JEE MAIN 2020 NEWS

NATIONAL LEVEL ONLINE TEST

    JEE Main Study Notes for Mathematical Reasoning: Important Notes,Short-cut Methods and Solved Questions

    Urvashi Singhal Urvashi Singhal
    Exams Prep Master

    Mathematical Reasoning has less weightage in JEE Main Mathematics section as compared to chapters like Coordinate Geometry and Integral Calculus. However, the questions asked from this chapter require lesser logical thinking and are very easy to score. Read JEE Main Mathematics Syllabus

    Around 1-2 questions are generally asked from this chapter which carry a total of 4 marks. Therefore, the total weightage of Mathematical Reasoning in JEE Main is 1-2%. Some important concepts tested under this chapter are Tautology, Fallacy, and various Mathematical Reasoning Laws. 

    Read the article to find notes on Mathematical Reasoning for JEE Main Preparation.

    Must Read:

    What is Mathematical Reasoning?

    What is Mathematical Reasoning?

    • A sentence is considered to be a mathematically reasonable one if it is either true or false, but not both. 
    • A sentence is not obligatory, probing, or exclamatory. 
    • An open statement is a declarative sentence containing variables if it is turned into a statement when the variables are substituted by some definite meaning. 
    • A compound statement is a statement that consists of two statements or more. Each of these declarations is called a compound statement. 
    • The word "and" (^) combines the compound statements to call the resulting statement a conjunction indicated as p q q. 
    • The compound statement with "And" is valid when all of its component statements are correct.
    • A truth table is an overview of truth values of the resulting statements for all the possible assignment of values to the variables displayed in a compound statement.

    The truth table given below displays the truth values of p ∧ q ( p and q) and q ∧ p ( q and p) -

    Truth Table (p ∨ q, q ∨ p)
    pqp ∧ qq ∧ p
    TTTT
    TFFF
    FTFF
    FFFF
    Rule: p ∧ q is true if and when p and q are true.
    • Compound statements p and q are merged through the connective 'OR' (v). Then, the compound statement indicated as p q q is termed as a disjunction. 

    The truth table of p ∨ q, q ∨ p is given below -

    Truth Table (p v q, q v p)
    pqp ∨ qq ∨ p
    TTTT
    TFTT
    FTTT
    FFFF
    Rule: p ∨ q is false if and when both p and q are false.
    • The denial of a sentence is known as the negation of the statement. 

    The truth table of ~p is given below -

    Truth Table (~p)
    p~p~ (~p)
    TFT
    FTF
    Rule: ~ is true if and when p is false.
    • Negation is not considered as a binary operation. It is, in fact, a unary operation, that is, a modifier.
    • There are three kinds of implications -
    • “If ….... then” 
    • “Only if”
    • “If and only if”
    • The “if …. then” kind of compound statement is known as a conditional statement. The statement ‘if p then q’ is expressed as p → q or as p ⇒ q.
    • p → q also indicates -
    • p is enough for q
    • q is required for p
    • p only if q
    • p leads to q
    • q if p
    • q when p
    • if p then q

    The truth table for p → q is given below -

    Truth Table (p → q, q → p)
    pqp → qq → p
    TTTT
    TFFT
    FTTF
    FFTT
    Rule: p → q is false if and when p is true and q is false.
    • If and when the kind of compound statement is known as biconditional, equivalence or double conditional, it is expressed as p ⇔ q or p ↔ q. This means -
    • p is a sufficient and essential condition for q
    • q is a sufficient and essential condition for p
    • If p then q and if q then p
    • q if and only if p

    The truth table for p ↔ q or q ↔ p is given below -

    Truth Table (p ↔ q, q ↔ p)
    pqp ↔ qq ↔ p
    TTTT
    TFFF
    FTFF
    FFTT
    Rule: p ↔ q is true if and when both p and q have the same truth value.
    • The contrapositive of p → q is ~ q → ~ p.
    • The converse of p → q is q → p.

    The truth table for p → q is given below - 

    Truth Table (p → q) 
    pqp → q~q → ~p (Contrapositive)q ↔ p (Converse)
    TTTTT
    TFFFT
    FTTTF
    FFTTT
    • The compound statements that are true for any truth value of their elements are known as tautologies.

    The truth table for a tautology ‘p ∨ ~p’, p being a logical statement is given below - 

    Truth Table (p ∨ ~p)
    p~pp ∨ ~p
    TFT
    FTT
    • The negation of a tautology is known as a fallacy or a contradiction.

