Quant Boosters - Swetabh Kumar - Set 2


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    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?

    (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

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

    (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

    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?

    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

    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.

    S(120) = 800
    A(100) = 600 so A=6 S=20/3
    so A/S = 800/d = 18/20
    so d=889. so Amit wins by 1000-889 = 111m

    In how many ways can 1,000,000 be written as sum of a square number and prime number?

    p+q^2 = 10^6
    p = (10^3+q)(10^3-q)
    since p is prime, 10^3-q=1 so q=(10^3-1)= 999
    so p = 1999
    hence only 1 way, 1999+(999)^2

    4 different digits are chosen and all possible positive 4 digit numbers are constructed out of them. The sum of the 4-digit numbers is 186,648. How many different sets of such 4 digits can be chosen?

    abcd so 24 arrangements.
    a (6000+600+60+6) +.... = 186648
    (a+b+c+d) = 186648/6666 = 28
    so (9,8,6,5), (9,8,4,7)

    The ratio of the factorial of a number x to the square of the factorial of another number, which when increased by 50% gives the required number , is 1.25. Find the number x.
    a) 6
    b) 5
    c) 9
    d) None

    x=6
    x and 2x/3
    so (x!) / (2x/3)^2 = 5/4
    x=6 pe 6!/(4!)^2 = 720/576 = 5/4 so 6

    A person has just sufficient money to buy either 30 guavas, 50 plums or 70 peaches. He spends 20% of the money on travelling, and buys 14 peaches, 'x' guavas and 'y' plums using rest of the money. If x, y > 0, what is the minimum value of the sum of x and y?

    30a =50b = 70c =5T
    4T = 14c + xa + yb
    4T = 14 * 5T/70 + x * 5T/30 + y * 5T/50
    4 = 1+ x/6 + y/10
    x/6+y/10 = 3
    5x+3y = 90
    so min. sum when x=15, y=5 so 20

    S = {1, 2, ..., 5n}, where n is a natural number. If R is a subset of S such that the average of no two elements of R is n or an integral multiple of n, then the maximum number of elements in R is:
    Options:
    a) 4n-1
    b) 2n
    c) 3n
    d) 3n-1

    3n-1
    n=1 so (1,2,3,4,5)
    so avg should not be integral multiple of 1, so avg should not be integral.
    so R=(1,2) no other number can come. so 2 elements= 3*1-1=2

    After travelling for 30mins , a train met with an accident and was stopped there for 45 mins. Due to accident it's speed reduces to 2/3 of its former speed and the train reached the station 1.5 hours late. Had the accident occurred 60km after the point it occurred earlier , the train would have reached reached 30mins earlier. What is the length of the journey?

    speed=60, distance=120
    a/v+3/4 + 3(d-a)/2v = d/v +3/2
    (a+60)/v + 3/4 + 3(d-a-60)/2v = d/v+1
    subtract.
    60/v - 90/v = -1/2
    30/v = 1/2 so v=60.
    so 1/2 +3d/2v = 2d/2v+3/2
    d/2v = 1 so d=2v = 2 * 60=120


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