Practice Questions

Q63.If ∑25r=0{(50Cr)(50−rC25−r)} = K(50C25) , then K is equal to (1) 225 (2) 225 −1 (3) 224 (4) (25)2 is 720, is

Q63.The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 + … upto 11th term is: (1) 945 (2) 916 (3) 946 (4) 915

Q63.A committee of 11 member is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then: (1) m = n = 68 (2) n = m –8 (3) m = n = 78 (4) m + n = 68 A is

Q63.If 5, 5r, 5r2 are the lengths of the sides of a triangle, then r can not be equal to: (1) 3 (2) 3 4 2 (3) 5 (4) 7 4 4

Q63.Let S = {1, 2, 3, … . , 100}, then number of non-empty subsets A of S such that the product of elements in A is even is : (1) 2100 −1 (2) 250 + 1 (3) 250(250 −1) (4) 250 −1

Q63.A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to (1) 24 (2) 27 (3) 25 (4) 28

Q63.There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is : (1) 11 (2) 12 (3) 7 (4) 9 is equal to 225K,

Q63.Let z0 be a root of quadratic equation, x2 + x + 1 = 0. If z = 3 + 6iz810 −3iz930 , then arg (z) is equal to: (1) 0 (2) π4 (3) π (4) π 6 3

Q63.If 19 th term of a non-zero A.P. is zero, then its (49th term): (29th term) is: (1) 4: 1 (2) 1: 3 (3) 3: 1 (4) 2: 1 2 n where q is a real number + + … +

Q63.Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys 𝐴 and 𝐵, who refuse to be the members of the same JEE Main 2019 (09 Jan Shift 1) JEE Main Previous Year Paper team, is: (1) 300 (2) 200 (3) 500 (4) 350

Q63.Suppose that 20 pillars of the same height have been erected along the boundary of circular stadium. If the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total number of beams is: (1) 170 (2) 180 (3) 210 (4) 190

Q63.The Number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is: (1) 220 (2) 221 (3) 220 + 1 (4) 220 - 1

Q63.The number of 6 digit number that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by 11 and no digit is repeated is: (1) 36 (2) 60 (3) 72 (4) 48

Q63.All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is (1) 175 (2) 162 (3) 180 (4) 160

Q63.Let a1, a2, … , a10 be a G.P. If a1a3 = 25, then a5a9 equals : (1) 54 (2) 4 (52) (3) 53 (4) 2 (52)

Q64.The number of four-digit numbers strictly greater than 4321 that can be formed using the digit 0,1, 2,3, 4,5 (repetition of digits is allowed) is: (1) 360 (2) 288 (3) 306 (4) 310 JEE Main 2019 (08 Apr Shift 2) JEE Main Previous Year Paper 20 1

Q64.Let Sn = 1 + q + q2 + … . +qn and Tn = 1 + ( q+12 ) ( q+12 ) ( q+12 ) and q ≠1. If 101C1 + 101C2 ⋅S1 + … . +101C101 ⋅S100 = αT100, then α is equal to : (1) 299 (2) 202 (3) 200 (4) 2100

Q64.Let the sum of the first n terms of a non-constant A. P. , a1, a2, a3, . . . . , an be 50n + n(n−7)2 A, where a constant. If d is the common difference of this A. P., then the ordered pair (d, a50) is equal to (1) (50, 50 + 46A) (2) (A, 50 + 45A) (3) (50, 50 + 45A) (4) (A, 50 + 46A) x is

Q64.If a1, a2, a3, . . . . are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P is: (1) 280 (2) 120 (3) 150 (4) 200

Q64.The sum of all natural numbers 𝑛 such that 100 < 𝑛< 200 and 𝐻. 𝐶. 𝐹. 91, 𝑛> 1 is (1) 3203 (2) 3221 (3) 3121 (4) 3303

Q64.The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is (1) 1356 (2) 1365 (3) 1256 (4) 1465 = 21k , then k equals

Q64.The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 27 . 19 Then the common ratio of this series is: JEE Main 2019 (11 Jan Shift 1) JEE Main Previous Year Paper (1) 1 (2) 2 3 3 (3) 2 (4) 4 9 9

Q64.If the sum of the first 15 terms of the series ( 43 ) 3 + (1 12 ) 3 + (2 14 ) 3 + 33 + (3 34 ) 3 + … then K is equal to : (1) 9 (2) 27 (3) 54 (4) 108

Q64.Consider three boxes, each containing 10 balls labelled 1, 2, … . , 10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni , the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is : (1) 240 (2) 82 (3) 120 (4) 164

Q64.If 𝑎, 𝑏 and 𝑐 be three distinct real numbers in G.P. and 𝑎+ 𝑏+ 𝑐= 𝑥𝑏, then 𝑥 cannot be: (1) -3 (2) 2 (3) 4 (4) -2