Quant Boosters - Swetabh Kumar - Set 1


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    (6-a)+(6-b)+(6-c) >15
    a+b+c < 3
    a+b+c=0 : 1 case
    a+b+c=1 : 3 cases
    a+b+c=2: 4C2=6 cases
    so 1+3+6=10 cases. so 10/216 = 5/108

    After a+b+c < 3, you can also use dummy variable to get rid of inequality.
    a+b+c+d=3 But min. d should be 1
    so a+b+c+(d+1) =3 so a+b+c+d=2 so 5C3 = 10 cases. so 10/216 = 5/108.


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    Q5) If the natural numbers starting from 1 are written one after the other to form a 121-digit number, then what will be the last digit of the resultant number?


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    1-9: 9 digits.
    112 more digits.
    so (112/2) = 56 numbers.
    so 10 to 65. so ends in 5


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    Q6) Find number of real valued satisfying [x] + 3 = 2x


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    2 values. (2.5 and 3)
    [x]+3= 2[x] + 2{x}
    0-2
    [x]>1 so [x]=2, 3.
    when [x]=2 5=4+2{x} so {x}=0.5 so 2.5
    when [x]=3 {x}=0 so 3


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    Q7) Find the number of zeroes at the end of 11^( (5!)! ) - 1


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    11^(120!) -1
    (10+1)^(120!)-1
    last term of binomial = 120! * 10
    number of zeroes = 28 + 1 = 29


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    Q8) A nine digit number N is formed using 0 to 9 at-most once.what's the probability that the no is divisible by 45


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    div. by 45 means div. by 5 and by 9. so unit digit is 0 or 5.
    case 1: unit digit 5: so rest 8 digits can be 1 to 8 so 8! (to make overall digit sum=9k) or they can be 0 to 8 so 8!-7! (1st digit cannot be 0 so subtracting the 7! cases)
    so 2 * 8!-7! = 15 * 7! here
    case 2: unit digit 0 : so rest 8 digits can only be 1 to 8 to give an overall 9k digit sum...so 8! here or 8 * 7!
    so total favourable= 15 * 7! +8 * 7! = 23 * 7!
    total cases = 9 * 9 * ... 2 * 1 = 9 * 9! = 9 * 9 * 8 * 7!=648 * 7!
    so probability = 23/648


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    Q9) Number of ways of distributing 20 chocolate among 4 boys so that each boy received atleast one chocolate and the number of chocolate with each boy is different


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    a+b+c+d = 20 so 19C3 = 969 total
    ONLY 2 equal:
    if a=b then 2a+c+d=20
    a=1 so c+d=18 so (2,16)...(16,2) so 14
    a=2 c+d=16 so (1,15)., (3,13)...(13,3),(15,1)..so 12
    a=3 c+d=14 so (1,13), (2,12)....(13,1) so 10
    a=4 c+d=12 so (1,11)...(11,1) so 8
    a=5 c+d=10 so (1,9)..(9,1) so 8
    a=6 c+d=8 so (1,7),..(8,8,) : 5
    a=7 c+d=6 so (1,5), (2,4),(4,2),(5,1) so 4
    a=8 c+d = 4 so (1,3),(3,1) so 2
    a=9 c+d= 2 0 cases here.
    so cases here: 63
    hence 66 * 4C2
    ONLY 3 equal:
    3a+d = 20 so (1,17)...(6,2) so 5 cases (excluding 555) so 4C3 * 5
    all 4 equal: (5555) so 1 case so 1 * 4C4=1
    so (969-66 * 4C2 - 4C3 * 5-1) ~ 552


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    Q10) What is the sum of all three digit palindromes which are multiples of 13 ?


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    aba
    101a+10b
    10a+10b (since 101=10 mod 13)
    10(a+b) so a+b =13
    so 494, 585, 676, 767, 858, 949 so sum=4329


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    Q11) Three biased coins A, B and C are such that the probability of getting a head when A is tossed is 1/4, the probability of getting a head when B is tossed is 3/4 and the probability of getting a head when C is tossed is 2/3. The three coins are tossed. If it is found that there are two heads and one tail, what is the probability that coin A shows a head?


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    (1/4 * 3/4 * 1/3 + 1/4 * 1/4 * 2/3) = 5/48
    (3/4 * 2/3 * 3/4 + 3/4 * 1/4 * 1/3 + 1/4 * 2/3 * 1/4)
    = 23/48 so 5/23


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    Q12) In how many ways can seven balls of different colours be put into 4 identical boxes such that none of the box is empty?


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    (4,1,1,1): 7C4 * 3 * 2 * 4 = 840
    (3,1,1,2): 7C3 * 4C2 * 2C1 * 4!/2! = 5040
    (2,2,2,1): 7C2 * 5C2 * 3C2 * 4 = 2520
    so(840+5040+2520)/4! = 350


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    Q13) A team is planning to participate in a Dahi-handi competition. On average, they succeed in breaking the handi on 2 out of every 11 attempts. How many attempts will they have to make to have at least a 50% chance of succeeding in breaking the handi?


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    2/11 + 9/11 * 2/11 + 9/11 * 9/11 * 2/11..n terms.
    r=9/11
    { 1-(9/11)^n} > = 0.5
    (9/11)^n < = 0.5
    so n=4


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    Q14) In a 1000 m race, when Sumit is given a head start of 20 seconds and 200 metres, he runs for 2 minutes and beats Amit by 400 metres. Who wins the race and by how much, if Amit gets a head start of 200 metres?
    a) Sumit, 91 m
    b) Amit, 91 m
    c) Sumit, 111 m
    d) Amit, 111 m.


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