Enroll to XAT 2023 Mocks & Sectionals <\/a><\/p>\n Question 1:\u00a0<\/b>The arithmetic mean of all the distinct numbers that can be obtained by rearranging\u00a0the digits in 1421, including itself, is<\/p>\n a)\u00a02222<\/p>\n b)\u00a02442<\/p>\n c)\u00a02592<\/p>\n d)\u00a03333<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The number of 4-digit numbers possible using 1,1,2, and 4 is\u00a0$\\frac{4!}{2!}=12$<\/p>\n Number of\u00a01’s, 2’s and 4’s in units digits will be in the ratio 2:1:1, i.e. 6 1’s, 3 2’s and 3 4’s.<\/p>\n Sum = 6(1) + 3(2) + 3(4) = 24<\/p>\n Similarly, in tens digit, hundreds digit and thousands digit as well.<\/p>\n Therefore, sum = 24 + 24(10) + 24(100) + 24(1000) = 24(1111)<\/p>\n Mean =\u00a0$\\frac{24\\left(1111\\right)}{12}=2222$<\/p>\n The answer is option A.<\/p>\n Question 2:\u00a0<\/b>The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is<\/p>\n a)\u00a01440<\/p>\n b)\u00a01200<\/p>\n c)\u00a01480<\/p>\n d)\u00a01420<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Case 1: 4-digit numbers<\/p>\n Given digits – 0, 1, 2, 3, 4, 5<\/p>\n _, _, _, _<\/p>\n As the numbers should be greater than 2000, first digit can be 2, 3, 4 and 5.<\/p>\n For remaining digits, we need to arrange 3 digits from the remaining 5 digits, i.e. 5*4*3 = 60 ways<\/p>\n Total number of possible 4-digit numbers = 4*60 = 240<\/p>\n Case 2: 5-digit numbers<\/p>\n _, _, _, _, _<\/p>\n First digit cannot be zero.<\/p>\n Therefore, total number of cases = 5*5*4*3*2 = 600<\/p>\n Case 3: 6-digit numbers<\/p>\n _, _, _, _, _, _<\/p>\n First digit cannot be zero.<\/p>\n Therefore, total number of cases = 5*5*4*3*2*1 = 600<\/p>\n Total number of integers possible = 600 + 600 + 240 = 1440<\/p>\n The answer is option A.<\/p>\n Question 3:\u00a0<\/b>The number of ways of distributing 20 identical balloons among 4 children such that each child gets some balloons but no child gets an odd number of balloons, is<\/p>\n 3)\u00a0Answer:\u00a084<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let the number of balloons each child received be 2a, 2b, 2c and 2d<\/p>\n 2a + 2b + 2c + 2d = 20<\/p>\n a + b + c + d = 10<\/p>\n Each of them should get more than zero balloons.<\/p>\n Therefore, total number of ways =\u00a0$(n-1)_{C_{r-1}}=(10-1)_{C_{4-1}}=9_{C_3}=84$<\/p>\n Question 4:\u00a0<\/b>A bag contains 6 red balls, 11 yellow balls and 5 pink balls. If two balls are drawn at a random from the bag, one after another. What is the probability that the first ball drawn was red and the second ball drawn was yellow in colour?<\/p>\n a)\u00a0$\\frac{3}{14}$<\/p>\n b)\u00a0$\\frac{5}{7}$<\/p>\n c)\u00a0$\\frac{1}{7}$<\/p>\n d)\u00a0$\\frac{1}{14}$<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n There are 6 red balls, 11 yellow balls and 5 pink balls<\/p>\n Total number of balls = 6 + 11 + 5 = 22<\/p>\n Probability of selecting red ball\u00a0=\u00a0$\\frac{6}{22}$<\/p>\n After selecting one ball, there are 21 balls left<\/p>\n Probability of selecting yellow ball\u00a0= $\\frac{11}{21}$<\/p>\n Therefore, probability =\u00a0$\\frac{6}{22}\\times\\ \\frac{11}{21}$ =\u00a0$\\frac{1}{7}$<\/p>\n Answer is option C.<\/p>\n Question 5:\u00a0<\/b>In a bag there are 15 red balls and 10 green balls. Three balls are selected at random. The probability of selecting 2 red balls and 1 green ball is :<\/p>\n a)\u00a0$\\frac{21}{46}$<\/p>\n b)\u00a0$\\frac{25}{117}$<\/p>\n c)\u00a0$\\frac{3}{25}$<\/p>\n d)\u00a0$\\frac{1}{50}$<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of ways of selecting 2 red balls and 1 green ball =\u00a0$15_{C_2}.