    The truth table for ‘p ∧ ~ p” which is a fallacy, p being a logical statement is given below -

    Truth Table (p ∧ ~p)
    p~pp ∧ ~p
    TFF
    FTF

    Must Read:


    Points to Remember for Mathematical Reasoning

    Points to Remember for Mathematical Reasoning

    • p ∨ q is true when at least one among p and q is true.
    • p ∧ q is true when both p and q are true.
    • A tautology is typically true
    • A fallacy is typically false.

    Given below are some laws that are satisfied by statements - 

    • Idempotent Laws: Suppose p is any statement, then, p ∨ p = p and p ∧ p = p.
    • Associative Laws: Suppose p, q, and r are any three statements, then, p ∨ (q ∨ r) = (p ∨ q) ∨ r and p ∧ (q ∧ r) = (p ∧ q) ∧ r.
    • Commutative Laws: Suppose p and q are any two statements, then, p ∨ q = q ∨ p and p ∧ q = q ∧ p.
    • Distributive Laws: Suppose p, q, and r are any three statements, then, p ∧ (q ∨ r) = (p ∧ q) ∨ (p ∧ r) and p ∨ (q ∧ r) = (p ∨ q) ∧ (p ∨ r).
    • Identity Laws: Suppose p is any statement, t is a tautology, and c is a contradiction, then, p ∨ t = t, p ∧ t = p, p ∨ c = p and p ∧ c = c.
    • Complement Laws: Suppose t is a tautology, c is a contradiction, and p is any statement, then, p ∨ (~p) = t, p ∧ (~p) = c, ~t = c and ~c = t.
    • Involution Law: Suppose p is any statement, then, ~(~p) = p.
    • De-Morgan’s Law: Suppose p and q are two statements, then, ~(p ∨ q) = (~p) ∧ (~q) and ~(p ∧ q) = (~p) ∨ (~q).
    • Two compound statements, S1 and S2, are said to duals of one another if one can be obtained from the other through the replacement of ∧ with ∨, or ∨ with ∧, that is, S∗ (p,q)=p∨q.

    Solved Examples on Mathematical Reasoning

    Solved Examples on Mathematical Reasoning

    Question 1: What is the contrapositive of the inverse of p ⇒ ~q?

    Solution: The inverse of p ⇒ ∼q is ∼ p ⇒ q.

    As the contrapositive of p ⇒ q is ∼ q ⇒ p,

    Therefore, the contrapositive of ∼ p ⇒ q is ∼ q ⇒ p. 

    Question 2: Which of the given statements is the contrapositive of, if two triangles are identical, then, these are similar?

    A) if two triangles are not similar, they are not identical

    B) If two triangles are not identical, they these are not similar

    C) If two triangles are not identical, they are similar

    D) If two triangles are not similar, they are identical

    Solution: Look at the given statements -

    p: Two triangles are identical.

    q: Two triangles are similar.

    Evidently, the given statement in representative form is p ⇒ q.

    Thus, its contrapositive is expressed as ∼ q ⇒ ∼ p

    Now,

    ∼p: two triangles are not identical.

    ∼q: two triangles are not similar.

    Therefore, ~ q ⇒ ~ p: If two triangles are not similar, they are not identical.

    Question 3: Suppose p is true and q is false, then which of the statements given below is not true?

    A) p ∨ q

    B) p ⇒ q

    C) p ∧ ( ~q)

    D) p ⇒ p

    Solution: If p is true and q is false, then p∨q is true, q ⇒ p is true and p ∧ (∼q) is true.

    Thus, both p and ∼q are true.

    Here, p ⇒ q is not true as a true statement. 

    Question 4: What is ~( p ∨ q) ∨ (~p ∧ q) is equivalent to?

    Solution: ∼ (p ∨ q) ∨ (∼p ∧ q) = (∼p ∧ ∼q) ∨ (∼p ∧ q) =∼q ∧ ∼(∼ (p ∧ q))

    Question 5: Which of the options given below is logically equivalent to ~( ~ p ⇒ q)?

    A) p ∧ q

    B) p ∧ ~q

    C) ~p ∧ q

    D) ~p ∧ ~q

    Solution: From the table given below, it is clear that ∼ (∼p ⇒ q) is equivalent to ∼p ∧ ∼q.

