (p ∨ q) ∧ q ∧ (r ∨ ~p).

(v)

Solution:

The symbolic form of the given circuit is : [p ∨ (~p ∧ ~q)] ∨ (p ∧ q).

(vi)

Solution:

The symbolic form of the given circuit is : (p ∨ q) ∧ (q ∨ r) ∧ (r ∨ p)

Question 2.

Construct the switching circuit of the following :

(i) (~p∧ q) ∨ (p∧ ~r)

Solution:

Let p : the switch S_{1} is closed

q : the switch S_{2} is closed

r : the switch S_{3} is closed

~p : the switch S_{1}‘ is closed or the switch S_{1} is open

~ q : the switch S_{2}‘ is closed or the switch S_{2} is open

~ r : the switch S_{3}‘ is closed or the switch S_{3} is open.

Then the switching circuits corresponding to the given statement patterns are :

(ii) (p∧ q) ∨ [~p ∧ (~q ∨ p ∨ r)]

Solution:

(iii) [(p ∧ r) ∨ (~q ∧ ~r)] ∧ (~p ∧ ~r)

Solution:

(iv) (p ∧ ~q ∧ r) ∨ [p ∧ (~q ∨ ~r)]

Solution:

(v) p ∨ (~p ) ∨ (~q) ∨ (p ∧ q)

Solution:

(vi) (p ∧ q) ∨ (~p) ∨ (p ∧ ~q)

Solution:

Question 3.

Give an alternative equivalent simple circuits for the following circuits :

(i)

Solution:

(i) Let p : the switch S_{1} is closed

q : the switch S_{2} is closed

~ p : the switch S_{1}‘ is closed or the switch Si is open Then the symbolic form of the given circuit is :

p ∧ (~p ∨ q).

Using the laws of logic, we have,

p ∧ (~p ∨ q)

= (p ∧ ~ p) ∨ (p ∧ q) …(By Distributive Law)

= F ∨ (p ∧ q) … (By Complement Law)

= p ∧ q… (By Identity Law)

Hence, the alternative equivalent simple circuit is :

(ii)

Let p : the switch S_{1} is closed

q : the switch S_{2} is closed

r : the switch S_{3} is closed

~q : the switch S_{2}‘ is closed or the switch S_{2} is open

~r : the switch S_{3}‘ is closed or the switch S_{3} is open.

Then the symbolic form of the given circuit is :

[p ∧ (q ∨ r)] ∨ (~r ∧ ~q ∧ p).

Using the laws of logic, we have

[p ∧ (q ∨ r)] ∨ (~r ∧ ~q ∧ p)

≡ [p ∧ (q ∨ r)] ∨ [ ~(r ∨ q) ∧ p] …. (By De Morgan’s Law)

≡ [p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)] … (By Commutative Law)

≡ p ∧ [(q ∨ r) ∨ ~(q ∨ r)) … (By Distributive Law)

≡ p ∧ T … (By Complement Law)

≡ p … (By Identity Law)

Hence, the alternative equivalent simple circuit is :

Question 4.

Write the symbolic form of the following switching circuits construct its switching table and interpret it.

i)

Solution:

Let p : the switch S_{1} is closed

q : the switch S_{2} is closed

~p : the switch S_{1}‘ is closed or the switch S_{1} is open

~ q : the switch S_{2}‘ is closed or the switch S_{2} is open.

Then the symbolic form of the given circuit is :

(p ∨ ~q) ∨ (~p ∧ q)

Since the final column contains all’ 1′, the lamp will always glow irrespective of the status of switches.

ii)

Solution:

Let p : the switch S_{1} is closed

q : the switch S_{2} is closed

~p : the switch S_{1} is closed or the switch S_{1} is open.

~q : the switch S_{2}‘ is closed or the switch S_{2} is open.

Then the symbolic form of the given circuit is : p ∨ (~p ∧ ~q) ∨ (p ∧ q)

Since the final column contains ‘0’ when p is 0 and q is ‘1’, otherwise it contains ‘1′.

Hence, the lamp will not glow when S_{1} is OFF and S_{2} is ON, otherwise the lamp will glow.

iii)

Solution:

Let p : the switch S_{1} is closed

q : the switch S_{2} is closed

r : the switch S_{3} is closed

~q : the switch S_{2}‘ is closed or the switch S_{2} is open

~r: the switch S_{3}‘ is closed or the switch S_{3} is open.

Then the symbolic form of the given circuit is : [p ∨ (~q) ∨ r)] ∧ [p ∨ (q ∧ r)]

From the switching table, the ‘final column’ and the column of p are identical. Hence, the lamp will glow which S_{1} is ‘ON’.

Question 5.

Obtain the simple logical expression of the following. Draw the corresponding switching circuit.

(i) p ∨ (q ∧ ~ q)

Solution:

Using the laws of logic, we have, p ∨ (q ∧ ~q)

≡ p ∨ F … (By Complement Law)

≡ p … (By Identity Law)

Hence, the simple logical expression of the given expression is p.

Let p : the switch S_{1} is closed

Then the corresponding switching circuit is :

(ii) (~p ∧ q) ∨ (~p ∧ ~q) ∨ (p ∧ ~q)]

Solution:

Using the laws of logic, we have,

(~p ∧ q) ∨ (~p ∨ ~q) ∨ (p ∧ ~q)

≡ [~p ∧ (q ∨ ~q)] ∨ (p ∧ ~ q)… (By Distributive Law)

≡ (~p ∧ T) ∨ (p ∧ ~q) … (By Complement Law)

≡ ~p ∨ (p ∧ ~q) … (By Identity Law)

≡ (~p ∨ p) ∧ (~p ∧~q) … (By Distributive Law)

≡ T ∧ (~p ∧ ~q) … (By Complement Law)

≡ ~p ∨ ~q … (By Identity Law)

Hence, the simple logical expression of the given expression is ~ p ∨ ~q.

Let p : the switch S_{1} is closed

q : the switch S_{2} is closed

~ p : the switch S_{1}‘ is closed or the switch S_{1} is open

~ q : the switch S_{2}‘ is closed or the switch S_{2} is open,

Then the corresponding switching circuit is :

(iii) [p (∨ (~q) ∨ ~r)] ∧ (p ∨ (q ∧ r)

Solution:

Using the laws of logic, we have,

[p ∨ (~ (q) ∨ (~r)] ∧ [p ∨ (q ∧ r)]

= [p ∨ { ~(q ∧ r)}] ∧ [p ∨ (q ∧ r)] … (By De Morgan’s Law)

= p ∨ [~(q ∧ r) ∧ (q ∧ r) ] … (By Distributive Law)

= p ∨ F … (By Complement Law)

= p … (By Identity Law)

Hence, the simple logical expression of the given expression is p.

Let p : the switch S_{1} is closed

Then the corresponding switching circuit is :

(iv) (p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) ∨ (p ∧ q ∧ r)

Question is Modified

(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r)∨ (p ∧ q ∧ r)

Solution:

Using the laws of logic, we have,

(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r)

= (p ∧ ~p ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Commutative Law)

= (F ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Complement Law)

= F ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Identity Law)

= (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Identity Law)

= (~ p ∨ p) ∧ (q ∧ r) … (By Distributive Law)

= T ∧ (q ∧ r) … (By Complement Law)

= q ∧ r … (By Identity Law)

Hence, the simple logical expression of the given expression is q ∧ r.

Let q : the switch S_{2} is closed

r : the switch S_{3} is closed.

Then the corresponding switching circuit is :