Quant Boosters - Hemant Malhotra - Set 10


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Number of Questions - 30
    Topic - Quant Mixed Bag
    Solved ? - Yes
    Source - Elite's Grid Prep Forum


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q1) What is the approximate probability that two integers a and b chosen at random from 0 to 100 are such that a^2 − 15a + 56 > 0 and |a− b| > 60?
    a) 0.15
    b) 0.20
    c) 0.23
    d) 0.25
    e) 0.29


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Total cases = 101 * 101
    a^2 − 15a + 56 > 0
    (a− 7)(a − 8) > 0
    so a< 7 or a > 8
    |a-b| > 60
    (a-b) > 60 or (a-b) < −60

    Case 1: (a-b) > 60
    a=61,b = 0
    a= 62, b =0, 1
    .
    .
    .
    a = 100, b= 0, 1, 2, …, 39

    Case 2: (a-b) < −60
    b= 61,a = 0
    b = 62, a= 0, 1
    For b = 63, a = 0, 1, 2
    For b = 64, a = 0, 1, 2, 3
    For b = 65, a = 0, 1, 2, 3, 4
    For b = 66, a = 0, 1, 2, 3, 4, 5
    For b = 67, a = 0, 1, 2, 3, 4, 5, 6
    For b = 68, a = 0, 1, 2, 3, 4, 5, 6 (a cannot be 7)
    For b = 69, a = 0, 1, 2, 3, 4, 5, 6 (a cannot be 7 and 8)
    For b= 70, a = 0, 1, 2, 3, 4, 5, 6, 9
    For b = 71, a= 0, 1, 2, 3, 4, 5, 6, 9, 10
    .
    .
    For b = 100, a = 0 to 6 and 9 to 39

    so Total number of favourable cases = (1 + 2 + 3 + … + 40) + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 7 + 7 + 8 + … + 38) = 1575
    so probability = 1575/10201 = 0.15


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q2) Two integers, x and y are chosen with replacement out of the set {0,1,2,3,....10). What is the probability that |x - y| > 5


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    x - y > 5 or x - y < -5 so y - x > 5
    now check
    x - y > 5 means x > 5 + y
    so when y = 0 then x > 5 so x = 6 to 10 so 5 cases
    when y = 1 then 4 cases
    when y = 2 then 3 cases
    when y = 3 then 2 cases
    when y = 4 then 1 case
    so 1+ 2 + 3 + 4 + 5 = 5 * 6/2 = 15 cases
    and same cases for y - x > 5 so total 30 cases
    and total =11^2
    so required probability = 30/11^2


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q3) Four numbers are multiplied together. The probability that the product is divisible by 5 or 10 is


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    When dividing a number by 5, the remainders can be 0,1,2,3,4 -> means 5 cases
    so when one number is div by 5 and rest are not then 4c1 * 4^3
    now all div by 5 : 1 case
    3 numbers div by 5 : 4c3 * 4
    2 numbers div by : 4c2 * 4^2
    Now (1 + 4c3 * 4 + 4c2 * 16 + 4c1 * 64)/5^4
    = (1 + 16 + 96 + 256)/625
    =369/625


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q4) What is the probability that there are exactly 5 cars between 2 particular cars if 10 cars are parked in a row?


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    A could be placed at 1st, 2nd, 3rd, 4th place only
    So A has 4 options and A and B can shuffle in 2! Ways and rest 8 can shuffle in 8! Ways
    so prob = 4 * 2! * 8!/10! = 4/45


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q5) The probabilities that a student passes in Mathematics, Physics and Chemistry are m,p and c respectively. Of these subjects a student has a 75% chance of passing in atleast one, a 50% chance of passing in atleast two and a 40% chance of passing in exactly two.Which of the following relations hold true
    a) p + m + c = 27/20
    b) p + m + c = 13/20
    c) pmc = 1/10
    d) both a and c


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    P(A U B U C) = 3/4
    P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C) = 3/4
    P(A ∩ B) + P(B ∩ C) + P(A ∩ C) - 2 * P(A ∩ B ∩ C) = 1/2
    P(A ∩ B) + P (B ∩ C) + P( C ∩ A) - 3 * P(A ∩ B ∩ C) = 2/5
    Solve 2nd and 3rd
    SO P (A ∩ B ∩ C) =1/2 - 2/5 = 1/10
    NOW P(A) * P(B) * P(C) = 1/10
    SO P(A) + P(B) + P(C) - (1/2 + 2/10) + 1/10 = 3/4
    SO P(A) + P(B) + P(C) = 27/20


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q6) When three dice are thrown simultaneously , what is the probability that the sum of the three numbers that turn up is equal to 12 ?
    a) 7/54
    b) 25/216
    c) 28/216
    d) None of the above


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Min value = 1
    so a + b + c = 9
    where a < = 5
    9 + 3 - 1c3 - 1 = 11c2 = 55
    now when a > 5
    so 6 + a' + b + c = 9
    so a' + b + c = 3
    so 5c2 = 10
    so 3 * 10 = 30
    so 55 - 30 = 25
    so prob = 25/216


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q7) A fair coin is tossed 10times. Let x/y be the probability that no two consecutive tosses show tails. where x & y are natural numbers such that HCF(x, y) = 1 find x+y.
    a) 53
    b) 83
    c) 73
    d) none of the above


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    10 things out of which we want to select r things in which no two are consecutive
    10 - r + 1 C r = 11 - r C r
    let
    11c0 + 10c1 + 9c2 + 8c3 + 7c4 + 6c5
    1 + 10 + 36 + 56 + 35 + 6 = 144
    so 144/2^10=9/64


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q8) The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is
    (a) 6
    (b) 7
    (c) 8
    (d) 9


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    at-least 2 = total - (no head + one head))
    probability of gettin no head = 1/2
    and in n times = (1/2)^n
    now prob of getting one head in n turns
    nc1 * (1/2)^n
    so 1-(1/2)^n - n * (1/2)^n > = 0.96
    1-((n+1)) * (1/2)^n > = 0.96
    (n+1) * (1/2)^n < = 0.04
    (n+1) * (1/2)^n < = 1/25
    so n = 8


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q9) If 3 consecutive letters are selected at random from the English alphabet, then the probability that all three are consonants is:
    a) 1/2
    b) 11/24
    c) 5/12
    d) 7/12


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    A , E , I , O , U
    now A will be in (1 consecutive)
    _ _ E or _ E _ or E _ _ ( 3consecutive)
    same in case of I ,O ,U so total consecutives which include vowels = 3 * 4 + 1 = 13
    so not include vowels wll be 24 - 13 = 11
    so OA=11/24


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Q10) A is a set consisting 20 elements . Find probability of selecting two subsets P and Q of set A such that Q = (not P)


Log in to reply
 

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.