REASONING-QUIZ-3

MATHEMATICS REASONING QUIZ-3

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background. 

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. 

Q1. The false statement in the following is

•  p∧(∼p) is a contradiction
•  (p⇒q)⇔(∼q⇒∼p) is a contradiction
•  ∼(∼p)⇔p is a tautology
•  p∨(∼p) is a tautology

Solution
p⇒q is logically equivalent to ∼q⇒∼p ∴ (p⇒q)⇔(∼q⇒∼p) Is a tautology but not a contradiction

Q2.The switching function for switching network is :

•  (x∧y^' )∨(y∧z^' )∨(z∧x^')
•  (x∧y∧z)∨(x^'∧y^'∧z^')
•  (x∨y^' )∧(y∨z^' )∧(z∨x^')
•  None of the above

Solution
∵ Switches x and y^' are connected parallel which is denoted by (x∧y)^') Similarly, y and z^'and z and x^' are also connected parallel Which are denoted by (y∧z^') and (z∧x^') respectively Now, x∧y^',y∧z^' and z∧x^' are connected in series. So, switching function of given network is (x∧y^' )∨(y∧z^' )∨(z∧x^')

Q3.  Some triangles are not isosceles. Identify the Venn diagram

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Solution

Q4. A compound sentence formed by two simple statements p and q using connective 'and' is called

•  Conjunction
•  Disjunction
•  Implication
•  None of these

Q5.Simplify (p∨q)∧(p∨~q)

•  p
•  T
•  F
•  q

Solution
 (p∨q)∧(p∨∼q) =p∨(q∧∼q) (distributive law) =p∨0 (complement law) =p (0 is identify for v)

Q6. The statement ~(p→q) is equivalent to

•  p∧(~q)
•  ~p∧q
•  p∧q
•  ~p∧~q

Solution

Q7.~(~p)↔p is

•  A tautology
•  A contradiction
•  Neither a contradiction nor a tautology
•  None of these

Solution
We have, ∼(∼p)=p ∴∼(∼p)↔p≅p↔p Hence, ∼(∼p)↔p is a tautology

Q8.If p and q are two simple propositions, then p↔~q is true when

•  p and q both are true
•  Both p and q are false
•  p is false and q is true
•  None of these

Solution
From the truth table of p↔q it is evident that p↔q is true when p and q both are true or both are false ∴p↔~q is true when p is false and ~q is false i.e.p is false and q is true

Q9.If p=∆ ABC is equilateral and q=each angle is 60°. Then, symbolic form of statement

•  p∨p
•  p∧q
•  p⇒q
•  p⇔q

Solution
The symbolic form of given statement is p⇔q

Q10. If p,q,r have truth values T,F,T respectively, which of the following is true?

•  (p→q)∧r
•  (p→q)∧∼r
•  (p∧q)∧(p∨r)
•  q→(p∧r)

Solution
Since p is true and q is false ∴p→q has truth value F Statement r has truth value T ∴(p→q)∧r has truth value F. Also, (p→q)∧∼r has truth value F p∧q has truth value F and p∨r has truth value T ∴(p∧q)∧(p∨r) has truth value F As p∧r has truth value T. Therefore, q→(p∧r) has truth value T