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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.

Q2.A class has 175 students. The following data shows the number of students opting one or
more subjects. Mathematics 100; Physics 70; Chemistry 40; Mathematics and Physics 30;
Mathematics and Chemistry 28; Physics and Chemistry 23; Mathematics, Physics and
Chemistry 18. Hoe many students have offered Mathematics alone?
•  35
•  48
•  60
•  22
Solution
Given, n(M)=100,n(P)=70, n(C)=40 n(M∩P)=30, n(M∩C)=28, n(P∩C)=23 and n(M∩P∩C)=18 ∴n(M∩P^'∩C^' )=n[M∩(P∩C')] =n(M)-n[M∩(P∩C) ] =n(M)-[n(M∩P)+n(M∩C)-n(M∩P∩C)] =100-[30+28-18=60]

Q3.  If R={(a,b):a+b=4} is a relation onN, then R is
•  Reflexive
•  Symmetric
•  AntiSymmetric
•  Transitive
Solution
Clearly, R={(1,3),(3,1),(2,2)} We observe that R is symmetric only

Q4. Let Z denote the set of all integers and A={(a,b):a2+3b2=28, a,b∈Z} and
B={(a,b):a>b,a,b∈Z}. Then, the number of elements in A∩B is
R={(x,x),(y,y),(z,z),(z,x),(z,y)} is
•  2
•  3
•  4
•  6

Q5.Let U be the universal set for sets A and B such that n(A)=200,n(B)=300 and n(A∩B)=100. Then,         n(A'∩B') is equal to 300, provided that n(U) is equal to
•  600
•  700
•  800
•  900
Solution
We have, n(A^'∩B^' )=n((A∪B)') ⇒n(A^'∩B^' )=n(U)-n(A∪B) ⇒n(A^'∩B^' )=n(U)-{n(A)+n(B)-n(A∩B) } ⇒300=n (U)-{200+300-100} ⇒n(U)=700

Q6.From 50 students taking examinations in Mathematics, Physics and Chemistry, 37 passed
Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, at
most 29 passed Mathematics and Chemistry and at most 20 passed Physics and Chemistry.
The largest possible number that could have passed all three examinations is
•  11
•  12
•  13
•  14
Solution
Let M,P and C be the sets of students taking examinations in Mathematics, Physics and Chemistry respectively. We have, n(M∪P∪C)=50,n(M)=37,n(P)=24,n(C)=43 n(M∩P)< 19,n(M∩C)≤29,n(P∩C)≤20 Now, n(M∪P∪C)=n(M)+n(P)+n(C)-n(M∩P) -n(M∩C)-n(P∩C)+n (M∩P∩C) ⇒50=37+24+43-{n(M∩P)+n(M∩C)+n(P∩C)} +n(M∩P∩C) ⇒n(M∩P∩C)=n(M∩P)+n(M∩C)+n(P∩C)-54 ⇒n(M∩P)+n(M∩C)+n(P∩C) =n(M∩P∩C)+54 …(i) Now, n(M∩P)≤19,n(M∩C)≤29,n(P∩C)≤20 ⇒n(M∩P)+n(M∩C)+n(P∩C)≤19+29+20 [Using (i)] ⇒n(M∩P∩C)+54≤68 ⇒n(M∩P∩C)≤14

Q7.For any three sets A1,A2,A3, let B1=A1,B2=A2-A1 and B3=A3-(A1∪A2 ), then which one of the             following statement is always true
•   A1∪A2∪A3⊂B1∪B2∪B3
•   A1∪A2∪A3=B1∪B2∪B3
•   A1∪A2∪A3⊃B1∪B2∪B3
•   All of the above
Solution

Q8.Given n(U)=20,n(A)=12,n(B)=9,n(A∩B)=4, where U is the universal set, A and B are subsets of U,     then n[(A∪B)c] equals to
• 10
•  9
•  11
•  3
Solution
n(A∪B)=n(A)+n(B)-n(A∩B) =12+9-4=17 Hence, n[(AUB)^c]=n(U)-n(A∪B) =20-17=3

Q9.The relation ‘is subset of’ on the power set P(A) of a set A is
•  Symmetric
•  AntiSymmetric
•  Equivalence relation
•  None of these
Solution
The relation is not symmetric, because A⊂B does not imply that B⊂A. But, it is anti-symmetric because A⊂B and B⊂A⇒A=B

Q10. In a battle 70% of the combatants lost one eye, 80% an ear, 75% an arm, 85% a leg, x% lost all the four limbs. The minimum value of x is
•  10
•  12
•  15
•  None of these
Solution
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