\\ 10_{C_1}$<\/p>\n Number of ways of selecting 3 balls =\u00a0$25_{C_3}$<\/p>\n Probability of selecting 2 red balls and 1 green ball =\u00a0$\\ \\frac{15_{C_2}.10_{C_1}}{25_{C_3}}$ =\u00a0$\\frac{21}{46}$<\/p>\n Answer is option A.<\/p>\n Take XAT 2023 Mock Tests<\/a><\/p>\n Question 6:\u00a0<\/b>The following five candidates have applied for a position in an organisation: Male 30\u00a0years, Male 32 years, Female 45 years, Female 20 years, and Male 40 years. What is the\u00a0probability that the candidate selected for the position will be either female or over 35\u00a0years?<\/p>\n a)\u00a04\/5<\/p>\n b)\u00a02\/5<\/p>\n c)\u00a01\/5<\/p>\n d)\u00a03\/5<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Female – A Answer is option D.<\/p>\n Question 7:\u00a0<\/b>In a class of 45 students, 25 are girls. In a test, 30 students scored above 90% marks\u00a0and 18 of them are girl students. A student is selected at random. The probability of\u00a0selecting a girl scoring above 90% marks is___________<\/p>\n a)\u00a018\/45<\/p>\n b)\u00a012\/45<\/p>\n c)\u00a018\/30<\/p>\n d)\u00a030\/45<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n It is given, Answer is option A.<\/p>\n Question 8:\u00a0<\/b>In a charity show, tickets numbered consecutively from 101 through 350 are placed in\u00a0a box. What is the probability that a ticket selected randomly will have a number with a\u00a0hundredth digit as two?<\/p>\n a)\u00a0100\/250<\/p>\n b)\u00a099\/249<\/p>\n c)\u00a0100\/249<\/p>\n d)\u00a099\/250<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n 350 = 101 + (n-1)1 Answer is option A.<\/p>\n Question 9:\u00a0<\/b>What is the probability that a leap year has 53 Sundays and 53 Mondays?<\/p>\n a)\u00a02\/7<\/p>\n b)\u00a01<\/p>\n c)\u00a00<\/p>\n d)\u00a01\/7<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number of days in leap year = 366 Answer is option D.<\/p>\n Question 10:\u00a0<\/b>There is a shooting event organised by the SBT youth Club to select the best candidate who will qualify to participate in Eklavya championship. The shooting board is designed by using 4 concentric circles of radii 2 inch, 3 inch, 5 inch and 9 inch. What is the probability that the participant will shoot only in the second ring [from the outside]\u00a0to qualify for Eklavya championship.<\/p>\n a)\u00a0$\\frac{16}{25}$<\/p>\n b)\u00a0$\\frac{20}{81}$<\/p>\n c)\u00a0$\\frac{20}{56}$<\/p>\n d)\u00a0$\\frac{16}{81}$<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The board is as follows.<\/p>\n The required probability =\u00a0$\\frac{\\pi\\left(5\\right)^2-\\pi\\left(3\\right)^2}{\\pi\\left(9\\right)^2}=\\frac{25-9}{81}=\\frac{16}{81}$<\/p>\n Hence, the answer is option D.<\/p>\n
\nabove 35 years – B
\nn(A $\\cup $ B ) = n(A)+n(B) – n (A $ \\cap $ B) = 2 + 2 – 1 = 3
\nProbability =\u00a0$\\frac{3}{5}$<\/p>\n
\nTotal number of students = 45
\nNumber of girls who scored above 90% marks = 18
\nTherefore, probability =\u00a0$\\frac{18}{45}\\ $<\/p>\n
\nn = 250
\nTotal number of tickets = 250
\nThere are 100 tickets(200 to 299) between the tickets 101 and 350 whose hundredth digit is 2.
\nProbability =\u00a0$\\ \\frac{100}{250}$<\/p>\n
\nThere are 2 odd days in a leap year. For a year to have 53 sundays and 53 mondays, 2 odd days should be sunday and monday.
\nThis is only possible when the first day of the year is sunday.
\nTherefore, the probability of the first day is sunday = $\\frac{1}{7}$<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot%20(469).png\")
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