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JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Morning)

  • question_answer
    In a class of 140 student numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is: [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

    A) 42                    

    B)      102

    C) 1                     

    D)                  38

    Correct Answer: D

    Solution :

    Maths = \[\{2,4,\text{ }6,\text{ }8,\text{ }......\text{ },\text{ }138,\text{ }140\}\] \[\Rightarrow \,\,\,n(M)=70\] Physics = \[\left\{ 3,\text{ }6,\text{ }9,\text{ }.......\text{ }138 \right\}\] \[\Rightarrow \,\,\,\,n\left( P \right)=46\] Maths & physics =\[\{6,12,\text{ }18,\text{ }......\text{ },\text{ }138\]; \[\Rightarrow \,\,\,n\left( M\cap P \right)=23\] Chem. = {5, 10, 15, 20, ...... , 140} \[\Rightarrow \,\,\,n\left( C \right)=28\] Maths & Chem = \[\left\{ 10,\text{ }20,\text{ }.......\text{ }140 \right\}\] \[\Rightarrow \,\,\,\,n\left( M\cap C \right)=14\] Phy & Chem = \[\left\{ 15,\text{ }30,\text{ }.......\text{ }135 \right\}\] \[\Rightarrow \,\,\,\,n\left( P\cap C \right)=9\] All three subject = \[\{30,\text{ }60,\text{ }......\text{ }120;=4\] \[\Rightarrow \,\,\,\,n\left( P\cap C\cap M \right)=4=n\left( \cup  \right)-n(P\cup C\cup M\text{ })\] Number of students without any subject opted \[=140-\left( 70 \right)-46-28+23+14\]\[+9\,\text{-}\,\text{4}=0=42-4=38\]


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