    Mathematical Reasoning Problems in JEE Main Exam

    Question 6: Suppose (p ∧ ~r) ∧ ( ~p / q) is false, write the truth values of p, q and r.

    Solution:

    Mathematical Reasoning Examples in JEE Main Exam

    Since (p ∧ ∼r) ⇒ (∼p ∨ q) is F.

    Then, p = T, q = F, r = F.

    Question 7: Suppose p and q are two statements, then what is (p ⇒ q) ⇔ ( ~q ⇒ ~p)?

    Solution:

    p⇒q∼p⇒∼qp⇒q⇔∼q⇒∼p
    TTT
    FFT
    TTT
    TTT

    Therefore, the (p ⇒ q) ⇔ ( ~q ⇒ ~p) is a tautology.

    Question 8: Suppose each of the given statements is true, then, P ⇒ ~q, q ⇒ r, ~r

    A) p is false

    B) p is true

    C) q is true

    D) None of these

    Solution: As ∼r is true, r is false. Also, q ⇒ r is true. Thus, q is false.

    Hence, a true statement cannot indicate a false one.

    p ⇒ q is also true.

    Therefore, p is false.

    Question 9: Find the negation of the compound proposition given below. 

    Suppose the examination is difficult, then I will pass if I study hard.

    Solution: Let p: Examination is difficult

    q: I shall pass

    r: I study hard

    Given result, P ⇒ (r ⇒ q)

    Now, ∼ (r ⇒ q) = r ∧ ∼q ∼(p ⇒ (r ⇒ q)) = p ∧ (r ∧ ∼q)

    The examination is difficult and I study hard but I will not pass.

    Question 10: What is the statement p → (p → q) equivalent to?

    Solution: p → (q → p) = −p (q → p) = ∼p ∨ (∼q ∨ p)

    Since p ∨∼p is always true,

     ∼p ∨ p ∨ q = p → (p ∨ q)


    Tricks to Solve Questions from Mathematical Reasoning

    Tricks to Solve Questions from Mathematical Reasoning

    Pro Tip - Use the elimination method whenever you feel stuck in a Mathematical reasoning problem. 

    Question 1: Take a look at the following statements -

    P: Suman is brilliant

    Q: Suman is rich

    R: Suman is honest

    How can the negation of the statement: Suman is brilliant and dishonest if and only if Suman is rich be expressed as?

    Trick to solve this question: Express “Suman is brilliant and dishonest” first. You will get P∧∼R. Then, express “Suman is brilliant and dishonest if and only if Suman is rich” using Q. You will get Q ↔ (P ∧ ∼R). Write the expression of negation from this.

    Solution: ∼(Q ↔ (P ∧ ∼R)

    Question 2: The statement ~(p ↔ ~q)is ________.

    Trick to solve this question: Use the truth table for easily identifying the answer. Find the nature of each element - p, q, and ~q. Then, find p ↔ ~q, ∼(p↔∼q), and p↔q. 

    Solution: 

    pq∼qp↔∼q∼(p↔∼q)p↔q
    TTFFTT
    TFTTFF
    FTFTFF
    FFTFTT

    Question 3: What is the Boolean Expression (p ∧ ~q) ∨ q ∨( ~p ∧ q) equivalent to?

    Trick to solve this question: Write [(p ∧ ∼q) ∨ q] ∨ (∼p ∧ q) and find its equivalent. Continue shortening the equivalent until you arrive at the correct solution. 

    Solution: p ∨ q


    Study Tips for JEE Main Mathematics

    Study Tips for JEE Main Mathematics

    1. Make your own formula book that you can refer to as often as possible. 

    2. Solve a minimum of 20 questions from Mathematics everyday. 

    3. Make use of logical thinking to solve JEE Main Mathematics questions. 

    4. Use NCERT, R. D Sharma, and Arihant books for learning and practice. 

    5. Solve as many mock tests and previous years’ questions as you can.

    The article above comprises all important notes for JEE Main Mathematical Reasoning. Adoption of study tips will help in acing the topic.

    Comments

    Comments



    No Comments To Show

    ×

    Related News & Articles

    July 03, 2020JEE-MAIN 2020

    JEE Main, JEE Advanced and NE ..

    July 03, 2020JEE-MAIN 2020

    HRD Minister Forms Panel To R ..

    July 01, 2020JEE-MAIN 2020

    NTA Looking Forward to Develo ..

    July 02, 2020JEE-MAIN 2020

    JEE Main 2020 July - Will it ..