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****MATHEMATICS REASONING QUIZ-10**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.

**MATHEMATICS REASONING QUIZ-10**

**Q1.**p∧(q∧r) is logically equivalent to

Solution

Clearly, (p∧q)∧r≅p∧(q∧r)

Clearly, (p∧q)∧r≅p∧(q∧r)

**Q2.**Let inputs of p and q be 1 and 0 respectively in electric circuit. Then, output of p∧q is

**Q4.**If p→(q∨r) is false, then the truth values of p,q,r are respectively

Solution

We know that p→q is false when p is true and q is false. So, p→(q∨r) is false when p is true and q∨r is false. But, q∨r is false when q and r both are false Hence, p→(q∨r) is false when p is true and q and r both are false

We know that p→q is false when p is true and q is false. So, p→(q∨r) is false when p is true and q∨r is false. But, q∨r is false when q and r both are false Hence, p→(q∨r) is false when p is true and q and r both are false

**Q5.**Dual of (x∧y)∨(x∧1)=x∧x∨y∧y is

**Q6.**Let R be the set of real numbers and x∈R. Then, x+3=8 is

**Q7.**Let p∧(q∨r)=(p∧q)∨(p∧r). Then, this law is known as

**Q8.**The property ∼(p∧q)≡∼p∨∼q is called

ax+by+bz=0

bx+ay+bz=0

bx+by+az=0

Will have a non-trivial solution, is

**Q9.**The contrapositive of p⇒∼q is

**Q10.**Which of the following is the inverse of the proposition : "If a number is a prime, then it is odd"?

Solution

Let p: A number is a prime q: It is odd Given proposition is p→q its inverse is ~q→~q. ie, If a number is not prime, then it is not odd.

Let p: A number is a prime q: It is odd Given proposition is p→q its inverse is ~q→~q. ie, If a number is not prime, then it is not odd.