Quant Boosters - Soumil Jain, CAT Quant 100 percentiler, IIM Calcutta - Set 2


  • IIM Calcutta | IIT Kanpur | CAT Quant 100 percentiler


    Region Q is defined by the equation 2x + y < 40. How many points (r, s) exist such that r is a natural number and s is a multiple of r?

    Yes When r = 1, s can take 37 values [371]
    When r = 2, s can take 17 values [352]
    When r = 3, s can take 11 values [333]
    When r = 4, s can take 7 values [314]
    When r = 5, s can take 5 values [295]
    When r = 6, s can take 4 values [276]
    When r = 7, s can take 3 values [257]
    When r = 8, s can take 2 values [238]
    When r = 9, s can take 2 values [219]
    When r = 10, s can take 1 values [1910]
    When r = 11, 12, 13 s can take one value each.
    Totally, there are 92 values possible.

    A boss decides to distribute Rs. 2000 between 2 employees. He knows X deserves more that Y, but does not know how much more. So he decides to arbitrarily break Rs. 2000 into two parts and give X the bigger part. What is the chance that X gets twice as much as Y or more?

    If he divides randomly 2000 into 2 parts then the larger part ranges uniformly from (1000,2000) we want the probability that the larger part lie in ( 1000*(4/3) , 2000 ) for X to get at least twice as much as Y. Therefore the required probability would be: (2000 - (4/3)*1000) / 1000 = 2 - (4/3) = 2/3 that is 66.67%

    A five-digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?

    If you assume that any digit is in fixed position, then the remaining four digits can be arranged in 4! = 24 ways.
    So, each of the 5 digit will appear in each of the five places 24 times. So, the sum of the digits in each position is 24(1+3+5+7+9) = 600
    The sum of all such numbers will be 600(1 + 10 + 100 + 1000 + 10000) = 6666600

    Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year.Unfortunately, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90 of the time. When it doesn't rain, he incorrectly forecasts rain 10 of the time. What is the probability that it will rain on the day of Marie's wedding?

    We want to know P(A1|B), the probability it will rain on the day of Marie's wedding, given a forecast for rain by the weatherman. The answer can be determined from Bayes' theorem, as shown below.
    P(A1|B)=P(A1)P(B|A1) / P(A1) P(B|A1)+P(A2) P(B|A2)
    P(A1|B)=(0.014×0.9 / ( 0.014×0.9+ 0.986×0.1) )
    P(A1|B)=0.111

    A swimmer jumps from abridge over a canal and swims 1 km upstream. after that first km he passes a floating cork. He continues swimming for half an hour and then, turns around and swims back to the bridge. The swimmer and the cork arrive at the same time. What is the speed of the water if the speed of the swimmer has been constant?

    If the swimmer is swimming away from the cork for half an hour (up stream), it will take him another half hour to swim back to the cork again. Because the swimmer is swimming with constant speed (constant relatively to the speed of the water!) you can look at it as if the water in the river doesn't move, the cork doesn't move, and the swimmer swims a certain time away from the cork and then back. So in that one hour time, the cork has floated from 1 kilometer up stream to the bridge.
    Conclusion: The water in the canal flows at a speed of 1 km/h.

    What is the digit at the ten's place of the number 6^11^7 ?
    a) 3
    b) 1
    c) 5
    d) 9

    The digit at the ten’s place of 6^2, 6^3, 6^4, 6^5 and 6^6 are 3, 1, 9, 7 and 5 respectively.
    Also, the digit at the ten’s place of 6^7 is 3.
    ⇒ The cyclicity of the ten’s digit of 6^N is 5. (N ≥ 2) The remainder when 11^7 is divided by 5 is 1.
    ⇒ The digit at the ten’s place of the given number N will be
    same as the digit at ten’s place of 6^6 which is 5.

    xy + zy = 37 and xz + zy = 72.Find the number of ordered triplets (x,y and z) such that x, y and z are positive integers.
    a) 0
    b) 1
    c) 2
    d) 3

    y(x+z) = 37=1×37
    It is obvious that x + z > 1, therefore ‘y’ has to be equal to 1. ⇒x+z = 37 and y=1
    ⇒xz + z =72
    ⇒ z(37 – z) + z = 72
    ⇒38z – z2 =72
    ⇒ z2 – 38z + 72=0 ⇒ (z – 2)(z – 36) = 0
    ⇒z= 2 or 36 So,there are 2 solutions: (x=35,y=1,z=2) and (x=1, y=1,z=36).

    A function f(x) is defined as f(x) = f(x – 2) + x(x – 2) for all the integer values of ‘x’. Given that f(1) + f(6) = 0. What is the value of f(1) + f(2) + f(3) + f(4) + f(5) + f(6)?

    Given that f(x)=f(x–2)+x(x–2)So,
    At x = 3, f(3) = f(1) + 3(1) = f(1) + 3
    and f(5) = f(3) +5(3) = f(3) +15 = f(1) + 18 Similarly,
    f(6)=f(4)+6(4)=f(4)+24 Hence, f(4) = f(6) – 24
    Also, f(4) = f(2) + 4(2) = f(2) + 8 Hence, f(2) = f(4) – 8 = f(6) – 32 Therefore,
    f(1) + f(2) + f(3) + f(4) + f(5) + f(6)
    = f(1) + {f(6) – 32} + {f(1) + 3} + {f(6) – 24} + {f(1) + 18} + f(6) =3{f(1)+f(6)}+ {–32+3–24+18}
    = 3(0) – 35
    = –35.

    How many divisors of 25200 can be expressed in the form 4n + 3, where n is a whole number?

    25200 = 2^4 ×3^2 ×5^2 ×7^1
    As the required divisors when divided by 4 leave remainder 3, the power of 2 in the divisors has to be 0. Therefore, any such divisor is of the form 3^a × 5^b ×7^c, which when divided by 4 leaves the remainder (–1)^a × 1^b × (–1)^c.
    For the remainder to be 3 i.e. –1, one of ‘a’ or ‘c’ must be even/0 and the other should be odd. Also, ‘b’ can take all the three possible values without making a difference to the remainder.
    The nine possibilities are listed below:
    a=0,b=0,c=1
    a=2,b=0,c=1
    a = 1, b = 0, c = 0
    a = 0, b = 1, c = 1
    a=2,b=1,c=1
    a = 1, b = 1, c = 0
    a=0,b=2,c=1
    a=2,b=2,c=1
    a = 1, b = 2, c = 0

    What is the total number of ways of selecting twenty balls from an infinite number of blue, green and yellow balls?

    Let the number of blue, green and yellow balls picked be x, y and z respectively.
    ∴ x + y + z = 20
    So the number of ways = (20+3–1)C(3–1)= 22C2 = 231


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