Mathematical Reasoning CBSE Questions & Answers
Mathematical Reasoning
This is Mathematics Class 11 Mathematical Reasoning CBSE Questions & Answers. There are 15 questions in this test with each question having around four answer choices.
Questions & Answers
1
The negation of the proposition “if a quadrilateral is a square, then it is a rhombus “ is
- Aif a quadrilateral is a square , then it is not a rhombus
- Ba quadrilateral is not a square and it is a rhombus
- Ca quadrilateral is a square and it is not a rhombusCorrect
- Dif a quadrilateral is not a square , then it is a rhombus
2
The contrapositive of \((p \vee q) \to r\) is
- A\( \sim r \to ( \sim p \wedge \sim q)\)
- B\( \sim r \to \sim (p \vee q)\)Correct
- C\(p \to (p \wedge q)\)
- D\( \sim r \to (p \wedge q)\)
3
The contrapositive of \(p \to ( \sim q \to \sim r)\) is
- A\(( \sim q \wedge r) \to \sim p\)Correct
- B\((q \wedge \sim r) \to \sim p\)
- C\((q \wedge \sim r) \to p\)
- D\((q \vee \sim r) \vee p\)
4
The contrapositive of the statement “ if \({2^2} = 5,\) then I get first class” is
- AIf I do not get a first class , then \({2^2} = 5,\)
- BIf I do not get a first class , then \({2^2} \ne 5,\)Correct
- CIf I get a first class , then \({2^2} = 5,\)
- Dnone of these
5
If x = 5 and y = - 2 , then x – 2y = 9 . The contrapositive of this proposition is
- AIf x – 2y = 9 , then x \( \ne \) 5 or y \( \ne \) - 2Correct
- Bx – 2y = 9 iff x = 5 and y = - 2
- CIf x – 2y = 9 , x \( \ne \) 5 and y \( \ne \) - 2
- Dnone of these
6
“The diagonals of a rhombus are perpendicular “. The contrapositive of the above statement is
- AIf the diagonals are not perpendicular, then the figure is a rhombus
- BIf the diagonals are perpendicular, then the figure is a rhombus
- CIf the figure is not a rhombus, then its diagonals are not perpendicular
- DIf the diagonals are not perpendicular, then the figure is not a rhombusCorrect
7
Which of the following statement is a tautology ?
- A\((p \wedge q) \wedge \left( { \sim \left( {p \wedge q} \right)} \right)\)
- B\(( \sim q \wedge p) \wedge (p \wedge \sim p)\)
- C\(( \sim q \wedge p) \wedge q\)
- D\(( \sim q \wedge p) \vee (p \vee \sim p)\)Correct
8
The statement \(p \to (q \to p)\) is equivalent to
- A\(p \to (p \vee q)\)Correct
- B\(p \to (p \to q)\)
- C\(p \to (p \leftrightarrow q)\)
- D\(p \to (p \wedge q)\)
9
The inverse of the proposition \((p \wedge \sim q) \to r\) is
- Anone of these
- B\( \sim r \to \sim p \vee q\)
- C\(r \to p \wedge \sim q\)
- D\(( \sim p \vee q) \to \sim r\)Correct
10
Logical equivalent proposition to the proposition \( \sim \left( {p \vee q} \right)\) is
- A\( \sim p \leftrightarrow \sim q\)
- B\( \sim p \to \sim q\)
- C\( \sim p \wedge \sim q\)Correct
- D\( \sim p \vee \sim q\)
11
Let p and q be two propositions. Then the inverse of the implication \(p \to q\) is
- A\( \sim p \to \sim q\)Correct
- B\( \sim q \to p\)
- C\(p \to \sim q\)
- D\( \sim p \to q\)
12
Let p and q be two propositions. Then the contrapositive of the implication \(p \to q\) is
- A\(p \leftrightarrow q\)
- B\( \sim p \to \sim q\)
- C\( \sim q \to p\)
- D\( \sim q \to \sim p\)Correct
13
Let p and q be two propositions. Then the implication \(p \to q\) is false ,when
- Ap is true and q is falseCorrect
- Bp is false and q is true
- Cboth p and q are false
- Dp is true and q is true
14
Let p and q be two propositions. Then the implication \(p \leftrightarrow \sim q\) is true ,when
- Aboth p and q are false
- Bp is false and q is trueCorrect
- Cp is true and q is true
- Dp is true and q is false
15
For any three propositions p , q , and r , the proposition \(\left( {p \wedge q} \right) \wedge \left( {q \wedge r} \right)\) is true , when
- Ap,q , r are all trueCorrect
- Bp, q are true and r is false
- Cp is true and q, r are false
- Dp, q , r are all false