Practice Questions

Q63.The total number of three-digit numbers, divisible by 3, which can be formed using the digits 1, 3, 5, 8, if repetition of digits is allowed, is (1) 21 (2) 20 (3) 22 (4) 18

Q63.Let s1, s2, s3. . . . , s10 respectively be the sum of 12 terms of 10 A. Ps whose first terms are 1, 2, 3, . . . . , 10 and the common differences are 1, 3, 5, . . . , 19 respectively. Then ∑10i=1 si is equal to (1) 7220 (2) 7360 (3) 7260 (4) 7380

Q63.The value of ∑𝑟=22 0 22𝐶𝑟· 23𝐶𝑟 is (1) 45𝐶23 (2) 44𝐶23 (3) 45𝐶24 (4) 44𝐶22

Q63.The number of triplets 𝑥, 𝑦, 𝑧 where 𝑥, 𝑦, 𝑧 are distinct non negative integers satisfying 𝑥+ 𝑦+ 𝑧= 15, is (1) 80 (2) 136 (3) 114 (4) 92

Q63.If the coefficient of 𝑥15 in the expansion of 𝑎𝑥3 + 1 is equal to the coefficient of 𝑥-15 in the expansion of 𝑏𝑥 3 1 15 1 𝑎𝑥 3 - , where 𝑎 and 𝑏 are positive real numbers, then for each such ordered pair 𝑎, 𝑏: 𝑏𝑥3 (1) 𝑎= 𝑏 (2) 𝑎𝑏= 1 (3) 𝑎= 3𝑏 (4) 𝑎𝑏= 3

Q63.The number of integers, greater than 7000 that can be formed, using the digits 3, 5, 6, 7, 8 without repetition is (1) 120 (2) 168 (3) 220 (4) 48 13+23+33......upto n terms

Q63.If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is 9 (1) 7 (2) 2 (3) 3 (4) 14

Q63.The letters of the word OUGHT are written in all possible ways and these words are arranged as in a dictionary, in a series. Then the serial number of the word TOUGH is : (1) 89 (2) 84 (3) 86 (4) 79

Q63.The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1, 3, 5, 7, 9 without repetition, is (1) 6 (2) 12 (3) 120 (4) 72

Q64.Let an be nth term of the series 5 + 8 + 14 + 23 + 35 + 50+. . . . . . .and Sn = ∑nk=1 ak . Then S30 −a40 is equal to (1) 11310 (2) 11260 (3) 11290 (4) 11280

Q64.The number of 4 -letter words, with or without meaning, each consisting of 2 vowels and 2 consonants, which can be formed from the letters of the word UNIVERSE without repetition is _____.

Q64.The value of 1 1 1 1 1 + + + … . + + is 1!50! 3!48! 5!46! 49!2! 51!1! (1) 250 (2) 250 50! 51! (3) 251 (4) 251 51! 50!

Q64.If n 1⋅3+2⋅5+3⋅7+....upto terms = 95 then the value of n is α is equal to

Q64.Five digit numbers are formed using the digits 1, 2, 3, 5, 7 with repetitions and are written in descending order with serial numbers. For example, the number 77777 has serial number 1 . Then the serial number of 35337 is

Q64.The sum to 20 terms of the series 2 ⋅22 −32 + 2 ⋅42 −52 + 2 ⋅62−. . . . . . . . . . . . is equal to __________.

Q64.The total number of six digit numbers, formed using the digits 4, 5, 9 only and divisible by 6, is _____ .

Q64.The coefficient of 𝑥301 in 1 + 𝑥500 + 𝑥1 + 𝑥499 + 𝑥21 + 𝑥498 + … . . + 𝑥500 is: (1) 501𝐶302 (2) 500𝐶301 (3) 500𝐶300 (4) 501𝐶200 1 1 1

Q64.Let 𝑎1, 𝑎2, 𝑎3, … . be a G.P. of increasing positive numbers. Let the sum of its 6th and 8th terms be 2 and the + 𝑎4𝑎4 + 𝑎6 is equal to product of its 3rd and 5th terms be 19. Then 6𝑎2 (1) 3 (2) 3√3 (3) 2 (4) 2√2

Q64.The number of ways, in which 5 girls and 7 boys can be seated at a round table so that no two girls sit together is (1) 720 (2) 126(5!)2 (3) 7(360)2 (4) 7(720)2

Q64.Let A1 and A2 be two arithmetic means and G1, G2 and G3 be three geometric means of two distinct positive numbers. Then G41 + G42 + G43 + G21G23 is equal to (1) (A1 + A2)2G1G3 (2) 2(A1 + A2)G1G3 (3) (A1 + A2)G21G23 (4) 2(A1 + A2)G21G23

Q64.If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of 4√2 + 4 1 is √3 √6: 1, then the third term from the beginning is: (1) 30√2 (2) 30√3 (3) 60√2 (4) 60√3

Q64.Let the number ( 22 2022 + ( 2022 22 leave the remainder α when divided by 3 and β when divided by 7 ) ) . Then (α2 + β2 ) is equal to (1) 20 (2) 13 (3) 5 (4) 10

Q64.Let a tangent to the curve 𝑦2 = 24𝑥 meet the curve 𝑥𝑦 = 2 at the points 𝐴 and 𝐵. Then the mid- points of such line segments 𝐴𝐵 lie on a parabola with the (1) directrix 4𝑥= 3 (2) directrix 4𝑥= - 3 3 (3) Length of latus rectum (4) Length of latus rectum 2 2 Q65. 1 1 1 1 sin2𝑡 𝑡→01lim sin 2𝑡+ 2 sin 2𝑡+ 3 sin 2𝑡. . . . . . 𝑛 sin 2𝑡 is equal to (1) 𝑛2 + 𝑛 (2) 𝑛 𝑛𝑛+ 1 (3) (4) 𝑛2 2

Q65.Let 0 < z < y < x be three real numbers such that x1 , 1y , 1z are in an arithmetic progression and x, √2y, z are in a geometric progression. If xy + yz + zx = 3 xyz, then 3(x + y + z)2 is equal to √2

Q65.The coefficient of 𝑥5 in the expansion of 2𝑥3 - 1 5 is 3𝑥2 (1) 80 (2) 9 9 (3) 8 (4) 26 3