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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.

Q1. There are five different green dyes, four different blue dyes and three different red dyes. The total number of combinations of dyes that can be chosen taking at least one green and one blue dye is
•  3255
•  212
•  3760
•  3720
Solution
(c) The total number of combinations which can be formed of five different green dyes, taking one or more of them is 25-1=31. Similarly, by taking one or more of four different red dyes 24-1=15 combinations can be formed. The number of combinations which can be formed of three different red dyes, taking none, one or more of them is 23=8 Hence, the required number of combinations of dyes =31×15×8=3720

Q2.If 1/( 4Cn )=1/( 5Cn )+1/( 6Cn ), then n is equal to
•  3
•  2
•  1
•  0
Solution
1/( 4Cn )=1/( 5Cn )+1/( 6Cn ) ⇒ (n!(4-n)!)/4!=(n!(5-n)!)/5!+(n!(6-n)!)/6! ⇒ ((4-n)!)/4!=((4-n)!(5-n))/(5×4!)+((6-n)(5-n)(4-n)!)/(6×5×4!) ⇒ 1=(5-n)/5+((6-n)(5-n))/(6×5) ⇒ n2-17n+30=0 ⇒ (n-15)(n-2)=0 ⇒ n=2 [∵ 4Cn is not meaningful for n=15]

Q3.  The number of ways in which 52 cards can be divided into 4 sets, three of them having 17 cards each and the fourth one having just one card
•   (52 !)/((17 !)3)
•  (51 !)/((17 !)3 3 !)
•  (51 !)/((17 !)3)
•  52!/((17 !)3 3 !)
Solution
Here, we have to divide 52 cards into 4 sets, three of them having 17 cards each and the fourth one having just one card. First we divide 52 cards into two groups of 1 card and 51 cards. This can be done in (52 !)/(1 ! 51 !) ways Now, every group of 51 cards can be divided into 3 groups of 17 each in (51 !)/((17 !)3 3 !) Hence, the required number of ways =(52 !)/(1 !51 !)∙(51 !)/((17 !)3 3 !)=(52 !)/((17 !)3 3 !)

Q4. If the total number of m elements subsets of the set A={a1,a2,a3,…an } is λ times the number of 3 elements subsets containing a4,then n is
•  (m-1)λ
•  0
•  (m+1)λ
•
Solution
(d) Total number of m-elements subsetcs of A= nCm …(i) and number of m-elements subsets of A each containing the element a4= n-1Cm-1 According to question, nCm=λ.n-1Cm-1 ⇒ n/m .n-1Cm-1=λ.n-1Cm-1 ⇒ λ=n/m or n=mλ

Q5. Consider the following statement: 1.The number of ways of arranging m different things taken all at a time in which p≤m perticular things are never together is m!-(m-p+1)!p! 2. A pack of 52 cards can be divided equally among four players in order in 52!/(13!4) ways Which of these is/are correct?
•  Both of these
•  Only(2)
•  Only(1)
•  None of these
Solution
(c) (1) Total number of ways of arranging m things =m! To find the number of ways in which p particular things are together, we consider p particular thing as a group. ∴ Number of ways in which p particular things are together =(m-p+1)!p! So, number of ways in which p particular things are not together =m!-(m-p+1)!p! (2) Each player shall receive 13 cards. Total number of ways =52!/(13!4) Hence, both statements are correct

Q6. Ramesh has 6 friends. In how many ways can he invite one or more of them at a dinner?
•  61
•  63
• 62
•  64
Solution
(b) Required number = 6C1+ 6C2+ 6C3+ 6C4+ 6C5+ 6C6=26-1=63

Q7. Let l1 and l2 be two lines intersecting at P. If A1,B1,C1 are points on l1 and A2,B2,C2,D2,E2 are points on l2 and if none of these coincides with P, then the number of triangles formed by these eight points, is
•  56
•  55
•  45
•  46
Solution
(c) If triangle is formed including point 'P' the other points must be one from l1 and other point from l2. Number of triangle formed with P=3C1×5C1=15 ways When P is not included. Number of triangle formed =3C2×5C1+3C1×5C2=15+15=30 Total number of triangles=15+30=45

Q8. The greatest possible number of points of intersection of 8 straight lines and 4 circle is
•  32
•  104
•  76
•  64
Solution
(b) The required number of points = 8C2×1+ 4C2×2+( 8C1× 4C1)×2 =28+12+32×2=104

Q9. The number of ways choosing a committee of 4 woman and 5 men from 10 women and 9 men, if Mr. A refuses to serve on the committee when Ms. B is a member of the committee, is
•  22030
•  21000
•  21580
•  20580
Solution
(d) The number of ways of choosing a committee if there is no restriction is 10C49C5=(10!)(4 !6 !)(9!)(4 !5 !)=26460 The number of ways of choosing the committee if both Mr.A and Ms. B are included in the committee is 9C38C4=5880 Therefore, the number of ways of choosing the committee when Mr. A and Ms. B are not together =26480-5880=20580

Q10. The products of any r consecutive natural numbers is always divisible by
•  r2
•  r!
•  rn
• None of these
Solution
(b) The product of r consecutive natural numbers =1.2.3.4…..r=r! The natural number will divided by r! ## Want to know more

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